Important Questions

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 16 Chemistry in Everyday Life Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 16 Chemistry in Everyday Life

Question 1.
Write a note on nutrients.
Answer:
Nutrients:

• Nutrients are obtained from food and are used as a source of energy by the body.
• The main nutrients obtained from food are carbohydrates, lipids, proteins, vitamins, minerals and water. Most nutrients are organic macromolecules.
• Along with providing energy, these nutrients also regulate various body functions like growth, repair of damaged body tissues, etc.

The following table consists of different types of nutrients and their major sources.

Type of nutrient	Sources
Carbohydrates	Grains, fruits, vegetables, etc.
Proteins	Meat, fish. eggs, dairy products, pulses, etc.
Lipids	Dairy products, vegetable oil, animal fats, etc.
Vitamins	Grains, fruits, vegetables, meat, fish, eggs, dairy products, pulses, etc.

Question 2.
What happens when proteins and carbohydrates present in foods are digested in presence of enzymes?
Answer:

• Proteins and carbohydrates are organic polymeric macromolecules.
• When food is digested in presence of enzymes, the polymeric carbohydrates and proteins break down into monomers, i.e., glucose and α-amino acids, respectively.

Question 3.
Quality of food changes on shelving. Explain.
Answer:

• Enzymes are naturally present in all food materials.
• Quality of food changes on shelving mostly due to enzyme action, chemical reactions with the environment and due to the action of microorganisms.

Question 4.
Give two beneficial effects of shelving of food in day to day life.
Answer:

• Setting of milk into curd.
• Raising flour dough to make bread.

[Note: These changes are brought about by action of microorganisms.]

Question 5.
How are the chemical reactions of foodstuff with the environment controlled during storage?
Answer:

• Primarily the oxygen and microorganisms in air are responsible for adverse effects on stored food.
• The exposure of stored food to atmosphere is minimized by storing them in air tight container, evacuation or filling the container with N2 gas.
• Rate of a chemical reaction decreases with lowering of temperature. Thus, refrigeration is useful for controlling chemical reaction of foodstuff with environment.
• The reactions of foodstuff with environment are catalysed by enzymes. Due to boiling, the enzymes become denatured and the reactions are controlled.

Question 6.
What is the main aim of food preservation and food processing methods?
Answer:
Food preservation and food processing methods aim at prevention of undesirable changes and attempt to bring about desirable changes in food.

Question 7.
The melting points of unsaturated fats are lower. Give reason.
Answer:

• The long carbon chains of unsaturated fatty acids contain one or more C=C double bonds which produces one or more ‘kinks’ in the chain. This prevent the molecules from packing closely together.
• Also, the van der Waals forces between the unsaturated chains are weak.

Hence, the melting points of unsaturated fats are lower.

Question 8.
i. What are natural fats? Comment on their melting points.
ii. Explain how unsaturation affects the melting point and crystalline nature of fats.
Answer:
i. Natural fats are mixtures of triglycerides. They do not have sharp melting points and melt over a range of temperatures.

ii. Effect of unsaturation:

• The more unsaturated the fat (i.e., the presence of C=C bonds), lower is its melting point.
• Also, they will be less crystalline in nature.

Note: Naturals fats and their physical states

Mainly saturated fats	Mainly monounsaturated fats	Mainly polyunsaturated fats
Coconut fat/oil, butterfat, lard, margarine, Vanaspati ghee	Olive oil, peanut oil, canola oil	Safflower oil, sunflower oil, soybean oil, com oil, fish oil
Solid	Liquid	Liquid

Note: Molecular shapes of fats (A schematic representation):

Question 9.
Write a note on cis and trans form of unsaturated fats.
Answer:
Due to the presence of C=C double bonds, unsaturated fats can have two geometrical isomers, i.e., cis form and transform.
i. Cis form:

• In the cis form of an unsaturated fatty acid, the two hydrogens on the two double bonded carbons are on the same side of the double bond.
• It is the most common form of unsaturated fats.
• Cis fats do not cause deposition of cholesterol in blood vessels and thus, decrease the chance of developing coronary heart disease.

ii. Transform:

• In the transform of an unsaturated fatty acid, the two hydrogens on the two double bonded carbons are on the opposite sides of the double bond.
• Trans fats are difficult to metabolize and may get deposited to dangerous levels in fatty tissues.
• Large amount of trans unsaturated fats, increase the tendency of cholesterol getting deposited in the blood vessels leading to increased risk of cardiovascular disease.

[Note: Transform occurs only in animal fats and processed unsaturated fats.]

Question 10.
In which form, fats are used to transport cholesterol in the body?
Answer:
Fats in the form of lipoprotein are used to transport cholesterol in the body.

Question 11.
Give reason: Excessive low-density lipoprotein (LDL) increases the risk of cardio vascular diseases.
Answer:
Excessive low-density lipoprotein (LDL) increases the risk of cardio vascular diseases because it causes deposition of cholesterol in blood vessels.

Question 12.
What does the term omega represent in unsaturated fatty acids? Explain with the help of an example.
Answer:
i. Omega denotes the last carbon of the carbon chain in unsaturated fatty acids.
ii. Depending upon the position of the double bond, there are several omega fatty acids such as omega-3 and omega-6 fatty acids.
iii. These names are given for the position of the double bond in a long carbon chain of the unsaturated fatty acid.
iv. Omega-3 fatty acids have C = C bond between the third and fourth carbon from the end of a carbon chain.
e.g. Linolenic acid (9,12,15-octadecatrienoic acid).

Question 13.
Do omega-3 and omega-6 fats have same effect on the body? Discuss the effects.
Answer:
No, omega-3 and omega-6 fats have different effects on the body.
Omega-3 fats are found to raise the high density lipoprotein, HDL (good cholesterol) level of blood whereas omega-6 fats are considered to increase the risk of high blood pressure.

Question 14.
State TRUE or FALSE. Correct the false statement.
i. Fats are triglycerides of fatty acids.
ii. Linolenic acid is an omega-3 fatty acid having four C = C double bonds in its structure.
iii. Omega-6 fats are considered to increase the risk of high blood pressure.
Answer:
i. True
ii. False
Linolenic acid is an omega-3 fatty acid having three C = C double bonds in its structure.
iii. True

Question 15.
Name few sources of omega-3 fatty acids.
Answer:
Foods like walnuts, flaxseeds, chia seeds, soybean, cod liver oil are rich source of omega-3 fatty acids.

Question 16.
Draw the structure of vitamin E (tocopherol).
Answer:
Structure of vitamin E (tocopherol):

Question 17.
Give the sources of vitamin E.
Answer:
Vitamin E can be obtained from foods such as wheat germ, nuts, seeds, green leafy vegetables and oils like safflower oil.

Question 18.
What are synthetic antioxidants? Give an example.
Answer:
Synthetic antioxidant:

• Synthetic antioxidants are chemicals that are synthesized in the laboratory and used as a substitute for natural antioxidants.
• They delay the onset of oxidant or slow down the rate of oxidation of foodstuff.
• They are added as additives to increase the shelf life of packed foods.
• Common structural units found in synthetic antioxidants are phenolic -OH group and tertiary butyl group.
  e.g. BHT, which is 3,5-di-tert-butyl-4-hydroxytoluene.

Question 19.
Write a short note on BHT.
Answer:
1. BHT (butylated hydroxytoluene) is a synthetic antioxidant.
2. It is used as an additive to increase the shelf life of packed foods.
3. Structure of BHT contains a phenolic – OH group (which is responsible for its antioxidant properties) and tertiary butyl group.

IUPAC name: 3,5-Di-tert-butyl-4-hydroxytoluene

Question 20.
Define the term: Drug
Answer:
A chemical which interacts with biomolecules such as carbohydrates, lipids, proteins and nuclei acids and produces a biological response is called drug.

Question 21.
i. Which type of drug is used as medicine?
ii. What does a medicine contain?
iii. What are medicines used for?
Answer:
i. A drug having therapeutic and useful biological response is used as medicine.
ii. A medicine contains a drug as its active ingredient. Besides, it contains some additional chemicals which make the drug suitable for its use as medicine.
iii. Medicines are used in diagnosis, prevention and treatment of a disease.
[Note: Drugs being foreign substances in a body, often give rise to undesirable, adverse side effects.]

Question 22.
Explain the following terms:
i. Drug design
ii. Generic medicines
Answer:
i. Drug design is an important branch of medicinal chemistry which aims at synthesis of new molecules having better biological response. Now-a-days, there is an increasing trend in drug design to take cognizance of traditional medical knowledge such as Ayurvedic medicine or natural materials to discover new drugs.

ii. The drug manufacturing companies usually have a patent for drugs which are sold with the brand name. After the expiry of patent, the drug can be sold in the name of its active ingredient. These are called generic medicines.

Question 23.
What are analgesics? Explain their mode of action.
Answer:
Analgesics:

• Drugs which give relief from pain are called analgesics.
  e.g. Aspirin, paracetamol
• Most of the analgesics are anti-inflammatory drugs, which kill pain by reducing inflammation or swelling.

Question 24.
Mention the medicinal properties of salicylic acid.
Answer:
Salicylic acid has pain-killing and fever reducing properties.

Question 25.
i. What is aspirin? Write its use.
ii. Mention its side effect.
Answer:
i. Aspirin is acetyl derivative of salicylic acid.
It is widely used as an analgesic.
ii. It has a fewer side effects than salicylic acid. However, it retains stomach irritating side effects of salicylic acid.

Question 26.
Draw the structure of salicylic acid and write its IUPAC name.
Answer:
Structure of salicylic acid:

IUPAC name: 2-Hydroxybenzoic acid

Question 27.
Draw structures of following analgesics and write their molecular formula,
i. Aspirin
ii. Paracetamol
Answer:
i. Structure and molecular formula of aspirin:

ii. Structure and molecular formula of paracetamol:

Question 28.
What are antimicrobial drugs?
Answer:
Any drug that inhibits or kills microbial cells that include bacteria, fungi and viruses, are called antimicrobial drugs.

Question 29.
Give a brief classification of antimicrobials.
Answer:
Antimicrobials are classified into the following three categories:
i. Disinfectants:

• Disinfectants are non-selective antimicrobials, which kill a wide range of microorganisms including bacteria.
• Disinfectants are used on non-living surfaces. For example, floors, instruments, sanitary ware, etc.

ii. Antiseptics:
Antiseptics are used to sterilise surfaces of living tissue when the risk of infection is very high, such as during surgery or on wounds.

iii. Antibiotics:
Antibiotics are a type of antimicrobial designed to target bacterial infections within or on the body.

Question 30.
Name the ingredients present in dettol.
Answer:
Chloroxylenol is the active ingredient of dettol. The other ingredients of dettol are isopropyl alcohol, pine oil, castor oil soap, caramel and water.

Question 31.
State whether the following statements are TRUE or FALSE. Correct the statement, if false.
i. A concentrated solution of boric acid is used as an antiseptic for eyes.
ii. Iodoform is a powerful antiseptic.
iii. The active ingredient present in dettol is chloroxylenol.
Answer:
i. False
A dilute aqueous solution of boric acid is used as an antiseptic for eyes.
ii. True
iii. True

Question 32.
Instead of phenol, it’s chloro derivatives are used as antiseptics. Explain.
Answer:

• A dilute aqueous solution of phenol has antiseptic properties but it is found to be corrosive in nature.
• Many chloro derivatives of phenol are more potent antiseptic and have less corrosive effects than phenol, if used in lower concentrations.

Thus, instead of phenol it’s chloro derivatives are used as antiseptics.

Question 33.
Draw the structures of the following compounds and name the class of antimicrobials to which they belong.
i. Thymol
ii. p-Chloro-o-benzylphenol
iii. 2,4,6-Trichlorophenol
Answer:
i. Thymol: It is an antiseptic.

ii. p-Chloro-o-benzylphenol: It is a disinfectant. OH

iii. 2,4,6-Trichlorophenol: It is an antiseptic.

Question 34.
What are antibiotics?
Answer:
Antibiotics are drugs which are purely synthetic or obtained from microorganisms like bacteria, fungi or moulds.
e.g. Salvarsan, Prontosil

Question 35.
Name the first effective drug used in treatment of syphilis.
Answer:
Salvarsan was the first effective drug used in treatment of syphilis.

Question 36.
Name the following:
i. An effective diazo antibacterial drug.
ii. One example of a sulpha drug.
Answer:
i. Prontosil
ii. Sulphapyridine

Question 37.
Name the diazo antibacterial, which gets converted to sulphanilamide in the body.
Answer:
Prontosil is an effective diazo antibacterial, which gets converted to a simpler compound, sulphanilamide, in the body.

Question 38.
Draw the structure of the following:
i. An azodye
ii. Prontosil
iii. Sulphapyridine
iv. Sulphanilamide
Answer:

Question 39.
Draw the general structure of penicillin.
Answer:
General structure of penicillin:

Question 40.
Draw the structure of chloramphenicol.
Answer:
Structure of chloramphenicol:

Question 41.
Give classification of antibiotics.
Answer:
Antibiotics can be of three types, which are as given below:

• Broad spectrum antibiotics: They are effective against wide range of bacteria.
• Narrow spectrum antibiotics: They are effective against one group of bacteria.
• Limited spectrum antibiotics: They are effective against a single organism.

[Note: Antibiotics can be synthetic, semisynthetic or of microbial origin.]

Question 42.
State the disadvantage of broad spectrum antibiotics.
Answer:
The disadvantage of broad spectrum antibiotics is that they also kill the useful bacteria in the alimentary canal.

Question 43.
Complete the following table.

Plant	Medicinal property	Active ingredient
Cinnamon	Antimicrobial for cold	————
————	————	Eugenol
Citrus fruits	Antioxidant	————
Wintergreen	————	————
Indian gooseberry (amla)	Antidiabetic, antimicrobial, antioxidant	Vitamin C, Gallic acid

Answer:

Plant	Medicinal property	Active ingredient
Cinnamon	Antimicrobial for cold	Cinnamaldehyde
Clove	Antimicrobial and analgesic	Eugenol
Citrus fruits	Antioxidant	Vitamin C (ascorbic acid)
Wintergreen	Analgesic	Methyl salicylate
Indian gooseberry (amla)	Antidiabetic, antimicrobial, antioxidant	Vitamin C, Gallic acid

Question 44.
Draw the structures of following:
i. Curcumin
ii. Methyl salicylate
iii. Cinnamaldehyde
iv. Eugenol
v. Vitamin C
vi. Gallic acid
Answer:

Question 45.
What are cleansing agents?
Answer:
Cleansing agents are substances which are used to remove stain, dirt or clutter on surfaces.

Question 46.
What are the different types of cleansing agents?
Answer:
Commercially cleansing agents are of the following two main types, depending on their chemical composition:

• Soaps
• Synthetic detergents

[Note: Cleansing agents may be natural or synthetically developed.]

Question 47.
What are soaps? How soaps are prepared?
Answer:
Soaps:
i. Soaps are sodium or potassium salts of long chain fatty acids.
ii. They are obtained by alkaline hydrolysis of natural oils and fats with NaOH or KOH. This is called saponification reaction.
iii. Chemically, oils are triesters of long chain fatty acids and propane-1,2,3-triol (commonly known as glycerol or glycerin).
iv. Saponification of oil produces soap and glycerol as shown in the reaction below:

Question 48.
Give reason: Potassium soaps can be used for bathing purpose.
Answer:

• The quality of soap depends upon the nature of oil and alkali used.
• Potassium soaps (toilet soaps) are prepared by using better grades of oil and KOH. Therefore, they are soft to skin.
• Also, care is taken to remove excess of alkali which may otherwise cause skin irritation.

Hence, potassium soaps can be used for bathing purpose.

Question 49.
Laundry soaps are made using which alkali?
Answer:
Laundry soaps are made using alkali NaOH (sodium hydroxide).

Question 50.
Give examples of fillers used in making of laundry soaps.
Answer:
Laundry soaps contain fillers like sodium rosinate (a lathering agent), sodium silicate, borax, sodium and trisodium phosphate.

Question 51.
Explain why soaps become inactive in hard water.
Answer:
i. Soaps form scum in hard water and become inactive.
ii. This is because, hard water contains dissolved salts of calcium and magnesium. Soaps react with these salts to form insoluble calcium and magnesium salts of fatty acids.
iii. This insoluble substance is termed as scum which sticks to the fabric.
iv. Reaction of soap with calcium salt (CaCl₂) from hard water is given below:

Question 52.
Which chemical can be used for softening of hard water? Why?
Answer:

• Washing soda (Na₂CO₃) can be used for softening of hard water.
• This is because, washing soda precipitates the dissolved calcium salts as carbonate and helps the soap action by softening of water.

Question 53.
i. What are synthetic detergents?
ii. Mention their different types.
Answer:
i. Synthetic detergents are manmade cleansing agents designed to use in soft water as well as in hard water.
ii. There are three types of synthetic detergents which are as follows:

• Anionic detergents
• Cationic detergents
• Nonionic detergents

Question 54.
Complete the following table:

Answer:

Question 55.
Give an example of detergent used as:
i. Additive in toothpaste
ii. Used as germicide
Answer:
i. Additive in toothpaste: Sodium lauryl sulphate, CH₃(CH₂)₁₀CH₂O\(\mathrm{SO}_{3}^{-}\)Na⁺
ii. Used as germicide: Ethyltrimethylammonium bromide, [CH₃(CH₅)₁₅ – N⁺(CH₃)₃]Br^–.

Question 56.
Explain cleansing mechanism of soaps and detergents.
Answer:
i. Soaps and detergents bring about cleansing of dirty, greasy surfaces by the same mechanism.
ii. Dirt is held at the surface by means of oily matter, and therefore cannot get washed with water.
iii. The molecules of soaps and detergent have two parts. One part is polar called head and the other part is long nonpolar chain of carbons called tail.
iv. The hydrophilic polar head can dissolve in water which is a polar solvent, while the hydrophobic nonpolar tail dissolve in oil/fat/grease.
v. The molecules of soap/detergent are arranged around the oily droplet such that the nonpolar tail points towards the central oily drop while the polar head is directed towards the water.
vi. Thus, micelles of soap/detergent are formed surrounding the oil drops, which are removed in the washing process.

Question 57.
Compound “X” having the following structure is used as synthetic antioxidant to increase the shelf life of packed foods.

i. What is the molecular formula of compound “X”?
ii. Identify the structural unit responsible for antioxidant activity of “X”.
iii. Give one example of a compound with structure, similar to compound “X”, which is commonly used as synthetic antioxidant.
iv. Give the IUPAC name of compound “X”.
Answer:
i. Molecular formula: C₁₁H₁₆O₂
ii. Structural unit responsible for antioxidant activity of compound “X” is phenolic -OH group.
iii. Butylated hydroxytoluene (BHT) is commonly used synthetic antioxidant similar to compound “X”.

iv. The IUPAC name of compound “X” is 2-tert-butyl-4-methoxyphenol.

Multiple Choice Questions

1. BHT as a food additive act as …………….
(A) antioxidant
(B) flavouring agent
(C) colouring agent
(D) emulsifier
Answer:
(A) antioxidant

2. The structure of antioxidant BHT is …………….


Answer:

3. The molecular formula of aspirin is …………….
(A) C₈H₈O₃
(B) C₉H₈O₄
(C) C₉H₁₀O₄
(D) C₉H₈O₃
Answer:
(B) C₉H₈O₄

4. Aspirin is a/an …………….
(A) antibiotic
(B) analgesic
(C) antimicrobial
(D) disinfectant
Answer:
(B) analgesic

5. The CORRECT structure of the drug paracetamol is …………….

Answer:

6. Which of the following is used as a weak antiseptic for eyes?
(A) Tincture of iodine
(B) Dilute solution of dettol
(C) Iodoform
(D) Dilute aqueous solution of boric acid
Answer:
(D) Dilute aqueous solution of boric acid

7. The structure of thymol is …………….

Answer:

8. Salvarsan is arsenic containing drug which was first used for the treatment of …………….
(A) syphilis
(B) typhoid
(C) ulcer
(D) dysentery
Answer:
(A) syphilis

9. The linkage present in salvarsan is …………….
(A) -N = N –
(B) – As = As –
(C) -S – S –
(D) – O – O –
Answer:
(B) – As = As –

10. Which of following contains – N = N – in its structure?
(A) Chloramphenicol
(B) Sulphapyridine
(C) Salvarsan
(D) Prontosil
Answer:
(D) Prontosil

11. Which of the following contains – As = As – linkage?
(A) Salvarsan
(B) Prontosil
(C) Sulphanilamide
(D) Sulphapyridine
Answer:
(A) Salvarsan

12. Which of the following element is NOT present in penicillin?
(A) O
(B) S
(C) P
(D) N
Answer:
(C) P

13. Methyl salicylate having analgesic properties is obtained from which of the following plant?
(A) Clove
(B) Indian gooseberry
(C) Wintergreen
(D) Cinnamon
Answer:
(C) Wintergreen

14. Hydrolysis of oil by aqueous alkali is called …………….
(A) esterification
(B) saponification
(C) acetylation
(D) carboxylation
Answer:
(B) saponification

15. Sodium lauryl sulphate is an example of …………….
(A) soap
(B) cationic detergent
(C) anionic detergent
(D) nonionic detergent
Answer:
(C) anionic detergent

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons

Question 1.
What are unsaturated and saturated hydrocarbons?
Answer:
Hydrocarbons which contain carbon-carbon multiple bond (C=C or C≡C) are called unsaturated hydrocarbons, whereas those which contain carbon-carbon single bond (C-C) are called saturated hydrocarbons.

Question 2.
How are hydrocarbons classified?
Answer:

Question 3.
Define alkanes. Write general formula of alkanes.
Answer:

1. Alkanes are aliphatic saturated hydrocarbons containing carbon-carbon and carbon-hydrogen single
   covalent bonds.
2. They have a general formula C_nH_2n+2 where, ‘n’ stands for number of carbon atoms in the alkane molecule.

Question 4.
Give information about isomerism in alkanes. Write the all possible structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.
Answer:
i. Alkanes with more than three carbon atoms generally exhibit, structural isomerism and in particular, the chain isomerism.
ii. The number of possible structural isomers increase rapidly with the number of carbon atoms.
iii. Structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.

Question 5.
Write all the possible structural isomers of a saturated hydrocarbon having molecular formula C₅H₁₂.
Answer:

Question 6.
Define sigma bond.
Answer:
A single covalent bond formed by the coaxial overlap of orbitals is called sigma (σ) bond.

Question 7.
i. Why do C – C bonds in alkanes undergo rotation?
ii. What are conformations?
Answer:
i. a. Alkanes have single covalent bonds (sigma bonds) formed by the coaxial overlap of orbitals.
b. As a direct consequence of coaxial overlap of orbitals, a sigma bond is cylindrically symmetrical and the extent of orbital overlap is unaffected by rotation about the single bond and therefore, C – C bonds undergo rotation.

ii. a. In alkanes, the atoms bonded to one carbon of a C – C single bond change their relative position with reference to the atoms on the other carbon of that bond on rotation of that C – C single bond.
b. The resulting arrangements of the atoms in space about the C – C single bond are called conformations or conformational isomers. Innumerable conformations result on complete rotation of a C – C single bond through 360°.

Question 8.
i. What is conformational isomerism?
ii. Name the two extreme conformations shown by ethane molecule.
Answer:
i. The phenomenon of existence of conformation is a type of stereoisomerism and is known as conformational isomerism.
ii. Ethane molecule shows the following two extreme conformations:

• Staggered conformation
• Eclipsed conformation

Question 9.
Draw structures representing staggered and eclipsed conformations of ethane using:
i. Sawhorse projection
ii. Newman projection
Answer:
i. Sawhorse projection of ethane:

ii. Newman projection of ethane:

Question 10.
How are alkanes obtained from crude oil?
Answer:
Alkanes are obtained by fractional distillation of crude oil in oil refineries.

Question 11.
How are alkanes obtained from alkenes and alkynes?
OR
How are alkanes obtained from catalytic hydrogenation of alkenes and alkynes?
Answer:
i. Catalytic hydrogenation of alkenes or alkynes with dihydrogen gas gives corresponding alkanes.
ii. Finely divided powder of platinum (Pt) or palladium (Pd) catalyse the hydrogenation of alkenes and alkynes at room temperature.
iii. Relatively high temperature and pressure are required with finely divided nickel as the catalyst.
e.g. a. Propene to propane:

b. Ethyne to ethane:

Question 12.
Write the general reactions for the catalytic hydrogenation of alkenes and alkynes.
Answer:
General reaction for catalytic hydrogenation of alkenes:

General reaction for catalytic hydrogenation of alkynes:

Question 13.
Write the structures of alkenes that on catalytic hydrogenation give n-butane.
Answer:

Question 14.
Explain the preparation of alkanes by reduction of alkyl halides with the help of an example.
Answer:
i. Alkanes can be prepared by reduction of alkyl halides using zinc and dilute hydrochloric acid.
ii. The reduction of alkyl halides is due to nascent hydrogen obtained from the reaction between reducing agent Zn and dilute HCl.
e.g. Reduction of methyl iodide to methane.

Question 15.
How are alkanes obtained by Wurtz reaction?
Answer:
Alkyl halides on treatment with reactive sodium metal in dry ether, gives higher alkanes having double the number of carbon atoms. This is called as Wurtz coupling reaction.

Question 16.
How will you convert ethyl chloride into n-butane?
Answer:
Ethyl chloride on heating with sodium metal in presence of dry ether gives n-butane.

Question 17.
Write chemical equations for reactions that take place on treating ethereal solutions of:
i. Methyl iodide with sodium metal
ii. Ethyl iodide with sodium metal
Answer:

Question 18.
Explain the preparation of Grignard reagents.
OR
What is Grignard reagent? Explain its preparation.
Answer:
Grignard reagent are alkyl magnesium halides obtained by treating alkyl halides with dry magnesium metal in the presence of dry ether.

General Reaction:

Question 19.
State the action of water on methyl magnesium bromide in dry ether with the help of a chemical reaction.
Answer:

Question 20.
Write the reagents involved in the following conversions.

Answer:

Question 21.
Straight chain alkanes have higher melting and boiling points as compared to branched isomeric alkanes. Give reason.
Answer:
i. The electronegativity of carbon and hydrogen is nearly the same. Therefore, C-H and C-C bonds are nonpolar covalent bonds and hence, alkanes are nonpolar.
ii. Alkane molecules are held together by weak intermolecular van der Waals forces.
iii. Larger the surface area of molecules, stronger are such intermolecular van der Waals forces.
iv. In straight chain alkane molecules, surface area is relatively larger as compared to branched chain alkanes and as a result, the intermolecular forces are relatively stronger in straight chain alkanes than in branched chain alkanes.

Hence, straight chain alkanes have higher melting and boiling points as compared to branched alkanes.

Question 22.
State physical properties of alkanes.
Answer:

• Alkanes are colourless and odourless.
• At room temperature, the first four alkanes are gases, alkanes having 5 to 17 carbon atoms are liquids while the rest all are solids.
• Alkanes are readily soluble in organic solvents such as chloroform, ether or ethanol while they are insoluble in water.
• Alkanes have low melting and boiling points which increases with an increase in the number of carbon atoms for straight chain molecules. But for branched chain molecules, more the number of branches, lower is the boiling/melting point.

Note: [Melting and boiling points of alkanes]

Question 23.
Define substitution reactions.
Answer:
The reactions in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms is called as substitution reactions.
e.g. Halogenation of alkanes.

Question 24.
i. What is halogenation of alkanes?
ii. Write the order of reactivity of halogens towards alkanes.
Answer:
i. Substitution of H atoms of alkanes by X (halogen, X = Cl, Br, I and F) atom is called halogenation of alkanes.
ii. The reactivity of halogens toward alkanes follows the order: F₂ > Cl₂ > Br₂ > I₂
[Note: The ease of replacement of hydrogen atoms from the carbon in alkanes is in the order: 3 > 2 > 1.]

Question 25.
Explain reactions involved in chlorination of methane.
Answer:
Alkanes react with chlorine gas in presence of UV light or diffused sunlight or at a high temperature (573-773 K) to give a mixture of alkyl halides.

Chlorination of methane:


Tetrachloromethane is a major product when excess of chlorine is used. Chloromethane is obtained as major product when excess of methane is employed.

Question 26.
Predict the products in the following set of reactions.

Answer:

Question 27.
What is the action of Cl₂ and Br₂ on 2-methylpropane?
Answer:

[Note: In bromination, there is high degree of selectivity as to which hydrogen atoms are replaced.]

Question 28.
Explain mechanism of halogenation of alkanes.
Answer:
i. Halogenation of alkanes follows the free radical mechanism.
ii. Homolysis of halogen molecule (X₂) generates halogen atoms, i.e., halogen free radicals.
iii. The mechanism of the first step of chlorination of methane is shown below:

Question 29.
Why are alkanes used as fuels?
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. Hence, alkanes are used as fuels.

Question 30.
What is combustion of alkanes? Write a general equation for alkane combustion.
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. This is known as combustion reaction of alkanes.

General representative equation for combustion is:

Question 31.
Write chemical equations for combustion of butane and methane.
Answer:
i. Combustion of butane:

ii. Combustion of methane:

Question 32.
Write a short note on pyrolysis of alkanes.
Answer:
Alkanes on heating at higher temperature in absence of air decompose to lower alkanes, alkenes and hydrogen, etc. This is known as pyrolysis or cracking.
e.g. Pyrolysis of hexane

Question 33.
Explain aromatization reaction of alkanes. Give its one application.
Answer:
i. Straight chain alkanes containing 6 to 10 carbon atoms are converted to benzene and its homologues, on heating under 10 to 20 atm pressure at about 773 K in the presence of V₂O₅, Cr₂O₃, MO₂O₃, etc. supported over alumina.

ii. The reaction involves simultaneous dehydrogenation and cyclization. This reaction is known as aromatization or reforming.

This process is used in refineries to produce high quality gasoline which is used in automobiles as fuel.

Question 34.
Collect the information on CNG and LPG with reference to the constituents and the advantages of CNG over LPG.
Answer:
Constituents of CNG (Compressed Natural Gas):
It mainly consists of methane compressed at a pressure of 200-248 bar.
Constituents of LPG (Liquefied Petroleum Gas):
It contains a mixture of propane and butane liquefied at 15°C and a pressure of 1.7 – 7.5 bar.

Advantages of CNG over LPG:

• CNG is cheaper and cleaner than LPG.
• CNG produces less pollutants than LPG.
• It does not evolve gases containing sulphur and nitrogen.
• Octane rating of CNG is high, hence thermal efficiency is more.
• Vehicles powered by CNG produces less carbon monoxide and hydrocarbon emission.

[Note: Students are expected to collect additional information on their own.]

Question 35.
Write the uses of alkane.
Answer:
Uses of alkanes:

• First four alkanes are used as a fuel mainly for heating and cooking purpose. For example, LPG and CNG.
• CNG, petrol and diesel are used as fuel for automobiles.
• Lower liquid alkanes are used as solvent.
• Alkanes with more than 35 C atoms (tar) are used for road surfacing.
• Waxes are high molecular weight alkanes. They are used as lubricants. They are also used for the preparation of candles and carbon black that is used in manufacture of printing ink, shoe polish, etc.

Question 36.
i. Write the general molecular formula of alkenes.
ii. Why are alkenes also known as olefins?
Answer:
i. Alkenes have general formula C_nH_2n, where, n = 2,3,4… etc.
ii. Alkenes are also known as olefins because the first member ethene/ethylene reacts with chlorine to form oily substance.
[Note: Alkenes with one carbon-carbon double bond, contain two hydrogen atoms less than the corresponding alkanes.]

Question 37.
Define alkadienes and alkatrienes. Give one example for each.
Answer:
i. The aliphatic unsaturated hydrocarbons containing two carbon-carbon double bonds are called as alkadienes.

ii. The aliphatic unsaturated hydrocarbons containing three carbon-carbon double bonds are called as alkatrienes.

Question 38.
Explain structural isomerism in alkenes by giving an example.
Answer:
Alkenes with more than three carbon atoms show structural isomerism.
e.g. Alkene with molecular formula C4H8 is butene. The structural formulae for C₄H₈ can be drawn in three different ways:

Question 39.
Draw structures of chain isomers of butene.
Answer:

Question 40.
Draw structures of position isomers of butene.
Answer:

Question 41.
Define geometrical isomerism.
Answer:
The isomerism which arises due to the difference in spatial arrangement of atoms or groups about doubly bonded carbon (C=C) atoms is called geometrical isomerism.

Question 42.
Explain geometrical isomerism using a general example.
Answer:
i. If the two atoms or groups bonded to each end of the C=C double bond are different, then the molecule can be represented by two different special arrangements of the groups as follows:

ii. In structure (A), two identical atoms or groups lie on the same side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called cis-isomer.
iii. In structure (B), two identical atoms or groups lie on the opposite side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called trans-isomer.
iv. Due to different arrangement of atoms or groups in space, these isomers differ in their physical properties like melting point, boiling point, solubility, etc.

Question 43.
i. Define cis- and trans-isomer.
ii. Draw geometrical isomers of but-2-ene.
Answer:
i. a. Cis isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called a cis-isomer.
b. Trans isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called a trans-isomer.

ii. Geometrical or cis-trans isomers of but-2-ene are represented as:

Question 44.
State whether the following alkenes can exhibit geometrical (or cis-trans) isomerism or not. Give reason for the answer.
i. CH₃ – CH₂ – CH₂ – CH = CH₂
ii. CH₃ – CH₂ – CH = C(CH₃)₂
Answer:
Both the alkenes (i) and (ii) cannot exhibit geometrical isomerism, since 1 alkene is a terminal alkene (containing two H-atoms on the same side of the double bond) while the 2nd alkene is a 1,1-disubstituted alkene (containing two identical alkyl groups on the same side of the double bond).

Question 45.
Write the general formulae of alkenes which exhibit cis-trans isomerism.
Answer:
Alkenes having the following general formulae exhibit cis-trans isomerism:
RCH=CHR, R₁R₂C=CR₁R₃, R₁CH=CR₁R₂, R₁CH=CR₂R₃, R₁CH=CHR₂ and R₁R₂C=CR₃R₄

Question 46.
Draw structures of cis-trans isomers for the following:

Answer:

Question 47.
Which of the following compounds will show geometrical isomerism?

Answer:
Compounds (III), (IV) and (V) will show geometrical isomerism as they have each of the doubly bonded carbon atoms in their structures, attached to different atoms/groups of atoms.

Question 48.
Alkenes can be obtained from which industrial sources?
Answer:

1. The most important alkenes for chemical industry are ethene, propene and buta-1,3-diene.
2. Alkenes containing up to four carbon atoms can be obtained in pure form from the petroleum products.
3. Ethene is produced from natural gas and crude oil by cracking.

Question 49.
What is β-elimination reaction? Explain in brief.
Answer:
The reactions in which two atoms or groups are eliminated from adjacent carbon atoms are called 1,2-elimination reactions. Since the atom/group is removed from β-carbon atom (β to the leaving group) it is called as β-elimination reaction.

The hybridization of each C in the reactant is sp³ while that in the product is sp². This means elimination reactions cause change in hybridization state while forming multiple bonds from single bond.

Question 50.
i. What is dehydrohalogenation reaction?
ii. How is it carried out. Explain with an example.
Answer:
i. a. The reactions in which there is removal of hydrogen (H) atom and halogen (X) atom from adjacent carbon atoms are known as dehydrohalogenation reactions.
b. The carbon carrying X is called α-carbon atom. The hydrogen atom from adjacent carbon called β-carbon atom, is removed and hence, the reaction is known as β-elimination.

ii. When an alkyl halide is boiled with a hot concentrated alcoholic solution of a strong base like KOH or NaOH, alkene is formed with removal of water molecule.

[Note: The ease of dehydrohalogenation of alkyl halides is in the order 3 > 2 > 1.]

Question 51.
State Saytzeff rule.
Answer:
In dehydrohalogenation the preferred product is the alkene that has the greater number of alkyl groups attached to doubly bonded carbon atoms.

Question 52.
Write and explain dehydrohalogenation reaction of 2-chlorobutane.
Answer:

In dehydrohalogenation of 2-chlorobutane, but-2-ene (disubstituted alkene) is the preferred product because it is formed faster than but-1-ene (monosubstituted alkene) which is in accordance with Saytzeff rule.

Question 53.
Write the CORRECT order of stability of alkenes with respect to Saytzeff rule.
R CH = CH₂, CH₂ = CH₂, R₂C = CH₂, R₂C = CR₂, RCH = CHR, R₂C = CHR
Answer:
R₂C = CR₂ > R₂C = CHR > R₂C = CH₂, RCH = CHR > RCH = CH₂ > CH₂ = CH₂

Question 54.
Explain dehydration reaction of alcohols.
Answer:
i. Alcohols on heating with sulphuric acid form alkenes with elimination of water molecule. The reaction is known as catalysed dehydration of alcohols.
ii. The exact conditions of dehydration depend upon the alcohol.
iii. Dehydration of alcohol is an example of β-elimination since -OH group from α-carbon along with H-atom from β-carbon is removed.

The ease of dehydration of alcohol is in the order 3° > 2° > 1°.

Question 55.
Write dehydration reaction of 1°, 2° and 3° alcohols giving one example for each.
Answer:
The ease of dehydration of alcohols is in the order 3° > 2° > 1°.

Question 56.
Explain isomerism with structure in the product obtained by acid catalysed dehydration of pentan-2-ol.
Answer:
i. Pentan-2-ol on acid catalysed dehydration, forms the following isomers.

ii. A and B are position isomers.
iii. Pent-2-ene has the following geometrical isomers:

Question 57.
What is dehalogenation? Write the general reaction for dehalogenation of vicinal dihalides.
Answer:
i. Removal of two halogen atoms from adjacent carbon atoms is called dehalogenation.
ii. The dihalides of alkane in which two halogen atoms are attached to adjacent carbon atoms are called vicinal dihalides.
iii. Vicinal dihalides on heating with zinc metal form an alkene.

Question 58.
How is propene obtained by dehalogenation reaction?
Answer:

Question 59.
How are geometrical isomers of alkenes obtained from alkynes?
Answer:
Alkenes are obtained by partial reduction of alkynes wherein C = C triple bond of alkynes is reduced to a C = C double bond by:
i. using calculated quantity of dihydrogen in presence of Lindlar’s catalyst (palladised charcoal deactivated partially with quinoline or sulphur compound) to give the cis-isomer of alkene.

ii. using sodium in liquid ammonia to give trans-isomer of alkene.

Question 60.
Write physical properties of alkenes.
Answer:

• Alkenes are nonpolar or weakly polar compounds that are insoluble in water, and soluble in nonpolar solvents like benzene, ether, chloroform.
• They are less dense than water.
• The boiling point of alkene rises with increasing number of carbons.
• Branched alkenes have lower boiling points than straight chain alkenes.
• The boiling point of alkene is very nearly the same as that of alkane with the same carbon skeleton.

Question 61.
Arrange the following alkenes in increasing order of their boiling points.
But-1-ene, 2,3-dimethylbut-2-ene, 2-methylpropene, propene, 2-methylbut-2-ene.
Answer:
Propene < 2-methylpropene < but-1-ene < 2-methylbut-2-ene < 2,3-dimethylbut-2-ene.
Note: Melting points and boiling points of alkenes.

Question 62.
What kind of reactions do alkenes undergo? Give reason.
Answer:
Alkenes undergo electrophilic addition reactions since they are unsaturated and contain pi (π) electrons.

Question 63.
Write a note on halogenation of alkenes.
OR
Explain the formation of vicinal dihalides from alkenes with the help of examples.
Answer:
Alkenes are converted into the corresponding vicinal dihalides by addition of halogens (X₂ = Cl₂ or Br₂).

Question 64.
How is carbon-carbon double bond in a compound detected by bromination?
Answer:
When an alkene like ethene is treated with bromine in presence of CCl₄, the red-brown colour of bromine disappears due to following reaction.

Hence, decolourisation of bromine is used to detect the presence of C = C bond in unknown compounds.

Question 65.
Explain the formation of alkyl halides from alkenes.
Answer:
i. Alkenes react with hydrogen halides (HX) like hydrogen chloride, hydrogen bromide and hydrogen iodide to give corresponding alkyl halides (haloalkanes). This reaction is known as hydrohalogenation of alkenes.


ii. The order of reactivity of halogen acids is HI > HBr > HCl.

Question 66.
State Markovnikov’s rule and explain it with the help of an example.
Answer:
i. Markovnikov’s rule: When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part (X-) of the reagent gets attached to the carbon atom which carries less number of hydrogen atoms.
ii. For example, addition of HBr to unsymmetrical alkenes yield two isomeric products.

iii. Experimentally it has been found that 2-Bromopropane is the major product.
[Note: Addition of HBr to symmetrical alkenes yields only one product.]

Question 67.
Explain Anti-Markovnikov’s addition or peroxide effect or Kharasch-Mayo effect.
Answer:
In 1933, M. S. Kharasch and F. R. Mayo discovered that the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.

Question 68.
Write the structure of major alkyl halide obtained by the action of HCl on pent-1-ene
i. in presence of peroxide
ii. in absence of peroxide.
Answer:
The structures of alkyl halides obtained by the action of hydrogen bromide on pent-1-ene are as follows:
i. In presence of peroxide:

ii. In absence of peroxide:

[Note: Presence/absence of peroxide has no effect on addition of HCl or HI.]

Question 69.
Explain the formation of alcohols from alkenes using conc. sulphuric acid with the help of an example.
Answer:
i. Alkenes react with cold concentrated sulphuric acid to form alkyl hydrogen sulphate (ROSO₃H). The addition takes place according to Markovnikov’s rule as shown in the following steps.


ii. If alkyl hydrogen sulphate is diluted with water and heated, then an alcohol having the same alkyl group as the original alkyl hydrogen sulphate is obtained.

iii. This is an excellent method for the large-scale manufacture of alcohols.

Question 70.
What is hydration of alkenes?
Answer:
i. Reactive alkenes on adding water molecules in the presence of concentrated sulphuric acid, form alcohol.
ii. The addition of water takes place according to Markovnikov’s rule. This reaction is known as hydration of alkenes.

Question 71.
Complete the following conversion.

Answer:

Question 72.
But-1-ene and 2-methylpropene are separately treated with following reagents. Predict the product/products. Indicate major/minor product,
i. HBr
ii. H₂SO₄ / H₂O
Answer:
i. HBr:

ii. H₂SO₄ / H₂O:

Question 73.
Explain: Ozonolysis
Answer:
i. The C = C double bond in alkenes, gets cleaved on reaction with ozone followed by reduction.
ii. The overall process of formation of ozonide by reaction of ozone with alkene in the first step and then decomposing it to the carbonyl compounds by reduction in the second step is called ozonolysis.
iii. When ozone gas is passed into solution of the alkene in an inert solvent like carbon tetrachloride, unstable alkene ozonide is obtained.
iv. This is subsequently treated with water in the presence of a reducing agent zinc dust to form carbonyl compounds, namely, aldehydes and/or ketones.

Question 74.
Write reactions for the ozonolysis of the following alkenes:
i. Ethene
ii. Propene
Answer:

Question 75.
What is the role of zinc dust, in ozonolysis reaction?
Answer:
In ozonolysis, the role of zinc dust is to prevent the formation of hydrogen peroxide which oxidizes aldehydes to corresponding acids.

Question 76.
State TRUE or FALSE. If false, correct the statement.
i. In the cleavage products of ozonide, a carbonyl group (C=O) is formed at each of the original doubly bonded carbon atoms.
ii. In ozonolysis, the structure of original alkene reactant cannot be identified by knowing the number and arrangement of carbon atoms in aldehydes and ketones produced.
iii. Ozonolysis reaction is used to locate the position and determine the number of double bonds in alkenes.
Answer:
i. True
ii. False
In ozonolysis, knowing the number and arrangement of carbon atoms in aldehydes and ketones produced, we can identify the structure of original alkene.
iii. True

Question 77.
Identify the alkene which produces a mixture of methanal and propanone on ozonolysis. Write the reactions involved.
Answer:
i. The structure of alkene which produces a mixture of methanol and propanone on ozonolysis is

ii. Reactions:

Question 78.
Explain the process of hydroboration-oxidation of alkenes.
Answer:
i. Alkenes with diborane in tetrahydrofuran (THF) solvent undergo hydroboration to form trialkylborane, which on oxidation with alkaline peroxide forms primary alcohol.
ii. The overall reaction gives anti-Markovnikov’s product from unsymmetrical alkenes.

Question 79.
Write reactions for the following conversion by hydroboration-oxidation reaction.
Ethene to ethanol
Answer:
Ethene to ethanol:

Question 80.
Define: Polymerization
Answer:
The process in which large number of small molecules join together and form very large molecules with repeating units is called polymerization.

Question 81.
What is the difference between monomer and polymer?
Answer:
The compound having very large molecules made of large number of repeating small units is called polymer while the simple compound forming the repeating units in the polymer is called monomer.

Question 82.
How is ethene converted to polyethene?
Answer:
Ethene at high temperature and under high pressure interacts with oxygen, and undergoes polymerization giving high molecular weight polymer called polyethene.

Here, n represents the number of repeating units and is a large number.

Question 83.
Explain the process of hydroxylation of alkenes.
OR
What is the action of alkaline KMnO₄ on alkenes?
Answer:
Alkenes react with cold and dilute alkaline potassium permanganate to form glycols.

Question 84.
Explain Baeyer’s test giving one example.
Answer:
i. During hydroxylation of alkenes the purple colour of KMnO₄ disappears.
ii. Hence, such reaction serves as a qualitative test for detecting the presence of double bond in the sample compound. This is known as Baeyer’s test.
e.g. As propene contains a double bond, it reacts with alkaline KMnO₄ to give colourless propane-1,2-diol as product. Therefore, the purple colour of alkaline KMnO₄ disappears.

Question 85.
What is the action of following reagents on but-1-ene and but-2-ene?
i. Bromine
ii. Cold and dilute alkaline KMnO₄.
Answer:
i. Action of Br₂:

ii. Action of cold and dilute alkaline KMnO₄:

Question 86.
Describe the action of acidic potassium permanganate on alkenes.
Answer:
Acidic potassium permanganate or acidic potassium dichromate oxidizes alkenes to ketones or acids depending upon the nature of the alkene and the experimental conditions. This is called oxidative cleavage of alkenes.
e.g.

Question 87.
Complete the following conversions.

Answer:

Question 88.
State some important uses of alkenes.
Answer:

• Alkenes are used as starting materials for preparation of alkyl halides, alcohols, aldehydes, ketones, acids, etc.
• Ethene and propene are used to manufacture polythene, polypropylene which are used in polyethene bags, toys, bottles, etc.
• Ethene is used for artificial ripening of fruits, such as mangoes.

Question 89.
What are alkynes? Write their general formula.
Answer:

• Alkynes are aliphatic unsaturated hydrocarbons containing at least one C = C.
• Their general formula is C_nH_2n-2.

Question 90.
Explain position isomerism in alkyne.
Answer:
Alkynes show position isomerism which is a type of structural isomerism.
e.g. But-1-yne and but-2-yne, both are represented by C₄H₆, however, both of them differ in position of triple bond in them.

[Note: 1-Alkynes are also called terminal alkynes.]

Question 91.
Draw the structural isomers of isomers of C₅H₈. Identify position isomers amongst them.
Answer:
i. Structural isomers of C₅H₁₀ (fourth member of homologous series of alkynes):

ii. The compounds pent-1-yne and pent-2-yne are position isomers of each other.

Question 92.
What are alkadiynes and alkatriynes? Give one example of each.
Answer:
The aliphatic unsaturated hydrocarbons containing two and three carbon-carbon triple bonds in their structure are called alkadiynes and alkatriynes, respectively.
e.g. CH ≡ C – CH₂ – C ≡ CH
Alkadiyne (Penta-1,4-diyne)

HC ≡ C- C ≡ C- C ≡ CH
Alkatriyne (Hexa-1,3,5-triyne)

Question 93.
Complete the following table:

Answer:

Question 94.
How is acetylene prepared from the following compounds?
i. Methane
ii. Calcium carbide
Answer:
i. From methane: Ethyne is industrially prepared by controlled, high temperature, partial oxidation of methane.

ii. From calcium carbide: Industrially, ethyne is prepared by reaction of calcium carbide with water.

Question 95.
How are alkynes prepared by dehydrohalogenation of vicinal dihalides? Write general reaction and explain it using an example.
Answer:
Vicinal dihalides react with alcoholic solution of potassium hydroxide to form alkenyl halide which on further treatment with sodamide forms alkyne.

Question 96.
Convert 1,2-dichloropropane to propyne.
Answer:

Question 97.
i. What are terminal alkynes?
ii. How are they converted to higher nonterminal alkynes? Give one example.
Answer:
i. Terminal alkynes are the compounds in which hydrogen atom is directly attached to triply bonded carbon atom.
ii. a. A smaller terminal alkyne first reacts with a very strong base like lithium amide to form metal acetylide (lithium amide is easier to handle than sodamide).
b. Higher alkynes are obtained by reacting metal acetylides (alkyn-1-yl lithium) with primary alkyl halides.

Question 98.
How is pent-2-yne prepared from propyne?
Answer:

Question 99.
Enlist physical properties of alkenes.
Answer:
The physical properties of alkynes are similar to those of alkanes and alkenes.

• They are less dense than water.
• They are insoluble in water and quite soluble in less polar organic solvents like ether, benzene, carbon tetrachloride.
• The melting points and boiling points of alkynes increase with an increase in molecular mass.

Question 100.
Lithium amide (LiNH₂) is very strong base and it reacts with terminal alkynes to form lithium acetylides with the liberation of hydrogen indicating acidic nature of terminal alkynes. Why is it so?
Answer:

• The hydrogen bonded to C ≡ C triple bond has acidic character.
• In terminal alkynes, hydrogen atom is directly attached to sp hybridized carbon atom.
• In sp hybrid orbital, the percentage of s-character is 50%. An electron in s-orbital is very close to the nucleus and is held tightly.
• The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp² carbon in ethene or the sp³ carbon in ethane.
• Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H⁺) to very strong base.

Question 101.
Give reason: Acidic nature of alkynes is used to distinguish between terminal and non-terminal alkynes.
Answer:

• Acidic alkynes react with certain heavy metal ions like Ag⁺ and Cu⁺ and form insoluble acetylides.
• On addition of acidic alkyne to the solution of AgNO₃ in alcohol, it forms a precipitate, which indicates that the hydrogen atom is attached to triply bonded carbon.

Hence, this reaction is used to differentiate terminal alkynes and non-terminal alkynes.

Question 102.
Predict the product in the following reactions.
\(\mathbf{H C} \equiv \mathbf{C H}+\mathbf{2 B r}_{2} \stackrel{\mathrm{CCl}_{4}}{\longrightarrow} ?\)
Answer:
Ethyne reacts with bromine in inert solvent such as carbon tetrachloride to give tetrabromoethane.

Question 103.
Write the general reaction for addition of halogens to alkynes.
Answer:

Question 104.
Explain the addition of hydrogen halides to alkynes using a general reaction.
Answer:
i. Hydrogen halides (HCl, HBr and HI) add to alkynes across carbon-carbon triple bond in two steps to form geminal dihalides (in which two halogen atoms are attached to the same carbon atom).
ii. The addition of HX in both the steps takes place according to Markovnikov’s rule as shown in below.
General reaction:

iii. The order of reactivity of hydrogen halides is HI > HBr > HCl.

Question 105.
State the action of HBr on acetylene and methyl acetylene.
Answer:

Question 106.
Explain reactions of alkynes with water using general reaction.
Answer:
Alkynes react with water in presence of 40% sulphuric acid and 1% mercuric sulphate to form aldehydes or ketones, i.e., carbonyl compounds.
General reaction:

Question 107.
Predict the products when ethyne and propyne are treated with 1% mercuric sulphate in H₂SO₄.
Answer:
i. Ethyne:

ii. Propyne:

Question 108.
Convert:
i. But-1-yne to butan-2-one
ii. Hex-3-yne to hexan-3-one
Answer:
i. But-l-yne to butan-2-one:

ii. Hex-3-yne to hexan-3-one:

Question 109.
What are products obtained on hydration of but-1-yne and but-2-yne? Are they same or different? Explain.
Answer:
i. Hydration of but-1-yne:

Hydration of but-2-yne:

The products obtained on hydration of but-1-yne and but-2-yne are same i.e., butan-2-onc since the addition of water to alkyncs takes place according to Markovnikovs rule.

Question 110.
How is ethylene converted into ethylidene dichloride?
Answer:

Question 111.
Write some important uses of acetylene.
Answer:

• Ethyne (acetylene) is used in preparation of ethanal (acetaldehyde), propanone (acetone), ethanoic acid (acetic acid).
• It is used in the manufacture of polymers, synthetic rubber, synthetic fibre, plastic, etc.
• For artificial ripening of fruits.
• In oxy-acetylene (mixture of oxygen and acetylene) flame for welding and cutting of metals.

Question 112.
Many organic compounds obtained from natural sources such as resins, balsams, oil of wintergreen, etc. have pleasant fragrance or aroma. Such compounds are named as aromatic compounds.
i. Name the simplest aromatic compound.
ii. Write the names of any two aromatic compounds.
Answer:
i. Benzene is the simplest aromatic hydrocarbon.
ii. Toluene and naphthalene

Question 113.
Draw structures of any four aromatic hydrocarbons.
Answer:

Question 114.
Write the molecular formula of benzene. Give its boiling point.
Answer:
The molecular formula for benzene is C₆H₆. Its boiling point is 353 K.

Question 115.
State TRUE or FALSE. Correct the false statement.
i. Aromatic hydrocarbons are also called as arenes.
ii. Toluene is a non-aromatic hydrocarbon.
iii. Benzene is colourless liquid having characteristic odour.
Answer:
i. True
ii. False
Toluene is an aromatic hydrocarbon.
iii. True

Question 116.
Name any two large-scale sources of benzene.
Answer:
Coal-tar and petroleum are the two large-scale sources of benzene.
[Note: Other aromatic compounds like toluene, phenol, naphthalene, etc. are also obtained from coal-tar and petroleum.]

Question 117.
Draw the structure of an aromatic compound that resembles benzene but does not have pleasant odour.
Answer:

Question 118.
Name and draw the structures of any three compounds that have pleasant odour but do not resemble benzene.
Answer:

Question 119.
Differentiate between aromatic and aliphatic compounds.
Answer:
Aromatic compounds:

• Aromatic compounds contain higher percentage of carbon.
• They bum with sooty flame.
• They are cyclic compounds with alternate single and double bonds.
• They are not attacked by normal oxidizing and reducing agents.
• They do not undergo addition reactions easily. They do not decolourise dilute alkaline aqueous KMnO₄ and Br₂ in CCl₄, though double bonds appear in their structure.
• They prefer substitution reactions.

Aliphatic compounds:

• Aliphatic compounds contain lower percentage of carbon.
• They bum with non-sooty flame.
• They are open chain compounds.
• They are easily attacked by oxidizing and reducing agents.
• Unsaturated aliphatic compounds undergo addition reactions easily. They decolourise dilute aqueous alkaline KMnO₄ and Br₂ in CCl₄.
• The saturated aliphatic compounds give substitution reactions.

Question 120.
Benzene cannot have open chain structure. Explain this statement.
Answer:

• The molecular formula of benzene is C₆H₆. This indicates high degree of unsaturation.
• Open chain or cyclic structure having double and triple bonds can be written for C₆H₆.
• However, benzene does not behave like alkenes or alkynes. This indicates that benzene cannot have the open chain structure.

Question 121.
Compare the reactivity of benzene and alkenes with the following reagents:
i. Dilute alkaline KMnO₄
ii. Br₆ in CCl₄
iii. H₆O in acidic medium
Answer:

Reagent	Alkenes	Benzene
Dilute alkaline aqueous KMnO4	Decolourisation of KMnO4	No decolourisation
Br2 in CCl4	Decolourisation of red brown colour of bromine	No decolourisation
H2O in acidic medium	Addition of H2O molecule	No reaction

Question 122.
Give the evidence for the cyclic structure of benzene.
Answer:
Evidence for the cyclic structure of benzene:
i. Benzene yields only one and no isomeric monosubstituted bromobenzene (C₆H₅Br) when treated with equimolar bromine in FeBr₃. This indicates that all six hydrogen atoms in benzene are identical.

ii. This is possible only if benzene has cyclic structure of six carbons bound to one hydrogen atom each.
iii. Benzene on catalytic hydrogenation gives cyclohexane.

This confirms the cyclic structure of benzene and three C = C in it.

Question 123.
Write a short note on the Kekule structure of benzene.
Answer:
Kekule structure of benzene:
i. August Kekule in 1865 suggested the structure for benzene having a cyclic planar ring of six carbon atoms with alternate single and double bonds and hydrogen atom attached to each carbon atom.

ii. The Kekule structure indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms would be attached to the doubly bonded carbon atoms whereas in the other, they would be attached to single bonded carbons.

iii. However, benzene was found to form only one ortho-disubstituted benzene. This problem was overcome by Kekule by suggesting the concept of oscillating nature of double bonds in benzene as given below.

iv. Even with this modification, Kekule structure of benzene failed to explain unusual stability and preference to substitution reactions rather than addition reactions, which was later explained by resonance.

Question 124.
Explain the resonance phenomenon with respect to benzene.
OR
Explain the resonance hybrid structure of benzene.
Answer:

• Benzene is a hybrid of various resonance structures. The two structures, (A) and (B) given by Kekule are the main contributing structures.
• The resonance hybrid is represented by inserting a circle or a dotted circle inscribed in the hexagon as shown in (C).
• The circle represents six electrons delocalized over the six carbon atoms of benzene ring.
• A double headed arrow between the resonance structures is used to represent the resonance phenomenon.

Question 125.
Why does benzene not prefer to undergo addition reactions?
Answer:

• Benzene is highly unsaturated molecule but despite of this feature, it does not give addition reaction.
• The actual structure of benzene is represented by the resonance hybrid which is the most stable form of benzene than any of its resonance structures.
• This stability due to resonance (delocalization of π electrons) is so high that π-bonds of the molecule becomes strong and thus, resist breaking.

Thus, benzene does not prefer to undergo addition reactions.

Question 126.
Explain the resonance structures of benzene using the orbital overlap concept.
Answer:
The structure of benzene can be better explained by the orbital overlap concept,
i. All six carbon atoms in benzene are sp² hybridized. Two sp² hybrid orbitals of each carbon atom overlap and form carbon-carbon sigma (σ) bond and the remaining third sp² hybrid orbital of each carbon overlaps with s orbital of a hydrogen atom to form six C – H sigma bonds.

ii. The unhybridized p orbitals of carbon atoms overlap laterally forming π bonds. There are two possibilities of forming three π bonds by overlap of p orbitals of C₁ – C₂, C₃ – C₄, C₅ – C₆ or C₂ – C₃, C₄ – C₅, C₆ – C₁, respectively, as shown in the following figure. Both the structures are equally probable.

According to resonance theory, these are two resonance structures of benzene.

Question 127.
Explain the structure of benzene with respect to molecular orbital theory.
Answer:
i. According to molecular orbital (MO) theory, the six p orbitals of six carbons give rise to six molecular orbitals of benzene.
ii. Shape of the most stable MO is as show in the figure below. Three of these π molecular orbitals lie above and the other below those of free carbon atom energies.

iii. The six electrons of the p orbitals cover all the six carbon atoms and are said to be delocalized. Delocalization of π electrons results in stability of benzene molecule.

Question 128.
Give the carbon-carbon bond length in benzene. Explain why benzene shows unusual behaviour.
Answer:
i. X-ray diffraction data indicate that all C – C bond lengths in benzene are equal (139 pm) which is an intermediate between C – C (154 pm) and C = C bond (133 pm).

ii. Thus, absence of pure double bond in benzene accounts for its reluctance to addition reactions under normal conditions, which explains unusual behaviour of benzene.

Question 129.
Write a short note on aromaticity.
Answer:
i. All aromatic compounds undergoes substitution reactions rather than addition reactions and this property is referred to as aromaticity or aromatic character.
ii. The aromatic character of benzene is correlated to its structure.
iii. Aromaticity is due to extensive cyclic delocalization of p electrons in the planar ring structure.
iv. Three rules of aromaticity that is used for predicting whether a particular compound is aromatic or non-aromatic are as follows:

• Aromatic compounds are cyclic and planar (all atoms in ring are sp² hybridized).
• Each atom in aromatic ring has a p orbital. The p orbitals must be parallel so that continuous overlap is possible around the ring.
• Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1, 2, 3, … etc.

Question 130.
State and explain the Huckel rule of aromaticity.
Answer:
Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1,2,3, … etc.

Explanation:
According to Huckel rule, a cyclic and planar compound is aromatic if it the number of π electrons is equal to (4n + 2), where n = integer 0, 1, 2, 3, … etc.

n	Number of π electrons
n = 0	(4 × 0) – 2 = 2
n = 1	(4 × 1) + 2 = 6
n = 2	(4 × 2) + 2 = 10

e.g. Consider benzene molecule:

Benzene has 6π electrons. According to Huckel rule, if n = 1, then (4n + 2)π = 6π electrons. Hence, benzene is aromatic.

Question 131.
By using the rules of aromaticity, explain whether the following compounds are aromatic or non-aromatic.
i. Benzene
ii. Naphthalene
iii. Cycloheptatriene
Answer:
i. Benzene:
a. It is cyclic and planar.
b. It has three double bonds and six π electrons.
c. It has a p orbital on each carbon of the hexagonal ring. Hence, a continuous overlap above and below the ring is possible.

d. According to Huckcl rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here, ‘n’ comes out to be an integer.
Hence, benzene is aromatic.

ii. Naphthalene:
a. It is cyclic and planar.
b. It has 5 double bonds and 10 n electrons.

c. It has a p orbital on each carbon atom of the ring. Hence, a continuous overlap around the ring is possible.
d. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 10,
∴ 4n = 10 – 2 = 8
n = 8/4 = 2, Here ‘n’ comes out to be an integer.
Hence, naphthalene is aromatic.

iii. Cycloheptatriene:
a. It is cyclic and planar.
b. It has three double bonds and 6 π electrons.

c. But one of the carbon atoms is saturated (sp³ hybridized) and it does not have a p orbital.
d. Hence, a continuous overlap around the ring is not possible in cycloheptatriene. Hence, it is non-aromatic.

Question 132.
How does Huckel rule help in determining the aromaticity of pyridine?
Answer:
i. Pyridine has three double bonds and 6 π electrons.
ii. The six p orbitals containing six electrons form delocalized π molecular orbital.
iii. The unused sp² hybrid orbital of nitrogen containing two non-bonding electrons is as it is.

iv. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here ‘n’ comes out to be an integer. Hence, pyridine is aromatic.

Question 133.
How is benzene prepared from ethyne/acetylene?
Answer:
From ethyne (By trimerization): Alkynes when passed through a red hot iron tube at 873 K, polymerize to form aromatic hydrocarbons. Ethyne when passed through a red hot iron tube at 873 K undergoes trimerization to form benzene.

Question 134.
How is benzene prepared from sodium benzoate?
OR
Explain preparation of benzene by decarboxylation.
Answer:
From sodium benzoate (by decarboxylation): When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.

[Note: This reaction is useful for decreasing the length of a carbon chain by one C-atom]

Question 135.
How will you convert phenol to benzene?
Answer:
From phenol (By reduction): When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.

Question 136.
Enlist physical properties of benzene.
Answer:
Physical properties of benzene:

• Benzene is a colourless liquid.
• Its boiling point is 353 K and melting point is 278.5 K.
• It is insoluble in water. It forms upper layer when mixed with water.
• It is soluble in alcohol, ether and chloroform.
• Its vapours are highly toxic which on inhalation lead to unconsciousness.

Question 137.
What is the action of chlorine on benzene in the presence of UV light?
Answer:
Addition of chlorine: When benzene is treated with chlorine in the presence of bright sunlight or UV light, three molecules of chlorine gets added to benzene to give benzene hexachloride.

Question 138.
Name the γ-isomer of benzene hexachloride which is used as insecticide.
Answer:
The γ-isomer of benzene hexachloride which is used as insecticide is called as gammaxene or lindane.

Question 139.
How will you convert benzene to cyclohexane?
Answer:
Addition of hydrogen: When a mixture of benzene and hydrogen gas is passed over heated catalyst nickel at 453 K to 473 K, cyclohexane is formed.

Question 140.
What is the action of ozone on benzene?
Answer:
Addition of ozone: When benzene is treated with ozone in the presence of an inert solvent carbon tetrachloride, benzene triozonide is formed, which is then decomposed by zinc dust and water to give glyoxal.

Question 141.
What are the different types of electrophilic substitution reactions of benzene?
Answer:
i. Benzene shows electrophilic substitution reactions, in which one or more hydrogen atoms of benzene ring are replaced by groups like – Cl, – Br, – NO₂, – SO₃H, -R, -COR, etc.
ii. Different types of electrophilic substitution reactions of benzene are as follows:

• Halogenation (chlorination and bromination)
• Nitration
• Sulphonation
• Friedel-Craft’s alkylation and
• Friedel-Craft’s acylation

Question 142.
Write a short note on chlorination reaction of benzene.
Answer:
Chlorination of benzene:
i. In chlorination reaction, hydrogen atom of benzene is replaced by chlorine atom.
ii. Chlorine reacts with benzene in dark in the presence of iron or ferric chloride or anhydrous aluminium chloride or red phosphorus as catalyst to give chlorobenzene (C₆H₅Cl).

iii. Electrophile involved in the reaction: Cl⁺, chloronium ion,
Formation of the electrophile: Cl – Cl + FeCl₃ → Cl⁺ + [FeCl₄]^–

Question 143.
Write a short note on bromination reaction of benzene.
Answer:
Bromination of benzene:
i. In bromination reaction, hydrogen atom of benzene is replaced by bromine atom.
ii. Bromine reacts with benzene in dark in presence of iron or ferric bromide or anhydrous aluminium bromide or red phosphorus as catalyst to give bromobenzene (C₆H₅Br).

iii. Electrophile involved in the reaction: Br⁺
Formation of the electrophile: Br – Br + FeBr₃ → Br⁺ + [FeBr₄]^–

Question 144.
Why direct iodination of benzene is not possible?
Answer:
Direct iodination of benzene is not possible as the reaction is reversible.

[Note: Iodination of benzene can be carried out in the presence of oxidising agents like HIO₃ or HNO₃.]

Question 145.
How will you convert benzene to hexachlorobenzene?
Answer:
When benzene is treated with excess of chlorine in presence of anhydrous aluminium chloride, it gives hexachlorobenzene.

Question 146.
State true or false. Correct the false statement.
i. In halogenation reaction, hydrogen atom of benzene ring is replaced by halogen atom.
ii. The molecular formula of hexachlorobenzene is C₆H₆Cl₆.
iii. Benzene forms the lower layer when mixed with water.
Answer:
i. True
ii. False
The molecular formula of hexachlorobenzene is C₆Cl₆
iii. False
Benzene forms the upper layer when mixed with water.

Question 147.
Explain the nitration reaction of benzene.
Answer:
Nitration of benzene:
i. When benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture) at about 313 K to 333 K, it gives nitrobenzene.

ii. Electrophile involved in the reaction: \(\mathrm{NO}_{2}^{+}\), nitronium ion
Formation of the electrophile: HO – NO₂ + 2H₂SO₄ ⇌ \(2 \mathrm{HSO}_{4}^{-}\) + H₃O⁺ + \(\mathrm{NO}_{2}^{-}\)

Question 148.
Write a short note on sulphonation of benzene.
Answer:
Sulphonation of benzene:
i. When benzene is heated with fuming sulfuric acid (oleum) at 373 K, it gives benzene sulfonic acid.

ii. Electrophile involved in the reaction: SO₃, free sulphur trioxide
Formation of the electrophile: 2H₂SO₄ → H₃O⁺ + \(\mathrm{HSO}_{4}^{-}\) + SO₃

Question 149.
Write a short note on Friedel-Craft’s alkylation reaction of benzene.
Answer:
Friedel-Craft’s alkylation reaction of benzene:
i. When benzene is treated with an alkyl halide like methyl chloride in the presence of anhydrous aluminium chloride, it gives toluene.

ii. Electrophile involved in the reaction: R⁺
Formation of the electrophile: R – Cl + AlCl₃ → R⁺ + \(\mathrm{AlCl}_{4}^{-}\)
iii. Friedel-Craft’s alkylation reaction is used to extend the chain outside the benzene ring.

Question 150.
Explain Friedel-Craft’s acylation reaction of benzene. Give example reactions.
Answer:
Friedel-craft’s acylation reaction of benzene:
i. When benzene is heated with an acyl halide or acid anhydride in the presence of anhydrous aluminium chloride, it gives corresponding acyl benzene.

ii. Electrophile involved in the reaction: R – C- = O, acylium ion
Formation of the electrophile: R – COCl + AlCl₃ → R – C⁺ = O + \(\mathrm{AlCl}_{4}^{-}\)

Question 151.
Write the general combustion reaction for hydrocarbons.
Answer:
General combustion reaction for any hydrocarbon (C_xH_y) can be represented as follows:

Question 152.
Write the combustion reaction of benzene.
Answer:
When benzene is heated in air, it bums with sooty flame forming carbon dioxide and water.
C₆H₆ + \(\frac {15}{2}\)O₂ → 6CO₂ + 3H₂O

Question 153.
Write a note on the directive influence of substituents (functional groups) in monosubstituted benzene.
Answer:
i. In benzene, all hydrogen atoms are equivalent and so, when it undergoes electrophilic substitution reactions, only one monosubstituted product is possible.
Monosubstituted benzene:

ii. When monosubstituted benzene undergoes further electrophilic substitution, the second substituent (electrophile, E) can occupy any of the five positions available and give three disubstituted products.
But these disubstituted products are not formed in equal amounts.

iii. The position of second substituent (E) is determined by the nature of substituent (S) already present in the benzene ring and not on the nature of second substituent (E).
iv. The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups. The groups which direct the incoming group to meta positions are called meta directing groups. Thus, depending on the nature of the substituent (S) either ortho and para products or meta products are formed as major products.

Question 154.
What are ortho and para directing groups? Enlist few ortho and para directing groups.
Answer:
The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups.
Ortho and para directing groups:

Question 155.
Explain the directive influence of ortho, para directing groups in monosubstituted benzene using suitable example.
OR
Explain the directive influence of -OH group in benzene.
Answer:
i. The directive influence of ortho, para directing groups can be explained with the help of inductive and resonance effects.
ii. phenol has the following resonating structures:

iii. It can be seen from the above resonating structures, that the ortho (o-) and para (p-) positions have a greater electron density than the meta positions.
iv. Therefore, -OH group activates the benzene ring for the attack of second substituent (E) at these electron rich centres. Thus, phenolic -OH group is activating and ortho, para-directing group.
v. In phenol, -OH group has electron withdrawing inductive (-I) effect which slightly decreases the electron density at ortho positions in benzene ring. Thus, resonance effect and inductive effect of -OH group act opposite to each other. However, the strong resonance effect dominates over inductive effect.

Question 156.
Explain the o, p-directive effect of methyl group.
Answer:

• All ortho and para directing groups possess nonbonding electron pair on the atom which is directly attached to the aromatic ring; however, methyl group is an exception.
  Methyl (or alkyl groups) is ortho and para directing, although it has no nonbonding electron pair on the key atom. This is explained on the basis of special type of resonance called hyperconjugation or no bond resonance.

Question 157.
Explain why halide group is an ortho and para directing group.
Answer:
i. In aryl halides, halogens are moderately deactivating. Because of their strong -I effect, overall electron density on the benzene ring decreases, which makes the electrophilic substitution difficult.
ii. However, halogens are ortho and para directing. This can be explained by considering resonance structures.
iii. e.g. Chlorobenzene has the following resonating structures:

iv. Due to resonance, the electron density on ortho and para positions is greater than meta positions and hence, -Cl is ortho and para directing.

Question 158.
What are meta directing groups? Enlist few of them.
Answer:
The groups which direct the incoming group to meta positions are called meta directing groups.

[Note: All meta directing groups have positive (or partial positive) charge on the atom which is directly attached to an aromatic ring.]

Question 159.
Explain the directive influence of nitro group in nitrobenzene.
OR
Explain why nitro group is a meta-directing group.
Answer:
i. Meta directing group withdraws electrons from the aromatic ring by resonance, making the ring electron-deficient. Therefore, meta groups are ring deactivating groups.
ii. Due to -I effect, -NO₂ group reduces electron density in benzene ring on ortho and para positions. So, the attack of incoming group becomes difficult at ortho and para positions. The incoming group can attack on meta positions more easily.
iii. The various resonance structures of nitrobenzene are as shown below:

iv. It is clear from the above resonance structures that the ortho and para positions have comparatively less electron density than at meta positions. Hence, the incoming group/electrophile attacks on meta positions.

Question 160.
What are polycyclic aromatic compounds? How are they produced?
Answer:

• Polycyclic aromatic compounds are the hydrocarbons containing more than two benzene rings fused together.
• They are produced by incomplete combustion of tobacco, coal and petroleum.

Question 161.
Write the harmful effects of benzene.
Answer:

• Benzene is both, toxic and carcinogenic (cancer causing).
• In fact, it might be considered “the mother of all carcinogens” as a large number of carcinogens have structures that include benzene rings.
• In liver, benzene is oxidized to an epoxide and benzopyrene is converted into an epoxy diol. These substances are carcinogenic and can react with DNA and thus, can induce mutation leading to uncontrolled growth of cancer cells.

Multiple Choice Questions

1. Alkanes are represented by the general formula ………….
(A) C_nH_2n-2
(B) C_nH_2n+2
(C) C_nH_2n
(D) C_nH_n
Answer:
(B) C_nH_2n+2

2. Which of the following compound is alkanes?
(A) C₅H₁₀
(B) C₁₀H₂₂
(C) C₁₅H₂₈
(D) C₉H₁₆
Answer:
(B) C₁₀H₂₂

3. Alkanes are commonly called …………
(A) arenes
(B) paraffins
(C) olefins
(D) acetylenes
Answer:
(B) paraffins

4. Every carbon atom in alkanes is …………..
(A) sp hybridized
(B) sp² hybridized
(C) sp³ hybridized
(D) sp³d hybridized
Answer:
(C) sp³ hybridized

5. Isomerism is the phenomenon in which two or more organic compounds have ………….
(A) same molecular formula but different structural formula
(B) same structural formula but different molecular formula
(C) same general formula, but different structural formula
(D) same empirical formula, same structural formula
Answer:
(A) same molecular formula but different structural formula

6. Pentane exhibits …………. chain isomers.
(A) two
(B) three
(C) four
(D) five
Answer:
(B) three

7. Which of the following is NOT an isomer of hexane?
(A) 2-Methylpentane
(B) 2,2-Dimethylbutane
(C) 2,2-Dimethylpentane
(D) 3-Methylpentane
Answer:
(C) 2,2-Dimethylpentane

8. Alkanes can be prepared by ………… of unsaturated hydrocarbons.
(A) hydrogenation
(B) oxidation
(C) hydrolysis
(D) cracking
Answer:
(A) hydrogenation

9. Catalytic hydrogenation of ethene or acetylene gives …………..
(A) ethane
(B) propylene
(C) methane
(D) propane
Answer:
(A) ethane

10. Ethyl iodide when reduced by zinc and dilute HCl, leads to the formation of …………..
(A) Methane
(B) Ethane
(C) Ethylene
(D) Butane
Answer:
(B) Ethane

11. The reaction of alkyl halides with sodium in dry ether to give higher alkanes is called ………..
(A) Wurtz reaction
(B) Kolbe’s reaction
(C) Frankland’s reaction
(D) Williamson’s reaction
Answer:
(A) Wurtz reaction

12. Methane is ………… molecule.
(A) polar
(B) nonpolar
(C) highly polar
(D) none of these
Answer:
(B) nonpolar

13. Alkanes are ………… in water.
(A) soluble
(B) sparingly soluble
(C) insoluble
(D) none of these
Answer:
(C) insoluble

14. As branching increases, boiling point of alkanes ………….
(A) increases
(B) decreases
(C) remains same
(D) None of these
Answer:
(B) decreases

15. Halogenation of alkane is an example of …………. reaction.
(A) dehydration
(B) substitution
(C) addition
(D) elimination
Answer:
(B) substitution

16. Order of reactivity of halogens in halogenation of alkanes is ………….
(A) F₂ > Cl₂ > Br₂ > I₂
(B) I₂ > Br₂ > Cl₂ > F₂
(C) Br₂ < I₂ < F₂ < Cl₂
(D) Cl₂ < I₂ < Br₂ < F₂
Answer:
(A) F₂ > Cl₂ > Br₂ > I₂

17. The thermal decomposition of alkanes in absence of air to give lower alkanes, alkenes and hydrogen is called ………….
(A) vapour phase nitration
(B) pyrolysis
(C) polymerisation
(D) combustion
Answer:
(B) pyrolysis

18. But-1-ene and But-2-ene are …………
(A) chain isomers
(B) position isomers
(C) geometrical isomers
(D) metamers
Answer:
(B) position isomers

19. Hex-2-ene and 2-Methylpent-2-ene exhibit …………
(A) chain isomerism
(B) position isomerism
(C) geometrical isomerism
(D) optical isomerism
Answer:
(A) chain isomerism

20. Which of the following shows position isomerism?
(A) Propene
(B) Ethene
(C) 2-Methylpropene
(D) Pent-2-ene
Answer:
(D) Pent-2-ene

21. When identical atoms or group of atoms are attached to the two carbon atoms on the same side of the double bond, the isomer is called ………… isomer.
(A) cis
(B) trans
(C) position
(D) chain
Answer:
(A) cis

22. Which of the following does NOT exhibit geometrical isomers?
(A) But-2-ene
(B) Pent-2-ene
(C) But-1-ene
(D) Hex-2-ene
Answer:
(C) But-1-ene

23. When ethyl bromide is heated with alcoholic KOH, ………… is formed.
(A) ethane
(B) ethanol
(C) ethene
(D) acetylene
Answer:
(C) ethene

24. Alkenes are insoluble in …………
(A) benzene
(B) water
(C) ether
(D) chloroform
Answer:
(B) water

25. Markownikov’s rule is applicable to …………
(A) symmetrical alkenes
(B) alkanes
(C) unsymmetrical alkenes
(D) alkynes
Answer:
(C) unsymmetrical alkenes

26. When propene is treated with HBr in the dark and in absence of peroxide, then the main product formed is …………
(A) 1-bromopropane
(B) 2-bromopropane
(C) 1,2-dibromopropane
(D) 1,3-dibromopropane
Answer:
(B) 2-bromopropane

27. The product formed by the addition of HCl to propene in presence of peroxide is …………..

Answer:

28. Propene reacts with HBr in presence of peroxide, to form …………..
(A) 2-bromopropane
(B) 1-bromopropane
(C) 3-bromopropane
(D) 1,2-dibromopropane
Answer:
(B) 1-bromopropane

29. Markovnikov’s rule is applicable for …………..
(A) CH₂ = CH₂
(B) CH₃CH = CHCH₃
(C) CH₃CH₂CH = CHCH₂CH₃
(D) (CH₃)₂C = CH₂
Answer:
(D) (CH₃)₂C = CH₂

30. The addition of HCl in presence of peroxide does not follow anti-Markownikov’s rule because …………..
(A) HCl bond is too strong to be broken homolytically
(B) Cl atom is not reactive enough to add on to a double bond
(C) Cl atom combines with H atom to form HCl
(D) HCl is a reducing agent
Answer:
(A) HCl bond is too strong to be broken homolytically

31. An alkene on ozonolysis produces a mixture of acetaldehyde and acetone. Identify the alkene.
(A) But-1-ene
(B) But-2-ene
(C) 2-Methylbut-1-ene
(D) 2-Methylbut-2-ene
Answer:
(D) 2-Methylbut-2-ene

32. The ozonolysis of (CH₃)₂C = C(CH₃)₂ followed by treatment with zinc and water will give ……………
(A) acetone
(B) acetone and acetaldehyde
(C) formaldehyde and acetone
(D) acetaldehyde
Answer:
(A) acetone

33. The compound which forms only acetaldehyde on ozonolysis is …………..
(A) ethene
(B) propyne
(C) but-1-ene
(D) but-2-ene
Answer:
(D) but-2-ene

34. Treatment of ethylene with ozone followed by decomposition of the product with Zn/H₂O gives two moles of ………….
(A) formaldehyde
(B) acetaldehyde
(C) formic acid
(D) acetic acid
Answer:
(A) formaldehyde

35. Ozonolysis of 2,3-Dimethylbut-1-ene followed by reduction with zinc and water gives ………….
(A) methanoic acid and 3-methylbutan-2-one
(B) methanal and 2-methylbutan-2-one
(C) methanal and 3-methylbutan-2-one
(D) methanoic acid and 2-methylbutan-2-one
Answer:
(C) methanal and 3-methylbutan-2-one

36. The reaction, CH₂ = CH₂ + H₂O + [O]

is called ……………
(A) hydroxylation
(B) decarboxylation
(C) hydration
(D) dehydration
Answer:
(A) hydroxylation

37. An alkene on vigorous oxidation with KMnO₄ gives only acetic acid. The alkene is …………..
(A) CH₃CH₂CH = CH₂
(B) CH₃CH = CHCH₃
(C) (CH₃)₂C = CH₂
(D) CH₃CH = CH₂
Answer:
(B) CH₃CH = CHCH₃

38. Ethylene reacts with Baeyer’s reagent to give a/an ………….
(A) glycol
(B) aldehyde
(C) acid
(D) alcohol
Answer:
(A) glycol

39. Baeyer’s reagent is ………….
(A) aqueous KMnO₄
(B) neutral KMnO₄
(C) alkaline KMnO₄
(D) aqueous bromine water
Answer:
(C) alkaline KMnO₄

40. Alkynes have general formula ………….
(A) C_nH_2n-2
(B) C_nH_2n
(C) C_nH_2n+2
(D) C_nH_2n+1
Answer:
(A) C_nH_2n-2

41. Aliphatic unsaturated hydrocarbons containing two carbon-carbon triple bonds in their structure are called as ………….
(A) alkadiynes
(B) alkatriynes
(C) alkynes
(D) alkanes
Answer:
(A) alkadiynes

42. Acetylene is prepared in the industry by the action of water on ………….
(A) calcium carbonate
(B) calcium carbide
(C) mercuric chloride
(D) calcium oxide
Answer:
(B) calcium carbide

43. The dihalogen derivatives of alkanes when heated with …………. form corresponding alkynes.
(A) alcoholic water
(B) sodamide
(C) zinc
(D) acids
Answer:
(B) sodamide

44. Alkynes readily undergo …………. reaction.
(A) addition
(B) substitution
(C) elimination
(D) rearrangement
Answer:
(A) addition

45. Liquid bromine reacts with acetylene to form ………….
(A) 1,2-dibromoethene
(B) 1,1,2,2-tetrabromoethane
(C) 1,1-dibromoethene
(D) methyl chloride
Answer:
(B) 1,1,2,2-tetrabromoethane

46. When acetylene is passed through dil H₂SO₄ in the presence of 1% mercuric sulphate, the compound formed is ………….
(A) ethanol
(B) acetone
(C) acetaldehyde
(D) acetic acid
Answer:
(C) acetaldehyde

47. The compounds which contain at least one benzene ring are ………….
(A) aliphatic compounds
(B) aromatic compounds
(C) cycloalkanes
(D) both (A) and (B)
Answer:
(B) aromatic compounds

48. Which of the following compounds does NOT contain any benzene rings in their structure?
(A) Benzaldehyde
(B) Benzoic acid
(C) Naphthalene
(D) Furan
Answer:
(D) Furan

49. Benzene undergoes ………….
(A) only addition reaction
(B) only substitution reaction
(C) both addition and substitution reactions
(D) nucleophilic substitution reactions
Answer:
(C) both addition and substitution reactions

50. If the substituents are on the adjacent carbon atoms in the benzene ring, it is called ………….
(A) meta
(B) para
(C) ortho
(D) beta
Answer:
(C) ortho

51. How many molecules of acetylene are required to form benzene?
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

52. Which of the following compound on reduction gives benzene?
(A) Sodium benzoate
(B) Acetylene
(C) Cyclohexane
(D) Phenol
Answer:
(D) Phenol

53. X-Ray diffraction reveals that benzene is a …………. structure.
(A) triangular
(B) planar
(C) co-planar
(D) 3D
Answer:
(B) planar

54. γ-isomer of BHC is known as ………….
(A) gammene
(B) gammaxane
(C) chlorobenzene
(D) hexachlorobenzene
Answer:
(B) gammaxane

55. Benzene when treated with ozone forms ………….
(A) glyoxal
(B) acetic acid
(C) formaldehyde
(D) benzaldehyde
Answer:
(A) glyoxal

56. …………. is formed as intermediate product in ozonolysis of benzene.
(A) Benzaldehyde
(B) Phenol
(C) Benzene triozonide
(D) Cyclohexane
Answer:
(C) Benzene triozonide

57. Electrophile in chlorination of benzene is ………….
(A) Cl
(B) Cl⁺
(C) Cl^–
(D) Cl₂
Answer:
(B) Cl⁺

58. Benzene when treated with fuming. H₂SO₄ at 373 K forms ………….
(A) ethylbenzene
(B) toluene
(C) benzene sulphonic acid
(D) acetophenone sulphonic acid
Answer:
(C) benzene sulphonic acid

59. Ethyl chloride reacts with benzene in presence of anhydrous aluminium chloride to form ………….
(A) ethyl benzene
(B) chlorobenzene
(C) toluene
(D) acetophenone
Answer:
(A) ethyl benzene

60. The electrophile in Friedel-Craft’s alkylation reaction is ………….
(A) R⁺
(B) R^–
(C) Cl⁺
(D) RCO⁺
Answer:
(A) R⁺

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 10 Electrostatics Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 10 Electrostatics

Question 1.
Explain: Atoms are electrically neutral.
Answer:

1. Matter is made up of atoms which in turn consists of elementary particles proton, neutron and electron.
2. A proton is considered to be positively charged and electron to be negatively charged.
3. Neutron is electrically neutral i.e., it has no charge.
4. An atomic nucleus is made up of protons and neutrons and hence is positively charged.
5. Negatively charged electrons surround the nucleus so as to make an atom electrically neutral.

Question 2.
What does the below diagrams show?

Answer:

1. Figure (a) shows insulated conductor.
2. Figure (b) shows that positive charge is neutralized by electron from Earth.
3. Figure (c) shows that earthing is removed, negative charge still stays in conductor due to positive charged rod.
4. Figure (d) shows that when rod is removed, negative charge is distributed over the surface of the conductor.

Question 3.
Explain concept of charging by conduction.
Answer:

1. When certain dissimilar substances, like fur and amber or comb and dry hair, are rubbed against each other, electrons get transferred to the other substance making them charged.
2. The substance receiving electrons develops a negative charge while the other is left with an equal amount of positive charge.
3. This can be called charging by conduction as charges are transferred from one body to another.

Question 4.
Explain concept of charging by induction.
Answer:

1. If an uncharged conductor is brought near a charged body, (not in physical contact) the nearer side of the conductor develops opposite charge to that on the charged body and the far side of the conductor develops charge similar to that on the charged body. This is called induction.
2. This happens because the electrons in a conductor are free and can move easily in presence of charged body.
3. A charged body attracts or repels electrons in a conductor depending on whether the charge on the body is positive or negative respectively.
4. Positive and negative charges are redistributed and are accumulated at the ends of the conductor near and away from the charged body.
5. In induction, there is no transfer of charges between the charged body and the conductor. So when the charged body is moved away from the conductor, the charges in the conductor are free again.

Question 5.
Explain the concept of additive nature of charge.
Answer:

1. Electric charge is additive, similar to mass. The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on different parts of the object.
2. It may be pointed out that while taking the algebraic sum, the sign (positive or negative) of the electric charges must be taken into account.
3. Thus, if two bodies have equal and opposite charges, the net charge on the system of the two bodies is zero.
4. This is similar to that in case of atoms where the nucleus is positively charged and this charge is equal to the negative charge of the electrons making the atoms electrically neutral.

Question 6.
State the analogy between the additive property of charge with that of mass.
Answer:

1. The masses of the particles constituting an object are always positive, whereas the charges distributed on different parts of the object may be positive or negative.
2. The total mass of an object is always positive whereas, the total charge on the object may be positive, zero or negative.

Question 7.
What is quantization of charge?
Answer:

1. Protons (+ve) and electrons (-ve) are the charged particles constituting matter, hence the charge on an object must be an integral multiple of ± e i.e., q = ± ne, where n is an integer.
2. Charge on an object can be increased or decreased in multiples of e.
3. It is because, during the charging process an integral number of electrons can be transferred from one body to the other body. This is known as quantization of charge or discrete nature of charge.

Question 8.
Explain with an example why quantization of charge is not observed practically.
Answer:
i. The magnitude of the elementary electric charge (e), is extremely small. Due to this, the number of elementary charges involved in charging an object becomes extremely large.

ii. For example, when a glass rod is rubbed with silk, a charge of the order of one µC (10^-6 C) appears on the glass rod or silk. Since elementary charge e = 1.6 × 10^-19 C. the number of elementary charges on the glass rod (or silk) is given by
n = \(\frac {10^{-6}C}{1.6×10^{-19}C}\) = 6.25 × 10¹²
Since, it is tremendously large number, the quantization of charge is not observed and one usually observes a continuous variation of charge.

Question 9.
The total charge of an isolated system is always conserved. Explain with an example.
Answer:

1. When a glass rod is rubbed with silk, it becomes positively charged and silk becomes negatively charged.
2. The amount of positive charge on glass rod is found to be exactly the same as negative charge on silk.
3. Thus, the systems of glass rod and silk together possesses zero net charge after rubbing.
   Hence, the total charge of an isolated system is always conserved.

Question 10.
Explain the conclusion when charges are brought close to each other.
Answer:

1. Unlike charges attract each other.
2. Like charges repel each other.

Question 11.
How much positive and negative charge is present in 1 g of water? How many electrons are present in it?
(Given: molecular mass of water is 18.0 g)
Answer:
Molecular mass of water is 18 gram, that means the number of molecules in 18 gram of water is 6.02 × 10²³
∴ Number of molecules in lgm of water = \(\frac {6.02×10^{23}}{18}\)
One molecule of water (H₂O) contains two hydrogen atoms and one oxygen atom. Thus, the number of electrons in ILO is sum of the number of electrons in H₂ and oxygen. There are 2 electrons in H₂ and 8 electrons in oxygen.
∴ Number of electrons in H₂O = 2 + 8 = 10
Total number of protons / electrons in one gram of water
= \(\frac {6.02×10^{23}}{18}\) × 10 = 3.344 × 10²³
Total positive charge
= 3.344 × 10²³ × charge on a proton
= 3.344 × 10²³ × 1.6 × 10^-19C
= 5.35 × 10⁴ C
This positive charge is balanced by equal amount of negative charge so that the water molecule is electrically neutral.
∴ Total negative charge = 5.35 × 10⁴C

Question 12.
Define point charge. Which law explains the interaction between charges at rest?
Answer:

1. A point charge is a charge whose dimensions are negligibly small compared to its distance from another bodies.
2. Coulomb’s law explains the interaction between charges at rest.

Question 13.
State and explain Coulomb’s law of electric charge in scalar form.
Answer:
Coulomb’s law:
The force of attraction or repulsion between two point charges at rest is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them. This force acts along the line joining the two charges.
Explanation:
i. Let q₁ and q₂ be the two point charges at rest with each other and separated by a distance r. F is the magnitude of electrostatic force of attraction or repulsion between them.

ii. According to Coulomb’s law.
F ∝ \(\frac {q_1q_2}{r^2}\)
∴ F = K\(\frac {q_1q_2}{r^2}\)
where, K is the constant of proportionality which depends upon the units of F, q₁, q₂, r and medium in which charges are placed.

Question 14.
State conditions for electrostatic force to be attractive or repulsive.
Answer:

1. The force between the two charges will be attractive, if the charges are unlike (one positive and one negative).
2. The force between the two charges will be repulsive, if the charges are similar (both positive or both negative).

Question 15.
Prove that relative permittivity is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.
Answer:
i. The force between the two charges placed in a medium is given by,
F_med = \(\frac {1}{4πε}\) (\(\frac {q_1q_2}{r^2}\)) …………. (1)
where, ε is called the absolute permittivity of the medium.

ii. The force between the same two charges placed in free space or vacuum at distance r is given by,
F_vac = \(\frac {1}{4πε_0}\) (\(\frac {q_1q_2}{r^2}\)) …………. (1)
Dividing equation (2) by equation (1),

Hence, relative permittivity is the ratio of the force between two point charges placed a certain distance apart in free space or vacuum to the force between the same two point charges when placed at the same distance in the given medium.

Question 16.
If relative permittivity of water is 80 then derive the relation between F_water and F_vacuum. What can be concluded from it?
Answer:
i. The force between two point charges q₁ and q₂ placed at a distance r in a medium of relative permittivity ε_r, is given by
F_med = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (1)
If the medium is vacuum,
F_vac = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) …………. (2)

ii. Dividing equation (2) by equation (1),
\(\frac{\mathrm{F}_{\mathrm{vac}}}{\mathrm{F}_{\text {med }}}\) = ε_r
For water, ε_r = 80 ……….. (given)
∴ F_water = \(\frac {F_{vac}}{80}\)

iii. This means that when two point charges are placed some distance apart in water, the force between them is reduced to (\(\frac {1}{80}\))^th of the force between the same two charges placed at the same distance in vacuum.

iv. Thus, it is concluded that a material medium reduces the force between charges by a factor of er, its relative permittivity.

Question 17.
Give conversions of micro-coulomb, nano-coulomb and pico-coulomb to coulomb.
Answer:
1 microcoulomb (µC) = 10^-6 C
1 nanocoulomb (nC) = 10^-9 C
1 picocoulomb (pC) = 10^-12 C

Question 18.
Explain Coulomb’s law in vector form.
Answer:
i. Let q₁ and q₂ be the two similar point charges situated at points A and B and let \(\vec{r}\)₁₂ be the distance of separation between them.

ii. The force \(\vec{F}\)₂₁ exerted on q₂ by q₁ is given by,
\(\overrightarrow{\mathrm{F}}_{21}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{12}\right|^{2}} \times \hat{\mathrm{r}}_{12}\)
where, \(\hat{r}\)₁₂ is the unit vector from A to B.
\(\vec{F}\)₂₁ acts on q₂ at B and is directed along BA, away from B.

iii. Similarly, the force \(\vec{F}\)₁₂ exerted on q₁ by q₂ is given by, \(\vec{F}\)₁₂ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{12}\right|^{2}} \times \hat{\mathrm{r}_{21}}\)
where, \(\hat{r}\)₂₁ is the unit vector from B to A. \(\vec{F}\)₁₂ acts on q₁ at A and is directed along BA, away from A.

iv. The unit vectors \(\hat{r}\)₁₂ and \(\hat{r}\)₂₁ are oppositely directed i.e., \(\hat{r}\)₁₂ = –\(\hat{r}\)₂₁
Hence, \(\vec{F}\)₂₁ = –\(\vec{F}\)₁₂
Thus, the two charges experience force of equal magnitude and opposite in direction.

v. These two forces form an action-reaction pair.

vi. As \(\vec{F}\)₂₁ and \(\vec{F}\)₁₂ act along the line joining the two charges, the electrostatic force is a central force.

Question 19.
State similarities and differences of gravitational and electrostatic forces.
Answer:
i. Similarities:
a. Both forces obey inverse square law:
F ∝ \(\frac{1}{r^2}\)
b. Both are central forces and they act along the line joining the two objects.

ii. Differences:
a. Gravitational force between two objects is always attractive while electrostatic force between two charges can be either attractive or repulsive depending on the nature of charges.
b. Gravitational force is about 36 orders of magnitude weaker than the electrostatic force.

Question 20.
Charge on an electron is 1.6 × 10^-19 C. How many electrons are required to accumulate a charge of one coulomb?
Answer:
1 electron = 1.6 × 10^-19 C
∴ 1 C = \(\frac{1}{1.6 \times 10^{-19}}\) electrons
= 0.625 × 10¹⁹ electrons
……. (Taking reciprocal from log table)
= 6.25 × 10¹⁸ electrons
Hence, 6.25 × 10¹⁸ electrons are required to accumulate a charge of one coulomb.

Question 21.
What is the force between two small charge spheres having charges of 2 × 10^-7 C and 3 × 10^-7 C placed 30 cm apart in air?
Answer:
Given: q₁ = 2 × 10^-7 C, q₂ = 3 × 10^-7 C
r = 30 cm = \(\frac {30}{100}\) m = 0.3 m
To find: Force (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation: From formula,
F = \(\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}\)
∴ F = 6 × 10^-3 N

Question 22.
The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge -0.8 µC in air is 0.2 N. (i) What is the distance between the two spheres? (ii) What is the force on the second sphere due to the first?
Answer:
i. Given: q₁ = 0.4 µC = 0.4 × 10^-6 C,
q₂ = -0.8 µC = -0.8 × 10^-6 C, F = 0.2 N
To find: i. Distance (r)
ii. Force on second sphere (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:
i. From formula,
r² = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{F}\)
r² = \(\frac{9 \times 10^{9} \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2}\)
= 0.0144
∴ r = \(\sqrt{0.0144}\) = 0.12 m
∴ r = 12 cm

ii. The force on the second sphere due to the first is also 0.2 N and is attractive in nature.

Question 23.
i. Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion, if the charge on each is 6.5 × 10^-7 C? The radii of A and B are negligible compared to the distance of separation,
ii. What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
Answer:
Given: q₁ = 6.5 × 10^-7 C q₂ = 6.5 × 10^-7 C
r = 50 cm = 0.50 m
To find: Force of repulsion (F)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:
From formula,
F = \(\frac{9 \times 10^{9} \times 6.5 \times 10^{-7} \times 6.5 \times 10^{-7}}{(0.50)^{2}}\)
F = 1.52 × 10^-2 N

ii. When each charge is doubled and the distance between them is reduced to half, then
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(2 q_{1}\right)\left(2 q_{2}\right)}{(r / 2)^{2}}\)
= 16 × \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\) = 16 × 1.52 × 10^-2
∴ F = 0.24 N

Question 24.
Calculate and compare the electrostatic and gravitational forces between two protons which are 10^-15 m apart. Value of G = 6.674 × 10^-11 m³ kg^-1 s^-2 and mass of the porton is 1.67 × 10^-27 kg.
Answer:
Given: G = 6.674 × 10^-11 m³ kg^-1 s^-2
m_p = 1.67 × 10^-27 kg.
q_p = 1.67 × 10^-19 C, r = 10^-15
To find:
i. Electrostatic Force (F_E)
ii. Gravitational Force (F_G)
Formula: i. F_E = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
ii. F_E = \(\frac {Gm_1m_2}{r^2}\)
Calculation:
From formula (i),
\(\mathrm{F}_{\mathrm{E}}=9 \times 10^{9} \times \frac{1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(10^{-15}\right)^{2}}\)
= 9 × 1.6 × 1.6 × 10
= 90 × 1.6 × 1.6
= antilog [log 90 + log 1.6 + log 1.6]
= antilog [1.9542 + 0.2041 + 0.2041]
= antilog [2.3624]
= 2.303 × 10² N
From formula (ii),
\(\mathrm{F}_{\mathrm{G}}=6.674 \times 10^{-11} \times \frac{1.67 \times 10^{-27} \times 1.67 \times 10^{-27}}{\left(10^{-15}\right)^{2}}\)
= 6.674 × 1.67 × 1.67 × 10^-35
= {antilog [log 6.674 + log 1.67 + log 1.67]} × 10^-35
= {antilog [0.8244 + 0.2227 + 0.2227]} × 10^-35
= {antilog [1.2698]} × 10^-35
= 1.861 × 10¹ × 10^-35
= 1.861 × 10^-34 N
Now,
\(\frac{\mathrm{F}_{\mathrm{E}}}{\mathrm{F}_{\mathrm{G}}}=\frac{2.303 \times 10^{2}}{1.861 \times 10^{-34}}\)
= {antilog [log 2.303 – log 1.861]} × 10³⁶
= {antilog [0.3623 – 0.2697]} × 10³⁶
= {antilog [0.0926]}
= 1.238 × 10³⁶
∴ F_E ≈ 10³⁶ × F_G

Question 25.
State and explain principle of superposition.
Answer:
Statement: When a number of charges are interacting, the resultant force on a particular charge is given by the vector sum of the forces exerted by individual charges.
Explanation:
i. Consider a number of point charges q₁, q₂, q₃ ……………… kept at points A₁, A₂, A₃ ………….. as shown in figure.

ii. The force exerted on the charge q₁ by q₂ is \(\vec{F}\)₁₂ The value of \(\vec{F}\)₁₂ is calculated by ignoring the presence of other charges. Similarly, force \(\vec{F}\)₁₃, \(\vec{F}\)₁₄ can be found, using the Coulomb’s law.

iii. Total force \(\vec{F}\)₁ on charge qi is the vector sum of all such forces.
\(\vec{F}\)₁ = \(\vec{F}\)₁₂ + \(\vec{F}\)₁₃ + \(\vec{F}\)₁₄ + …………..
\( =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\left|\mathrm{r}_{21}\right|^{2}} \times \hat{\mathrm{r}}_{21}+\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{\left|\mathrm{r}_{31}\right|^{2}} \times \hat{\mathrm{r}}_{31}+\ldots .\right]\)

where \(\hat{r}\)₂₁, \(\hat{r}\)₃₁ are unit vectors directed to q₁ from q₂, q₃ respectively and r₂₁, r₃₁, r₄₁ are the distances from q₁ to q₂, q₃ respectively.

iv. If q₁, q₂, q₃ ……., q_n are the point charges then the force \(\vec{F}\) exerted by these charges on a test charge q₀ is given by,
\(\vec{F}\)_test = \(\vec{F}\)₁ = \(\vec{F}\)₂ + \(\vec{F}\)₃ + …. + \(\vec{F}\)_n
= \(\sum_{\mathrm{n}=1}^{\mathrm{n}} \mathrm{F}_{\mathrm{n}}=\frac{1}{4 \pi \varepsilon_{0}} \sum_{\mathrm{n}=1}^{\mathrm{n}} \frac{\mathrm{q}_{0} \mathrm{q}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}_{\mathrm{n}}\)
Where, \(\hat{r}\)_n, is a unit vector directed from the nth charge to the test charge q₀ and r₂ is the
separation between them, \(\vec{r}\)_n = r_n \(\hat{r}\)_n

Question 26.
Three charges of 2 µC, 3 µC and 4 µC are placed at points A, B and C respectively, as shown in the figure. Determine the force on A due to other charges.
(Given: AB = 4 cm, BC = 3 cm)

Answer:

Using pythagoras theorem
AC = \(\sqrt {AB^2+BC^2}\)
= \(\sqrt {4^2+3^2}\)
AC = 5 cm
Magnitude of force \(\vec{F}\)_AB on A due to B is,

In ∆ABC 4
cos θ = \(\frac {4}{5}\)
θ = cos^-1 (\(\frac {4}{5}\)) = 36.87°
Forces acting points A are

= 59.36 N
Direction of resultant force is 36.87° (north of west)

[Note: The question given above is modified considering minimum requirement of data needed to solve the problem.]

Question 27.
There are three charges of magnitude 3 pC, 2 pC and 3 pC located at three corners A, B and C of a square ABCD of each side measuring 2 m. Determine the net force on 2 pC charge.
Answer:
Given: q₁ = 3 µC, q₂ = 2 µC, q₃ = 3 µC, r = 2 m
To find: Net force on q₂ (R)
Formula: F = \(\frac {1}{4πε_0}\) \(\frac {q_1q_2}{r^2}\)
Calculation:

From the formula,
Force on q₂ because of q₁

Net force on q₂ is the resultant force of \(\vec{F}\)₂₁ and \(\vec{F}\)₂₃ which is given by,
R = \(\sqrt{\mathrm{F}_{21}^{2}+\mathrm{F}_{23}^{2}}\)
= \(\sqrt{\left(1.35 \times 10^{-2}\right)^{2}+\left(1.35 \times 10^{-2}\right)^{2}}\)
∴ R = 1.91 × 10^-2 N

Question 28.
Explain the concept of electric field.
Answer:

1. The space around a charge gets modified when a test charge is brought in that region, it experiences a coulomb force. The region around a charged object in which coulomb force is experienced by another charge is called electric field.
2. Mathematically, electric field is defined as the force experienced per unit charge.
3. The coulomb force acts across an empty space (vacuum) and does not need any intervening medium for its transmission.
4. The electric field exists around a charge irrespective of the presence of other charges.
5. Since the coulomb force is a vector, the electric field of a charge is also a vector and is directed along the direction of the coulomb force, experienced by a test charge.

Question 29.
Define electric field. State its SI unit and dimensions.
Answer:

1. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
   \(\vec{E}\) = \(\lim _{q \rightarrow 0} \frac{\vec{F}}{q}\)
2. SI unit: newton per coulomb (N/C) or volt per metre (V/m).
3. Dimensions: [L M T^-3 A^-1]

Question 30.
Establish relation between electric field intensity and electrostatic force.
Answer:
i. Let Q and q be two charges separated by a distance r.
The coulomb force between them is given by \(\vec{F}\) = \(\frac {1}{4πε_0}\) \(\frac {Qq}{r^2}\) \(\hat{r}\)
where, \(\hat{r}\) is the unit vector along the line joining Q to q.

ii. Therefore, electric field due to charge Q is given \(\vec{F}\) = \(\frac{\vec{F}}{\mathrm{q}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)

iii. Electric field at a point is useful to estimate the force experienced by a charge at that point.

Question 31.
State an expression for electric field on the surface of the sphere due to a positive point charge placed at its centre.
Answer:
The magnitude of electric field at a distance r from a point charge Q is same at all points on the surface of a sphere of radius r as shown in figure.

ii. Magnitude of electric field is given by,
E = \(\frac {1}{4πε_0}\) \(\frac {Q}{r^2}\)
iii. Its direction is along the radius of the sphere, pointing away from its centre if the charge is positive.

Question 32.
Derive expression for electric field intensity due to a point charge in a material medium.
Answer:
i. Consider a point charge q placed at point O in a medium of dielectric constant K as shown in figure.

ii. Consider the point P in the electric field of point charge at distance r from q. A test charge q0 placed at the point P will experience a force which is given by the Coulomb’s law,
\(\vec{F}\) = \(\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{qq}_{0}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)
where \(\hat{r}\) is the unit vector in the direction of force i.e., along OP.

iii. By the definition of electric field intensity,
\(\vec{F}\) = \(\frac{\vec{F}}{\mathrm{q}_{0}}=\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{q}}{\mathrm{r}^{2}} \hat{\mathrm{r}}\)
The direction of \(\vec{E}\) will be along OP when q is positive and along PO when q is negative.

iv. The magnitude of electric field intensity in a medium is given by, E = \(\frac{1}{4 \pi \varepsilon_{0} \mathrm{~K}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)

v. For air or vacuum, K = 1 then
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)

Question 33.
Show graphical representation of variation of coulomb force and electric field due to point charge with distance.
Answer:
Electrostatic force: F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\)
Electric field: E : \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
The coulomb force (F) between two charges and electric field (E) due to a charge both follow the inverse square law.
(F ∝ 1/r², E ∝ 1/r²)

Question 34.
What is non-uniform electric field?
Answer:
A field whose magnitude and direction is not the same at all points.
For example, field due to a point charge. In this case, the magnitude of field is same at distance r from the point charge in any direction but the direction of the field is not same.

Question 35.
Derive relation between electric field (E) and electric potential (V).
Answer:
i. A pair of parallel plates is connected as shown in the figure. The electric field between them is uniform

ii. A potential difference V is applied between two parallel plates separated by a distance ‘d’.

iii. The electric field between them is directed from plate A to plate B.

iv. A charge +q placed between the plates experiences a force F due to the electric field.

v. If the charge is moved against the direction of field, i.e., towards the positive plate, some amount of work is done on it.

vi. If the charge is moved +q from the negative plate B to the positive plate A, then the work done against the field is W = Fd; where ‘d’ is the separation between the plates.

vii. The potential difference V between the two plates is given by W = V_q,
but W = Fd
∴ Vq = Fd
∴ \(\frac {F}{q}\) = \(\frac {V}{d}\) = E
∴ Electric field can be defined as E = V/d.

Question 36.
What are electric lines of force?
Answer:
i. An electric line of force is an imaginary curve drawn in such a way that the tangent at any given point on this curve gives the direction of the electric field at that point.

ii. If a test charge is placed in an electric field it would be acted upon by a force at every point in the field and will move along a path.

iii. The path along which the unit positive charge moves is called a line of force.

iv. A line of force is defined as a curve such that the tangent at any point to this curve gives the direction of the electric field at that point.

v. The density of field lines indicates the strength of electric fields at the given point in space.

Question 37.
State the characteristics of electric lines of force.
Answer:

1. The lines of force originate from a positively charged object and end on a negatively charged object.
2. The lines of force neither intersect nor meet each other, as it will mean that electric field has two directions at a single point.
3. The lines of force leave or terminate on a conductor normally.
4. The lines of force do not pass through conductor i.e., electric field inside a conductor is always zero, but they pass through insulators.
5. Magnitude of the electric field intensity is proportional to the number of lines of force per unit area of the surface held perpendicular to the field.
6. Electric lines of force are crowded in a region where electric intensity is large.
7. Electric lines of force are widely separated from each other in a region where electric intensity is small
8. The lines of force of an uniform electric field are parallel to each other and are equally spaced.

Question 38.
Find the distance from a charge of 4 µC placed in air which produces electric field of intensity 9 × 10³ N/C.
Answer:
Given: K = 1, E = 9 × 10³ N/C
q = 4 µC = 4 × 10^-6
To Find: Distance (r)
Formula: E = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q}{r^{2}}\)
Calculation from formula
9 × 10³ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{4 \times 10^{-6}}{r^{2}}\)
∴ 9 × 10³ = 9 × 10⁹ \(\frac{4 \times 10^{-6}}{\mathrm{r}^{2}}\)
∴ r² = 4
∴ r = 2 m

Question 39.
What is the magnitude of a point charge chosen so that the electric field 50 cm away has magnitude 2.0 N/C?
Answer:
Given: r = 50 cm – 0.5 m, E = 2 N/C,
To find: Magnitude of charge (q)
Formula: E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}\)
Calculation from formula

Question 40.
Three point charges are placed at the vertices of a right angled isosceles triangle as shown in the given figure. What is the magnitude and direction of the resultant electric field at point P which is the mid point of its hypotenuse?

Answer:
Electric field at P due to the charges at A, B and C are shown in the figure.

Let \(\vec{E}\)_A be the field at P due to charge at A and \(\vec{E}\)_c be the field at P due to charge at C.
Since P is the midpoint of AC and the fields at A and C are equal in magnitudes and are opposite in direction, E_A = – E_C .
i.e., \(\vec{E}\)_A + \(\vec{E}\)_C = 0.
Thus, the field at P is only to the charge at B and is given by,

Question 41.
A simplified model of hydrogen atom consists of an electron revolving about a proton at a distance of 5.3 × 10^-11 m. The charge on a proton is +1.6 × 10^-19 C. Calculate the intensity of the electric field due to proton at this distance. Also find the force between electron and proton.
Answer:
Given: r = 5.3 × 10^-11 m
q = 1.6 × 10^-19 C
To Find: i. Intensity of electric field (E)
ii. Force (F)
Formula: i. E = \(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
ii. E = \(\frac {F}{q}\)
Calculation from formula (i)
E = 9 × 10⁹ × \(\frac{1.6 \times 10^{-19}}{\left(5.3 \times 10^{-11}\right)^{2}}\)
= 5.126 × 10¹¹ N/C
Force between electron and proton,
Force between electron and proton,
F = E × q_e ….[From formula (ii)]
= 5.126 × 10^-11 × -1.6 × 10^-19
= -8.201 × 10⁸ N

Question 42.
The force exerted by an electric field on a charge of +10 µC at a point is 16 × 10^-4 N. What is the intensity of the electric field at the point?
Answer:
Given: q = 10 µC= 10 × 10^-6 C, F = 16 × 10^-4 N
To find: Electric field intensity (E)
Formula: E = \(\frac {F}{q}\)
Calculation: From formula,
E = \(\frac {16×10^{-4}}{10×10^{-6}}\) = 160 N/C

Question 43.
What is the force experienced by a test charge of 0.20 µC placed in an electric field of 3.2 × 10⁶ N/C?
Answer:
Given: q₀ = 0.20 µC = 0.2 × 10⁶ C,
E = 3.2 × 10⁶ N/C
To find: Force (F)
Formula: E = \(\frac {F}{q_0}\)
Calculation: From formula,
F = Eq₀
∴ F = 3.2 × 10⁶ × 0.2 × 10^-6 = 0.64 N

Question 44.
Gap between two electrodes of the spark-plug used in an automobile engine is 1.25 mm. If the potential of 20 V is applied across the gap, what will be the magnitude of electric field between the electrodes?
Answer:
Given: V = 20 V
d = 1.25 mm = 1.25 × 10^-3 m
To Find: Magnitude of electric field (E)
Formula: E = \(\frac {V}{d}\)
Calculation: From formula,
E = \(\frac {20}{1.25×10^{-3}}\)
= 1.6 × 10⁴ V/m

Question 45.
If 100 joules of work must be done to move electric charge equal to 4 C from a place, where potential is -10 volt to another place where potential is V volt, find the value of V.
Answer:
Given: q₀ = 4 C,
V_A = -10 volt,
V_B = V volt,
W_AB = 100 J
To Find: Potential (V)
Formula: V_B – V_A = \(\frac {W_{AB}}{q_0}\)
Calculation: From formula,
V – (-10) = \(\frac {100}{4}\) = 25
∴ V + 10 = 25
∴ V = 15 volt

Question 46.
Find the work done when a point charge of 2.0 pC is moved from a point at a potential of -10 V to a point at which the potential is zero.
Answer:
V_A = -10V,
V_B = 0,
q = 2 × 10^-6 C
To Find: Work done (W)
Formula: V_BA = \(\frac {W}{q}\)
Calculation: From formula,
W = V_BA × q
= (V_B – V_A) × q
= (0 + 10) × 2 × 10^-6
= 20 × 10^-6 J
∴ W = 2 × 10^-5 J

Question 47.
Explain the term: Electric flux
Answer:
i. The number of lines of force per unit area is the intensity of the electric field \(\vec{E}\).

ii. When the area is inclined at an angle θ with the direction of electric field, the electric flux can be calculated as follows.
Let the angle between electric field \(\vec{E}\), and area vector \(\vec{dS}\) be θ, then the electric flux passing through are dS is given by
dø = (component of dS along \(\vec{E}\)) × (area of \(\vec{dS}\))
dø = EdS cos θ
dø = \(\vec{E}\) .\(\vec{dS}\)
Total flux through the entire surface .
ø = ∫dø = \( \int_{S} \vec{E} \cdot d \vec{S}=\vec{E} \cdot \vec{S}\)

iii. The SI unit of electric flux can be calculated using,
ø = \(\vec{E}\). \(\vec{S}\) = (V/m) m² = V m
[Note: Area vector is a vector whose magnitude is equal to area and is directed normal to its surface]

Question 48.
The electric flux through a plane surface of area 200 cm² in a region of uniform electric field 20 N/C is 0.2 N m²/C. Find the angle between electric field and normal to the surface.
Answer:
Given: ds = 200 cm² = 2 × 10^-2 m², E = 20 N/C,
ø = 0.2 N m²/C
To find: Angle between electric field and normal (θ)
Formula: ø = Eds cos θ
Calculation:
From formula,
cos θ = \(\frac {ø}{Eds}\) = \(\frac {0.2}{20×2×10^{-2}}\) = \(\frac {1}{2}\)
∴ θ = cos^-1 (\(\frac {1}{2}\))
∴ θ = 60°

Question 49.
A charge of 5.0 C is kept at the centre of a sphere of radius 1 m. What is the flux passing through the sphere? How will this value change if the radius of the sphere is doubled?
Answer:
Given: q = 5C, r = 1 m
To find: Flux (ø)
Formulae: i. E = \(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{\mathrm{q}}{\mathrm{r}^{2}}\)
ii. ø = E × A = E (4πr²)
Calculation: From formula (i),
E = 9 × 10⁹ × \(\frac {5}{1^2}\)
= 4.5 × 10¹⁰ N/C
From formula (ii),
ø = E × 4 π r²
= 4.5 × 10¹⁰ × 4 × 3.14 × 1²
ø = 5.65 × 10¹¹ Vm
This value of flux will not change if radius of sphere is doubled. Though radius of sphere will increase, increased distance will reduce the electric field intensity. As E ∝ \(\frac {1}{r^2}\) and A × r² net variation in total flux will not be observed.

Question 50.
State and prove Gauss’ law of electrostatics.
Answer:
Statement:
The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by Eo.
\(\int \vec{E} \cdot \overrightarrow{\mathrm{dS}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)
where Q is the total charge within the surface.

Proof:
i. Consider a closed surface of any shape which encloses number of positive electric charges.

ii. Imagine a small charge +q present at a point O inside closed surface. Imagine an infinitesimal area dS of the given irregular closed surface.

iii. The magnitude of electric field intensity at point P on dS due to charge +q at point O is, E = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}}{\mathrm{r}^{2}}\right)\) ………… (1)

iv. The direction of E is away from point O. Let θ be the angle subtended by normal drawn to area dS and the direction of E

v. Electric flux passing through area (dø)
= Ecosθ dS
= \(\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\) cosθ dS ………….. (from 1)
= \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\right)\left(\frac{\mathrm{d} \mathrm{S} \cos \theta}{\mathrm{r}^{2}}\right)\)
But, dω = \(\frac {dS cos θ}{r^2}\)
where, dco is the solid angle subtended by area dS at a point O.
∴ dø = \(\left(\frac{\mathrm{q}}{4 \pi \varepsilon_{0}}\right)\) dω …………. (2)

vi. Total electric flux crossing the given closed surface can be obtained by integrating equation (2) over the total area.
\(\phi_{\mathrm{E}}=\int_{\mathrm{s}} \mathrm{d} \phi=\int_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\int \frac{\mathrm{q}}{4 \pi \varepsilon_{0}} \mathrm{~d} \omega=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}} \int \mathrm{d} \omega\)

vii. But ∫dω = 4π = solid angle subtended by entire closed surface at point O.
Total Flux = \(\frac {q}{4πε_0}\) (4π)
∴ ø_E = \(\int_{\mathrm{s}} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=\frac{+\mathrm{q}}{\varepsilon_{0}}\)

viii. This is true for every electric charge enclosed by a given closed surface.
Total flux due to charge q₁, over the given closed surface = + \(\frac {q_1}{ε_0}\)
Total flux due to charge q₂, over the given closed surface = + \(\frac {q_2}{ε_0}\)
Total flux due to charge q_n, over the given closed surface = +\(\frac {q_n}{ε_0}\)

ix. According to the superposition principle, the total flux c|> due to all charges enclosed within the given closed surface is
\(\phi_{\mathrm{E}}=\frac{\mathrm{q}_{1}}{\varepsilon_{0}}+\frac{\mathrm{q}_{2}}{\varepsilon_{0}}+\frac{\mathrm{q}_{3}}{\varepsilon_{0}}+\ldots+\frac{\mathrm{q}_{\mathrm{n}}}{\varepsilon_{0}}=\sum_{\mathrm{i}=1}^{\mathrm{i}=\mathrm{n}} \frac{\mathrm{q}_{\mathrm{i}}}{\varepsilon_{0}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\)

Question 51.
With a help of diagram, state the direction of flux due to positive charge, negative charge and charge outside a closed surface.
Answer:

Positive sign indicates that the flux is directed outwards, away from the charge.

If the charge is negative, the flux will be is directed inwards.

If a charge is outside the closed surface, the net flux through it will be zero.

Question 52.
Explain: Electric flux is independent of shape and size of closed surface.
Answer:
i. The net flux crossing an enclosed surface is equal to \(\frac {q}{ε_0}\) where q is the net charge inside the closed surface.

ii. Consider a charge +q at the centre of concentric circles as shown in figure below.

As the charge inside the sphere is unchanged, the flux passing through a sphere of any radius is the same.

iii. Thus, if the radius of the sphere is increased by a factor of 2, the flux passing through is surface remains unchanged.

iv. As shown in figure same number of lines of force cross both the surfaces.
Hence, total flux is independent of shape of the closed surface radius of the sphere and size of closed surface.

Question 53.
Define the following terms with the help of a diagram.
i. Electric dipole
ii. Dipole axis
iii. Axial line
iv. Equatorial line
Answer:
i. Electric dipole: A pair of equal and opposite charges separated by a finite distance is called an electric dipole.

ii. Dipole axis: Line joining the two charges is called the dipole axis.

iii. Axial line: A line passing through the dipole axis is called axial line.

iv. Equatorial line: A line passing through the centre of the dipole and perpendicular to the axial line is called the equatorial line.

AB : Electric dipole Line joining
AB: Dipole axis
X-Y : Axial line
P-Q : Equatorial line

Question 54.
What are polar molecules? Explain with examples.
Answer:

1. Polar molecules are the molecules in which the centre of positive charge and the negative charge is naturally separated.
2. Molecules of water, ammonia, sulphur dioxide, sodium chloride etc. have an inherent separation of centres of positive and negative charges. Such molecules are called polar molecules.

Question 55.
What are non-polar molecules? Explain with examples.
Answer:
i. Non-polar molecules are the molecules in which the centre of positive charge and the negative charge is one and the same. They do not have a permanent electric dipole. When an external electric field is applied to such molecules, the centre of positive and negative charge are displaced and a dipole is induced.

ii. Molecules such as H₂, CI₂, CO₂, CH₄, etc., have their positive and negative charges effectively centred at the same point and are called non-polar molecules.

Question 56.
Derive expression for couple acting on an electric dipole in a uniform electric field.
Answer:
i. Consider an electric dipole placed in a uniform electric field E. The axis of electric dipole makes an angle θ with the direction of electric field.

ii. The force acting on charge – q at A is \(\vec{F}\)_A = -q\(\vec{E}\) in the direction of\(\vec{E}\) and the force acting on charge +q at B is \(\vec{F}\)_B = + q \(\vec{E}\) in the direction opposite to \(\vec{E}\).

iii. Since \(\vec{F}\)_A = –\(\vec{F}\)_B, the two equal and opposite forces separated by a distance form a couple.

iv. Moment of the couple is called torque and is defined by \(\vec{τ}\) = \(\vec{d}\) × \(\vec{F}\) where, d is the perpendicular distance between the two equal and opposite forces.

v. Magnitude of Torque = Magnitude of force × Perpendicular distance
∴ Torque on the dipole (\(\vec{τ}\)) = \(\vec{BA}\) × q\(\vec{E}\)
= 2lqE sin θ
but p = q2l
∴ τ = pEsin θ
∴ In vector form \(\vec{τ}\) = \(\vec{d}\) × \(\vec{E}\)

vi. If θ = 90° sin θ = 1, then τ = pE
When the axis of electric dipole is perpendicular to uniform electric field, torque of the couple acting on the electric dipole is maximum, i.e., τ = pE.

vii. If θ = 0 then τ = 0, this is the minimum torque on the dipole. Torque tends to align its axis along the direction of electric field.

Question 57.
Derive expression for electric intensity at a point on the axis of an electric dipole.
Answer:
i. Consider an electric dipole consisting of two charges -q and +q separated by a distance 2l.

ii. Let P be a point at a distance r from the centre C of the dipole.

iii. The electric intensity \(\vec{E}\)_a at P due to the dipole is the vector sum of the field due to the charge -q at A and +q at B.

iv. Electric field intensity at P due to the charge -q at A = \(\vec{E}\)_A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(-q)}{(r+l)^{2}} \hat{\mathrm{u}}_{\mathrm{pD}}\),
where, \(\hat{u}\)_PD is unit vector directed along \(\vec{PD}\)

v. Electric intensity at P due to charge +q at B
\(\vec{E}\)_B = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{(\mathrm{r}-l)^{2}} \hat{\mathrm{u}}_{\mathrm{PQ}}\)
where, \(\hat{u}\)_PQ is a unit vector directed along \(\vec{PQ}\)
The magnitude of \(\vec{E}\)_B is greater than that of \(\vec{E}\)_A since BP < AP

vi. Resultant field \(\vec{E}\)_a at P on the axis, due to the dipole is
\(\vec{E}\)_a = \(\vec{E}\)_B + E\(\vec{E}\)_A

vii. The magnitude of \(\vec{E}\)_a is given by

ix. |\(\vec{E}\)_a| is directed along PQ, which is the direction of the dipole moment \(\vec{p}\) i.e., from the negative to the positive charge, parallel to the axis.

x. If r >> l, l² can be neglected compared to r²,
|\(\vec{E}\)_a| = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\)

The field will be along the direction of the dipole moment \(\vec{p}\).

Question 58.
Drive expression for electric intensity at a point on the equator of an electric dipole.
Answer:
i. Electric field at point P due to charge -q at A is \(\vec{E}\)_A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(-\mathrm{q})}{(\mathrm{AP})^{2}} \hat{\mathrm{u}}_{\mathrm{PA}}\)
where, \(\hat{u}\)_PA is a unit vector directed along \(\vec{PA}\)

ii. Similarly, electric field at P due to charge +q at B is
\(\vec{E}\)_A = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{(\mathrm{BP})^{2}} \hat{\mathrm{u}}_{\mathrm{BP}}\)
where \(\hat{u}\)_BP is a unit vector directed along \(\vec{BP}\)

iii. Electric field at P is the sum of E_A and E_B
∴ \(\vec{E}\)_eq = \(\vec{E}\)_A + \(\vec{E}\)_B

iv. Consider ∆ACP
(AP)² = (PC)² + (AC)² = r² + l² = (BP)²

v. The resultant of fields \(\vec{E}\)_A and \(\vec{E}\)_B acting at point P can be calculated by resolving these vectors E\(\vec{E}\)_A and E\(\vec{E}\)_B along the equatorial line and along a direction perpendicular to it.

vi. Let the Y-axis coincide with the equator of the dipole X-axis will be parallel to dipole axis and the origin is at point P as shown.

vii. The Y-components of E_A and E_B are E_Asin θ and E_B sin θ respectively. They are equal in magnitude but opposite in direction and cancel each other. There is no contribution from them towards the resultant.

viii. The X-components of E_A and E_B are E_Acos θ and E_Bcos θ respectively. They are of equal magnitude and are in the same direction.
∴ |\(\vec{E}\)_eq| = E_A cos θ + E_B cos θ From equation (3),
|\(\vec{E}\)_eq| = 2E_A cos θ

x. The direction of this field is along –\(\vec{P}\) (anti-parallel to \(\vec{P}\)).

Question 59.
An electric dipole of length 2.0 cm is placed with its axis making an angle of 30° with a uniform electric field of 10⁵ N/C as shown in figure. If it experiences a torque of 10√3 N m, calculate the magnitude of charge on dipole.

Answer:
Given: 2l = 2 cm = 2 × 10² m
E = 10⁵ N/C, τ = 10√3 Nm, θ = 30°
To find: Charge (q)
Formula: τ = q E 2 l sin θ
Calculation: From Formula.
q = \(\frac{τ}{\mathrm{E} \times 2 l \times \sin \theta}\)
= \(\frac{10 \sqrt{3}}{10^{5} \times 2 \times 10^{-2} \times \sin 30^{\circ}}\)
= 1.732 × 10^-2 C

Question 60.
Explain the concept of continuous charge distribution.
Answer:
i. A system of charges can be considered as a continuous charge distribution, if the charges are located very close together, compared to their distances from the point where the intensity of electric field is to be found out.

ii. Thus, the charge distribution is said to be continuous for a system of closely spaced charges. It is treated equivalent to a total charge which is continuously distributed along a line or a surface or a volume.

Question 61.
Explain linear charge density.
Answer:
Consider charge q uniformly distributed along a linear conductor of length l, then the linear charge density (λ) is given as,
λ = \(\frac {q}{l}\)
For example, charge distributed uniformly on a straight thin rod or a thin nylon thread. If the charge is not distributed uniformly over the length of thin conductor then charge dq on small element of length dl can be written as dq = λ dl.

Question 62.
Explain surface charge density.
Answer:
i. Consider a charge q uniformly distributed over a surface of area A then the surface charge density c is given as
σ = \(\frac {q}{A}\)
For example, charge distributed uniformly on a thin disc or a synthetic cloth. If the charge is not distributed uniformly over the surface of a conductor, then charge dq on small area element dA can be written as dq = σ dA.

ii. SI unit of σ is (C / m²)

Question 63.
Explain volume charge density.
Answer:
i. Consider a charge q uniformly distributed throughout a volume V, then the volume charge density ρ is given as
ρ = \(\frac {q}{V}\)
For example, charge on a plastic sphere or a plastic cube. If the charge is not distributed uniformly over the volume of a material, then charge dq over small volume element dV can be written as dq = ρ dV.

ii. S.I. unit of p is (C/m³)
[Note: Electric field due to a continuous charge distribution can be calculated by adding electric fields due to all these small charges.]

Question 64.
Explain the concept of static charge.
Answer:

1. Static charges can be created whenever there is a friction between an insulator and other object.
2. For example, when an insulator like rubber or ebonite is rubbed against a cloth, the friction between them causes electrons to be transferred from one to the other.
3. This property of insulators is used in many applications such as photocopier, inkjet printer, painting metal panels, electrostatic precipitation/separators etc.

Question 65.
Explain the disadvantage of static charge.
Answer:

1. When charge transferred from one body to other is very large, sparking can take place. For example, lightning in sky.
2. Sparking can be dangerous while refuelling your vehicle.
3. One can get static shock, if charge transferred is large.
4. Dust or dirt particles gathered on computer or TV screens can catch static charges and can be troublesome.

Question 66.
State the precautions against static charge.
Answer:

1. Home appliances should be grounded.
2. Avoid using rubber soled footwear.
3. Keep your surroundings humid (dry air can retain static charges).

Question 67.
Two charged particles having charge 3 × 10^-8 C each are joined by an insulating string of length 2 m. Find the tension in the string when the system is kept on a smooth horizontal table.
Answer:
Tension (T) in the string is the force of repulsion (F) between the two charges.
According to Coulomb’s law,
F = \(\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}}\)
= \(\frac{9 \times 10^{9} \times 3 \times 10^{-8} \times 3 \times 10^{-8}}{2^{2}}\)
F = 2.025 × 10^-6 N
Hence, tension in the string is 2.025 × 10^-6 N.

Question 68.
A free pith ball of mass 5 gram carries a positive charge of 0.6 × 10^-7 C. What is the nature and magnitude of charge that should be given to second ball fixed 6 cm vertically below the former pith ball so that the upper pith bath is stationary?
Answer:
Let +q₂ be the charge on lower pith ball.
Now, the upper pith ball become stationary only when its weight acting downward is balanced by the upward force of repulsion between two pith balls,
i.e., F_E = mg
∴ \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\) = mg
∴ \(\frac{9 \times 10^{9} \times 0.6 \times 10^{-7} \times \mathrm{q}_{2}}{\left(6 \times 10^{-2}\right)^{2}}\) = 5 × 10^-3 × 9.8
∴ q₂ = 3.27 × 10^-7C
Hence, the second pith ball carries a positive charge of 3.27 × 10^-7C.

Question 69.
A water drop of mass 11.0 mg and having a charge of 1.6 × 10^-6 C stays suspended in a room. What will be the magnitude and direction of electric Held in the room?
Answer:
As the drop is suspended,
Force (F) due to electric field balances the weight of the drop.
∴ F = mg ………….. (1)
Here, m = 11.0 mg
= 11 × 10^-6 kg,
q = 1.6 × 10^-6 C
Electric field is given by,
E = \(\frac {F}{q}\)
= \(\frac {mg}{q}\)
= \(\frac {11×10^{-6}×9.8}{1.6×10^{-6}}\)
E = 67.4 N/C
As upward force balances the weight, hence direction of electric field must be vertically upwards.

Question 70.
A charged metallic sphere A is suspended by a nylon thread. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm, as shown in figure (a). The resulting repulsion of A is noted (for example, by shining a beam of light and measuring the deflection of its shadow on a screen.) Spheres A and B are touched by uncharged spheres C and D respectively, as shown in figure (b). C and D are then removed and B is brought closer to A to a distance of 5.0 cm between their centres, as shown in figure (c). What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B comparison to the separation between their centres.

Answer:
Let the original charge on sphere A be q and that on B be q’. At a distance r between their centres, the magnitude of the electrostatic force on each is given by
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{qq}^{\prime}}{\mathrm{r}^{2}}\)

Neglecting the sizes of spheres, A and B in comparison to r. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2. Similarly, after D touches B, the redistributed charge on each is q’/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is
F’ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(\mathrm{q} / 2)\left(\mathrm{q}^{\prime} / 2\right)}{(\mathrm{r} / 2)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\mathrm{qq}^{\prime}\right)}{\mathrm{r}^{2}}\) = F

Thus, the electrostatic force on A, due to B, remains unaltered.

Multiple Choice Questions

Question 1.
Force between two charges separated by a certain distance in air is F. If each charge is doubled and the distance between them is also doubled, force would be
(A) F
(B) 2 F
(O’ 4 F
(D) F/4
Answer:
(A) F

Question 2.
For what order of distance is Coulomb7 s law true?
(A) For all distances.
(B) Distances greater than 10^-13 m.
(C) Distances less than 10^-13 m.
(D) Distance equal to 10^-13 m.
Answer:
(B) Distances greater than 10^-13 m.

Question 3.
The permittivity of medium is 26.55 × 10^-12 C²/Nm². The dielectric constant of the medium will be
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

Question 4.
A glass rod when rubbed with a piece of fur acquires a charge of magnitude 3.2 µC. The number of electrons transferred is
(A) 2 × 10^-13 from fur to glass
(B) 5 × 10¹² from glass to fur
(C) 2 × 10¹³ from glass to fur
(D) 5 × 10¹² from fur to glass
Answer:
(A) 2 × 10^-13 from fur to glass

Question 5.
Choose the correct answer.
(A) Total charge present in the universe is constant.
(B) Total positive charge present in the universe is constant.
(C) Total negative charge present in the universe is constant.
(D) Total number of charged particles present in the universe is constant.
Answer:
(A) Total charge present in the universe is constant.

Question 6.
If a charge is moved against the Coulomb force of an electric field,
(A) work is done by the electric field
(B) energy is used from some outside source
(C) the strength of the field is decreased
(D) the energy of the system is decreased
Answer:
(B) energy is used from some outside source

Question 7.
Two point charges +4 µC and +2 µC repel each other with a force of 8 N. If a charge of -4 µC is added to each of these charges, the force would be
(A) zero
(B) 8 N
(C) 4 N
(D) 12 N
Answer:
(A) zero

Question 8.
The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is
(A) 0 V
(B) 2.5 V
(C) 250 V
(D) 1000 V
Answer:
(D) 1000 V

Question 9.
The dimensional formula of electric field intensity is
(A) [M¹E¹T^-2A^-1]
(B) [M¹L¹T^-3A^-1]
(C) [M^-1L²T^-3A^-1]
(D) [M¹L²T^-3A^-2]
Answer:
(B) [M¹L¹T^-3A^-1]

Question 10.
A force of 2.25 N acts on a charge of 15 × 10^-4C. Calculate the intensity of electric field at that point.
(A) 1500 NC^-1
(B) 150 NC^-1
(C) 15000NC^-1
(D) 2500 NC^-1
Answer:
(A) 1500 NC^-1

Question 11.
A point charge q produces an electric field of magnitude 2 N C^-1 at a point distant 0.25 m from it. What is the value of charge?
(A) 1.39 × 10^-11 C
(B) 1.39 × 10¹¹ C
(C) 13.9 × 10^-11 C
(D) 13.9 × 10¹¹ C
Answer:
(A) 1.39 × 10^-11 C

Question 12.
The electric intensity in air at a point 20 cm from a point charge Q coulombs is 4.5 × 10⁵ N/ C. The magnitude of Q is
(A) 20 µC
(B) 200 µC
(C) 10 µC
(D) 2 µC
Answer:
(D) 2 µC

Question 13.
The charge on the electron is 1.6 × 10^-19 C. The number of electrons need to be removed from a metal sphere of 0.05 m radius so as to acquire a charge of 4 × 10^-15 C is
(A) 1.25 × 10⁴
(B) 1.25 × 10³
(C) 2.5 × 10³
(D) 2.5 × 10⁴
Answer:
(D) 2.5 × 10⁴

Question 14.
Electric lines of force about a positive point charge and negative point charge are respectively .
(A) circular, clockwise
(B) radially outward, radially inward
(C) radially inward, radially outward
(D) circular, anticlockwise
Answer:
(B) radially outward, radially inward

Question 15.
Which of the following is NOT the property of equipotential surfaces?
(A) They do not intersect each other.
(B) They are concentric spheres for uniform electric field.
(C) Potential at all points on the surface has constant value.
(D) Separation of equipotential surfaces increases with decrease in electric field.
Answer:
(B) They are concentric spheres for uniform electric field.

Question 16.
In a uniform electric field, a charge of 3 C experiences a force of 3000 N. The potential difference between two points 1 cm apart along the electric lines of force will be
(A) 10 V
(B) 3 V
(C) 0.1 V
(D) 20 V
Answer:
(A) 10 V

Question 17.
Gauss’ law helps in
(A) determination of electric field due to symmetric charge distribution.
(B) determination of electric potential due to symmetric charge distribution.
(C) determination of electric flux.
(D) situations where Coulomb’s law fails.
Answer:
(A) determination of electric field due to symmetric charge distribution.

Question 18.
The electric flux over a sphere of radius 1.0 m is ø. If the radius of the sphere is doubled without changing the charge, the flux will be
(A) 4ø
(B) 2ø
(C) ø
(D) 8ø
Answer:
(C) ø

Question 19.
Gauss’ theorem states that total normal electric induction over a closed surface in an electric field is equal to
(A) \( \frac{1}{\varepsilon} \sum \mathrm{q}_{\mathrm{n}}\)
(B) ^εΣ q_n
(C) Σ q_n
(D) q₁ × q₂ × q₃ × ……… q_n
Answer:
(C) Σ q_n

Question 20.
Number of lines of induction starting from a conductor holding + q charge surrounded by a medium of permittivity ε is
(A) q and they leave the surface in normal direction.
(B) q and they leave the surface in any direction.
(C) q/ε and they leave the surface normally at every point.
(D) q/ε and they leave the surface in any direction.
Answer:
(C) q/ε and they leave the surface normally at every point.

Question 21.
An electric dipole of moment p is placed in the position of stable equilibrium in a uniform electric field of intensity E. The torque required to rotate, when the dipole makes an angle 0 with the initial position is
(A) pE cosθ
(B) pE sinθ
(C) pE tanθ
(D) pE cotθ
Answer:
(B) pE sinθ

Question 22.
Four coulomb charge is uniformly distributed on 2 km long wire. Its linear charge density is
(A) 2 C/m
(B) 4 C/m
(C) 4 × 10³ C/m
(D) 2 × 10^-3 C/m
Answer:
(D) 2 × 10^-3 C/m

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 14 Basic Principles of Organic Chemistry Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 14 Basic Principles of Organic Chemistry

Question 1.
How is the structural formula of a molecule represented? Give an example.
Answer:
Structural formula:
i. Structural formula of a molecule shows all the constituent atoms denoted with their respective chemical symbols and all the covalent bonds therein represented by a dash joining mutually bonded atoms.
ii. Structural formula of methane is:

Question 2.
Write a note on Lewis structures with the help of an example.
Answer:
Lewis structures:
i. The electron dot structures are called as Lewis structures,
e. g. The Lewis structure of methane is shown below.

ii. All the valence electrons of carbon and hydrogen are shown as dots around them. Two dots drawn between two atoms indicate one covalent bond between them. The covalent bond can be represented by a dash joining mutually bonded atoms.
iii. The dash formula represents simplified Lewis formula of the molecule.
e.g. Dash formula of methane:

Question 3.
How is the condensed formula of an organic molecule written?
Answer:
The complete structural formula is further simplified by hiding some or all the covalent bonds and indicating the number of identical groups attached to an atom by a subscript. The resulting formula of a compound is known as condensed formula.
e.g.

1. The condensed formula of ethane is written as CH₃-CH₃ or CH₃CH₃.
2. The condensed formula of n-pentane is written as CH₃CH₂CH₂CH₂CH₃ or CH₃(CH₂)₃CH₃.

Question 4.
What do you understand by the term bond-line formula?
Answer:
Bond-line or zig-zag formula:
i. The condensed formula is simplified into bond-line formula, which is also known as zig-zag formula.
ii. In this representation of an organic molecule, the symbols of carbon and hydrogen atoms are not written. The carbon-carbon bonds are represented by lines drawn in a zig-zag manner
iii. The terminals of the zig-zag line indicate methyl groups and the intersection of lines denote a carbon atom bonded to appropriate number of hydrogen atoms which satisfy the tetravalency of the carbon atom.
e.g. Propane is represented by bond-line or zig-zag formula:

iv. If a compound contains heteroatom(s) or H-atom(s) bonded to heteroatom(s), then they are represented by their symbols.
e.g. Ethanol is represented by bond-line or zig-zag formula:

Question 5.
Name the different methods used to represent three-dimensional structure of a molecule on the paper.
Answer:
Four different methods are used to represent three-dimensional structure of a molecule on the paper:

1. Wedge formula
2. Fischer projection formula or cross formula
3. Newman projection formula
4. Sawhorse or andiron or perspective formula

Question 6.
Write a short note on: Wedge formula.
Answer:
Wedge formula:
i. The three-dimensional (3-D) structure of organic molecules can be represented on plane paper by using solid and dashed wedges and normal line (-) for single bonds.
ii. In this formula, the solid wedge is used to indicate a bond projecting up from the plane of paper, towards the reader (observer), whereas the dashed wedge is used to depict a bond going backward, below the paper away from the reader.
iii. The bonds lying in plane of the paper are depicted by using a normal line (-).
iv. Wedge formula of methane molecule is shown below:

Question 7.
How is Fischer projection formula of a molecule drawn? Explian by giving an example.
Answer:
Fischer projection (cross) formula:

• In this representation, a three dimensional molecule is projected on plane of paper.
• Fischer projection formula can be drawn by visualizing the molecule with its main carbon chain vertical.
• Each carbon on the vertical chain is represented by a cross. By convention, the horizontal lines of the cross represent bonds projecting up from the carbon and the vertical lines represent the bonds going below the carbon.

Fischer projection formula of a molecule along with its wedge formula is represented below:

[Note: Fischer projection formula is more commonly used in carbohydrate chemistry.]

Question 8.
Write the Fischer projection and wedge formula for 2-chloro-propan-2-ol.
Answer:
2-Chloropropan-2-ol has formula CH₃C(Cl)(OH)CH₃.
Fischer projection and wedge formula for 2-chloropropan-2-ol can be given as:

Question 9.
Convert the following wedge formula to Fischer projection formula:

Answer:

Question 10.
Explain how will you represent the Newman projection formula and Sawhorse formula of ethane molecule?
Answer:
i. Newman projection formula of ethane molecule:
a. A Newman projection views the carbon-carbon single bond directly head-on. The front carbon atom is represented by a point while the rear carbon atom is represented by a circle. The point is drawn at the centre of the circle.
b. Bonds attached to the front carbon atom are represented by three lines drawn at an angle of 120° to each other from the centre of the circle and bonds attached to the rear carbon atom are represented by three lines drawn at an angle of 120° to each other from the circumference of the circle.
c. Newman projections of ethane molecule is represented in adjacent diagram.

ii. Sawhorse (or andiron or perspective) formula of ethane molecule:
a. In this representation, a C-C single bond is represented by a long slanting line. The lower end of the line represents the front carbon and the upper end represents the rear carbon.
b. The remaining three bonds at the two carbons are shown to radiate from the respective carbons. (As the central C-C bond is drawn rather elongated the bonds radiating from the front and rear carbons do not intermingle.)
c. Sawhorse formula of ethane molecule is represented in adjacent diagram.

Question 11.
Explain the classification of organic compounds based on carbon skeleton.
Answer:
On the basis of their carbon skeleton, organic compounds are classified into two main groups:
i. Acyclic or aliphatic or open chain compounds:
a. Organic compounds in which carbon atoms are joined to form an open chain are called aliphatic compounds.
b. Their structure may consist of straight chains (in which carbon atoms are bonded to one or two other carbon atoms) or branched chains (in which at least one carbon atom is bonded to three or four other carbon atoms).
e.g.

ii. Cyclic or closed chain or ring compounds:
a. Organic compounds in which carbon atoms are joined to form one or more closed rings with or without hetero atom are called cyclic compounds.
b. They are further divided into two types: Homocyclic and heterocyclic compounds.
1. Homocyclic or carbocyclic compounds: The cyclic organic compounds which have a ring made up of only carbon atoms are called as homocyclic or carbocyclic compounds.
They are further divided into:
i. Alicyclic compounds: These are cyclic compounds (ring of 3 or more C-atoms) exhibiting properties similar to those of aliphatic compounds.

ii. Aromatic compounds: These compounds have special stability.
Aromatic compounds are further classified as benzenoid and non-benzenoid aromatics.
a. Benzenoid aromatics contain at least one benzene ring in the structure.

b. Non-benzenoid aromatics contain an aromatic ring, other than benzene.

2. Heterocyclic compounds: Cyclic organic compounds which contain one or more heteroatoms (such as O, N, S, etc.) in the ring are called heterocyclic compounds.
They are further divided into:
i. Heterocyclic aromatic compounds: Aromatic compounds which contain at least one heteroatom in the ring are called heterocyclic aromatic (hetero-aromatic) compounds.

ii. Heterocyclic non-aromatic compounds: Alicyclic compounds, which contain at least one heteroatom in the ring are called heterocyclic non-aromatic compounds (hetero-alicyclic) compounds.

Question 12.
What is a functional group? Give two examples.
Answer:
Functional group:
i. A part of an organic molecule which undergoes change as a result of a reaction is called functional group.
OR
An atom or a group of atoms in the organic molecule which determines its characteristic chemical
properties is called functional group.
e.g. a. The functional group in alcohols is -OH group.
b. The functional group in aldehydes is -CHO group.

ii. There are a large variety of functional groups in organic compounds. Hence, organic compounds can be classified based on the nature of functional group present in them.
iii. The resulting individual class of compounds is called a family and is named after the constituent functional group.
e.g. Family of alcohols, which includes organic compounds having -OH functional group.

Note: Functional groups in organic compounds:

Question 13.
Indicate all the functional groups present in the following compounds.

Answer:

Question 14.
Identify the functional group in the following compounds:
i. n-Butyl alcohol
ii. Propanone
iii. Acetylene
Answer:

Question 15.
Write the name of the family of the following organic compounds:
i. CH₃(CH₂)₃CH₂Cl
ii. CH₃CH₂CH₂NH₂
iii. CH₃CH₂COCH₃
iv. CH₃CH₂OCH₃
Answer:

Question 16.
Write a note on homologous series.
Answer:
Homologous series:

• A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (-CH₂-) in its molecular and structural formula is called as homologous series.
• The individual members of the series are called homologues and they can be represented by a same general formula.
• Two successive homologues differ by one – CH₂ (methylene) unit (i.e., molecular weight of each successive member differs by 14 units).
• Homologues show similar chemical properties.
• Physical properties (like melting point, boiling point, density, solubility, etc.) of the homologues show a gradual change with increase in the molecular weight of the member.

Note: Consider the homologous series of straight chain aldehydes. The boiling point increases down the series as molecular weight increases.

Name	Molecular formula	Boiling point
Formaldehyde	HCHO	-21 °C
Acetaldehyde	CH3CHO	21 °C
Propionaldehyde	C2H5CHO	48 °C
Butyraldehyde	C3H7CHO	75 °C
Valeraldehyde	C4H9CHO	103 °C

Question 17.
Alkanes constitute a homologous series of straight chain saturated hydrocarbons. Write down the structural formulae of the first five homologues of this series. Write their molecular formulae and deduce the general formula of such homologous series.
Answer:
The first five homologues are generated by adding one – CH₂ – at a time, starting with the first homologue, methane (CH₄).

By counting carbon and hydrogen atoms in the five homologues, we get their molecular formulae as CH₄, C₂H₆, C₃H₈, C₄H₁₀ and C₅H₁₂.
Comparing these molecular formulae and assigning the number of carbon atoms as ‘n’, the following general formula is deduced: C_nH_2n+2.

Question 18.
Write down structural formulae of (i) the third higher and (ii) the second lower homologue of CH₃CH₂COOH.
Answer:
i. Structural formula of the third higher homologue is obtained by adding three – CH₂ – units to the carbon chain of the given structure.

ii. Structural formula of the second lower homologue is obtained by removing two – CH₂ – units from the carbon chain of the given structure.

Question 19.
Write the general formula of homologous series of alcohols.
Answer:
General formula of homologous series of alcohols can be represented as, C_nH_2n+1OH (where n = 1, 2, 3, …).

Question 20.
Write the name and molecular formulae of the first three higher homologues of propyl chloride.
Answer:
General formula: C_nH_2n+1Cl (where n = 1, 2, 3, …)

No. of carbon atoms	Molecular formula	Name
n = 3	C3H7Cl	Propyl chloride
n = 4	C4H9Cl	Butyl chloride
n = 5	C5H11Cl	Pentyl chloride
n = 6	C6H13Cl	Hexyl chloride

Question 21.
What is the molecular formula of:
i. first higher homologue of propionic acid?
ii. first lower homologue of propionic acid?
Answer:
i. First higher homologue of propionic acid:
(Addition of 1-CH₃ group to CH₃CH₂COOH)
Butyric acid: C₃H₇COOH

ii. First lower homologue of propionic acid:
(1-CH₃ group less from CH₃CH₃COOH)
Acetic acid: CH₃COOH

Question 22.
How are the saturated (sp³) carbon atoms in a molecule classified based on the number of other carbon atoms bonded to it? Give an example that has all the four types of carbon atoms.
Answer:
i. The saturated (sp³) carbons in a molecule are classified as primary, secondary, tertiary and quaternary in accordance with the number of other carbons bonded to it by single bonds.

• Primary carbon atom (1°): This carbon atom is bonded to only one other carbon atom. Terminal carbon atoms are always 1° carbon atoms.
• Secondary carbon atom (2°): This carbon atom is bonded to two other carbon atoms.
• Tertiary carbon atom (3°): This carbon atom is bonded to three other carbon atoms.
• Quaternary carbon atom (4°): This carbon atom is bonded to four other carbon atoms.

ii. An example molecule having all the four types of carbon atoms:

Thus, in 2,2,5-trimethylhexane, there are five primary, two secondary, one tertiary and one quaternary carbon atoms.
[Note: Hydrogen atoms attached to primary’, secondary and tertiary carbon atoms are referred to as primary, secondary and tertiary H-atoms respectively.]

Question 23.
Give common name/trivial name of the following compounds.

Answer:
i. Lactic acid
ii. Glycine
iii. Glycerol
iv. Chloroform

Question 24.
Give a basic idea about IUPAC nomenclature system and comment on IUPAC names of straight chain alkanes.
Answer:
i. International Union of Pure and Applied Chemistry (IUPAC) was founded (in 1919) and a systematic method of nomenclature for organic compounds was developed under its banner.
ii. This was done because of growing number of organic compounds with increasingly complicated structures and it was difficult to name them. To simplify and avoid confusions, IUPAC system is accepted and widely used all over the world today. According to this system, a unique name is given to each organic compound.

Following things are taken into consideration while naming a particular organic compound:

• To arrive at the IUPAC name of an organic compound, its structure is considered to be made of three main parts: parent hydrocarbon, branches and functional groups.
• The IUPAC names of a compound are obtained by modifying the name of its parent hydrocarbon further incorporating names of the branches and functional groups as prefix and suffix.

IUPAC names of straight chain alkanes:
a. The homologous series of straight chain alkanes forms the parent hydrocarbon part of the IUPAC names of aliphatic compounds.
b. The IUPAC name of a straight chain alkane is derived from the number of carbon atoms it contains.
c. IUPAC names of the first twenty alkanes are mentioned in the following table:

Question 25.
Match the following:

	Column – I		Column – II
i.	C19H40	a.	Undecane
ii.	C12H26	b.	Nonadecane
iii.	C11H24	c.	Dodecane
		d.	Nonane

Answer:
i – b,
ii – c,
iii – a

Question 26.
Explain the following with two examples:
i. straight chain alkyl groups
ii. branched chain alkyl group
Answer:
i. Straight chain alkyl group: It is obtained by removing one H-atom from the terminal carbon of an alkane molecule.
ii. It is named by replacing ‘ane’ of the alkane by ‘yl’.

iii. Branched chain alkyl group: It is obtained by removing a H-atom from any one of the non-terminal carbons of an alkane or any H-atom from a branched alkane.

Note: Straight chain alkyl groups

Trivial names of small branched alkyl groups

Question 27.
Write names of following groups.
i. C₆H₅–
ii. (CH₃)₃C-
Answer:
i. Phenyl group
ii. tert-Butyl group

Question 28.
State the rules to assign IUPAC nomenclature of a branched chain alkane.
Answer:
i. Select the longest continuous chain of carbon atoms to be called the parent chain. All other carbon atoms not included in this chain constitute, side chains or branches or alkyl substituents. For example:

Parent chain has five carbon atoms and -CH3 group is alkyl substituent.

Parent chain has six carbon atoms and methyl group is the alkyl substituent.
If two chains of equal length are located, then the one with maximum number of substituents is selected as the parent chain.

Parent chain hexane with one alkyl substituent is the incorrect chain.

ii. The parent chain is numbered from one end to the other to locate the position, called locant number of the alkyl substituent. The numbering is done in that direction which will result in lowest possible locant numbers.

iii. Names of the alkyl substituents are added as prefix to the name of the parent alkane. Different alkyl substituents are listed in alphabetical order with each substituent name preceded by the appropriate locant number. The name of the substituent is separated from the locant number by a hyphen.

The name is 4-ethyl-3-methylheptane and not 3-methyl-4-ethylheptane.
iv. When both the numberings give the same set of locants, that numbering is chosen which gives smaller locant to the substituent having alphabetical priority.

The name is 3-ethyl-4-methylhexane and not 3-methyl-4-ethylhexane.

v. If two or more identical substituents are present the prefix di (for 2), tri (for 3), tetra (for 4) and so on, are used before the name of the substituent to indicate how many identical substituents are there. The locants of identical substituents are listed together, separated by commas.

There must be as many numbers in the name as the substituents. A digit and an alphabet are separated by hyphen. The prefixes di, tri, tetra, sec and tert are ignored in alphabetizing the substituent names. Substituent and parent hydrocarbon names are joined into one word.

vi. Branched alkyl group having no accepted trivial name is named with the longest continuous chain beginning at the point of attachment as the base name. Carbon atom of this group attached to parent chain is numbered as ‘1’. The name of such substituent is enclosed in bracket.

Question 29.
Complete the following table.

Answer:

Question 30.
Explain the rules for IUPAC nomenclature of unsaturated hydrocarbons (Alkenes and Alkynes).
Answer:
While writing IUPAC names of alkenes and alkynes following rules are to be followed in addition to rules for alkanes.
i. The longest continuous chain must include carbon-carbon multiple bond. Thus, the longest continuous chains in 1 and II contain four and six carbons, respectively.

ii. Numbering of this chain must be done such that carbon-carbon multiple bond has the lowest possible locant number.

iii. The ending ‘ane’ of alkane is replaced by ‘ene’ for an alkene and ‘yne’ for an alkyne.
iv. Position of carbon atom from which multiple bond starts is indicated by smaller locant number of two multiple bonded carbons before the ending ‘ene’ or ‘yne’. e.g.

v. If the multiple bond is equidistant from both the ends of a selected chain, then carbon atoms are numbered from that end, which is nearer to first branching.

vi. If the parent chain contains two double bonds or two triple bonds, then it is named as diene or diyne. In all these cases ‘a’ of ‘ane’ (alkane) is retained.

vii. If the parent chain contains both double and triple bond, then carbon atoms are numbered from that end where multiple bond is nearer. Such systems are named by putting ‘en’ ending first followed by ‘yne’. The number indicating the location of multiple bond is placed before the name.

viii. If there is a tie between a double bond and a triple bond, the double bond gets the lower number.

Question 31.
Give IUPAC rules for naming simple monocyclic hydrocarbons.
Answer:
i. A saturated monocyclic hydrocarbon is named by attaching prefix ‘cyclo’ to the name of the corresponding open chain alkane.

ii. An unsaturated monocyclic hydrocarbon is named by substituting ‘ene’, ‘yne’, etc. for ‘ane’ in the name of corresponding cycloalkane.

iii. If side chains are present then the numbering of the ring carbon is started from a side chain.

iv. If alkyl groups contain greater number of carbon atoms than the ring, the compound is named as derivative of alkane. Ring is treated as substituent.

Question 32.
Give the IUPAC names of the following compounds:

Answer:
i. 1-Ethyl-1-methyl-2-propylcyclohexane
ii. 1,2-Dimethylcyclobutane
iii. Cyclopentene
iv. 3-Cyclopropylhex-1-yne

Question 33.
Explain in short how naming of monofunctional compound is done.
Answer:
Naming of monofunctional compounds: When a molecule contains only one functional group, the longest carbon chain containing that functional group is identified as the parent chain and numbered so as to give the smallest locant number to the carbon bearing the functional group. The parent name is modified by applying appropriate suffix. Location of the functional group is indicated where necessary and when it is NOT numbered ‘1’.

When the functional group cannot be used as suffix, and can be only the prefix, the molecule is named as parent alkane carrying the functional group as substituent at specified carbon.

Question 34.
Complete the following.

Answer:

Question 35.
Give examples of functional groups which can appear only as prefix?
Answer:
Functional groups which can appear only as prefix are as follows:
i. Nitro group (-NO₂)
ii. Halides (-X): Represented by prefix “halo” (like fluoro, chloro, bromo, iodo).
iii. Alkoxy group (-OR): Groups like methoxy (-OCH₃), ethoxy (-OC₂H₅), etc.

Note: Functional groups appearing as prefix and suffix

Functional Group	Prefix	Suffix
-COOH	Carboxy	– oic acid
-COOR	alkoxycarbonyl	– oate
-COCl	Chlorocarbonyl	– oyl chloride
-CONH2	Carbamoyl	– amide
-CN	Cyano	– nitrile
-CHO	Formyl	– al
-CO-	Oxo	– one
-OH	Hydroxy	– ol
-NH2	Amino	– amine

Question 36.
Write a note on principal functional group.
Answer:
i. The organic compounds possessing two or more functional groups (same or different) in their molecules are called polyfunctional compounds.
ii. When there are two or more different functional groups, one of them is selected as the principal functional group and the others are considered as substituents.
iii. The principal functional group is used as suffix of the IUPAC name while the other substituents are written with appropriate prefixes. The principal functional group is decided on the basis of the following order of priority:

Question 37.
Explain the rules for naming mono or polyfunctional compounds.
Answer:

• Identification of parent chain: The longest carbon chain containing the single or the principal functional group is identified as parent chain.
  e.g. Ethers are named as alkoxyalkane. While naming it, the larger alkyl group is chosen as parent chain.
• Numbering of parent chain: It is done so as to give the lowest possible locant numbers to the carbon atom of this functional group.
• Suffix: The name of the parent hydrocarbon is modified adequately with appropriate suffix in accordance with the single/principal functional group.
• Names of the other functional groups (if any) are attached to this modified name as prefixes. The locant numbers of all the functional groups are indicated before the corresponding suffix/prefix.

[Note: The carbon atom in -COOR, -COCl, -CONH₂, -CN and -CHO is C – 1 by rule and therefore, is not mentioned in the IUPAC name.]

Question 38.
Write IUPAC names for the following structures:

Answer:

Here, the principal functional group, ketone is located at the C-3 on the five carbon chain. The -OH group, the hydroxyl substituent is at C-2. Therefore, the IUPAC name is 2-hydroxypentan-3-one.

Here, the principal functional group is carboxylic acid. The amino substituent is located at C-3 on four carbon chain. Therefore, the IUPAC name 3-aminobutanoic acid.

Here, two same functional groups are present at C-1 and C-2 position. They are indicated by using the term ‘di’ before the class suffix. Therefore, the IUPAC name is propane-1,2-diol.
iv. CH₂ = CH – CH = CH₂
Here, the parent chain contains two double bonds at C-1 and C-3, hence it is named as diene. Therefore, the IUPAC name is buta-1,3-diene.

Question 39.
Give IUPAC rules for naming substituted benzene.
Answer:
i. Monosubstituted benzene : The IUPAC name of a monosubstituted benzene is obtained by placing the name of substituent as prefix to the parent skeleton which is benzene.

ii. Some monosubstituted benzenes have trivial names which may show no resemblance with the name of the attached substituent group. For example, methylbenzene is known as toluene, aminobenzene as aniline, hydroxybenzene as phenol and so on. The common names written in the bracket are also used universally and accepted by IUPAC.

iii. If the alkyl substituent is larger than benzene ring (7 or more carbon atoms) the compound is named as phenyl-substituted alkane.

iv. Benzene ring can as well be considered as substituent when it is attached to an alkane with a functional group.

v. Disubstituted benzene derivatives:
Common names of the three possible isomers of disubstitued benzene derivatives are given using one of the prefixes ortho (o-), meta (m-) or para (p-).
IUPAC system, however, uses numbering instead of prefixes, o-, m-, or p-.

vi. If two substituents are different, then they enter in alphabetical order.

vii. If one of the two groups gives special name to the molecule then the compound is named as derivative of the special compound.

viii. Trisubstituted benzene derivatives : If more than two substituents are attached to benzene ring, numbers are used to indicate their relative positions following the alphabetical order and lowest locant rule. In some cases, common name of benzene derivatives is taken as parent compound.

Question 40.
Write the structural formula of following derivatives of benzene.
i. 2,4,6-Trinitrotoluene
ii. 1-Chloro-2,4-dinitrobenzene
iii. 4-Broniobenzaldehyde
iv. 1-Iodo-3-phenylpentane
v. 2-Hydroxybenzaldehyde
Answer:

Question 41.
Write the IUPAC names of the following compounds.

Answer:
i. 5-Phenylpent-1-ene
ii. 2-Hydroxybenzoic acid

Question 42.
Define the terms:
i. Isomerism
ii. Isomers
Answer:
i. Isomerism: The phenomenon of existence of two or more compounds possessing the same molecular formula is known as isomerism.

ii. Isomers: Two or more compounds having the same molecular formula are called as isomers of each other. [Note: The isomers are different compounds having same molecular formula and therefore they exhibit different physical and chemical properties.]

Question 43.
Define: Structural isomerism
Answer:
Structural isomerism: When two or more compounds have same molecular formula but different structural formulae, they are said to be structural isomers of each other and the phenomenon is known as structural isomerism.

Question 44.
Define: Stereoisomerism
Answer:
When different compounds have the same structural formula but different relative arrangement of groups/atoms in space, that is, different spatial arrangement of groups/atoms, it is called as stereoisomerism.

Question 45.
Give different types of structural isomerism that organic compounds can exhibit.
Answer:
Different types of structural isomerism that organic compounds may exhibit are as follows:

• Chain isomerism
• Position isomerism
• Functional group isomerism
• Metamerism
• Tautomerism

Question 46.
Explain chain isomerism in alkanes with two suitable examples.
Answer:
Chain isomerism: When two or more compounds have the same molecular formula but different parent chain or different carbon skeletons, it is referred to as chain isomerism and such isomers are known as chain isomers.
e.g.
i. Butane (C₄H₁₀) exists in two isomeric forms:

Here, n-butane contains longest chain of four carbon atoms whereas isobutane contains longest chain of three carbon atoms. Such isomers having different carbon skeletons are called as chain isomers.
[Note: Methylpropane has no other branched isomers, hence locant (2) can be dropped.]

ii. Pentene (C₅H₁₂) exists in three isomeric forms:

[Note: The numbers of chain isomers increase with the increase in the number of carbon atoms in the molecule.]

Question 47.
Write a note on position isomerism.
Answer:
i. The phenomenon in which diffèrent compounds having the same functional group at different positions on the parent chain is known as position isomerism.
ii. e.g. But-1-ene and but-2-ene are position isomers of each other as they have the same molecular formula (C₄H₈) and the sanie carbon skeleton hut the double bonds are located at different positions.

Question 48.
Define: Functional group isomerism
Answer:
Different compounds having the same molecular formula but different functional groups are called as futictional group isomers and the phenomenon is called as junctional group isomerism.
e.g. CH₃ – O – CH₃ (Dimethyl ether) and C₂H₅ – OH (ethyl alcohol) have same molecular formula (C₂H₆O) but former has ether (-O-) functional group and the latter has alcoholic (-OH) functional group.

Question 49.
Explain: Metamerism
Answer:
i. Metamerism may be defined as a type of isomerism in which different compounds have same molecular formula and the same functional group but have unequal distribution of carbon atoms on either side of the functional group. Such isomers are known as metamers.

ii. e.g. Ether with molecular formula C₄H₁₀O has three metamers. They have same functional group as ether but have different distribution of carbon atoms attached to etheral oxygen. These metamers are:

Question 50.
Explain: Tautomerism
Answer:
When same compound exists as mixture of two or more structurally distinct molecules which are in rapid equilibrium with each other, then the phenomenon is referred to as tautomerism. Such interconverting isomers are called tautomers.
i. In nearly all the cases, it is the proton which shifts from one atom to another atom in the molecule to form its tautomer.
ii. Keto-enol tautomerism is very common form of tautomerism.
iii. Here, a hydrogen atom shifts reversibly from the a-carbon of the keto form to oxygen atom of the enol. This type of isomerism is known as keto-enol tautomerism.

Question 51.
Explain the terms substrate, reagent and byproduct in an organic reaction.
Answer:

• Organic molecules primarily contain various types of covalent bonds between the constituent atoms. During an organic reaction, molecules of the reactant undergo change in their structure. A covalent bond at a carbon atom in the reactant is broken and a new covalent bond is formed at it, giving rise to the product.
• The reactant that provides carbon to the new bond is called substrate. In other words, substrate is a chemical species which reacts with reagent to give corresponding products.
• The other reactant which brings about this change is called reagent.
• Apart from the product of interest, some other products are also formed in an organic reaction. These are called byproducts.

e.g. In following reaction, methane is the substrate and chlorine is the reagent. The product of interest is methyl chloride and the byproduct is HCl.

Question 52.
Explain: Organic reactions are often a multi-step process.
Answer:

• Organic molecules contain covalent bonds, which are made of valence electrons of the constituent atoms.
• During an organic reaction, molecules of the reactant undergo change in their structure due to redistribution of valence electrons of constituent atoms.
• This results in the bond breaking or bond forming processes as organic reaction proceeds. However, these processes are usually not instantaneous.
• As a result of this, the overall organic reaction occurs by the formation of one or more unstable species called intermediates.

Thus, organic reactions are often a multi-step process.

Question 53.
What do you mean by reaction mechanism? Give importance of reaction mechanism.
Answer:
i. Mechanism of an organic reaction is the complete step by step description of exactly which bonds break and which bonds form, in what manner and in what order to give the observed products.

ii. In general, reaction mechanism is a sequential account of:

• the electron movement taking place during each step
• the bond cleavage and/or bond formation
• accompanying changes in energy and shapes of various species and
• rate of the overall reaction.

The individual steps, constitute the reaction mechanism.

iii. Importance of reaction mechanism:
The knowledge of mechanism of a reaction is useful for understanding the reactivity of the concerned organic compounds and, in turn, helpful for planning synthetic strategies.

Question 54.
What are the different ways in which a covalent bond fission can takes place?
Answer:
The covalent bond fission/cleavage takes place in two ways:

1. Homolytic fission
2. Heterolytic fission

Question 55.
Explain homolytic cleavage of a bond with suitable example.
Answer:
Homolytic cleavage:
i. A covalent bond consists of two electrons (i.e., a bond pair of electrons) shared between the two bonded atoms.
ii. In homolytic cleavage of a covalent bond, one of the two electrons go to one of the bonded atoms and the other is bound to the other atom.
iii. This type of cleavage gives rise to two neutral species carrying one unpaired electron each. Such a species with single unpaired electron is called as free radical.
iv. The free radicals are short lived (transitory) and unstable. Therefore, they are very reactive, having tendency to seek an electron for pairing.
v. Homolytic cleavage can be represented as follows:

where movement of a single electron is represented by a half-headed curved arrow or fish hook,
vi. Thus, the symmetrical breaking of a covalent bond between two atoms such that each atom retains one electron of the shared pair forming free radicals is known as homolytic cleavage (homolysis).

Question 56.
What conditions favour homolytic cleavage?
Answer:
Homolytic cleavage is favoured in the presence of UV radiation or in presence of catalyst such as peroxides (H₂O₂) or at high temperatures.

Question 57.
Write a short note on free radical.
Answer:
Free radical:
i. A species with unpaired electron is called free radical.
OR
An uncharged species which is electrically neutral and which contains a single electron is called free radical.
ii. A free radical is highly reactive, unstable and therefore has a transitory existence (short-lived).
iii. Free radicals are formed as reaction intermediate which subsequently react with another radical/molecule to restore stable bonding pair.
iv. In a carbon free radical, the carbon atom having unpaired electron is sp hybridized and has planar trigonal geometry.

v. The alkyl free radicals are classified as primary, secondary or tertiary depending upon the number of carbon atoms attached to the C-atom carrying the unpaired electron.

vi. Stability of alkyl free radicals decreases in the order 3° > 2° > 1° > methyl free radical.

Question 58.
Explain heterolytic cleavage with suitable example.
Answer:
Heterolytic cleavage:
i. In hetcrolytic cleavage of a covalent bond, both shared electrons go to one of the two bonded atoms.
ii. This type of cleavage gives rise to two charged species, one with negative charge (anion) and the other with positive charge (cation).
iii. The negatively charged species has the more electronegative atom which has taken away the shared pair of electrons with it.
iv. Heterolytic cleavage can be represented as follows:

Where B is more electronegative than A and the movement of an electron pair is represented by a curved arrow.
v. Thus, the unsymmetrical breaking of a covalent bond between two atoms in such a way that the more electronegative atom acquires both the electrons of the shared pair. thereby fòrming charged ions is known as heterolytic fission or heterolysis.

Question 59.
What is carbocation? Explain with the help of an example and comment on the stability of carbocation.
Answer:
Carbocation:
i. A carbon atom having sextet of electrons and a positive charge is called a carbocation.
ii. They are unstable and highly reactive species formed as intermediates in many organic reactions.
iii. In a carbocation, the central carbon atom is sp² hybridized and has trigonal planar geometry.
e. g. In a methyl carbocation C If, the positively charged carbon atom is covalently bonded to three hydrogen atoms. It is planar with H-C-H bond angle of 120°.
The unhybridized p_z orbital is vacant and lies perpendicular to the plane containing the three sigma C-H bonds.

iv. Carbocation are classified as primary (1°), secondary (2°) and tertiary (3°).
v. The stability of carbocations decreases in the order:

Question 60.
Write a short note on carbanion.
Answer:
Carbanion:
i. Carbanion is a species with a negatively charged carbon atom having complete octet (eight electrons) in its valence shell.
ii. It is formed due to heterolytic bond fission when carbon atom is bonded to the more electropositive atom.

(Where Z is more electropositive than C)
iii. Carbanions are unstable and highly reactive species formed as intermediates in many organic reactions.

Question 61.
Give the types of reagents used to carry out polar organic reactions.
Answer:
The polar organic reactions are brought about by two types of reagents.
Depending upon the ability to accept or donate electrons from or to the substrate, reagents are classified as

1. Electrophiles (E⁺)
2. Nucleophiles (Nu:)

Question 62.
Explain the term electrophile. Give examples.
Answer:
Electrophiles:
i. The species which accept electron pairs from the substrate during the reaction are called electrophiles.
ii. The electrophiles are electron seeking (or electron loving) species because they themselves are electron deficient.
iii. e.g. a. Positively charged/cationic electrophiles:

b. Neutral species with vacant orbitals or incomplete octet of electrons in the outermost orbit: AlCl₃, BF₃, FeCl₃, SO₂, BeCl₂, ZnCl₂, PCl₅, etc.
iv. A polyatomic electrophile has an electron deficient atom in it called the electrophilic centre.
e.g. The electrophilic centre of the electrophile AlCl₃ is AlCl₃ which has only 6 valence electrons.

Question 63.
Explain the term nucleophile. Give examples.
Answer:
Nucleophiles:
i. The species which donate (give away) electron pairs to the substrate during the reaction are called nucleophiles.
ii. Since, nucleophiles are electron rich species, they donate a pair of electrons to acceptor atoms and thus, they are nucleus seeking (or nucleus loving) species.
iii. e.g. a. Negatively charged nucleophiles: OH^–, CN^–, Cl^–, Br^–, etc.
b. Neutral species containing at least one lone pair of electrons:
H₂O, NH₃, H₂S, R – OH, R – NH₂, R – OR, etc.
iv. A polyatomic nucleophile has an electron rich atom in it called the nucleophilic centre.
e.g. The nucleophilic centre of the nucleophile H₂O is ‘O’ which has two lone pairs of electrons.

Question 64.
Identify the nucleophile and electrophile from NH₃ and \(\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}\). Also indicate the nucleophilic and electrophilic centres in them. Justify.
Answer:
The structural formulae of two reagents showing all the valence electrons are:

Thus, NH₃ contains N with a lone pair of electrons which can be given away to another species. Therefore, NH₃ is a nucleophile and ‘N’ in it is the nucleophilic centre.
The \(\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}\) is a positively charged electron deficient species having a vacant orbital on the carbon. It is an electrophile and the ‘C’ in it is the electrophilic centre.

Question 65.
What is the difference between nucleophilic reaction and electrophilic reaction. Give one example.
Answer:
In nucleophilic reaction nucleophile attacks the electrophilic centre in the substrate and brings about a nucleophilic reaction whereas, in electrophilic reaction an electrophile attacks a nucleophilic centre in the substrate and brings about an electrophilic reaction.

Here, the nucleophilic centre N: in the nucleophile NH₃ attacks the electrophilic centre ‘B’ in the electrophile BF₃ to form the product.
[Note: Given reaction is not an organic reaction.]

Question 66.
How electrophilic or nucleophilic centre is generated in a neutral substrate?
Answer:

• The displacement of valence electrons resulting in polarization of an organic molecule is called electronic effect.
• Polarization can be either due to the presence of an atom or substituent group, or due to the influence of certain atornattacking reagent or due to the certain structural feature present in the molecule.
• Such polarization results in the formation of electrophilic or nucleophilic centre in the neutral organic molecule.

Question 67.
Explain the difference between permanent electronic effect and temporary electronic effect.
Answer:
i. Permanent electronic effect:
The electronic effect that occurs in a substrate in the ground state is a permanent effect.
e.g. Inductive effect and resonance effect are two examples of permanent electronic effect.

ii. Temporary electronic effect:
The electronic effect that occurs in a substrate due to approach of the attacking reagent is a temporary effect. This type of electronic effect is called as electromeric effect or polarizability effect.

Question 68.
Define: Inductive effect
Answer:
Inductive effect: When an organic molecule has a polar covalent bond in its structure, polarity is induced in adjacent carbon-carbon single bonds too. This effect is called as inductive effect.

Question 69.
Describe inductive effect in detail.
Answer:
Inductive effect:
i. When an organic molecule has a polar covalent bond in its structure, polarity is induced in adjacent carbon- carbon single bonds too. This effect is called as inductive effect.

ii. For example, in chloroethane molecule, the covalent bond between ‘C’ and ‘Cl’ is a polar covalent bond whereas C-2 and C-1 bond (C-C bond) is expected to be nonpolar covalent bond. But, this bond acquires some polarity as chlorine is more electronegative than carbon. Chlorine pulls the bonding pair of electrons towards itself. Thus, the chlorine atom acquires a fractional negative charge, while the C-1 carbon atom acquires a fractional positive charge. As C-1 is further bonded to C-2, the positive polarity of C-1 pulls the shared pair of electrons of the C-2 – C-l bond more towards itself. As a result, a smaller positive charge is developed on C-2. Thus, the electron density gets displaced towards the chlorine atom not only along the [C-1 – Cl] bond, but also along the [C-2 – C-1] bond due to the inductive effect of Cl. This is represented as follows:

iii. The arrow head shown in the centre of the bond represents inductive effect. The direction of the arrow head indicates the direction of the permanent electron displacement along the sigma bond in the ground state.
iv. The inductive effect of an influencing group is transmitted along a chain of C-C bonds. However, this effect decreases rapidly with the increase in the number of intervening C-C single bonds and it becomes negligible beyond three C-C bonds.

v. The direction of the inductive effect of a bonded group depends upon whether electron density of the bond is withdrawn from the bonded carbon or donated by the bonded carbon. On the basis of this ability, the groups/substituents are classified as either electron withdrawing (accepting) or electron donating (releasing) groups with respect to hydrogen.
e.g. In chloroethane, Cl withdraws electron density from the carbon chain and is electron withdrawing. Therefore, chlorine is said to exert an electron withdrawing inductive effect or negative inductive effect (-I effect) on the carbon chain.

vi. a. Substituents or groups that shows -I effect: -Cl, -NO₂, -CN, -COOH, -COOR, -OAr, etc.
b. Substituents or groups that shows +I effect: Alkyl groups such as -CH₃, -CH₂CH₃, etc.

Question 70.
Consider the following molecules and answer the questions:
CH₃ – CH₂ – CH₂ – Cl, CH₃ – CH₂ – CH₂ – Br, CH₃ – CH₂ – CH₂ – I.
i. What type of inductive effect is expected to operate in these molecules?
ii. Identify the molecules from these three, having the strongest and the weakest inductive effect.
Answer:
i. The groups responsible for inductive effect in these molecules are -Cl, -Br and -I, respectively. All these are halogen atoms which are more electronegative than carbon. Therefore, all of them exert -I effect, that is, electron withdrawing inductive effect.
ii. The -I effect of halogens is due to their electronegativity. A decreasing order of electronegativity in these halogens follows Cl > Br > I. Therefore, the strongest -I effect is expected in CH₃ – CH₂ – CH₂ – Cl, while the weakest -I effect is expected for CH₃ – CH₂ – CH₂ – I.

Question 71.
Which of the CH₃ – CHCl₂ and CH₃CH₂Cl is expected to have stronger -I effect?
Answer:
The group exerting -I effect is -Cl. In CH₃CH₂Cl, there is only one -Cl atom while in CH₃ – CHCl₂ there are two -Cl atoms. Therefore, CH₃ – CHCl₂ is expected to have strong -I effect.

Question 72.
Give an account of expected and observed values of carbon-carbon bond lengths in benzene.
Answer:

• In cyclic structure of benzene, three alternating C – C single bonds and C=C double bonds are present.
• Expected values of bond length of the C – C bond and C = C are 154 pm and 133 pm respectively.
• Experimental measurements show that benzene has a regular hexagonal shape and all the six carbon-carbon bonds have the same bond length of 138 pm, which is intermediate between C – C single bond and C=C double bond.
• This means that all the six carbon-carbon bonds in benzene are equivalent.

Note: Structure of benzene

Question 73.
What do you understand by the term conjugated system of π bonds?
Answer:
When Lewis structure of a compound has two or more multiple bonds alternating with single bonds, it is called a conjugated system of π bonds.
e.g. Benzene molecule
[Note: In such a system or in species having an atom carrying p orbital attached to a multiple bond, resonance theory is applicable.]

Question 74.
Identify the species that contains a conjugated system of π bonds. Explain your answer,
i. CH₂ = CH – CH₂ – CH = CH₂
ii. CH₂ = CH – CH = CH – CH₃
Answer:
i. It does not contain conjugated system of π bonds, as the two C = C double bonds are separated by two C – C single bonds.
ii. It contains a conjugated system of π bonds, as the two C = C double bonds are separated by only one C – C single bond.

Question 75.
Explain in detail the important points of resonance theory.
Answer:
Resonance theory:
i. The π electrons in conjugated system of π bonds are not localized to a particular π bond.
ii. For a compound having a conjugated system of π bonds (or similar other systems), two or more Lewis structures are written by showing movement of π electrons (that is, delocalization of π electrons) using curved arrows.
The Lewis structures so generated are linked by double headed arrow and are called resonance structures or contributing structures or cononical structures of the species. Thus, two resonance structures can be drawn for benzene by delocalizing or shifting the π electrons :

iii. The positions of the carbon atoms in the conjugated system of π bonds remain unchanged, but the positions of π electrons are different in different resonance structures.
e.g. In the resonance structure I of benzene there is a single bond between C1 and C2 while in the resonance structure II there is a double bond between C1 and C2.

iv. Any resonance structure is hypothetical and does not by itself represent any real molecule and can explain all the properties of the compound. The real molecule has, however, character of all the resonance structures those can be written. The real or actual molecule is said to be the resonance hybrid of all the resonance structures.
e.g. An actual benezene molecule is the resonance hybrid of structures I and II and exhibit character of both these structures. Its approximate representation can be shown as a dotted.circle inscribed in a regular hexagon. Thus, each carbon-carbon bond in benzene has single as well as double bond character and the ring has a regular hexagonal shape.

v. Hypothetical energy of an individual resonance structure can be calculated using bond energy values. The energy of actual molecule is, however, lower than that of any one of the resonance structures. In other words, resonance hybrid is more stable than any of the resonance structures. The difference in the actual energy and the lowest calculated energy of a resonance structure is called resonance stabilization energy or just resonance energy. Thus, resonance leads to stabilization of the actual molecule.

Question 76.
State the rules to be followed for writing resonating structures.
Answer:
Rules to be followed for writing resonating structures:

1. Resonance structures can be written only when all the atoms involved in the n conjugated system lie in the same place.
2. All the resonance structures must have the same number of unpaired electrons.
3. Resonance structures contribute to the resonance hybrid in accordance to their energy or stability. More stable (having low energy) resonance structures contribute largely and thus are important.

Question 77.
What are the important points considered while selecting the most stable resonance structure if there are several contributing/resonance structures for a compound?
Answer:
When several resonance structures are compared, then the resonance structure is considered to be more stable if it has:

• more number of covalent bonds,
• more number of atoms with complete octet or duplet,
• less separation, if any, of opposite charges,
• negative charge, if any, on more electronegative atom and positive charge, if any, on more electropositive atom and
• more dispersal of charge.

[Note: When all the resonance structures of a species are equivalent to each other, the species is highly resonance stabilized. For example, R – COO-, \(\mathrm{CO}_{3}^{2-}\)]

Question 78.
Write resonance structures of H – COO^– and comment on their relative stability.
Answer:
i. First the detailed bond structure of H – COO^– showing all the valence electron is drawn and then other resonance structures are generated using curved arrow to show movement of π-electrons.
ii. Two resonance structures are written for H – COO^–.

Both the resonance structures I and II are equivalent to each other, and therefore, are equally stable.

Question 79.
Identify the species which has resonance stabilization. Justify your answer.

Answer:
i. The bond structure shows that there is no π bond. Therefore, no resonance and no resonance stabilization.
ii.

N = O double b5itd is attached to ‘O’ which carries lone pair of electrons in a p orbital.
Therefore, resonance structures can be written as shown and species is resonance stabilized.
iii.

The Lewis structure shows two C = C double bonds alternating with a C – C single bond.
Therefore, resonance structures can be written as shown and the species is resonance stabilized.

Question 80.
Write three resonance structures for CH₃ – CH = CH – CHO. Indicate their relative stabilities and explain.
Answer:
Three resonance structures are:

Stability order: I > II > III
I: Contains more number of covalent bonds, each carbon atom and oxygen atom has complete octet, and involves no separation of opposite charges. Therefore, the most stable resonance structure.

II: Contains one covalent bond less than in I, one carbon (C⁺) has only 6 valence electrons, involves separation of opposite charges; the resonance structure II has -ve charge on more electronegative ‘O’ and +ve charge on more electropositive ‘C’. It has intermediate stability.

III: Contains one covalent bond less than in I, oxygen has only 6 valence electrons, involves separation of opposite charge, has -ve charge on the more electropositive ‘C’ and +ve charge on more electronegative ‘O’. All these factors are unfavourable for stability. Therefore, it is the least stable.

Question 81.
Define: Resonance effect.
Answer:
The polarity produced in the molecule by the interaction between conjugated n bonds (or that between n bond and p orbital on attached atom) is called the resonance effect or mesomeric effect.

Question 82.
Explain in short:
i. Positive resonance (+R) effect
ii. Negative resonance (-R) effect
Answer:
i. Positive resonance (+R) effect or electron donating/releasing resonance effect:
a. If the substituent group has a lone pair of electrons to donate to the attached K bond or conjugated system of π bonds, the effect is called +R effect.
b. The +R effect increases electron density at certain positions in a molecule.
e.g. +R effect in aniline increases the electron density at ortho and para positions.
c. Halogen, -OH, -OR, -O^–, -NH₂, -NHR, -NR₂, – NHCOR, -OCOR, etc. are the groups which show +R effect.

ii. Negative resonance (-R) effect:
a. If the substituent group has a tendency to withdraw electrons from the attached π bond or conjugated system of π bonds towards itself the effect is called -R effect.
b. The -R effect results in developing a positive polarity at certain positions in a molecule.
e.g. -R effect in nitrobenzene develops positive polarity at ortho and para positions.
c. -COOH, -CHO, – CO -, -CN, -NO₂, -COOR, etc., are the groups which represent -R effect.

Question 83.
Draw resonance structures showing +R effect in aniline.
Answer:
The following resonance structures can be drawn for aniline:

Question 84.
Draw resonance structures showing -R effect in nitrobenzene.
Answer:
The following resonance structures can be drawn for nitrobenzene:

Question 85.
Write a note on electromeric effect.
Answer:
Electromeric effect:
i. This is a temporary electronic effect exhibited by multiple-bonded groups in the excited state in the presence of a reagent.
ii. When a reagent approaches a multiple bond, the electron pair gets completely shifted to one of the multiply, bonded atoms, giving a charge separated structure.
iii.

This effect is temporary and disappears when the reagent is removed from the reacting system.

Question 86.
Explain the term hyperconjugation in short.
Answer:
Hyperconjugation:
i. Hyperconjugation is a permanent electronic effect.
ii. It explains the stability of a carbocation, free radical or alkenes.
iii. It involves delocalization of sigma electrons of a C – H bond of an alkyl group directly attached to a carbon atom, which is part of an unsaturated system or has an empty p orbital or a p orbital with an unpaired electron.
iv. Following species are stabilized by resonance:

Question 87.
Explain hyperconjugation in ethyl carbocation.
Answer:
i. In ethyl cation \(\mathrm{CH}_{3} \stackrel{+}{\mathrm{CH}}_{2}\), positively charged carbon atom is attached to a methyl group.
ii. The positively charged carbon atom has six electrons; it is sp² hybridized and has an empty p orbital available for hyperconjugation.
iii. One of the C – H bonds of the methyl group can align in plane of the empty p orbital. The sigma electrons constituting the C – H bond can be delocalized into this empty p orbital.
iv. Therefore, hyperconjugation arises due to the partial overlap of a C-H bond with the empty p orbital of an adjacent positively charged carbon atom. Thus, hyperconjugation is a σ-π conjugation.
v. Hyperconjugation structures in ethyl carbocation can be represented as:

vi. In the contributing structures, there is no covalent bond shown between the carbon and one of the α-hydrogens. Hence, hyperconjugation is also called as ‘no bond resonance’.
vii. This type of overlap stabilizes the cation, because the electron density from the adjacent a bond helps in dispersing the positive charge.

Question 88.
Explain the stability of tert-butyl cation, isopropyl cation, ethyl cation and methyl cation on the basis of hyperconjugation.
Answer:
i. Greater the number of alkyl groups attached to a positively charged carbon atom, more is the number of α-hydrogens, more is the hyperconjugation structures and more is the stability of the cation.

ii. Thus, the relative stability of the cations decreases in the order:
3° carbocation > 2° carbocation > 1° carbocation > Methyl cation

Question 89.
Explain hyperconjugation in propene.
Answer:
i. In propene, CH₃ – CH = CH₂, one of the sp² hybridized carbon atom of the double bond is attached to sp³ hybridized carbon atom of methyl group.
ii. One of the C-H bonds of the methyl group can align in plane of the p orbital of sp² hybridized C-atom and the electrons constituting the C-H bond in plane with this p orbital can then be delocalized into the p orbital.
iii. Therefore, hyperconjugation arises due to the partial overlap of a C-H bond with the p orbital of an adjacent sp² hybridized carbon atom.

iv. Hyperconjugation (no bond resonance) structures for propene can be represented as:

Question 90.
Write the Lewis dot structures of but-1-ene and but-2-ene? Also, write the bond line formula of both the compounds.
Answer:

Question 91.
Due to contamination by viruses, the hospital authorities had asked Ranjan, the ward boy, to keep cleaning the hospital lobby using some antiseptic. Ranjan would wipe the floor by adding Dettol to water and would always keep the premises clean. One of the active ingredients in Dettol is chloroxylenol (4-chloro-3,5-dimethylphenol). Ranjan was also actively associated with an NGO, which was involved in Swachh Bharat campaign. Based on this passage, answer the following questions.
i. Which functional groups are present in chloroxylenol?
ii. Write the bond line and molecular formula of chloroxylenol.
iii. Identify one group each in chloroxylenol which show +I and -I effect, respectively.
Answer:
i. chloroxylenol is

Functional groups present in chloroxylenol are chloro (-Cl) and phenolic -OH group.

ii. The bond line formula of chloroxylenol can be shown as,

Its molecular formula is C₈H₉OCl or C₈H₈ClOH

iii. Group which shows +I effect = -CH₃; group which shows -I effect = -Cl

Multiple Choice Questions

1. Which of the following method can be used to represent 3-D structure of organic molecules?
i. Wedge formula
ii. Fischer projection formula
iii. Newman projection formula
iv. Sawhorse formula
(A) Only ii and iii.
(B) Only i and iii.
(C) Only iii and iv.
(D) All of the above
Answer:
(D) All of the above

2. Which one is the INCORRECT statement?
(A) Open chain compounds are called aliphatic compounds.
(B) Unsaturated compounds contain multiple bonds in them.
(C) Saturated hydrocarbons are called alkenes.
(D) Aromatic compounds possess a characteristic aroma.
Answer:
(C) Saturated hydrocarbons are called alkenes.

3. Choose the INCORRECT statement from the following.
(A) Cyclohexane is an alicyclic compound.
(B) Pyridine is a heterocyclic compound.
(C) Piperidine is an aromatic compound.
(D) Tropone is a non-benzenoid compound.
Answer:
(C) Piperidine is an aromatic compound.

4. Which of the following is NOT a cyclic compound?
(A) Anthracene
(B) Pyrrole
(C) Phenol
(D) Neopentane
Answer:
(D) Neopentane

5. Which of the following is a cycloalkane?

Answer:

6. Which one of the following could be a cyclic alkane?
(A) C₅H₅
(B) C₃H₆
(C) C₄H₆
(D) C₂H₆
Answer:
(B) C₃H₆

7. Which of the following is a heterocyclic compound?
(A) Naphthalene
(B) Thiophene
(C) Phenol
(D) Aniline
Answer:
(B) Thiophene

8. Which of the following is NOT aromatic?
(A) Benzene
(B) Toluene
(C) Cyclopentane
(D) Phenol
Answer:
(C) Cyclopentane

9. Cyclohexene is …………….
(A) aromatic
(B) alicyclic
(C) benzenoid
(D) aliphatic
Answer:
(B) alicyclic

10. An organic compound ‘X’ (molecular formula C₆H₇O₂N) has six carbons in a ring system, two double bonds and also a nitro group as a substituent, ‘X’ is …………..
(A) homocyclic and aromatic
(B) homocyclic but not aromatic
(C) heterocyclic
(D) aromatic but not homocyclic
Answer:
(B) homocyclic but not aromatic

11. Which of the following structure represents an aldehyde?

Answer:

12. A member of a homologous series differs from immediate above or below member by …………… group.
(A) – CH₃
(B) – CH₂ –
(C) – CH₂CH₃
(D) – C₆H₅ –
Answer:
(B) – CH₂ –

13. Which of the following is NOT a branched chain alkyl group?
(A) Isobutyl group
(B) n-Butyl group
(C) sec-Butyl group
(D) tert-Butyl group
Answer:
(B) n-Butyl group

14. In IUPAC nomenclature, the number which indicates the position of the substituent is called ………….
(A) locant
(B) delocant
(C) prefix
(D) suffix
Answer:
(A) locant

15. The IUPAC name of the following compound is …………..
(A) 1,1 -dimethyl-2-ethylcyclohexane
(B) 2-ethyl-1,1 -dimethylcyclohexane
(C) 1 -ethyl-2,2-dimethylcyclohexane
(D) 2,2-dimethyl-1-ethylcyclohexane
Answer:
(B) 2-ethyl-1,1 -dimethylcyclohexane

16. Which is the CORRECT name of ………….

(A) Propyl ethanoate
(B) Ethyl propanoate
(C) Methyl butanoate
(D) Butyl methanoate
Answer:
(C) Methyl butanoate

17. Homolytic fission is NOT favourable in presence of …………..
(A) UV light
(B) catalyst like peroxide
(C) polar solvent
(D) high temperature
Answer:
(C) polar solvent

18. The total number of electrons in the carbon atom of methyl free radical is ………….
(A) six
(B) seven
(C) eight
(D) nine
Answer:
(B) seven

19. The most unstable carbocation amongst the following is ……………
(A) (CH₃)₃C⁺
(B) (CH3)₂CH⁺
(C) CH₃ – CH₂⁺
(D) CH₃⁺
Answer:
(D) CH₃⁺

20. Which of the following represents a pair of electrophiles?
(A) BF₃, H₂O
(B) AlCl₃, NH₃
(C) CN^–, ROH
(D) BF₃, AlCl₃
Answer:
(D) BF₃, AlCl₃

21. This group shows +I effect.
(A) -Br
(B) -CN
(C) -COOH
(D) -CH₂CH₃
Answer:
(D) -CH₂CH₃

22. Which of the following group shows negative resonance effect?
(A) -O-
(B) -COOH
(C) -NHCOR
(D) -NH₂
Answer:
(B) -COOH

23. Resonance is NOT exhibited by ………….
(A) phenol
(B) aniline
(C) nitrobenzene
(D) cyclohexane
Answer:
(D) cyclohexane

24. All bonds in benzene are equal due to ………….
(A) tautomerism
(B) metamerism
(C) resonance
(D) isomerism
Answer:
(C) resonance

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 11 Electric Current Through Conductors

Question 1.
Define current. State its formula and SI unit.
Answer:

1. Current is defined as the rate of flow of electric charge.
2. Formula: I = \(\frac {q}{t}\)
3. SI unit: ampere (A)

Question 2.
Derive an expression for a current generated due to flow of charged particles
Answer:
i. Consider an imaginary gas of both negatively and positively charged particles moving randomly in various directions across a plane P.

ii. In a time interval t, let the amount of positive charge flowing in the forward direction be q⁺ and the amount of negative charge flowing in the forward direction be q^–. Thus, the net charge flowing in the forward direction is q = q⁺ – q^–

iii. Let I be the current varying with time. Let ∆q be the amount of net charge flowing across the plane P from time t to t + At, i.e. during the time interval ∆t.

iv. Then the current is given by
I(t) = \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \mathrm{q}}{\Delta \mathrm{t}}\)
Flere, the current is expressed as the limit of the ratio (∆q/∆t) as ∆t tends to zero.

Question 3.
Match the amount of current generated A given in column – II with the sources given in column -I.

Column I	Column II
1. Lightening	a. Few amperes
2. House hold circuits	b. 10000 A
	c. Order of µA

Answer:

Column I	Column II
1. Lightening	b. 10000 A
2. House hold circuits	a. Few amperes

Question 4.
Which are the most common units of current used in semiconductor devices?
Answer:

1. milliampere (mA)
2. microampere (µA)
3. nanoampere (nA)

Question 5.
Six ampere current flows through a bulb. Find the number of electrons that should flow through the bulb in a time of 4 hrs.
Answer:
Given: I = 6 A, t = 4 hrs = 4 × 60 × 60 s
To find: Number of electrons (N)
Formula: I = \(\frac {q}{t}\) = \(\frac {Ne}{t}\)
Calculation: As we know, e = 1.6 × 10^-19 C
From formula,
N = \(\frac {It}{e}\) = \(\frac {6×4×60×60}{1.6×10^{-19}}\) 6x4x60x60 = 5.4 × 10²³

Question 6.
Explain flow of current in different conductor.
Answer:

1. A current can be generated by positively or negatively charged particles.
2. In an electrolyte, both positively and negatively charged particles take part in the conduction.
3.  In a metal, the free electrons are responsible for conduction. These electrons flow and generate a net current under the action of an applied electric field.
4. As long as a steady field exists, the electrons continue to flow in the form of a steady current.
5. Such steady electric fields are generated by cells and batteries.

Question 7.
State the sign convention used to show the flow of electric current in a circuit.
Answer:
The direction of the current in a circuit is drawn in the direction in which positively charged particles would move, even if the current is constituted by the negatively charged particles, (electrons), which move in the direction opposite to that the electric field.

Question 8.
Explain the concept of drift velocity with neat diagrams.
Answer:
i. When no current flows through a copper rod, the free electrons move in random motion. Therefore, there is no net motion of these electrons in any direction.

ii. If an electric field is applied along the length of the copper rod, a current is set up in the rod. The electrons inside rod still move randomly, but tend to ‘drift’ in a particular direction.

iii. Their direction is opposite to that of the applied electric field.

iv. The electrons under the action of the applied electric field drift with a drift speed v_d.

Question 9.
What is current density? State its SI unit.
Answer:
i. Current density at a point in a conductor is the amount of current flowing per unit area of the conductor.
Current density, J = \(\frac {I}{A}\)
where, I = Current
A = Area of cross-section

ii. SI unit: A/m²

Question 10.
A metallic wire of diameter 0.02 m contains 10 free electrons per cubic metre. Find the drift velocity for free electrons, having an electric current of 100 amperes flowing through the wire.
(Given: charge on electron = 1.6 × 10^-19C)
Answer:
Given: e = 1.6 × 10^-19 C, n = 10²⁸ electrons/m³,
D = 0.02 m, r = D/2 = 0.01 m,
I = 100 A
To find: Drift velocity (v_d)
Formula: v_d = \(\frac {I}{nAe}\)
Calculation: From formula,
v_d = \(\frac {I}{nπr^2e}\)
∴ v_d = \(\frac {100}{10^{28}×3.142×10^{-4}×1.6×10^{-19}}\)
= \(\frac {10^{-3}}{3.142×1.6}\)
= 1.989 × 10^-4 m/s

Question 11.
A copper wire of radius 0.6 mm carries a current of 1 A. Assuming the current to be uniformly distributed over a cross sectional area, find the magnitude of current density. Answer:
Given: r = 0.6 mm = 0.6 × 10^-3 m, I = 1 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {1}{3.142×(0.6)^2×10^{-6}}\)
= 0.884 × 10⁶ A/m²

Question 12.
A metal wire of radius 0.4 mm carries a current of 2 A. Find the magnitude of current density if the current is assumed to be uniformly distributed over a cross sectional area.
Answer:
Given: r = 0.4 mm = 0.4 × 10^-3 m, I = 2 A
To find: Current density (J)
Formula: J = \(\frac {I}{A}\)
Calculation: From formula,
J = \(\frac {2}{3.142×(0.4)^2×10^{-6}}\)
= 3.978 × 10⁶ A/m²

Question 13.
State and explain ohm’s law.
Answer:
Statement: The current I through a conductor is directly proportional to the potential difference V applied across its two ends provided the physical state of the conductor is unchanged.
Explanation:
According to ohm’s law,
I ∝ V
∴ V = IR or R = \(\frac {V}{I}\)
where, R is proportionality constant and is called the resistance of the conductor.

Question 14.
Draw a graph showing the I-V curve for a good conductor and ideal conductor.
Answer:

Question 15.
Define one ohm.
Answer:
If potential difference of 1 volt across a conductor produces a current of 1 ampere through it, then the resistance of the conductor is one ohm.

Question 16.
Define conductance. State its SI unit.
Answer:

1. Reciprocal of resistance is called conductance.
   C = \(\frac {I}{R}\)
2. S.I unit statement or Ω^-1

Question 17.
Explain the concept of electrical conduction in a conductor.
Answer:

1. Electrical conduction in a conductor is due to mobile charge carriers (electrons).
2. These conduction electrons are free to move inside the volume of the conductor.
3. During their random motion, electrons collide with the ion cores within the conductor. Assuming that electrons do not collide with each other these random motions average out to zero.
4. On application of an electric field E, the motion of the electron is a combination of the random motion of electrons due to collisions and that due to the electric field \(\vec{E}\).
5. The electrons drift under the action of the field \(\vec{E}\) and move in a direction opposite to the direction of the field \(\vec{E}\). In this way electrons in a conductor conduct electricity.

Question 18.
Derive expression for electric field when an electron of mass m is subjected to an electric field (E).
Answer:
i. Consider an electron of mass m subjected to an electric field E. The force experienced by the electron will be \(\vec{F}\) = e\(\vec{E}\).

ii. The acceleration experienced by the electron will then be
\(\vec{a}\) = \(\frac {e\vec{E}}{m}\) …………. (1)

iii. The drift velocities attained by electrons before and after collisions are not related to each other.

iv. After the collision, the electron will move in random direction, but will still drift in the direction opposite to \(\vec{E}\).

v. Let τ be the average time between two successive collisions.

vi. Thus, at any given instant of time, the average drift speed of the electron will be,
v_d = a τ = \(\frac {eEτ}{m}\) ………………(From 1)
v_d = \(\frac {eEτ}{m}\) = \(\frac {J}{ne}\) ……………(2) [∵ v_d = \(\frac {J}{ne}\)]

vii. Electric field is given by,
E = (\(\frac {m}{e^2nτ}\))J ………… (from 2)
= ρJ = [∵ ρ = \(\frac {m}{ne^2τ}\)]
where, ρ is resistivity of the material.

Question 19.
A Flashlight uses two 1.5 V batteries to provide a steady current of 0.5 A in the filament. Determine the resistance of the glowing filament.
Answer:
Given: For each battery, V₁ = V₂ = 1.5 volt,
I = 0.5 A
To find: Resistance (R)
Formula: V = IR
Calculation: Total voltage, V = V₁ + V₂ = 3 volt
From formula,
R = \(\frac {V}{I}\) = \(\frac {3}{0.5}\) = 6.0 Ω

Question 20.
State an expression for resistance of non-ohmic devices and draw I-V curve for such devices.
Answer:
i. Resistance (R) of a non-ohmic device at a particular value of the potential difference V is given by,
R = \(\lim _{\Delta I \rightarrow 0} \frac{\Delta V}{\Delta I}=\frac{d V}{d I}\)
where, ∆V = potential difference between the
two values of potential V – \(\frac {∆V}{2}\) to V + \(\frac {∆V}{2}\),
and ∆I = corresponding change in the current.

Question 21.
Derive an expression for decrease in potential energy when a charge flows through an external resistance in a circuit.
Answer:
i. Consider a resistor AB connected to a cell in a circuit with current flowing from A to B.

ii. The cell maintains a potential difference V between the two terminals of the resistor, higher potential at A and lower at B.

iii. Let Q be the charge flowing in time ∆t through the resistor from A to B.

iv. The potential difference V between the two points A and B, is equal to the amount of work (W) done to carry a unit positive charge from A to B.
∴ V = \(\frac {W}{Q}\)

v. The cell provides this energy through the charge Q, to the resistor AB where the work is performed.

vi. When the charge Q flows from the higher potential point A to the lower potential point B, there is decrease in its potential energy by an amount
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

Question 22.
Prove that power dissipated across a resistor is responsible for heating up the resistor. Give an example for it.
OR
Derive an expression for the power dissipated across a resistor in terms of its resistance R.
Answer:
i. When a charge Q flows from the higher potential point to the lower potential point, its potential energy decreases by an amount,
∆U = QV = I∆tV
where I is current due to the charge Q flowing in time ∆t.

ii. By the principle of conservation of energy, this energy is converted into some other form of energy.

iii. In the limit as ∆t → 0, \(\frac {dU}{dt}\) = IV
Here, \(\frac {dU}{dt}\) is power, the rate of transfer of energy ans is given by p = \(\frac {dU}{dt}\) = IV
Hence, power is transferred by the cell to the resistor or any other device in place of the resistor, such as a motor, a rechargeable battery etc.

iv. Due to the presence of an electric field, the free electrons move across a resistor and their kinetic energy increases as they move.

v. When these electrons collide with the ion cores, the energy gained by them is shared among the ion cores. Consequently, vibrations of the ions increase, resulting in heating up of the resistor.

vi. Thus, some amount of energy is dissipated in the form of heat in a resistor.

vii. The energy dissipated per unit time is actually the power dissipated which is given by,
P = \(\frac {V^2}{R}\) = I²R
Hence, it is the power dissipation across a resistor which is responsible for heating it up.

viii. For example, the filament of an electric bulb heats upto incandescence, radiating out heat and light.

Question 23.
Calculate the current flowing through a heater rated at 2 kW when connected to a 300 V d. c. supply.
Answer:
Given: P = 2 kW = 2000 W, V = 300 V
To find: Current (I)
Formula: P = IV
Calculation: From formula,
I = \(\frac {P}{V}\) = \(\frac {2000}{300}\) = 6.67 A

Question 24.
An electric heater takes 6 A current from a 230 V supply line, calculate the power of the heater and electric energy consumed by it in 5 hours.
Answer:
Given: I = 6 A, V = 230 V, t = 5 hours
To find: Power (P), Energy consumed
Formulae: i. P = IV
ii. Energy consumed = power × time
Calculation: From formula (i),
P = 6 × 230
= 1380 W = 1.38 kW
From formula (ii),
Energy consumed = 1.38 × 5 = 6.9 kWh
= 6.9 units

Question 25.
When supplied a voltage of 220 V, an electric heater takes 6 A current. Calculate the power of heater and electric energy consumed by its in 2 hours?
Answer:
Given: I = 6 A, V = 220 volt, t = 2 hour
To find: i. Power of heater (P)
ii. Electric energy consumed (E)
Formulae: i. P = IV
ii. Electric energy consumed
= Power × time
Calculation: From formula (i),
P = 6 × 220 = 1320 W = 1.32 kW
From formula (ii),
Electric energy consumed
= 1.32 × 2 = 2.64 kWh = 2.64 units

Question 26.
Explain the colour code system for resistors with an example.
Answer:
i. In colour code system, resistors has 4 bands on it.

ii. In the four band resistor, the colour code of the first two bands indicate two numbers and third band often called decimal multiplier.

iii. The fourth band separated by a space from the three value bands, indicates tolerance of the resistor.

iv. Following table represents the colour code of carbon resistor.

v. Example:
Let the colours of the rings of a resistor starting from one end be brown, red and orange and gold at the other end. To determine resistance of resistor we have,
x = 1, y = 2, z = 3 (From colour code table)
∴ Resistance = xy × 10^z Ω ± tolerance
= 12 × 10³ Ω ± 5%
= 12 kΩ ± 5%
[Note: To remember the colours in order learn the Mnemonics: B.B. ROY of Great Britain had Very Good Wife]

Question 27.
Explain the concept of rheostat.
Answer:

1. A rheostat is an adjustable resistor used in applications that require adjustment of current or resistance in an electric circuit.
2. The rheostat can be used to adjust potential difference between two points in a circuit, change the intensity of lights and control the speed of motors, etc.
3. Its resistive element can be a metal wire or a ribbon, carbon films or a conducting liquid, depending upon the application.
4. In hi-fi equipment, rheostats are used for volume control.

Question 28.
Explain series combination of resistors.
Answer:
i. In series combination, resistors are connected in single electrical path. Hence, the same electric current flows through each resistor in a series combination.

ii. Whereas, in series combination, the supply voltage between two resistors R₁ and R₂ is divided into V₁ and V₂ respectively.

iii. According to Ohm’s law,
R₁ = \(\frac {V_1}{I}\), R₂ = \(\frac {v_2}{I}\)
Total Voltage, V = V₁ + V₂
= I(R₁ + R₂)
∴ V = I R_s
Thus, the equivalent resistance of the series circuit is, R_s = R₁ + R₂

iv. When a number of resistors are connected in series, the equivalent resistance is equal to the sum of individual resistances.
For ‘n’ number of resistors,
R_s = R₁ + R₂ + R₂ + ………….. + R_n = \(\sum_{i=1}^{i=n} R_{i}\)

Question 29.
Explain parallel combination of resistors.
Answer:
i. In parallel combination, the resistors are connected in such a way that the same voltage is applied across each resistor.

ii. A number of resistors are said to be connected in parallel if all of them are connected between the same two electrical points each having individual path.

iii. In parallel combination, the total current I is divided into I, and I2 as shown in the circuit diagram.

iv. Since voltage V across them remains the same,
I = I₁ + I₂
where I₁ is current flowing through R₁ and I₂ is current flowing through R₂.

v. When Ohm’s law is applied to R₁,
V = I₁R₁
i.e. I₁ = \(\frac {V}{R_1}\) ………(1)
When Ohm’s law applied to R2,
V = I₂R₂
i.e., I₂ = \(\frac {V}{R_2}\) …………(2)

vi. Total current is given by,
I = I₁ + I₂
∴ I = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\) ………[From (1) and (2)]
Since, I = \(\frac {V}{R_p}\)
∴ \(\frac {V}{R_p}\) = \(\frac {V}{R_1}\) + \(\frac {V}{R_2}\)
∴ \(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\)
Where, R_p is the equivalent resistance in parallel combination.

vii. If ‘n’ number of resistors R₁, R₂, R₃, ………….. R_n are connected in parallel, the equivalent resistance of the combination is given by
\(\frac {1}{R_p}\) = \(\frac {1}{R_1}\) + \(\frac {1}{R_2}\) + \(\frac {1}{R_3}\) ……….. + \(\frac {1}{R_n}\) = \(\sum_{i=1}^{\mathrm{i}=\mathrm{n}} \frac{1}{\mathrm{R}}\)
Thus, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of individual resistances.

Question 30.
Colour code of resistor is Yellow-Violet- Orange-Gold. Find its value.
Answer:

	Yellow (x)	Violet (y)	Orange (z)	Gold (T%)
Value	4	7	3	± 5

Value of resistance: xy × 10^z Ω ± tolerance
∴ Value of resistance = 47 × 10³ Ω ± 5%
= 47 kΩ ± 5%

Question 31.
From the given value of resistor, find the colour bands of this resistor.
Value of resistor: 330 Ω
Answer:
Value = 330 Ω = 33 × 10¹ Ω = xy × 10^z Ω

Value	3	3	1
Colour	Orange (x)	Orange (y)	Broen(z)

ii. Given: Green – Blue – Red – Gold

Question 32.
Evaluate resistance for the following colour-coded resistors:
i. Yellow – Violet – Black – Silver
ii. Green – Blue – Red – Gold
ill. Brown – Black – Orange – Gold
Answer:
i. Given: Yellow – Violet – Black – Silver
To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%)Ω

Colour	Yellow (x)	Violet (y)	Black (z)	Sliver (T%)
Code	4	7	0	±10

Hence x = 4, y = 7, z = 0, T = 10%
Value of resistance = (xy ×10^z ± T%) Ω
= (47 × 10° ± 10%) Ω
Value of resistance = 47 Ω ± 10%

To find: Value of resistance
Formula: Value of resistance
= (xy × 10^z ± T%) Ω
Calculation:

Colour	Green (x)	Blue (y)	Red (z)	Gold (T%)
Code	5	6	2	±5

Hence x = 5, y = 6, z = 2, T = 5%
Value of resistance = (xy × 10^z ± T%) Q
= 56 × 102 Ω ± 5%
= 5.6 k Ω ± 5%

iii. Given: Brown – Black – Orange – Gold
To find: Value of the resistance
Formula: Value of the resistance
= (xy × 10^z ± T%) Ω
Calculation:

Colour	Brown (x)	Black (y)	Orange (z)	Gold (T%)
Code	1	0	3	±5

Hence x = 1, y = 0, z = 3, T = 5%
Value of resistance = (xy × 10^z ± T%) Ω
= 10 × 10³ Ω ± 5%
= 10 kΩ ± 5%

Question 33.
Calculate
i. total resistance and
ii. total current in the following circuit.
R₁ = 3 Ω, R₂ = 6 Ω, R₃ = 5 Ω, V = 14 V

Answer:
i. R₁ and R₂ are connected in parallel. This combination (R_p) is connected in series with R₃.
∴ Total resistance, R_T = R_p + R₃
R_p = \(\frac {R_1R_2}{R_1+R_2}\) = \(\frac {3×6}{3+6}\) = 2 Ω
∴ R_T = 2+ 5 = 7 Ω

ii. Total current: I = \(\frac {V}{R_T}\) = \(\frac {14}{7}\) = 2 A

Question 34.
State the factors affecting resistance of a conductor.
Answer:
Factors affecting resistance of a conductor:

1. Length of conductor
2. Area of cross-section
3. Nature of material

Question 35.
Derive expression for specific resistance of a material.
Answer:
At a particular temperature, the resistance (R) of a conductor of uniform cross section is
i. directly proportional to its length (l),
i.e., R ∝ l ……….. (1)

ii. inversely proportional to its area of cross section (A),
R ∝ \(\frac {1}{A}\) ……….. (1)
From equations (1) and (2),
R = ρ\(\frac {l}{A}\)
where ρ is a constant of proportionality and it is called specific resistance or resistivity of the material of the conductor at a given temperature.

iii. Thus, resistivity is given by,
ρ = \(\frac {RA}{l}\)

Question 36.
State SI unit of resistivity.
Answer:
SI unit of resistivity is ohm-metre (Ω m).

Question 37.
What is conductivity? State its SI unit.
Answer:
i. Reciprocal of resistivity is called as conductivity of a material.
Formula: σ = \(\frac {1}{ρ}\)
ii. SI unit: (\(\frac {1}{ohm m}\)) or siemens/metre

Question 38.
Explain the similarities between R = \(\frac {V}{I}\) and ρ = \(\frac {E}{J}\)
Answer:

1. Resistivity (ρ) is a property of a material, while the resistance (R) refers to a particular object.
2. The electric field \(\vec{E}\) at a point is specified in a material with the potential difference across the resistance and the current density \(\vec{J}\) in a material is specified instead of current I in the resistor.
3. For an isotropic material, resistivity is given by ρ = \(\frac {E}{J}\)
   For a particular resistor, the resistance R given by, R = \(\frac {V}{I}\)

Question 39.
State expression for current density in terms of conductivity.
Answer:
Current density, \(\vec{J}\) = \(\frac {1}{ρ}\) \(\vec{E}\) = σ \(\vec{E}\)
where, ρ = resistivity of the material
E = electric field intensity
σ = conductivity of the material

Question 40.
Calculate the resistance per metre, at room temperature, of a constantan (alloy) wire of diameter 1.25 mm. The resistivity of constantan at room temperature is 5.0 × 10^-7 Ωm.
Answer:
Given: ρ = 5.0 × 10^-7 Ω m, d = 1.25 × 10^-3 m,
∴ r = 0.625 × 10^-3 m
To find: Resistance per metre (\(\frac {R}{l}\))
Formula: ρ = \(\frac {RA}{l}\)
Calculation:
From formula,
\(\frac{\mathrm{R}}{l}=\frac{\rho}{\mathrm{A}}=\frac{\rho}{\pi \mathrm{r}^{2}}\)
= \(\frac{5 \times 10^{-7}}{3.142 \times\left(0.625 \times 10^{-3}\right)^{2}}\)
= \(\frac{5}{3.142 \times 0.625^{2}} \times 10^{-1}\)
= { antilog [log 5 – log 3.142 -2 log 0.625]} × 10^-1
= {antilog [ 0.6990 – 0.4972 -2(1.7959)]} × 10^-1
= {antilog [0.2018- 1.5918]} × 10^-1
= {antilog [0.6100]} × 10^-1
= 4.074 × 10^-1
∴ \(\frac {R}{l}\) ≈ 0.41 Ω m^-1

Question 41.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6 × 10^-7 m², and its resistance is measured to be 5 Ω. What is the resistivity of the material at the temperature of the experiment?
Answer:
Given: l = 15 m, A = 6.0 × 10^-7 m², R = 5 Ω
To find: Resistivity (ρ)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: From formula,
ρ = \(\frac {5×6×10^{-7}}{15}\)
∴ ρ = 2 × 10^-7 Ω m

Question 42.
A constantan wire of length 50 cm and 0.4 mm diameter is used in making a resistor. If the resistivity of constantan is 5 × 10^-7m, calculate the value of the resistor.
Answer:
Given: l = 50 cm = 0.5 m,
d = 0.4 mm = 0.4 × 10^-3 m,
r = 0.2 × 10^-3 m, p = 5 × 10^-7 Ωm
To Find: Value of resistor (R)
Formula: ρ = \(\frac {RA}{l}\)
Calculation: from formula,

Question 43.
The resistivity of nichrome is 10^-6 Ωm. What length of a uniform wire of this material and of 0.2 mm diameter will have a resistance of 200 ohm?
Answer:
Given: ρ = 10^-6 Ω m, d = 0.2 mm,
∴ r = 0.1 mm = 0.1 × 10^-3 m, R = 200 Ω
To find: Length (l)
Formula: R = \(\frac {ρl}{A}\) = \(\frac {ρl}{πr^2}\)
Calculation: From formula,
l = \(\frac {πr^2}{ρ}\)
∴ l = \(\frac{200 \times 3.142 \times\left(0.1 \times 10^{-3}\right)^{2}}{10^{-6}}\) = 6 284 m

Question 44.
A wire of circular cross-section and 30 ohm resistance is uniformly stretched until its new length is three times its original length. Find its resistance.
Answer:
Given: R₁ = 30 ohm,
l₁ = original length, A₁ = original area,
l₂ = new length, A₂ = new area
l₂= 3l₁
To find: Resistance (R₂)
Formula: R= ρ\(\frac {l}{A}\)
Calculation: From formula,

The volume of wire remains the same in two cases, we have

Question 45.
Define temperature coefficient of resistivity. State its SI unit.
Answer:
i. The temperature coefficient of resistivity is defined as the increase in resistance per unit original resistance at the chosen reference temperature, per degree rise in temperature.
α = \(\frac{\rho-\rho_{0}}{\rho_{0}\left(T-T_{0}\right)}\)
= \(\frac{\mathrm{R}-\mathrm{R}_{0}}{\mathrm{R}_{0}\left(\mathrm{~T}-\mathrm{T}_{0}\right)}\)
For small difference in temperatures,
α = \(\frac {1}{R_0}\) \(\frac {dR}{dT}\)

ii. SI unit: °C^-1 (per degree Celsius) or K^-1 (per kelvin).

Question 46.
Give expressions for variation of resistivity and resistance with temperature. Represent graphically the temperature dependence of resistivity of copper.
Answer:
i. Resistivity is given by,
ρ = ρ₀ [1 + α (T – T₀)] where,
T₀ = chosen reference temperature
ρ₀ = resistivity at the chosen temperature
α = temperature coefficient of resistivity
T = final temperature

ii. Resistance is given by,
R = R₀ [1+ α (T – T₀)]
Where,
T₀ = chosen reference temperature
R₀ = resistance at the chosen temperature
α = temperature coefficient of resistance
T = final temperature

iii. For example, for copper, the temperature dependence of resistivity can be plotted as shown:

Question 47.
What is super conductivity?
Answer:

1. The resistivity of a metal decreases as the temperature decreases.
2. In case of some metals and metal alloys, the resistivity suddenly drops to zero at a particular temperature (T_c), this temperature is called critical temperature.
3. Super conductivity is the phenomenon where resistivity of a material becomes zero at particular temperature.
4. For example, mercury loses its resistance completely to zero at 4.2 K.

Question 48.
A piece of platinum wire has resistance of 2.5 Ω at 0 °C. If its temperature coefficient of resistance is 4 × 10^-3/°C. Find the resistance of the wire at 80 °C.
Answer:
Given: R₀ = 2.5 Ω
α = 4 × 10^-3/°C = 0.004/°C
T = 80 °C
To find: Resistance at 80 °C (R_T)
Formula: R_T = R₀(l + α T)
Calculation: From formula,
R_T = 2.5 [1+ (0.004 × 80)]
= 2.5(1 + 0.32)
R_T = 2.5 × 1.32
R_T = 3.3 Ω

Question 49.
The resistance of a tungsten filament at 150 °C is 133 ohm. What will be its resistance at 500 °C? The temperature coefficient of resistance of tungsten is 0.0045 per °C.
Answer:
Given: Let resistance at 150 °C be R₁ and resistance at 500 °C be R₂
Thus,
R₁= 133 Ω, α = 0.0045 °C^-1
To find: Resistance (R₂)
Formula: R_T = R₀ (1 + α∆T)
Calculation:
From formula,
R₁ = R₀ (1 + α × 150)
∴ 133 = R₀(1 + 0.0045 × 150) ……….(i)
R₂ = R₀ (1 + α × 500)
∴ R₂ = R₀(1 + 0.0045 × 500) ………(ii)
Dividing equation (ii) by (i), we get
\(\frac{\mathrm{R}_{2}}{133}=\frac{1+(0.0045 \times 500)}{1+(0.0045 \times 150)}=\frac{3.25}{1.675}\)
∴ R₂ = \(\frac {3.25}{1.675}\) × 133 = 258 Ω

Question 50.
A silver wire has resistance of 2.1 Ω at 27.5 °C. If temperature coefficient of silver is 3.94 × 10^-3/°C, find the silver wire resistance at 100 °C.
Answer:
Given: R₁ = 2.1 Ω, T₁ = 27.5 °C,
α = 3.94 × 10^-3/°C, T₂ = 100 °C
To find: Resistance (R₂)
Formula: R_T = R_o (1 + αT)
Calculation:
From the formula,
R₁ = R₀(1 + α × 27.5) ……….. (i)
R₂ = R₀(l + α × 100) ………….. (ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+\left(3.94 \times 10^{-3} \times 27.5\right)}{1+\left(3.94 \times 10^{-3} \times 100\right)}\)
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1.10835}{1.394}\) = 0.795
∴ R₂ = \(\frac{\mathrm{R}_{1}}{0.795}=\frac{2.1}{0.795}\) = 2.641 Ω

Question 51.
At what temperature would the resistance of a copper conductor be double its resistance at 0 °C?
(a for copper = 3.9 × 10^-3/°C)
Answer:
Given: Let the resistance of the conductor at 0°C be R₀
R₁ = R₀ at T₁ = 0°C
R₂ = 2R₀ at T₂ = T
To find: Final temperature (T)
Formula: α = \(\frac {R_2-R_1}{R_1(T_2-T_1)}\)
Calculation: From formula,
α = \(\frac {2R_0-R_0}{R_1(T_2-T_1)}\) = \(\frac {1}{T}\)
∴ T = \(\frac {1}{α}\) = \(\frac {1}{3.9×10^{-3}}\) ≈ 256 °C

Question 52.
A conductor has resistance of 15 Ω at 10 °C and 18 Ω at 400 °C. Find the temperature coefficient of resistance of the material.
Answer:
Given: R₁ = 15 Ω, T₁ = 10 °C, R₂ = 18 Ω,
T₂ = 400 °C
To find: Temperature coefficient of resistance (α)
Formula: R_T = R₀ (1 + αT)
Calculation:
From formula,
R₁ = R₀ (1 + α × 10) ……..(i)
R₂ = R₀ (1 + α × 400) …….(ii)
Dividing equation (i) by (ii), we get,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{1+(\alpha \times 10)}{1+(\alpha \times 400)}\)
∴ \(\frac{15}{18}=\frac{1+10 \alpha}{1+400 \alpha}\)
∴ 18 + 180 α = 15 + 6000 α
∴ 5820 α = 3
∴ α = \(\frac {3}{5820}\) = 5.155 × 10^-4/°C

Question 53.
Write short note on e.m.f. devices.
Answer:

1. When charges flow through a conductor, a potential difference get established between the two ends of the conductor.
2. For a steady flow of charges, this potential difference is required to be maintained across the two ends of the conductor.
3. There is a device that does so by doing work on the charges, thereby maintaining the potential difference. Such a device is called an emf device and it provides the emf E.
4. The charges move in the conductor due to the energy provided by the emf device. This energy is supplied by the e.m.f. device on account of its work done.
5. Power cells, batteries, Solar cells, fuel cells, and even generators, are some examples of emf devices.

Question 54.
Explain working of a circuit when connected to emf device.
Answer:
i. A circuit is formed with connecting an emf device and a resistor R. Flere, the emf device keeps the positive terminal (+) at a higher electric potential than the negative terminal (-)

ii. The emf is represented by an arrow from the negative terminal to the positive terminal.

iii. When the circuit is open, there is no net flow of charge carriers within the device.

iv. When connected in a circuit, the positive charge carriers move towards the positive terminal which acts as cathode inside the emf device.

v. Thus, the positive charge carriers move from the region of lower potential energy, to the region of higher potential energy.

vi. Consider a charge dq flowing through the cross section of the circuit in time dt.

vii. Since, same amount of charge dq flows throughout the circuit, including the emf device. Hence, the device must do work dW on the charge dq, so that the charge enters the negative terminal (low potential terminal) and leaves the positive terminal (higher potential terminal).

viii. Therefore, e.m.f. of the emf device is,
E = \(\frac {dW}{dq}\)
The SI unit of emf is joule/coulomb (J/C).

Question 55.
What is an ideal e.m.f. device?
Answer:

1. In an ideal e.m.f. device, there is no internal resistance to the motion of charge carriers.
2. The emf of the device is then equal to the potential difference across the two terminals of the device.

Question 56.
What is a real e.m.f. device?
Answer:

1. In a real emf device, there is an internal resistance to the motion of charge carriers.
2. If such a device is not connected in a circuit, there is no current through it.

Question 57.
Derive an expression for current flowing through a circuit when an external resistance is connected to a real e.m.f. device.
Answer:

i. If a current (I) flows through an emf device, there is an internal resistance (r) and the emf (E) differs from the potential difference across its two terminals (V).
V = E – Ir ……… (1)

ii. The negative sign is due to the fact that the current I flows through the emf device from the negative terminal to the positive terminal.

iii. By the application of Ohm’s law,
V = IR …….(2)
From equations (1) and (2),
IR = E – Ir
∴ \(\frac {E}{R+r}\)

Question 58.
Explain the conditions for maximum current.
Answer:

1. Current in a circuit is given by, I = \(\frac {E}{R+r}\)
2. Maximum current can be drawn from the emf device, only when R = 0, i.e.
   I_max = \(\frac {E}{R}\)
3. I_max is the maximum allowed current from an emf device (or a cell) which decides the maximum current rating of a cell or a battery.

Question 59.
A network of resistors is connected to a 14 V battery with internal resistance 1 Q as shown in the circuit diagram.
i. Calculate the equivalent resistance,
ii. Current in each resistor,
iii. Voltage drops V_AB, V_BC and V_DC.

Answer:

For equivalent resistance (R_eq):
R_AB is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\)
∴ R_AB = 2 Ω
R_BC = R₃ = 1 Ω
Also, R_CD is given as,
\(\frac{1}{\mathrm{R}_{\mathrm{CD}}}=\frac{1}{\mathrm{R}_{4}}+\frac{1}{\mathrm{R}_{5}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}\)
∴ R_CD = 3 Ω
∴ R_eq = R_AB + R_BC + R_CD
= 2 + 1 + 3 = 6Ω

ii. Current through each resistor:
Total current, I = \(\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eq}}+\mathrm{r}}\) = \(\frac {14}{6+1}\) = 2 A
Across AB, as, R₁ = R₂
V₁ = V₂
∴I₁ × 4 = I₂ × 4
∴ I₁ = I₂
But, I₁ + I₂ = I
∴ 2I₁ = I
∴ I₁ = I₂ =1 A ….(∵I = 2 A)
Similarly, as R₄ = R₅
I₃ = I₄ = 1 A
Current through resistor BC is same as I.
∴ I_BC = 2 A

iii. Voltage drops across AB, BC and CD:
V_AB = IR_AB = 2 × 2 = 4 V
V_BC = IR_BC = 2 × 1 = 2 V
V_CD = IR_CD = 2 × 3 = 6 V

Question 60.
i. Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery.
Answer:
Given: R₁ = 2Ω, R₂ = 4 Ω, R₃ = 5 Ω,
V = 20 V
To Find: i. Total resistance (R)
ii. Current through each resistor (I₁, I₂, I₃ respectively)
iii. Total current (I)
Formulae:
i. \(\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}\)
ii. V = IR
iii. Total current, I = I₁ +I₂ + I₃
Calculation
From formula (i):
\(\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}\)
∴ R = \(\frac {20}{19}\) Ω
From formula (ii):

From formula (iii):
I = 10 + 5 + 4
∴ I = 19 A

Question 61.
i. Three resistors 1 Ω, 2 Ω and 3 Ω are combined in series. What is the total resistance of the combination?
ii. If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:
Given: R₁ = 1Ω, R₂ = 2 Ω, R₃ = 3 Ω,
V = 12 V
To Find: i. Total resistance (R)
ii. P.D Across R₁, R₂, R₃ (V₁, V₂, V₃ respectively)
Formulae:
i. R_s = R₁ + R₂ + R₃
ii. V = IR
Calculation
From formula (i):
Rs = l + 2 + 3 = 6 Ω
From formula (ii),
1 = \(\frac {V}{R}\) = \(\frac {12}{6}\) = 2A
∴ V₁ = IR₁ = 2 × 1 = 2 V
∴ V₂ = IR₂ = 2 × 2 = 4 V
∴ V₃ = IR₃ = 2 × 3 = 6 V

Question 62.
A voltmeter is connected across a battery of emf 12 V and internal resistance of 10 Ω. If the voltmeter resistance is 230 Ω, what reading will be shown by the voltmeter? Answer:
Given: E = 12 volt, r = 10 Ω, R = 230 Ω
To find: Reading shown by voltmeter (V)
Formula: i. I = \(\frac {E}{R+r}\)
ii. V = E – Ir
Calculation
From formula (i),
I = \(\frac{12}{230+10}=\frac{12}{240}=\frac{1}{20} \mathrm{~A}\)
From formula (ii),
V= 12 – \(\frac {1}{20}\) × 10 = 12 – 0.5
= 11.5 volt

Question 63.
A battery of e.m.f. 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer:
Given: E = 10 V, r = 3 Ω, I = 0.5 A
To find: i. Resistance of resistor (R)
ii. Terminal voltage of battery (V)
Formula: I = \(\frac {E}{R+r}\)
Calculation: From formula, R = \(\frac {E}{I}\) – r
∴ R = \(\frac {10}{0.5}\)– 3 = 17 Ω
∴ V = IR = 0.5 × 17 = 8.5 volt

Question 64.
How many cells each of 1.5 V/500 mA rating would be required in series-parallel combination to provide 1500 mA at 3 V?
Answer:
21 = ………… = 1.5 V (given)
I₁ = I₂ = …………… = 500 mA (given)
1500 mA at 3 V is required.
To determine required number of cells:
For series V = V₁ + V₂ + ………….., and current remains same.
For parallel I = I₁ + I₂ + ………, and voltage remains same.
To achieve battery output of 3V, the cells should be connected in series.
If n are the number of cells connected in series, then
V = V₁ + V₂ + …………. + V_n
∴ V = nV₁
∴ 3 = n × 1.5
∴ n = 2 cells in series
The series combination of two cells in series will give a current 500 mA.
To achieve output of 1500 mA, the number of batteries (n) connected in parallel, each one having output 3V is,
I = I₁ + I₂ + ………. + I_n
∴ I = nI₁
∴ 1500 = n × 500
∴ n = 3 batteries each of two cells
∴ No of cells required are 2 × 3 = 6 .
∴ Number of cells = 6
The six cells must be connected as shown

Question 65.
Explain the concept of series combination of cells.
Answer:
i. In a series combination, cells are connected in single electrical path, such that the positive terminal of one cell is connected to the negative terminal of the next cell, and so on.

ii. The terminal voltage of batteiy/cell is equal to the sum of voltages of individual cells in series. Example: Given figure shows two 1.5 V cells connected in series. This combination provides total voltage,
V = 1.5 V + 1.5 V = 3 V.

iii. The equivalent emf of n number of cells in series combination is the algebraic sum of their individual emf.
\(\sum_{i} \mathrm{E}_{\mathrm{i}}\) = E₁ + E₂ + E₂+ …….. + E_n

iv. The equivalent internal resistance of n cells in a series combination is the sum of their individual internal resistance.
\(\sum_{i} \mathrm{r}_{\mathrm{i}}\) = r₁ + r₂ + r₃ + ……… + r_n

Question 66.
State advantages of cells in series.
Answer:

1. The cells connected in series produce a larger resultant voltage.
2. Cells which are damaged can be easily identified, hence can be easily replaced.

Question 67.
Explain combination of cells in parallel. Ans:
Answer:
i. Consider two cells which are connected in parallel. Here, positive terminals of all the cells are connected together and the negative terminals of all the cells are connected together.

ii. In parallel connection, the current is divided among the branches i.e. I₁ and I₂ as shown in figure.

iii. Consider points A and B having potentials V_A and V_B, respectively.

iv. For the first cell the potential difference across its terminals is, V = V_A – V_B = E₁ – I₁ r₁
∴ I₁ = \(\frac {E_1V}{r_1}\) ………. (1)

v. Point A and B are connected exactly similarly to the second cell.
Hence, considering the second cell,
V = V_A – V_B = E₂ – I₂r₂
∴ I₂ = \(\frac {E_2V}{r_2}\) ………. (2)

vi. Since, I = I₁ + I₂ ………….. (3)
Combining equations (1), (2) and (3),

viii. If we replace the cells by a single cell connected between points A and B with the emf E_eq and the internal resistance r_eq then,
V = E_eq– Ir_eq
From equations (4) and (5),

ix. For n number of cells connected in parallel with emf E₁, E₂, E₃, ………….., E_n and internal resistance r₁, r₂, r₃, …………, r_n
\(\frac{1}{\mathrm{r}_{\mathrm{rq}}}=\frac{1}{\mathrm{r}_{1}}+\frac{1}{\mathrm{r}_{2}}+\frac{1}{\mathrm{r}_{3}}+\ldots \ldots \ldots+\frac{1}{\mathrm{r}_{\mathrm{n}}}\)
and \(\frac{\mathrm{E}_{\mathrm{eq}}}{\mathrm{r}_{\mathrm{rq}}}=\frac{\mathrm{E}_{1}}{\mathrm{r}_{1}}+\frac{\mathrm{E}_{2}}{\mathrm{r}_{2}}+\ldots \ldots \ldots+\frac{\mathrm{E}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}\)

Question 68.
State advantages and disadvantages of cells in parallel.
Answer:
Advantages:
For cells connected in parallel in a circuit, the circuit will not break open even if a cell gets damaged or open.

Disadvantages:
The voltage developed by the cells in parallel connection cannot be increased by increasing number of cells present in circuit.

Question 69.
State the basic categories of electrical cells.
Answer:
Electrical cells can be divided into several categories like primary cell, secondary cell, fuel cell, etc.

Question 70.
Write short note on primary cell.
Answer:

1. A primary cell cannot be charged again. It can be used only once.
2. Dry cells, alkaline cells are different examples of primary cells.
3. Primary cells are low cost and can be used easily. But these are not suitable for heavy loads.

Question 71.
Write short note on secondary cell.
Answer:

1. The secondary cells are rechargeable and can be reused.
2. The chemical reaction in a secondary cell is reversible.
3. Lead acid cell and fuel cell are some examples of secondary cells.
4. Lead acid battery is used widely in vehicles and other applications which require high load currents.
5. Solar cells are secondary cells that convert solar energy into electrical energy.

Question 72.
Write short note on fuel cells vehicles.
Answer:

1. Fuel cells vehicles (FCVs) are electric vehicles that use fuel cells instead of lead acid batteries to power the vehicles.
2. Hydrogen is used as a fuel in fuel cells. The by product after its burning is water.
3. This is important in terms of reducing emission of greenhouse gases produced by traditional gasoline fuelled vehicles.
4. The hydrogen fuel cell vehicles are thus more environment friendly.

Question 73.
What can be concluded from the following observations on a resistor made up of certain material? Calculate the power drawn in each case.

Case	Current (A)	Voltage (V)
A	0.2	1.6
B	0.4	3.2
C	0.6	4.8
d	0.8	6.4

Answer:
i. As the ratio of voltage and current different readings are same, hence ohm’s is valid i.e., V = IR.

ii. Electric power is given by, P = IV
∴ (a) P₁ = 0.2 × 1.6 = 0.32 watt
(b) P₂ = 0.4 × 3.2 = 1.28 watt
(c) P₃ = 0.6 × 4.8 = 2.88 watt
(d) P₄ = 0.8 × 6.4 = 5.12 watt

Question 74.
Answer the following questions from the circuit given below. [S₁, S₂, S₃, S₄, S₅ ⇒ Switches]. Calculate the current (I) flowing in the following cases:
i. S₁, S₄ → open; S₂, S₃, S₅ → closed.
ii. S₂, S₅ → open; S₁, S₃, S₄ → closed.
iii. S₃ → open; S₁, S₂, S₄, S₅ → closed.

Answer:
i. Here, the circuit can be represented as,

ii. Here, the circuit can be represented as,

iii. Here, the circuit can be represented as,

∴ As switch S3 is open, no current will flow in the circuit.

Question 75.
An electric circuit with a carton resistor and an electric bulb (60 watt, 300 Ω) are connected in series with a 230 V source.

i. Calculate the current flowing through the circuit.
ii. If the electric bulb of 60 watt is replaced by an electric bulb (80 watt, 300 Ω), will it glow? Justify your answer.
Answer:
Resistance of carbon resistor (R₁)
= 16 × 10 Ω = 160 Ω ….(using colour code)
Resistance of bulb (R₂) = 300 Ω
∴ Current through the circuit = \(\frac{V}{R_{1}+R_{2}}\)
∴ I = \(\frac{230}{(160+300)}=\frac{230}{460}\) = 0.5 A

ii. Power drawn through electric bulb
= I²R₂ = (0.5)² × 300 = 75 watt
Hence, if the bulb is replaced by 80 watt bulb, it will not glow.

Question 76.
From the graph given below, which of the two temperatures is higher for a metallic wire? Justify your answer.

Answer:
As R = \(\frac {V}{I}\)

For constant V,
I₂ > I₁
∴ R₁ > R₂
Now, for metallic wire,
R ∝ T
∴ T₁ > T₂
T₁ is greater than T₂.

Question 77.
If n identical cells, each of emf E and internal resistance r, are connected in series, write an expression for the terminal p.d. of the combination and hence show that this is nearly n times that of a single cell.
Answer:
i. Let n identical cells, each of emf E and internal resistance r, be connected in series. Let the current supplied by this combination to an external resistance R be I.

ii. The equivalent emf of the combination,
E_eq = E + E + …….. (n times) = nE

iii. The equivalent internal resistance of the combination,
r_eq= r + r + … (n times)
= nr

iv. The terminal p.d. of the combination is
V = E_eq – Ir_eq = nE – Inr = n (E – Ir)
∴ V = n × terminal p.d. of a single cell
Thus, the terminal p.d. of the series combination is n times that of a single cell.

Question 78.
If n identical cells, each of emf E and internal resistance r, are connected in parallel, derive an expression for the current supplied by this combination to external resistance R. Prove that the combination supplies current almost n times the current supplied by a single cell, when the external resistance R is much smaller than the internal resistance of the parallel combination of the cells.
Answer:
i. Consider n identical cells, each of emf E and internal resistance r, connected in parallel.

ii. Let the current supplied by the combination to the external resistance R be I.
In this case, the equivalent emf of the combination is E.

iii. The equivalent internal resistance r’ of the combination is,
\(\frac{1}{\mathrm{r}^{\prime}}=\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}\) + …………. (n terms)
∴ \(\frac{1}{\mathrm{r}^{\prime}}=\frac{\mathrm{n}}{\mathrm{r}} \Rightarrow \mathrm{r}^{\prime} \frac{\mathrm{n}}{\mathrm{r}}\)

iv. But V = IR is the terminal p.d. across each cell.

v. Hence, the current supplied by each cell,
I = \(\frac {E-V}{r}\)

vi. This gives the current supplied by the combination to the external resistance as
I = \(\frac {E-V}{r}\) + \(\frac {E-V}{r}\) + …….. (n terms) = n(\(\frac {E-V}{r}\))
Thus, current I = n × current supplied by a single cell
This proves that, the current supplied by the combination is n times the current supplied by a single cell.

Multiple Choice Questions

Question 1.
The drift velocity of the free electrons in a conductor is independent of
(A) length of the conductor.
(B) cross-sectional area of conductor.
(C) current.
(D) electric charge.
Answer:
(A) length of the conductor.

Question 2.
The direction of drift velocity in a conductor is
(A) opposite to that of applied electric field.
(B) opposite to the flow of positive charge.
(C) in the direction of the flow of electrons,
(D) all of these.
Answer:
(D) all of these.

Question 3.
The drift velocity of free electrons in a conductor is vd, when the current is flowing in it. If both the radius and current are doubled, the drift velocity will be
(A) \(\frac {v_d}{8}\)
(B) \(\frac {v_d}{4}\)
(C) \(\frac {v_d}{2}\)
(D) v_d
Answer:
(C) \(\frac {v_d}{2}\)

Question 4.
The drift velocity vd of electrons varies with electric field strength E as
(A) v_d ∝ E
(B) v_d ∝ \(\frac {1}{E}\)
(C) v_d ∝ E^1/2
(D) v_d × E^{\(\frac {1}{1/2}\)}
Answer:
(A) v_d ∝ E

Question 5.
When a current I is set up in a wire of radius r, the drift speed is v_d. If the same current is set up through a wire of radius 2r, then the drift speed will be
(A) v_d/4
(B) v_d/2
(C) 2v_d
(D) 4v_d
Answer:
(A) v_d/4

Question 6.
When potential difference is applied across an electrolyte, then Ohm’s law is obeyed at
(A) zero potential
(B) very low potential
(C) negative potential
(D) high potential.
Answer:
(D) high potential.

Question 7.
A current of 1.6 A is passed through an electric lamp for half a minute. If the charge on the electron is 1.6 × 10^-19 C, the number of electrons passing through it is
(A) 1 × 10¹⁹
(B) 1.5 × 10²⁰
(C) 3 × 10¹⁹
(D) 3 × 10²⁰
Answer:
(D) 3 × 10²⁰

Question 8.
The SI unit of the emf of a cell is
(A) V/m
(B) V/C
(C) J/C
(D) C/J
Answer:
(C) J/C

Question 9.
The unit of specific resistance is
(A) Ω m^-1
(B) Ω^-1 m^-1
(C) Ω m
(D) Ω m^-2
Answer:
(C) Ω m

Question 10.
If the length of a conductor is halved, then its conductivity will be
(A) doubled
(B) halved
(C) quadrupled
(D) unchanged
Answer:
(D) unchanged

Question 11.
The resistance of a metal conductor increases with temperature due to
(A) change in current carriers.
(B) change in the dimensions of the conductor.
(C) increase in the number of collisions among the current carriers.
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.
Answer:
(D) increase in the rate of collisions between the current carriers and the vibrating atoms of the conductor.

Question 12.
The resistivity of Nichrome is 10^-6 Ω-m. The wire of this material has radius of 0.1 mm with resistance 100 Ω, then the length will be
(A) 3.142 m
(B) 0.3142 m
(C) 3.142 cm
(D) 31.42 m
Answer:
(A) 3.142 m

Question 13.
Given a current carrying wire of non-uniform cross-section. Which of the following is constant throughout the length of the wire?
(A) Current, electric field and drift speed
(B) Drift speed only
(C) Current and drift speed
(D) Current only
Answer:
(D) Current only

Question 14.
A cell of emf E and internal resistance r is connected across an external resistance R (R >> r). The p.d. across R is A 1
(A) \(\frac {E}{R+r}\)
(B) E(I – \(\frac {r}{R}\))
(C) E(I + \(\frac {r}{R}\))
(D) E (R + r)
Answer:
(B) E(I – \(\frac {r}{R}\))

Question 15.
The e.m.f. of a cell of negligible internal resistance is 2 V. It is connected to the series combination of 2 Ω, 3 Ω and 5 Ω resistances. The potential difference across 3 Ω resistance will be
(A) 0.6 V
(B) 10 V
(C) 3 V
(D) 6 V
Answer:
(A) 0.6 V

Question 16.
A P.D. of 20 V is applied across a conductance of 8 mho. The current in the conductor is
(A) 2.5 A
(B) 28 A
(C) 160 A
(D) 45 A
Answer:
(C) 160 A

Question 17.
If an increase in length of copper wire is 0.5% due to stretching, the percentage increase in its resistance will be
(A) 0.1%
(B) 0.2%
(C) 1 %
(D) 2 %
Answer:
(C) 1 %

Question 18.
If a certain piece of copper is to be shaped into a conductor of minimum resistance, its length (L) and cross-sectional area A shall be respectively
(A) L/3 and 4 A
(B) L/2 and 2 A
(C) 2L and A2
(D) L and A
Answer:
(A) L/3 and 4 A

Question 19.
A given resistor has the following colour scheme of the various strips on it: Brown, black, green and silver. Its value in ohm is
(A) 1.0 × 104 ± 10%
(B) 1.0 × 105 ± 10%
(C) 1.0 × 106 + 10%
(D) 1.0 × 107 ± 10%
Answer:
(C) 1.0 × 106 + 10%

Question 20.
A given carbon resistor has the following colour code of the various strips: Orange, red, yellow and gold. The value of resistance in ohm is
(A) 32 × 104 ± 5%
(B) 32 × 104 ± 10%
(C) 23 × 105 ± 5%
(D) 23 × 105 ± 10%
Answer:
(A) 32 × 104 ± 5%

Question 21.
A typical thermistor can easily measure a change in temperature of the order of
(A) 10^-3 °C
(B) 10^-2 °C
(C) 10² °C
(D) 10³ °C
Answer:
(A) 10^-3 °C

Question 22.
Thermistors are usually prepared from
(A) non-metals
(B) metals
(C) oxides of non-metals
(D) oxides of metals
Answer:
(D) oxides of metals

Question 23.
On increasing the temperature of a conductor, its resistance increases because
(A) relaxation time decreases.
(B) mass of the electron increases.
(C) electron density decreases.
(D) all of the above.
Answer:
(A) relaxation time decreases.

Question 24.
Which of the following is used for the formation of thermistor?
(A) copper oxide
(B) nickel oxide
(C) iron oxide
(D) all of the above
Answer:
(D) all of the above

Question 25.
Emf of a cell is 2.2 volt. When resistance R = 5 Ω is connected in series, potential drop across the cell becomes 1.8 volt. Value of internal resistance of the cell is
(A) 10/9 Ω
(B) 7/12 Ω
(C) 9/10 Ω
(D) 12/7 Ω
Answer:
(A) 10/9 Ω

Question 26.
A strip of copper, another of germanium are cooled from room temperature to 80 K. The resistance of
(A) copper strip decreases germanium decreases. and that of
(B) copper strip decreases germanium increases. and that of
(C) Both the strip increases.
(D) copper strip increases germanium decreases. and that of
Answer:
(B) copper strip decreases germanium increases. and that of

Question 27.
The terminal voltage of a cell of emf E on short circuiting will be
(A) E
(B) \(\frac {E}{2}\)
(C) 2E
(D) zero
Answer:
(D) zero

Question 28.
If a battery of emf 2 V with internal resistance one ohm is connected to an external circuit of resistance R across it, then the terminal p.d. becomes 1.5 V. The value of R is
(A) 1 Ω
(B) 1.5 Ω
(C) 2 Ω
(D) 3 Ω
Answer:
(D) 3 Ω

Question 29.
A hall is used 5 hours a day for 25 days in a month. It has 6 lamps of 100 W each and 4 fans of 150 W. The total energy consumed for the month is
(A) 1500 kWh
(B) 150 kWh
(C) 15 kWh
(D) 1.5 kWh
Answer:
(B) 150 kWh

Question 30.
The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell will be
(A) 0.5 V
(B) 1.5 V
(C) 1.95 V
(D) 2 V
Answer:
(C) 1.95 V

Question 31.
The emf of a cell is 12 V. When it sends a current of 1 A through an external resistance, the p.d. across the terminals reduces to 10 V. The internal resistance of the cell is
(A) 0.1 Ω
(B) 0.5 Ω
(C) 1 Ω
(D) 2 Ω
Answer:
(D) 2 Ω

Question 32.
Three resistors, 8 Ω, 4 Ω and 10 Ω connected in parallel as shown in figure, the equivalent resistance is

(A) \(\frac {19}{40}\) Ω
(B) \(\frac {40}{19}\) Ω
(C) \(\frac {80}{19}\) Ω
(D) \(\frac {34}{23}\) Ω
Answer:
(B) \(\frac {40}{19}\) Ω

Question 33.
A potential difference of 20 V is applied across the ends of a coil. The amount of heat generated in it is 800 cal/s. The value of resistance of the coil will be
(A) 12 Ω
(B) 1.2 Ω
(C) 0.12 Ω
(D) 0.012 Ω
Answer:
(C) 0.12 Ω

Question 34.
In a series combination of cells, the effective internal resistance will
(A) remain the same.
(B) decrease.
(C) increase.
(D) be half that of the 1st cell.
Answer:
(C) increase.

Question 35.
The terminal voltage across a cell is more than its e.m.f., if another cell of
(A) higher e.m.f. is connected parallel to it.
(B) less e.m.f. is connected parallel to it.
(C) less e.m.f. is connected in series with it.
(D) higher e.m.f. is connected in series with it.
Answer:
(A) higher e.m.f. is connected parallel to it.

Question 36.
A 100 W, 200 V bulb is connected to a 160 volt supply. The actual’ power consumption would be
(A) 64 W
(B) 125 W
(C) 100 W
(D) 80 W
Answer:
(A) 64 W

Balbharti Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors Important Questions and Answers.

Maharashtra State Board 11th Physics Important Questions Chapter 14 Semiconductors

Question 1.
What are the factors on which electrical conductivity of any solid depends?
Answer:
Electrical conduction in a solid depends on its temperature, the number of charge carriers, how easily these carries can move inside a solid (mobility), its crystal structure, types and the nature of defects present in a solid.

Question 2.
Why are metals good conductor of electricity?
Answer:
Metals are good conductors of electricity due to the large number of free electrons (≈ 10²⁸ per m³) present in them.

Question 3.
Give the formula for electrical conductivity of a solid and give significance of the terms involved.
Answer:
Electrical conductivity (σ) of a solid is given by a = nqµ,
where, n = charge carrier density (number of charge carriers per unit volume)
q = charge on the carriers
µ = mobility of carriers

Question 4.
Explain in brief temperature dependence of electrical conductivity of metals and semiconductors with the help of graph.
Answer:
i. The electrical conductivity of a metal decreases with increase in its temperature.

ii. When the temperature of a semiconductor is increased, its electrical conductivity also increases

Question 5.
Mention the broad classification of semiconductors along with examples.
Answer:
A broad classification of semiconductors can be:

1. Elemental semiconductors: Silicon, germanium
2. Compound Semiconductors: Cadmium sulphide, zinc sulphide, etc.
3. Organic Semiconductors: Anthracene, doped pthalocyanines, polyaniline etc.

Question 6.
What are some electrical properties of semiconductors?
Answer:

1. Electrical properties of semiconductors are different from metals and insulators due to their unique conduction mechanism.
2. The electronic configuration of the elemental semiconductors plays a very important role in their electrical properties.
3. They are from the fourth group of elements in the periodic table.
4. They have a valence of four.
5. Their atoms are bonded by covalent bonds. At absolute zero temperature, all the covalent bonds are completely satisfied in a single crystal of pure semiconductor like silicon or germanium.

Question 7.
Explain in detail the distribution of electron energy levels in an isolated atom with the help of an example.
Answer:

1. An isolated atom has its nucleus at the centre which is surrounded by a number of revolving electrons. These electrons are arranged in different and discrete energy levels.
2. Consider the electronic configuration of sodium (atomic number 11) i.e, 1s², 2s², 2p⁶, 3s¹. The outermost level 3s can take one more electron but it is half filled in sodium,
3. The energy levels in each atom are filled according to Pauli’s exclusion principle which states that no two similar spin electrons can occupy the same energy level.
4. That means any energy level can accommodate only two electrons (one with spin up state and the other with spin down state)
5. Thus, there can be two states per energy level.
6. Figure given below shows the allowed energy levels of a sodium atom by horizontal lines. The curved lines represent the potential energy of an electron near the nucleus due to Coulomb interaction.
   

Question 8.
Explain formation of energy bands in solid sodium with neat labelled energy band diagrams.
Answer:
i. For an isolated sodium atom (atomic number 11) the electronic configuration is given as 1s², 2s², 2p⁶, 3s¹. The outermost level 3s is half filled in sodium.

ii. The energy levels are filled according to Pauli’s exclusion principle.

iii. Consider two sodium atoms close enough so that outer 3s electrons can be considered equally to be part of any atom.

iv. The 3s electrons from both the sodium atoms need to be accommodated in the same level.

v. This is made possible by splitting the 3 s level into two sub-levels so that the Pauli’s exclusion principle is not violated. Figure given below shows the splitting of the 3 s level into two sub levels.

vi. When solid sodium is formed, the atoms come close to each other such that distance between them remains of the order of 2 – 3 Å. Therefore, the electrons from different atoms interact with each other and also with the neighbouring atomic cores.

vii. The interaction between the outer most electrons is more due to overlap while the inner most electrons remain mostly unaffected. Each of these energy levels is split into a large number of sub levels, of the order of Avogadro’s number due to number of atoms in solid sodium is of the order of this number.

viii. The separation between the sublevels is so small that the energy levels appear almost continuous. This continuum of energy levels is called an energy band. The bands are called 1 s band, 2s band, 2p band and so on. Figure shows these bands in sodium metal.

Question 9.
Explain concept of valence band and conduction band in solid crystal.
Answer:
A. Valence band (V.B):

1. The topmost occupied energy level in an atom is the valence level. The energy band formed by valence energy levels of atoms in a solid is called the valence band.
2. In metallic conductors, the valence electrons are loosely attached to the nucleus. At ordinary room temperature, some valence electrons become free. They do not leave the metal surface but can move from atom to atom randomly.
3. Such free electrons are responsible for electric current through conductors.

B. Conduction band (C.B):

1. The immediately next energy level that electrons from valence band can occupy is called conduction level. The band formed by conduction levels is called conduction band.
2. It is the next permitted energy band beyond valence band.
3. In conduction band, electrons move freely and conduct electric current through the solids.
4. An insulator has empty conduction band.

Question 10.
Draw neat labelled diagram showing energy bands in sodium. Why broadening of higher bands is different than that of the lower energy bands?
Answer:

Broadening of valence and higher bands is more since interaction of these electrons is stronger than the inner most electrons.

Question 11.
State the conditions when electrons of a semiconductor can take part in conduction.
Answer:

1. All the energy levels in a band, including the topmost band, in a semiconductor are completely occupied at absolute zero.
2. At some finite temperature T, few electrons gain thermal energy of the order of kT, where k is the Boltzmann constant.
3. Electrons in the bands between the valence band cannot move to higher band since these are already occupied.
4. Only electrons from the valence band can be excited to the empty conduction band, if the thermal energy gained by these electrons is greater than the band gap.
5. Electrons can also gain energy when an external electric field is applied to a solid. Energy gained due to electric field is smaller, hence only electrons at the topmost energy level gain such energy and participate in electrical conduction.

Question 12.
Define 1 eV.
Answer:
1 eV is the energy gained by an electron while it overcomes a potential difference of one volt. 1 eV= 1.6 × 10^-19 J.

Question 13.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors?
Answer:

1. The 4 valence electrons of C, Si or Ge lie respectively in the second, third and fourth orbit.
2. Energy required to take out an electron from these atoms (i.e., ionisation energy E_g) will be least for Ge, followed by Si and highest for C.
3. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for

Question 14.
What is intrinsic semiconductor?
Answer:
A pure semiconductor is blown as intrinsic semiconductor.

Question 15.
Explain characteristics and structure of silicon using a neat labelled diagram.
Answer:

1. Silicon (Si) has atomic number 14 and its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p².
2. Its valence is 4.
3. Each atom of Si forms four covalent bonds with its neighbouring atoms. One Si atom is surrounded by four Si atoms at the comers of a regular tetrahedron as shown in the figure.
   

Question 16.
Describe in detail formation of holes in ii. intrinsic semiconductor.
Answer:
i. In intrinsic semiconductor at absolute zero temperature, all valence electrons are tightly bound to respective atoms and the covalent bonds are complete.

ii. Electrons are not available to conduct electricity through the crystal because they cannot gain enough energy to get into higher energy levels.

iii. At room temperature, however, a few covalent bonds are broken due to heat energy produced by random motion of atoms. Some of the valence electrons can be moved to the conduction band. This creates a vacancy in the valence band as shown in figure.

iv. These vacancies of electrons in the valence band are called holes. The holes are thus absence of electrons in the valence band and they carry an effective positive charge.

Question 17.
How does electric conduction take place inside a pure silicon?
Answer:

1. There are two different types of charge carriers in a pure semiconductor. One is the electron and the other is the hole or absence of electron.
2. Electrical conduction takes place by transportation of both carriers or any one of the two carriers in a semiconductor.
3. When a semiconductor is connected in a circuit, electrons, being negatively charged, move towards positive terminal of the battery.
4. Holes have an effective positive charge, and move towards negative terminal of the battery. Thus, the current through a semiconductor is carried by two types of charge carriers moving in opposite directions.
5. Figure given below represents the current through a pure silicon.
   

Question 18.
Why do holes not exist in conductor?
Answer:

1. In case of semiconductors, there is one missing electron from one of the covalent bonds.
2. The absence of electron leaves an empty space called as hole; each hole carries an effective positive charge.
3. In case of an conductor, number of free electrons are always available for conduction. There is no absence of electron in it. Hence holes do not exist in conductor.

Question 19.
What is the need for doping an intrinsic semiconductor?
Answer:
The electric conductivity of an intrinsic semiconductor is very low at room temperature; hence no electronic devices can be fabricated using them. Addition of a small amount of a suitable impurity to an intrinsic semiconductor increases its conductivity appreciably. Hence, intrinsic semiconductors are doped with impurities.

Question 20.
Explain what is doping.
Answer:

1. The process of adding impurities to an intrinsic semiconductor is called doping.
2. The impurity atoms are called dopants which may be either trivalent or pentavalent. The parent atoms are called hosts.
3. The dopant material is so selected that it does not disturb the crystal structure of the host.
4. The size and the electronic configuration of the dopant should be compatible with that of the host.
5. Doping is expressed in ppm (parts per million), i.e., one impurity atom per one million atoms of the host.
6. Doping significantly increases the concentration of charge carriers.

Question 21.
What is extrinsic semiconductors?
Answer:
The semiconductor with impurity is called a doped semiconductor or an extrinsic semiconductor.

Question 22.
Draw neat diagrams showing schematic electronic structure of:
i. A pentavalent atom [Antimony (Sb)]
ii. A trivalent atom [Boron (B)]
Answer:

[Note: Electronic structure of antimony is drawn as per its electronic configuration in accordance with Modern Periodic Table.]

Question 23.
With the help of neat diagram, explain the structure of n-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony we get n-type semiconductor.

ii. When a dopant atom of 5 valence electrons occupies the position of a Si atom in the crystal lattice, 4 electrons from the dopant form bonds with 4 neighbouring Si atoms and the fifth electron from the dopant remains very weakly bound to its parent atom

iii. To make this electron free even at room temperature, very small energy is required. It is 0.01 eV for Ge and 0.05 eV for Si.

iv. As this semiconductor has large number of electrons in conduction band and its conductivity is due to negatively charged carriers, it is called n-type semiconductor.

v. The n-type semiconductor also has a few electrons and holes produced due to the thermally broken bonds.

vi. The density of conduction electrons (n_e) in a doped semiconductor is the sum total of the electrons contributed by donors and the thermally generated electrons from the host.

vii. The density of holes (n_h) is only due to the thermally generated holes of the host Si atoms.

viii. Thus, the number of free electrons exceeds the number of holes (n_e >> n_h). Thus, in n-type semiconductor electrons are the majority carriers and holes are the minority carriers.

Question 24.
What are some features of n-type semiconductor?
Answer:

1. These are materials doped with pentavalent impurity (donors) atoms.
2. Electrical conduction in these materials is due to majority charge carriers i.e., electrons.
3. The donor atom loses electrons and becomes positively charged ions.
4. Number of free electrons is very large compared to the number of holes, n_e >> n_h. Electrons are majority charge carriers.
5. When energy is supplied externally, negatively charged free electrons (majority charges carries) and positively charged holes (minority charges carries) are available for conduction.

Question 25.
With the help of neat diagram, explain the structure of p-type semiconductor in detail.
Answer:
i. When silicon or germanium crystal is doped with a trivalent impurity such as boron, aluminium or indium, we get a p-type semiconductor.

ii. The dopant trivalent atom has one valence electron less than that of a silicon atom. Every trivalent dopant atom shares its three electrons with three neighbouring Si atoms to form covalent bonds. But the fourth bond between silicon atom and its neighbour is not complete.

iii. The incomplete bond can be completed by another electron in the neighbourhood from Si atom.

iv. The shared electron creates a vacancy in its place. This vacancy or the absence of electron is a hole.

v. Thus, a hole is available for conduction from each acceptor impurity atom.

vi. Holes are majority carriers and electrons are minority carriers in such materials. Acceptor atoms are negatively charged ions and majority carriers are holes. Therefore, extrinsic semiconductor doped with trivalent impurity is called a p-type semiconductor.

vii. For a p-type semiconductor, n_h >> n_e^.

Question 26.
What are some features of p-type semiconductors?
Answer:

1. These are materials doped with trivalent impurity atoms (acceptors).
2. Electrical conduction in these materials is due to majority charge carriers i.e., holes.
3. The acceptor atoms acquire electron and become negatively charged-ions.
4. Number of holes is very large compared to the number of free electrons. n_h >> n_e. Holes are majority charge carriers.
5. When energy is supplied externally, positively charged holes (majority charge carriers) and negatively charged free electrons (minority charge carriers) are available for conduction.

Question 27.
What are donor and acceptor impurities?
Answer:

1. Every pentavalent dopant atom which donates one electron for conduction is called a donor impurity.
2. Each trivalent atom which can accept an electron is called an acceptor impurity.

Question 28.
Explain the energy levels of both donor and acceptor impurities with a schematic band structure.
Answer:
i. The free electrons donated by the donor impurity atoms occupy energy levels which are in the band gap and are close to the conduction band.

ii. The vacancies of electrons or the extra holes are created in the valence band due to addition of acceptor impurities. The impurity levels are created just above the valence band in the band gap.

Question 29.
Distinguish between p-type and n-type semiconductor.
Answer:

p-type semiconductor	n-type semiconductor
1. The impurity of some trivalent element like B, Al, In, etc. is mixed with semiconductor.	The impurity of some pentavalent element like P, As, Sb, etc. is mixed
2. The impurity atom accepts one electron hence the impurities	The impurity atom donates – one electron, hence the impurities added are known as donor impurities.
3. The holes are majority charge carriers and electrons are minority charge carriers.	The electrons are j majority charge carriers and holes are minority charge carriers.
4. The acceptor energy level is close to the valence band and far away from conduction band.	Donor energy level is close to the conduction band and far away from valence band.

Question 30.
What is the charge on a p-type and n-type semiconductor?
Answer:
n-type as well as p-type semiconductors are electrically neutral.

Question 31.
Explain the transportation of holes inside a p-type semiconductor.
Answer:
i. Consider a p-type semiconductor connected to terminals of a battery as shown.

ii. When the circuit is switched on, electrons at 1 and 2 are attracted to the positive terminal of the battery and occupy nearby holes at x and y. This creates holes at the positions 1 and 2 previously occupied by electrons.

iii. Next, electrons at 3 and 4 move towards the positive terminal and create holes in their previous positions.

iv. But, the holes are captured at the negative terminal by the electrons supplied by the battery.

v. In this way, holes are transported from one place to other and density of holes is kept constant so long as the battery is working.

Question 32.
A pure Si crystal has 4 × 10²⁸ atoms m^-3. It is doped by 1 ppm concentration of antimony. Calculate the number of electrons and holes. Given n₁ = 1.2 × 10¹⁶/m³.
Answer:
As, the atom is doped with 1 ppm concentration of antimony (Sb).
1 ppm = 1 parts per one million atoms. = 1/10⁶
∴ no. of Si atoms = \(\frac {Total no. of Si atoms}{10^6}\)
= \(\frac {4×10^{28}}{10^6}\) = 4 × 10²² m^-3
i.e., total no. of extra free electrons (n_e)
= 4 × 10²² m^-3
n_i² = n_e n_h
∴ n_h = \(\frac {n_i^2}{n_e}\) = \(\frac {(1.2×10^{16})^2}{4×10^{22}}\)
= \(\frac {144×10^{30}{4×10^{22}}\)
= 36 × 10^-8
= 3.6 × 10⁹ m^-3.

Question 33.
A pure silicon crystal at temperature of 300 K has electron and hole concentration 1.5 × 10¹⁶ m^-3 each. (n_e = n_h). Doping bv indium increases n_h to 4.5 × 10²² m^-3. Calculate ne for the doped silicon crystal.
Answer:
Given: At 300 K, n_i = n_e = n_h = 1.5 × 10¹⁶ m^-3
After doping n_h = 4.5 × 10²² m^-3
To find: Number density of electrons (n_e)
Formula: n_i² = n_e nn_h
Calculation From formula:
n_e = \(\frac {n_i^2}{n_h}\) = \(\frac {(1.5×10^{16})^2}{4×10^{22}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 5 × 10^-9 m^-3.

Question 34.
A Ge specimen is doped with A/. The concentration of acceptor atoms is ~10²¹ atoms/m³. Given that the intrinsic concentration of electron-hole pairs is ~10 ¹⁹/m³, calculate the concentration of electrons in the specimen.
Answer:
Given: At room temperature,
n_i = n_e = n_h = 10¹⁹ m^-3
After doping n_h = 10²¹ m^-3
To find: Number density of electrons (nc)
Formulae: n_i² = n_en_h
Calculation: From formula,
n_e = \(\frac {n_i^2}{n_h}\) = \(\frac {(10^{19})^2}{10^{21}}\)
= \(\frac {255×10^{30}{45×10^{21}}\)
= 10¹⁷ m^-3.

Question 35.
A semiconductor has equal electron and hole concentration of 2 × 10⁸ m^-3. On doping with a certain impurity, the electron concentration increases to 4 × 10¹⁰ m^-3, then calculate the new hole concentration of the semiconductor.
Answer:
Given: n_i = 2 × 10⁸ m^-3, n = 4 × 10¹⁰ m^-3
After doping n_h = 10²¹ m^-3
To find: Number density of holes (n_h)
Formulae: n_i ²= n_en_h
Calculation: From formula.
n_h = \(\frac {n_i^2}{n_e}\) = \(\frac {(2×10^{8})^2}{4×10^{10}}\) = 10⁶ m^-3

Question 36.
What is a p-n junction?
Answer:
When n-type and p-type semiconductor materials are fused together, the junction formed is called as p-n junction.

Question 37.
Explain the process of diffusion in p-n junction.
Answer:
i. The transfer of electrons and holes across the p-n junction is called diffusion.

ii. When an n-type and a p-type semiconductor materials are fused together, initially, the number of electrons in the n-side of a junction is very large compared to the number of electrons on the p-side. The same is true for the number of holes on the p-side and on the n-side.

iii. Thus, a large difference in density of carriers exists on both sides of the p-n junction. This difference causes migration of electrons from the n-side to the p-side of the

iv. They fill up the holes in the p-type material and produce negative ions.

v. When the electrons from the n-side of a junction migrate to the p-side, they leave behind positively charged donor ions on the n- side. Effectively, holes from the p-side migrate into the n-region.

vi. As a result, in the p-type region near the junction there are negatively charged acceptor ions, and in the n-type region near the junction there are positively charged donor ions.

vii. The extent up to which the electrons and the holes can diffuse across the junction depends on the density of the donor and the acceptor ions on the n-side and the p-side respectively, of the junction.

Question 38.
Define potential barrier.
Answer:
The diffusion of carriers across the junction and resultant accumulation of positive and negative charges across the junction builds a potential difference across the junction. This potential difference is called the potential barrier.

Question 39.
Draw neat labelled diagrams for potentials barrier and depletion layer in a p-n junction.
Answer:

Question 40.
Explain in brief electric field across a p-n junction with a neat labelled diagram.
Answer:

1. When p-type semiconductor is fused with n-type semiconductor, a depletion region is developed across the junction.
2. The n-side near the boundary of a p-n junction becomes positive with respect to the p-side because it has lost electrons and the p-side has lost holes.
3. Thus, the presence of impurity ions on both sides of the junction establishes an electric field across this region such that the n-side is at a positive voltage relative to the p-side.
   

Question 41.
What is the need of biasing a p-n junction?
Answer:

1. Due to potential barrier across depletion region, charge carriers require extra energy to overcome the barrier.
2. A suitable voltage needs to be applied to the junction externally, so that these charge carriers can overcome the potential barrier and move across the junction.

Question 41.
Explain the mechanism of forward biased p-n junction.
Answer:

1. In forward bias, a p-n junction is connected in an electric circuit such that the p-region is connected to the positive terminal and the n-region is connected to the negative terminal of an external voltage source.
2. The external voltage effectively opposes the built-in potential of the junction. The width of potential barrier is thus reduced.
3. Also, negative charge carriers (electrons) from the n-region are pushed towards the junction.
4. A similar effect is experienced by positive charge carriers (holes) in the p-region and they are pushed towards the junction.
5. Both the charge carriers thus find it easy to cross over the barrier and contribute towards the electric current.
   

Question 42.
Explain the mechanism of reverse biased p-n junction.
Answer:
i. In reverse biased, the p-region is connected to the negative terminal and the n-region is connected to the positive terminal of the external voltage source. This external voltage effectively adds to the built-in potential of the junction. The width of potential barrier is thus increased.

ii. Also, the negative charge carriers (electrons) from the n-region are pulled away from the junction.

iii. Similar effect is experienced by the positive charge carriers (holes) in the p-region and they are pulled away from the junction.

iv. Both the charge carriers thus find it very difficult to cross over the barrier and thus do not contribute towards the electric current.

Question 43.
State some important features of the depletion region.
Answer:

1. It is formed by diffusion of electrons from n-region to the p-region. This leaves positively charged ions in the n-region.
2. The p-region accumulates electrons (negative charges) and the n-region accumulates the holes (positive charges).
3. The accumulation of charges on either sides of the junction results in forming a potential barrier and prevents flow of charges.
4. There are no charges in this region.
5. The depletion region has higher potential on the n-side and lower potential on the p-side of the junction.

Question 44.
What is p-n junction diode? Draw its circuit symbol.
Answer:
A p-n junction, when provided with metallic connectors on each side is called a junction diode

Question 45.
Explain asymmetrical flow of current in p-n junction diode in detail.
Answer:

i. The barrier potential is reduced in forward biased mode and it is increased in reverse biased mode.

ii. Carriers find it easy to cross the junction in forward bias and contribute towards current because the barrier width is reduced and they are pushed towards the junction and gain extra energy to cross the junction.

iii. The current through the diode in forward bias is large and of the order of a few milliamperes (10^-3 A) for a typical diode.

iv. When connected in reverse bias, width of the potential barrier is increased and the carriers are pushed away from the junction so that very few carriers can cross the junction and contribute towards current.

v. This results in a very small current through a reverse biased diode. The current in reverse biased diode is of the order of a few microamperes (10^-6 A).

vi. When the polarity of bias voltage is reversed, the width of the depletion layer changes. This results in asymmetrical current flow through a diode as shown in figure.

Question 46.
What is knee voltage?
Answer:
In forward bias mode, the voltage for which the current in a p-n junction diode rises sharply is called knee voltage.

Question 47.
What is a forward current in case of zero biased p-n junction diode?
Answer:
When the diode terminals are shorted together, some holes (majority carriers) in the p-side have enough thermal energy to overcome the potential barrier. Such carriers cross the barrier potential and contribute to current. This current is known as the forward current.

Question 48.
Define reverse current in zero biased p-n junction diode.
Answer:
When the diode terminals are shorted together some holes generated in the n-side (minority carriers), move across the junction and contribute to current. This current is known as the reverse current.

Question 49.
Explain the I-V characteristics of a reverse biased junction diode.
Answer:
i. The positive terminal of the external voltage is connected to the cathode (n-side) and negative terminal to the anode (p-side) across the diode.

ii. In case of reverse bias the width of the depletion region increases and the p-n junction behaves like a high resistance.

iii. Practically no current flows through it with an increase in the reverse bias voltage. However, a very small leakage current does flow through the junction which is of the order of a few micro amperes, (µA).

iv. When the reverse bias voltage applied to a diode is increased to sufficiently large value, it causes the p-n junction to overheat. The overheating of the junction results in a sudden rise in the current through the junction. This is because covalent bonds break and a large number of carries are available for conduction. The diode thus no longer behaves like a diode. This effect is called the avalanche breakdown.

v. The reverse biased characteristic of a diode is shown in figure.

Question 50.
Explain zero biased junction diode.
Answer:
i. When a diode is connected in a zero bias condition, no external potential energy is applied to the p-n junction.

li. The potential barrier that exists in a junction prevents the diffusion of any more majority carriers across it. However, some minority carriers (few free electrons in the p-region and few holes in the n-region) drift across the junction.

iii. An equilibrium is established when the majority carriers are equal in number (n_e = n_h) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

iv. The minority carriers are continuously generated due to thermal energy.

v. When the temperature of the p-n junction is raised, this state of equilibrium is changed.

vi. This results in generating more minority carriers and an increase in the leakage current. An electric current, however, cannot flow through the diode because it is not connected in any electric circuit

Question 51.
What is dynamic equilibrium?
Answer:
An equilibrium is established when the majority carriers are equal in number (n_e = n_h) and both moving in opposite directions. The net current flowing across the junction is zero. This is a state of‘dynamic equilibrium’.

Question 52.
Draw a neat diagram and state I-V characteristics of an ideal diode.
Answer:
An ideal diode offers zero resistance in forward biased mode and infinite resistance in reverse biased mode.

Question 53.
What do you mean by static resistance of a diode?
Answer:
Static (DC) resistance:

1. When a p-n junction diode is forward biased, it offers a definite resistance in the circuit. This resistance is called the static or DC resistance (R_g) of a diode.
2. The DC resistance of a diode is the ratio of the DC voltage across the diode to the DC current flowing through it at a particular voltage.
3. It is given by, R_g = \(\frac {V}{I}\)

Question 54.
Explain dynamic resistance of a diode.
Answer:

1. The dynamic (AC) resistance of a diode, r_g, at a particular applied voltage, is defined as
   r_g = \(\frac {∆V}{∆I}\)
2. The dynamic resistance of a diode depends on the operating voltage.
3. It is the reciprocal of the slope of the characteristics at that point.

Question 55.
Draw a graph representing static and dynamic resistances of a diode.
Answer:

Question 56.
Refer to the figure as shown below and find the resistance between point A and B when an ideal diode is (i) forward biased and (ii) reverse biased.

Answer:
We know that for an ideal diode, the resistance is zero when forward biased and infinite when reverse biased.
i. Figure (a) shows the circuit when the diode is forward biased. An ideal diode behaves as a conductor and the circuit is similar to two resistances in parallel.

∴ R_AB = (30 × 30)/(30 +30) = 900/60 = 15 Ω

ii. Figure (b) shows the circuit when the diode is reverse biased.

It does not conduct and behaves as an open switch along path ACB. Therefore, R_AB = 30 Ω. the only resistance in the circuit along the path ADB.

Question 57.
State advantages of semiconductor devices.
Answer:

1. Electronic properties of semiconductors can be controlled to suit our requirement.
2. They are smaller in size and light weight.
3. They can operate at smaller voltages (of the order of few mV) and require less current (of the order of pA or mA), therefore, consume lesser power.
4. Almost no heating effects occur, therefore these devices are thermally stable.
5. Faster speed of operation due to smaller size.
6. Fabrication of ICs is possible.

Question 58.
State disadvantages of semiconductor devices.
Answer:

1. They are sensitive to electrostatic charges.
2. Not very useful for controlling high power.
3. They are sensitive to radiation.
4. They are sensitive to fluctuations in temperature.
5. They need controlled conditions for their manufacturing.
6. Very few materials are semiconductors.

Question 59.
Explain applications of semiconductors.
Answer:
i. Solar cell:

1. It converts light energy into electric energy.
2. t is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photo resistor: It changes its resistance when light is incident on it.

iii. Bi-polar junction transistor:

1. These are devices with two junctions and three terminals.
2. A transistor can be a p-n-p or n-p-n transistor.
3. Conduction takes place with holes and electrons.
4. Many other types of transistors are designed and fabricated to suit specific requirements.
5. They are used in almost all semiconductor devices.

iv. Photodiode: It conducts when illuminated with light.

v. LED (Light Emitting Diode):

1. It emits light when current passes through it.
2. House hold LED lamps use similar technology.
3. They consume less power, are smaller in size and have a longer life and are cost effective.

vi. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

vii. Integrated Circuits (ICs): A small device having hundreds of diodes and transistors performs the work of a large number of electronic circuits.

Question 60.
Explain any four application of p-n junction diode.
Answer:
1. Solar cell:

1. It converts light energy into electric energy.
2. It is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photodiode: It conducts when illuminated with light.

iii. LED (Light Emitting Diode):

1. It emits light when current passes through it.
2. House hold LED lamps use similar technology.
3. They consume less power, are smaller in size and have a longer life and are cost effective.

iv. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

Question 61.
What is thermistor?
Answer:
Thermistor is a temperature sensitive resistor. Its resistance changes with change in its temperature.

Question 62.
What are different ty pes of thermistor and what are their applications?
Answer:
There are two types of thermistor:
i. NTC (Negative Thermal Coefficient) thermistor: Resistance of a NTC thermistor decreases with increase in its temperature. Its temperature coefficient is negative. They are commonly used as temperature sensors and also in temperature control circuits.

ii. PTC (Positive Thermal Coefficient) thermistor: Resistance of a PTC thermistor increases with increase in its temperature. They are commonly used in series with a circuit. They are generally used as a reusable fuse to limit current passing through a circuit to protect against over current conditions, as resettable fuses.

Question 63.
How are thermistors fabricated?
Answer:
Thermistors are made from thermally sensitive metal oxide semiconductors. Thermistors are very sensitive to changes in temperature.

Question 64.
Enlist any two features of thermistor.
Answer:

1. A small change in surrounding temperature causes a large change in their resistance.
2. They can measure temperature variations of a small area due to their small size.

Question 65.
Write a note on:
i. Electric devices
ii. Electronic devices
Answer:
i. Electric devices:

1. These devices convert electrical energy into some other form.
2. Examples: Fan, refrigerator, geyser etc. Fan converts electrical energy into mechanical energy. A geyser converts it into heat energy.
3. They use good conductors (mostly metals) for conduction of electricity.
4. Common working range of currents for electric circuits is milliampere (mA) to ampere.
5. Their energy consumption is also moderate to high. A typical geyser consumes about 2.0 to 2.50 kW of power.
6. They are moderate to large in size and are costly.

ii. Electronic devices:

1. Electronic circuits work with control or sequential changes in current through a cell.
2. A calculator, a cell phone, a smart watch or the remote control of a TV set are some of the electronic devices.
3. Semiconductors are used to fabricate such devices.
4. Common working range of currents for electronic circuits it is nano-ampere to µA.
5. They consume very low energy. They are very compact, and cost effective.

Question 66.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction?
Answer:

1. No. Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å).
2. Hence, continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 67.
What is Avalanche breakdown and zener breakdown?
Answer:
i. Avalanche breakdown: In high reverse bias, minority carriers acquire sufficient kinetic energy and collide with a valence electron. Due to collisions the covalent bond breaks. The valence electron enters conduction band. A breakdown occurring in such a manner is avalanche breakdown. It occurs with lightly doped p-n junctions.

ii. Zener breakdown: It occurs in specially designed and highly doped p-n junctions, viz., zener diodes. In this case, covalent bonds break directly due to application of high electric field. Avalanche breakdown voltage is higher than zener voltage.

Question 68.
Indicators on platform, digital clocks, calculators make use of seven LEDs to indicate a number. How do you think these LEDs might be arranged?
Answer:
i. The indicators on platforms, digital clocks, calculators are made using seven LEDs arranged in such a way that when provided proper signal they light up displaying desired alphabet or number.

ii. This arrangement of LEDs is called Seven Segment Display.

Multiple Choice Questions

Question 1.
The number of electrons in the valence shell of semiconductor is ……………
(A) less than 4
(B) equal to 4
(C) more than 4
(D) zero
Answer:
(B) equal to 4

Question 2.
If the temperature of semiconductor is increased, the number of electrons in the valence band will ……………….
(A) decrease
(B) remains same
(C) increase
(D) either increase or decrease
Answer:
(A) decrease

Question 3.
When N-type semiconductor is heated, the ……………..
(A) number of electrons and holes remains same.
(B) number of electrons increases while that of holes decreases.
(C) number of electrons decreases while that of holes increases.
(D) number of electrons and holes increases equally.
Answer:
(D) number of electrons and holes increases equally.

Question 4.
In conduction band of solid, there is no electron at room temperature. The solid is ……………
(A) semiconductors
(B) insulator
(C) conductor
(D) metal
Answer:
(B) insulator

Question 5.
In the crystal of pure Ge or Si, each covalent bond consists of …………..
(A) 1 electron
(B) 2 electrons
(C) 3 electrons
(D) 4 electrons
Answer:
(B) 2 electrons

Question 6.
A pure semiconductor is ……………..
(A) an extrinsic semiconductor
(B) an intrinsic semiconductor
(C) p-type semiconductor
(D) n-type semiconductor
Answer:
(B) an intrinsic semiconductor

Question 7.
For an extrinsic semiconductor, the valency of the donor impurity is …………..
(A) 2
(B) 1
(C) 4
(D) 5
Answer:
(D) 5

Question 8.
In a semiconductor, acceptor impurity is
(A) antimony
(B) indium
(C) phosphorous
(D) arsenic
Answer:
(B) indium

Question 9.
What are majority carriers in a semiconductor?
(A) Holes in n-type and electrons in p-type.
(B) Holes in n-type and p-type both.
(C) Electrons in n-type and p-type both.
(D) Holes in p-type and electrons in n-type.
Answer:
(D) Holes in p-type and electrons in n-type.

Question 10.
When a hole is produced in P-type semiconductor, there is ……………….
(A) extra electron in valence band.
(B) extra electron in conduction band.
(C) missing electron in valence band.
(D) missing electron in conduction band.
Answer:
(C) missing electron in valence band.

Question 11.
The number of bonds formed in p-type and n-type semiconductors are respectively
(A) 4,5
(B) 3,4
(C) 4,3
(D) 5,4
Answer:
(B) 3,4

Question 12.
The movement of a hole is brought about by the valency being filled by a ………………..
(A) free electrons
(B) valence electrons
(C) positive ions
(D) negative ions
Answer:
(B) valence electrons

Question 13.
The drift current in a p-n junction is
(A) from the p region to n region.
(B) from the n region to p region.
(C) from n to p region if the junction is forward biased and from p to n region if the junction is reverse biased.
(D) from p to n region if the junction is forward biased and from n to p region if the junction is reverse biased.
Answer:
(B) from the n region to p region.

Question 14.
If a p-n junction diode is not connected to any circuit, then
(A) the potential is same everywhere.
(B) potential is not same and n-type side has lower potential than p-type side.
(C) there is an electric field at junction direction from p-type side to n-type side.
(D) there is an electric field at the junction directed from n-type side to p-type side.
Answer:
(D) there is an electric field at the junction directed from n-type side to p-type side.

Question 15.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(A) free electrons in the n-region attract them.
(B) they move across the junction by the potential difference.
(C) hole concentration in p-region is more as compared to n-region.
(D) all the above.
Answer:
(C) hole concentration in p-region is more as compared to n-region.

Question 16.
The width of depletion region ……………
(A) becomes small in forward bias of diode
(B) becomes large in forward bias of diode
(C) is not affected upon by the bias
(D) becomes small in reverse bias of diode
Answer:
(A) becomes small in forward bias of diode

Question 17.
For p-n junction in reverse bias, which of the following is true?
(A) There is no current through P-N junction due to majority carriers from both regions.
(B) Width of potential barriers is small and it offers low resistance.
(C) Current is due to majority carriers.
(D) Both (B) and (C)
Answer:
(A) There is no current through P-N junction due to majority carriers from both regions.

Question 18.
In the circuit shown below Di and D2 are two silicon diodes. The current in the circuit is …………….

(A) 2 A
(B) 2 mA
(C) 0.8 mA
(D) very small (approx 0)
Answer:
(D) very small (approx 0)

Question 19.
For an ideal junction diode,
(A) forward bias resistance is infinity.
(B) forward bias resistance is zero.
(C) reverse bias resistance is infinity.
(D) both (B) and (C).
Answer:
(D) both (B) and (C).