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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 1.
Explain the phenomenon of adsorption with the help of examples.
Answer:
Consider the following two examples:

• Example 1: When a metal spoon is dipped in milk and taken out, it is observed that a film of milk particles covers the spoon surface.
• Example 2: If a cold water bottle is taken out from the refrigerator and kept on a table for a while, water vapour is seen to condense on the outer surface of the bottle, forming droplets or a film.
• In the above examples, the milk particles or the water molecules from the air get adsorbed on the surface of the spoon and the bottle, respectively.
• Similarly, surfaces of many objects around us are exposed to the atmosphere. Water molecules as well as other gas molecules such as N₂, O₂, from the air form an invisible multimolecular film on these objects.
  This is known as the phenomenon of adsorption.

Question 2.
Why does adsorption occur?
Answer:

• The adsorption phenomenon is caused by dispersion forces (also known as London dispersion forces or van der Waals forces) which are short range and additive. Adsorption force is the sum of all interactions between all the atoms.
• The pulling interactions cause the surface of a liquid to tighten like an elastic film.
• A measure of the elastic force at the surface of a liquid is called surface tension.
• There is a tendency to have minimum surface tension, i.e., decrease of free energy, which leads to adsorption.

Question 3.
Define surface tension.
Answer:
A measure of the elastic force at the surface of a liquid is called surface tension.
OR
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.

Question 4.
Define the following terms.
i. Adsorbent
ii. Adsorbate
Answer:
i. Adsorbent: The material or substance present in the bulk, on the surface of which adsorption takes place is called adsorbent.
ii. Adsorbate: The substance getting adsorbed on the adsorbent is called as adsorbate.

Question 5.
Give some examples of adsorption.
Answer:
Following are some examples of adsorption:

• Adsorption of gases like hydrogen and oxygen by finely divided metals, namely, platinum, palladium, copper, nickel, etc.
• Adsorption of gases like nitrogen and carbon dioxide by activated charcoal.
• Removal of colouring matter like an organic dye, for example, methylene blue. When charcoal is added to methylene blue solution and shaken, it becomes colourless after some time as dye molecules accumulate on the surface of charcoal.

Question 6.
What is desorption?
Answer:
The process of removal of an adsorbed substance from a surface on which it was adsorbed is called desorption.

Question 7.
Define sorption.
Answer:
When both adsorption and absorption occur simultaneously, it is known as sorption.
e.g. When a chalk is dipped in ink, the ink molecules are adsorbed at the surface of the chalk while the solvent of the ink goes deeper into the chalk due to absorption.

Question 8.
What is physisorption? State its characteristics.
Answer:
When the adsorbent such as gas molecules are accumulated at the surface of a solid on account of weak van der Waals forces, the adsorption is termed as physical adsorption or physisorption.

Characteristics:

• The van der Waals forces involved in physical adsorption are similar to forces causing condensation of gas into liquid. Thus, heat is released in physisorption.
• The heat released during physisorption is of the same order of magnitude as heat of condensation.
• Due to weak nature of van der Waals forces, physisorption is weak in nature.
• The adsorbed gas forms several layers of molecules at high pressures.
• The extent of adsorption is large at low temperatures.
• The equilibrium is attained rapidly.
• Physisorption is readily reversed by lowering of pressure of gas or by raising temperature.

Question 9.
Define chemisorption, Write its main features.
Answer:
When the gas molecules accumulate at the surface of a solid or adsorbate by means of chemical bonds (covalent or ionic), the adsorption is termed as chemical adsorption or chemisorption.
Features of chemical adsorption:

• Chemisorption is specific in nature.
• Chemisorption involving the gas-solid as the adsorbate and adsorbent is usually exothermic i.e., heat is released during this process (Exception: The adsorption of hydrogen on glass is endothermic).
• The heat evolved in chemisorption per mole of adsorbate is nearly the same order of magnitude as that accompanying chemical bonding.
• Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.
• Chemisorption increases with increase in temperature in the beginning, as a greater number of molecules can have activation energy. But after certain temperature chemisorption decreases with increase in temperature as the chemical bonds break.
• Sometimes at low’ temperature, physisorption occurs which passes into chemisorption as the temperature is raised.
• Chemisorption is dependent on surface area of the adsorbent.

[Note: Chemisorption was first investigated in 1916 by American Chemist, Irving Langmuir (1881-1957).]

Question 10.
Why is chemisorption also known as activated adsorption?
Answer:
Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.

Question 11.
Give reason: Adsorption of hydrogen on glass is an endothermic process.
Answer:
Adsorption of hydrogen on glass is an endothermic process because heat is absorbed during the process due to dissociation of hydrogen.

Question 12.
Explain graphically the effect of the following factors on the adsorption of gases by solids.
i. Temperature of the adsorbent surface
ii. Pressure of the gas (adsorbate)
Answer:
i. Temperature of the adsorbent surface:

• Adsorption is an exothermic process.
• According to Te Chatelier’s principle, it is favoured at low temperature.
• Therefore, the amount of gas adsorbed is inversely proportional to the temperature.
• The graph given below shows plots of volume of N? adsorbed per unit mass of adsorbent against the pressure of a gas at different temperatures.
• As temperature increases from 193 K to 273 K at a constant pressure ‘P’, the amount of gas adsorbed decreases.

ii. Pressure of the gas:

• At any temperature, the extent of gas adsorbed increases with an increase in pressure.
• The extent of adsorption is directly proportional to pressure of the gas.
• At high pressures extent of adsorption becomes independent of the pressure. The surface of adsorbent is then almost fully covered by adsorbed gaseous molecules.

Question 13.
What are the applications of adsorption?
Answer:
Following are the various applications of adsorption:
i. Catalysis (Heterogeneous catalysis):

• The solid catalysts are used in many industrial manufacturing processes.
• For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H₂SO₄ (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

• It is a device which consists of activated charcoal or mixture of adsorbents.
• It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.

iv. Production of high vacuum:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

v. Adsorption indicators: The adsorption is used to detect the end point of precipitation titrations. Dyes such as eosin, fluorescein are used as indicators.
e.g.
a. A solution of sodium chloride containing a small amount of fluorescein is titrated against silver nitrate solution.

b. When chloride ions are over, fluorescein is adsorbed on white silver chloride precipitate and hence, red colour is developed.
c. Thus, colour changes from pale yellow to reddish pink at the end point.

vi. Separation of inert gases:

• In a mixture of noble gases, different gases adsorb to different extent.
• Due to selective adsorption principle, gases can be separated on coconut charcoal.

vii. Froth floatation process:

• A low-grade sulphide ore is concentrated by separating it from silica and other earthy matter using pine oil as frothing agent.
• Hydrophobic pine oil preferentially adsorbs sulphide ore which is taken up in the froth.

viii. Chromatographic analysis:

• It is based on selective adsorption of ions from solution using powdered adsorbents such as silica or alumina gel.
• It has several industrial and analytical applications. Other applications include surface area determination, purification of water, etc.

Question 14.
Explain how high vacuum can be obtained by adsorption.
Answer:

• Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
• A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question 15.
State whether TRUE or FALSE. Correct if false.
i. The rate of adsorption of gases on charcoal powder decreases on lowering of temperature at a given pressure.
ii. Noble gases can be separated from their mixture using the principle of selective adsorption as they adsorb to different extent.
iii. Pine oil is used as frothing agent in froth floatation process.
Answer:
i. False
The rate of adsorption of gases on charcoal powder increases on lowering of temperature at a given pressure.
ii. True
iii. True

Question 16.
Match the following.

	Column A		Column B
i.	Iron	a.	Hydrogenation of oils
ii.	Nickel	b.	Production of sulphuric acid
iii.	Platinum	c.	Synthesis of ammonia

Answer:
i – c,
ii – a,
iii – b

Question 17.
What is a catalyst?
Answer:
A catalyst is a substance which when added to a reacting system, increases the rate of a reaction without itself undergoing any permanent chemical change.

Question 18.
Explain the importance of catalysts in chemical industries.
Answer:

• A large number of the chemicals manufactured in industries make use of catalysts to obtain specific products.
• The use of catalyst lowers the reaction temperature as well as energy costs significantly.
  Due to these advantages, catalysts are of great importance in chemical industry.

Question 19.
Name two types of catalysis.
Answer:

1. Homogeneous catalysis
2. Heterogeneous catalysis

Question 20.
Define homogeneous catalysis and give any two examples.
Answer:
When the reactants and the catalyst are in the same phase, it is said to be homogeneous catalysis.
e.g.
i. Iodide ion (I^–) is used as homogeneous catalyst in decomposition of aqueous hydrogen peroxide because both I^– and H₂O₂ are present in the same aqueous phase.
ii. Hydrolysis of sugar is catalysed by H⁺ ions furnished by sulphuric acid.

All reactants and catalyst are in same solution phase.
[Note: Enzyme catalysis is also an important type of homogeneous catalysis.]

Question 21.
Justify: Lead chamber process is an example of homogeneous catalysis.
Answer:
i. In the lead chamber process, sulphur dioxide is oxidized to sulphur trioxide with dioxygen (O₂) in the presence of nitric oxide as catalyst.

ii. Since all the reactants as well as the catalyst is present in gaseous state. i.e., in same phase, it is a homogeneous catalysis reaction.
Hence, lead chamber process is an example of homogeneous catalysis.

Question 22.
Describe heterogeneous catalysis with the help of one example.
Answer:
i. When the reactants and catalyst are in different phase, it is said to be heterogeneous catalysis.
ii. The heterogeneous catalyst is generally a solid and the reactants may either be gases or liquids.
iii. When the solid catalyst is added to the reaction mixture, it does not dissolve in the reacting system and the reaction occurs on the surface of the solid catalyst.
e.g. Dinitrogen (N₂) and dihydrogen (H₂) combine to form ammonia in Haber process in the presence of finely divided iron along with K₂O and Al₂O₃.

b. In the above reaction, Al₂O₃ and K₂O are promoters of the Fe catalyst. Al₂O₃ is added to prevent the fusion of Fe particles. K₂O causes chemisorption of nitrogen atoms. Molybdenum is also used as promoter.
c. Since the reactants are present in gaseous phase while the catalyst used is in solid phase, it represents heterogeneous catalysis.

Question 23.

i. State whether the given reaction is an example of heterogeneous or homogeneous catalysis.
ii. What is the role of Fe, K₂O and Al₂O₃ in this reaction?
Answer:
i. This reaction is an example of heterogeneous catalysis.
ii. Fe is used as a catalyst while K₂O and Al₂O₃ are promoters of the Fe catalyst. Al₂O₃ is used to prevent the fusion of Fe particles while K₂O causes chemisorption of nitrogen atoms.

Question 24.
Describe hydrogenation reaction of vegetable oils.
Answer:
i. Hydrogenation reaction of vegetable oils used in food industry to produce solid fats. The reaction is as follows:

ii. The reaction is catalysed by finely divided metals like Ni, Pd or Pt.
iii. Vegetable oil contains one or more carbon-carbon double bonds (C = C) in its structure.
iv. On hydrogenation, a solid product (which contains only carbon-carbon single bonds) is formed. It is called Vanaspati ghee.
v. The hydrogenation reaction of vegetable oils is an example of heterogeneous catalysis as the reactant and the catalyst are not present in the same phase.

Question 25.
i. Explain the role of catalytic converters in automobile exhaust.
ii. Why do automobiles with catalytic converter require unleaded petrol?
Answer:
i. a. An important application of heterogeneous catalysts is in automobile catalytic converters.
b. In automobile exhaust, large number of air pollutants such as carbon monoxide, nitric oxide, etc. are present.
c. The catalytic converter transforms these air pollutants into carbon dioxide, water, nitrogen and oxygen.
ii. The catalyst used in the catalytic converter gets poisoned by the adsorption of lead (Pb) present in the petrol. Hence, the automobiles with catalytic converter requires unleaded petrol.

Question 26.
What are inhibitors? Explain with an example.
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.

When 2% ethanol is added to chloroform, the formation of COCl₂ is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Question 27.
Write decomposition reaction of hydrogen peroxide. Suggest how this decomposition can be prevented.
Answer:
i. Hydrogen peroxide decomposes as,

ii. The reaction can be inhibited by addition of dilute acid or glycerol as they act as inhibitors.

Question 28.
Explain why 2% ethanol is added to chloroform?
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.

When 2% ethanol is added to chloroform, the formation of COCl₂ is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Question 29.
Describe the steps involved in heterogeneous catalysis by solid catalyst.
OR
Explain the mechanism involved in catalytic action of a heterogeneous catalyst.
Answer:
The catalytic action of a heterogeneous catalyst occurs on the surface of a catalyst.
The mechanism involves the following five steps.
i. Diffusion of reactants towards the surface of the catalyst.
ii. Adsorption of reactant molecules on the surface of the catalyst.
iii. Occurrence of chemical reaction on the catalyst surface and formation of an intermediate.
iv. Formation of the products.
v. Desorption of reaction products from the catalyst surface. Products leave the catalyst surface in the following steps.
Steps involved in desorption of reaction products:
Diffusion → Adsorption → Intermediate formation → Product formation → Desorption
vi. Fresh reactant molecules can replace the products to start the cycle again as in first step.
vii. This is why catalyst remains unchanged in mass and chemical composition at the end of the reaction.

Question 30.
Write a short note on catalytic activity.
Answer:

• The catalytic activity of a catalyst depends on the strength of chemisorption.
• If large number of reactant molecules (gas or liquid) are strongly adsorbed on the surface of solid catalyst, the catalyst is said to be active.
• However, the adsorption of reactant molecules on the surface, that is, the bond formed between adsorbate and adsorbent surface should not be very strong so that they are not immobilized.
• d-block metals such as Fe, V and Cr tend to be strongly active towards O₂, C₂H₂, C₂H₄, CO, H₂, CO₂, N₂, etc.
• Mn and Cu are unable to adsorb N₂ and CO₂.
• The metals Mg and Li adsorb O₂ selectively.

Question 31.
Explain catalytic selectivity with suitable examples.
Answer:
i. Some solid catalysts are selective in their action.
ii. The same gaseous reactants produce different products when different catalysts are used.
e.g.
a. The gaseous ethylene and O₂ react to produce different products with different catalysts.

b. The gaseous carbon monoxide and H₂ produce different products by using different catalysts.

Question 32.
i. What are zeolites?
ii. Zeolites are shape selective catalysts. Explain.
iii. What is the use of a zeolite catalyst ZSM-5 in petroleum industry?
Answer:
i. a. Zeolites are aluminosilicates with three-dimensional network of silicates.
b. Some silicon atoms in this network are replaced by aluminium atoms giving Al – O – Si framework which results in microporous structure.

ii. a. The reactions in zeolites are dependent on the size and shape of reactant or products, b. It also depends on the pores and cavities of zeolites.
b. Therefore, zeolites are shape selective catalysts.

iii. In petroleum industry, zeolite catalyst ZSM-5 converts alcohols directly to gasoline (petrol) by dehydration which gives a mixture of hydrocarbons.

Question 33.
State the importance of colloids in day-to-day life.
Answer:

• Colloid chemistry is the chemistry of everyday life.
• A number of substances we use in our day-to-day life are colloids. For example, milk, butter, jelly, whipped cream, mayonnaise.
• Knowledge of colloid chemistry is essential for understanding about many useful materials like cement, bricks, pottery, porcelain, glass, enamels, oils, lacquers, rubber, celluloid and other plastics, leather, paper, textiles, filaments, crayons, inks, road construction material, etc.
• In many daily processes like cooking, washing, dyeing, painting, ore floatation, water purification, sewage disposal, smoke prevention, photography, pharmacy, use of colloids is important.

Question 34.
What are colloids? Explain.
Answer:
i. Colloids are heterogeneous mixtures.
ii. The component of colloid present in the largest proportion is called dispersion medium and the other components are called dispersed phase.
iii. The particles of the dispersed phase are larger than the size of a molecule and smaller than the particles which we can see with naked eye.
e.g.

• Observe the formation of solution of salt and water. Salt dissolves completely in water and forms homogeneous system.
• On the other hand, ground coffee or tea leaves with milk form suspension.
• Between the two extremes of solution and suspension exists a large group of systems called colloidal dispersions or simply colloids.

Question 35.
State the differences between colloids and solutions.
Answer:
Colloids:

1. Colloids contain particles of dispersed phase with diameters in the range of 2 to 500 nm.
2. They are translucent to light.
3. e.g. Milk, fog, etc.

Solutions:

1. Solutions contain solute particles with diameters in the range of 0.1 to 2 nm.
2. They are transparent or may be coloured.
3. e.g. NaCl solution

Question 36.
Explain: Natural phenomena of colloids observed in daily life.
Answer:
Following are some examples of colloids observed in daily life.
i. Blue colour of the sky: The sky appears blue to us because minute dust particles along with minute water droplets dispersed in air scatter blue light which reaches our eyes.
ii. Blood: It is a colloidal dispersion of plasma proteins and antibodies in water arid at the same time blood is also a suspension of blood cells and platelets in water.
iii. Soils: Fertile soils are colloidal in nature where humus acts as a protective colloid. Soil adsorbs moisture and nourishing materials due to its colloidal nature.
iv. Fog, mist and rain:

• Mist is caused by small droplets of water dispersed in air.
• Fog is formed whenever there is temperature difference between ground and air.
• A large portion of air containing dust particles gets cooled below its dew point, the moisture from the air condenses on the surface of these particles which form fine droplets, which are colloidal particles and float in the air as fog or mist.

Question 37.
State different ways to classify colloids.
Answer:
Colloids can be classified in three different ways:

• Physical states of dispersed phase and dispersion medium
• Interaction or affinity of phases
• Molecular size

Question 38.
Name the types of colloids based on the physical states of dispersed phase and dispersion medium. Give two examples of each.
Answer:
There are eight types of colloids based on the physical states of dispersed phase and dispersion medium as given below.

Sr. No.	Type of Colloids	Examples
i.	Solid sol (solid dispersed in solid)	Coloured glasses, gemstones
ii.	Sols and gels (solid in liquid)	Gelatin, muddy water
iii.	Aerosol (solid in gas)	Smoke, dust
iv.	Gel (liquid in solid)	Cheese, jellies
v.	Emulsion (liquid in liquid)	Milk, hair cream
vi.	Aerosol (liquid in gas)	Fog, mist
vii.	Solid sol (gas in solid)	Foam rubber, plaster
viii.	Foam (gas in liquid)	Froth, soap lather

Question 39.
Complete the following chart.

Answer:

[Note: Students can write any one example of the given type of colloids.]

Note: Types of colloids based on the physical states of dispersed phase and dispersion medium.

Question 40.
Describe classification of colloids based on the interaction or affinity of phases.
Answer:
On the basis of interaction or affinity of phases, a colloidal solution is classified as lyophilic and lyophobic.
i. Lyophilic colloids:

• A colloidal solution in which the particles of dispersed phase have a great affinity for the dispersion medium are lyophilic colloids.
• If the lyophilic sol is evaporated, the dispersed phase separates. However, if it is remixed with the medium, the sol. can be formed again and hence, such sols are called reversible sols.
• They are stable and difficult to coagulate.

ii. Lyophobic colloids:

• Colloidal solution in which the particles of the dispersed phase have no affinity for the dispersion
  medium are called lyophobic colloids.
• The common examples are Ag, Au, hydroxides like Al(OH)₃, Fe(OH)₃, metal sulphides.
• Once precipitated or coagulated they have little tendency or no tendency to revert back to colloidal state.

[Note: Lyo means liquid and philic means loving whereas phobic means fearing and hence liquid hating. If water is the dispersion medium, the terms hydrophilic and hydrophobic are used.]

Question 41.
Give reason: Lyophilic sols are called reversible sols.
Answer:

• When lyophilic sol is evaporated, the dispersed phase separates.
• However, if the dispersed phase is remixed with the medium, the sol can be formed again.

Hence, lyophilic sols are called reversible sols.

Question 42.
How are colloids classified based on their molecular size?
Answer:
Colloids are classified into three types based on their molecular size as described below.
i. Multimolecular colloids:

• In multimolecular colloids, the individual particles consist of an aggregate of atoms or small molecules with size less than 10³ pm.
  e.g. Gold sol consists of particles of various sizes having several gold atoms.
• Colloidal solution in which particles are held together with van der Waals force of attraction is called multimolecular colloid.
  e.g. S₈ sulphur molecules

ii. Macromolecular colloids: In this type of colloids, the molecules of the dispersed phase are sufficiently large in size (macro) to be of colloidal dimensions.
e.g. Starch, cellulose, proteins, polythene, nylon, plastics.

iii. Associated colloids or micelles:

• The substances behave as normal electrolytes at low concentration and associated in higher concentration forming a colloidal solution.
• The associated particles are called micelles, e.g. Soaps and detergents

Question 43.
How can be colloids prepared by chemical methods?
Answer:
i. Colloidal dispersions can be prepared by chemical reactions leading to formation of molecules by double decomposition, oxidation, reduction or hydrolysis.
ii. Molecules formed in these reactions are water-insoluble and thus, they aggregate leading to the formation of colloids.
e.g.

Question 44.
Describe the process involved in peptization?
Answer:

• During peptization a precipitate is converted into colloidal sol by shaking with dispersion medium in the presence of a small amount of an electrolyte. The electrolyte used is known as peptizing agent.
• During the process, the precipitate adsorbs one of the ions of the electrolyte on its surface and as a result, positive or negative charge is developed on the precipitate which finally breaks up into small particles of colloidal size.

[Note: This method is generally applied to convert a freshly prepared precipitate into a colloidal sol.]

Question 45.
Why is it necessary to purify colloidal solutions?
Answer:

• Colloidal solution generally contains excessive amount of electrolytes and some other soluble impurities.
• A small quantity of an electrolyte is necessary for the stability of colloidal solution, however, a large quantity of electrolyte may result in coagulation.
• It is also necessary to reduce soluble impurities.

Hence, it is necessary to purify colloidal solutions.

Question 46.
i. What is purification of colloidal solution?
ii. How can a colloidal solution be purified using the method of dialysis?
Answer:
i. The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution.

ii. a. Dialysis is a process of removing a dissolved substance from a colloidal solution by diffusion through a suitable membrane.
b. Purification of colloidal solution can be carried out using dialysis by the following method.

• The apparatus used is dialyser.
• A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing.
• The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

Question 47.
What are the general properties exhibited by colloidal dispersions?
Answer:
General properties exhibited by colloidal dispersions:

• Colloidal system is heterogeneous and consists of two phases, dispersed phase and dispersion medium.
• The dispersed phase particles pass slowly through parchment paper or animal membrane, but readily pass through ordinary filter paper.
• Colloidal particles are usually not detectable by powerful microscope.

Question 48.
Discuss the factors that influence the colour of colloidal solutions.
Answer:

• Colour of colloidal solution depends on the wavelength of light scattered by dispersed particles.
• The colour of colloidal dispersion also changes with the manner in which the observer receives the light.
  e.g. Mixture of a few drops of milk and large amount of water appears blue when viewed by the scattered light and red when viewed by transmitted light.
• It also depends on size of colloidal particles.
  e.g. Finest gold sol is red in colour whereas with increase in size it appears purple.

Question 49.
Give three examples each:
i. Positively charged sols
ii. Negatively charged sols
Answer:
i. Positively charged sols: Al₂O₃. xH₂O, haemoglobin, TiO₂ sol
ii. Negatively charged sols: Au sols, Congo red sol, clay

Note: Some common sols with the nature of charge on the particles are listed in the table below.

Positively charged sols	Negatively charged sols
Hydrated metallic oxides: Al2O3.xH2O, CrO3.xH2O, Fe2O3.xH2O.	Metals: Cu, Ag. Au sols Metallic sulphides: As2S3, Sb2S3, CdS
Basic dye stuff, methylene blue sols	Acid dye stuff, eosin, Congo red sol
Haemoglobin (blood)	Sols of starch, gum
Oxides: TiO2 sol	Gelatin, clay, gum sols

Question 50.
Explain the term electroosmosis.
Answer:

• Movement of dispersed particles can be prevented by suitable means such as use of membrane.
• On doing so, it is observed that the dispersion medium begins to move in an electric field. This is known as electroosmosis.

Question 51.
What is coagulation?
Answer:
The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.

Question 52.
How can we bring about precipitation of lyophobic colloids?
Answer:

• The charge on the colloidal particles is due to the preferential adsorption of ions on their surface.
• Hence, lyophobic colloids can be precipitated out by removing the charge on the colloidal particles (dispersed phase).

Question 53.
Discuss various methods that are used to bring about coagulation of lyophobic sols.
Answer:
Coagulation of the lyophobic sols can be carried out in the following ways.

• By electrophoresis: The colloidal particles move towards oppositely charged electrodes, get discharged and precipitate.
• By mixing two oppositely charged sols: Oppositely charged sols when mixed in almost equal proportions neutralize their charges and get precipitated.
  e. g. Mixing of hydrated ferric oxide (positive sol) and arsenious sulphide (negative sol) brings them in the precipitated forms. This type of coagulation is called mutual coagulation.
• By boiling: When a sol is boiled, the adsorbed layer is disturbed as a result of increased collisions with molecules in the dispersion medium. This reduces charge on the particles and subsequently particles settle down as a precipitate.
• By persistent dialysis: On prolonged dialysis, traces of the electrolyte present in the sol are removed almost completely. The colloids then become unstable and finally precipitate.
• By addition of electrolytes: When excess of an electrolyte is added, the colloidal particles are precipitated.

Question 54.
Write Hardy-Schulze rule.
Answer:
Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.

Question 55.
Differentiate between oil in water and water in oil emulsions.
Answer:
Oil in water:

1. Oil is the dispersed phase and water is the dispersion medium.
2. If water is added, it will be miscible with the emulsion.
3. Addition of small amount of an electrolyte makes the emulsion conducting.
4. Continuous phase is water.
5. Basic metal sulphates, water soluble alkali metal soaps are used as emulsifiers.

Water in oil:

1. Water is the dispersed phase and oil is the dispersion medium.
2. If oil is added, it will be miscible with the emulsion.
3. Addition of small amount of an electrolyte has no effect on conducting power.
4. Continuous phase is oil.
5. Water insoluble soaps such as those of Zn, Al, Fe, alkaline earth metals are used as emulsifiers.

Question 56.
What are the properties of emulsion?
Answer:
Properties of emulsion:

• Emulsion can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed forms a separate layer.
• The droplets in emulsions are often negatively charged and can be precipitated by electrolytes.
• Emulsions show Brownian movement and Tyndall effect.
• The two liquids in emulsions can be separated by heating, freezing, centrifuging, etc.

Question 57.
Give applications of colloids.
Answer:
Applications of colloids:
i. Electrical precipitation of smoke:

• Smoke is a colloidal solution of solid particles of carbon, arsenic compound, dust, etc. in the air.
• When smoke is allowed to pass through chamber containing charged plates, smoke particles lose their charge and get precipitated. The particles then settle down on the floor of the chamber.
• The precipitator used is called Cottrell precipitator.

ii. Purification of drinking water:

• Water obtained from natural sources contains colloidal impurities.
• By addition of alum to such water, colloidal impurities get coagulated and settle down. This makes water potable.

iii. Medicines:

• Usually medicines are colloidal in nature.
• Colloidal medicines are more effective owing to large surface area to volume ratio of a colloidal particle and easy assimilation.
  e.g. Argyrol is a silver sol used as an eye lotion. Milk of magnesia, an emulsion is used in stomach disorders.

iv. Rubber industry: Rubber is obtained by coagulation of latex.
v. Cleansing action of soaps and detergents.
vi. Photographic plates, films, and industrial products like paints, inks, synthetic plastics, rubber, graphite lubricants, cement, etc. are colloids.

Question 58.
Match column A with column B.

	Column A		Column B
i.	Tyndall effect	i.	Kinetic property
ii.	Electrophoresis	ii.	Argyrol
iii.	Silver sol	iii.	Optical property
iv.	Brownian motion	iv.	Coagulation

Answer:
i – c,
ii – d,
iii – b,
iv – a

Question 59.
In drinking water treatment, often alum is added for the complete removal of suspended impurities. On complete dissolution, alum produces positive charge which neutralizes the charge on the suspended particles and thus, impurities are easily removed.
i. Name and define the process involved due to which charge on particles get neutralized.
ii. What is the role of alum in the above mentioned process?
Answer:
i. a. Charge on particles get neutralized due to coagulation.
b. The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.
ii. Alum acts as a reagent that helps in coagulation of the suspended particles by the removal of the charge associated with these particles.

Multiple Choice Questions

1. Which of the following is responsible for adsorption phenomenon?
(A) Hydrogen bonding
(R) Dipole-dipole forces
(C) Ion-dipole forces
(D) Dispersion forces
Answer:
(D) Dispersion forces

2. A substance which adsorbs another substance on its surface is called ……………..
(A) adsorbate
(B) absorbate
(C) adsorbent
(D) absorbent
Answer:
(C) adsorbent

3. During adsorption, the molecules of the substance which gets adsorbed are termed as
(A) adsorbent
(B) adsorbate
(C) absorbent
(D) absorbate
Answer:
(B) adsorbate

4. in adsorption of acetic acid on charcoal, acetic acid is ……………
(A) adsorhate
(B) adsorbent
(C) absorbent
(D) absorbate
Answer:
(A) adsorhate

5. The process of removal of an adsorbed substance from the surface is known as
(A) sorption
(B) oxidation
(C) reduction
(D) desorption
Answer:
(D) desorption

6. ………….. is the process in which adsorbate molecules are held on the surface of the adsorbent by weak van der Waals forces.
(A) Chemisorption
(B) Absorption
(C) Physisorption
(D) Biosorption
Answer:
(C) Physisorption

7. Which of the following is an example of physical adsorption?
(A) Adsorption of acetic acid in solution by charcoal
(B) Adsorption of O₂ on tungsten
(C) Adsorption of N₂ on Fe
(D) Adsorption of H₂ on Ni
Answer:
(A) Adsorption of acetic acid in solution by charcoal

8. Chemisorption is a slow process because …………….
(A) it forms multimolecular layer
(B) it is reversible
(C) it takes place at normal temperature
(D) it requires high activation energy
Answer:
(D) it requires high activation energy

9. The number of layer(s) formed on adsorbent in chemical adsorption is …………….
(A) one
(B) two
(C) three
(D) many
Answer:
(A) one

10. Which of the following statements is CORRECT regarding chemical adsorption?
(A) It is highly specific in nature.
(B) It is relatively strong.
(C) It involves the formation of monolayer of adsorbed particles.
(D) All of these.
Answer:
(D) All of these.

11. Which of the following is adsorbed to maximum extent on charcoal?
(A) H₂
(B) N₂
(C) Cl₂
(D) O₂
Answer:
(C) Cl₂

12. The relation between the amount of substance adsorbed by an adsorbent and the equilibrium pressure or …………. at any constant temperature is called adsorption isotherm.
(A) surface area
(B) volume
(C) circumference
(D) concentration
Answer:
(D) concentration

13. For equilibrium pressure (P), the mass of gas adsorbed (x) and mass of adsorbent (m) may be expressed as Freundlich adsorption isotherm as ……………

Answer:
(B) \(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{\frac{1}{\mathrm{n}}}\)

14. When log x/m is plotted against log P, the intercept obtained …………..
(A) on Y axis is equal to log K
(B) on Y axis is equal to K
(C) on X axis is equal to log K
(D) on X axis is equal to K
Answer:
(A) on Y axis is equal to log K

15. The adsorption isotherm tends to saturate at ………….. pressure.
(A) low
(B) moderate
(C) all of these
(D) high
Answer:
(D) high

16. In Haber process for manufacture of NH₃, the catalyst used is ……………
(A) iron
(B) copper
(C) vanadium pentoxide
(D) nickel
Answer:
(A) iron

17. A substance that decreases the rate of a chemical reaction is called ……………
(A) inhibitor
(B) prohibitor
(C) promoter
(D) reactor
Answer:
(A) inhibitor

18. Whether a given mixture forms a true solution or a colloidal dispersion depends on the …………….
(A) charge of solute particles
(B) size of solvent particles
(C) size of solute particles
(D) charge of solvent particles
Answer:
(C) size of solute particles

19. An aerosol is a dispersion of a ……………
(A) gas in a solid
(B) liquid in a gas
(C) solid in a gas
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

20. The dispersed phase in Pumice stone is ……………
(A) solid
(B) liquid
(C) gas
(D) none of these
Answer:
(C) gas

21. Colloidal solution in which the dispersed phase has little affinity for the dispersion medium is called ………………
(A) lyophobic colloids
(B) lyophilic colloids
(C) hydrophilic colloids
(D) emulsions
Answer:
(A) lyophobic colloids

22. Which of the following is NOT an example of macromolecular colloid?
(A) Starch
(B) Proteins
(C) S₈ molecules
(D) Nylon
Answer:
(C) S₈ molecules

23. Tyndall effect is useful ……………….
(A) to identify colloidal dispersions
(B) to count number of particles in colloidal dispersion.
(C) to determine the size of the colloidal particles
(D) all of these
Answer:
(D) all of these

24. Brownian movement is a ……………… type of property of the colloidal sol.
(A) electrical
(B) optical
(C) kinetic
(D) colligative
Answer:
(C) kinetic

25. The migration of colloidal particles under the influence of an electric field is called …………….
(A) catalysis
(B) Brownian movement
(C) electrophoresis
(D) Tyndall effect
Answer:
(C) electrophoresis

26. The capacity of an ion to coagulate a colloidal solution depends on ……………….
(A) its shape
(B) its valency
(C) the sign of charge
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

27. ……………… is an example of water in oil type of emulsion.
(A) Milk
(B) Cod liver oil
(C) Vanishing cream
(D) Paint
Answer:
(B) Cod liver oil

28. Which of the following has highest precipitation power to precipitate negative sol?
(A) Al³⁺
(B) Mg²⁺
(C) Na⁺
(D) K⁺
Answer:
(A) Al³⁺

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 12 Chemical Equilibrium

Question 1.
Explain irreversible reaction.
Answer:
Irreversible reaction:
i. Reactions which occur only in one direction, namely, from reactant to products are called irreversible reactions.
ii. They proceed in only a single direction until one of the reactants is exhausted.
iii. The direction in which an irreversible reaction occurs is indicated by an arrow (→) pointing towards the products in the chemical equation.
e.g. a. \(\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \stackrel{\text { Burn }}{\longrightarrow} \mathrm{CO}_{2(\mathrm{~g})}\)
b. \(2 \mathrm{KClO}_{3(\mathrm{~s})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{KCl}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})}\)

Question 2.
What is a closed system?
Answer:
A system in which there is no exchange of matter with the surroundings is called a closed system.

Question 3.
What is an open system?
Answer:
A system in which exchange of both matter and heat occurs with the surroundings is called an open system.

Question 4.
Why was calcium oxide used in theatre lighting?
Answer:
Calcium oxide (CaO) on strong heating glows with a bright white light. Hence, CaO was used in theatre lighting, which gave rise to the phrase ‘in the limelight’.

Question 5.
Explain liquid-vapour equilibrium with an example.
Answer:
Liquid-vapour equilibrium:
i. Consider reversible physical process of evaporation of liquid water into water vapour in a closed vessel. Initially, there is practically no water vapour in the vessel.

ii. When the liquid evaporates in the closed container, the liquid molecules escape from the liquid surface into vapour phase building up vapour pressure. They also condense back into liquid state because the container is closed.

iii. In the beginning the rate of evaporation is high and the rate of condensation is low. But with time, as more and more vapour is formed, the rate of evaporation goes down and the rate of condensation increases. Eventually the two rates become equal. This gives rise to a constant vapour pressure. This state is known as an ‘equilibrium state’.
In this state, the rate of evaporation is equal to the rate of condensation.
It may be represented as: H₂O_(l) ⇌ H₂O_(Vapour)

iv. At equilibrium, the pressure exerted by the gaseous water molecules at a given temperature remains constant, known as the equilibrium vapour pressure of water (or saturated vapour pressure of water or aqueous tension). The saturated vapour pressure increases with increase of temperature.

[Note: The saturated vapour pressure of water at 100 °C is 1 atm (1.013 bar). Hence, water boils at 100 °C when pressure is 1 atm.]

Question 6.
What is meant by the term ‘normal boiling point’ of a liquid?
Answer:
For any pure liquid at 1 atm pressure, the temperature at which its saturated vapour pressure equals to atmospheric pressure is called the normal boiling point of that liquid.
e.g. The boiling point of ethyl alcohol is 78 °C i.e., the saturated vapour pressure of ethyl alcohol at 78 °C is 1 atm (1.013 bar).

Question 7.
Give an example of solid-liquid equilibrium.
Answer:
A mixture of ice and water in a perfectly insulated thermos flask at 273 K is an example of solid-liquid equilibrium.
H₂O_(s) ⇌ H₂O_(l)

Question 8.
Identify the type of equilibrium in the following physical processes:
i. Camphor_(s) ⇌ Camphor_(g)
ii. Ammonium chloride_(s) ⇌ Ammonium chloride_(g)
iii. Carbon dioxide gas ⇌ Dry ice
iv. Water ⇌ Ice
Answer:
i. Solid – vapour equilibrium
ii. Solid – vapour equilibrium Solid
iii. Solid – vapour equilibrium
iv. Solid – liquid equilibrium

Question 9.
Name two substances that undergoes sublimation.
Answer:
Camphor, ammonium chloride.

Question 10.
Write a short note on chemical equilibrium.
Answer:
Chemical equilibrium:

• If a reaction takes place in a closed system so that the products and reactants cannot escape, we often find that reaction does not give a 100% yield of products. Instead some reactants remain after the concentrations stop changing.
• When there is no further change in concentration of reactant and product, the chemical reaction has attained equilibrium, with the rates of forward and reverse reactions being equal.
• Chemical equilibrium at a given temperature is characterized by constancy of measurable properties such as pressure, concentration, density, etc.
• Chemical equilibrium can be approached from either side of the chemical reaction.

Question 11.
Explain the law of mass action and give its mathematical representation.
Answer:
Statement: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.
Explanation: A rate equation can be written for a reaction by applying the law of mass action as follows: Consider a reaction, A + B → C
Here A and B are the reactants and C is the product. The concentrations of chemical species are expressed in mol L^-1 and denoted by putting the formula in square brackets. On applying the law of mass action to this
reaction, a proportionality expression can be written as: Rate ∝ [A] [B]
This proportionality expression is transformed into an equation by introducing a proportionality constant, k, as follows:
Rate = k [A] [B]
This equation is called the rate equation and the proportionality constant, k, is called the rate constant of the reaction.

Question 12.
Write the rate equation for the following reactions:
i. C + O₂ → CO₂
ii. 2KClO₃ → 2KCl + 3O₂
Answer:
The rate equation is written by applying the law of mass action.
i. The reactants are C and O₂
Rate ∝ [C] [O₂]
∴ Rate = k [C] [O₂]
ii. The reactant is KClO₃ and its 2 molecules appear in the balanced equation.
∴ Rate ∝ [KClO₃]²
∴ Rate = k [KClO₃]²

Question 13.
Derive the expression of equilibrium constant, K_C for the reaction:
A + B ⇌ C + D
Answer:
Consider a hypothetical reversible reaction A + B ⇌ C + D.
Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:
Rate_forward ∝ [A][B]
∴ Rate_forward = k_f [A] [B] …… (1)
∴ Rate_reverse ∝ [C] [D]
∴ Rate_reverse = k_r [C] [D] …. (2)
At equilibrium, the rates of forward and reverse reactions are equal. Thus,
Rate_forward = Rate_reverse
∴ k_f [A] [B] = k_r [C] [D]
∴ \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}=\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\) …….. (3)
K_C is called the equilibrium constant.

Question 14.
Show that the equilibrium constant of the reverse chemical reaction (K_C) is the reciprocal of the equilibrium constant (K_C).
Answer:
Consider a reversible chemical reaction:
aA + bB ⇌ cC + dD
The equilibrium constant, K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Consider the reverse reaction:
cC + dD ⇌ aA + bB.
The equilibrium constant, K_C is:
K_C = \(\frac{[\mathrm{A}]^{a}[\mathrm{~B}]^{\mathrm{b}}}{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}=\frac{1}{\mathrm{~K}_{\mathrm{C}}}\)
Thus, equilibrium constant of the reverse chemical reaction (K_C) is the reciprocal of the equilibrium constant K_C.

Question 15.
Write equilibrium constant expressions for both forward and reverse reaction for the synthesis of ammonia by the Haber process.
Answer:
Synthesis of ammonia by Haber process:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)

Question 16.
How are the equilibrium constants of the following pair of equilibrium reactions related?

Answer:

ii. K_C = \(\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{N}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{N}_{2} \mathrm{O}\right]}\)

Question 17.
Write KP expression for the reaction:
aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)
Answer:
For the given reaction,
K_P = \(\frac{\left(P_{c}\right)^{c}\left(P_{D}\right)^{d}}{\left(P_{A}\right)^{a}\left(P_{B}\right)^{b}}\)

Question 18.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Write expressions for K_P and substitute expressions for P_N2, P_H2 and P_NH3 using ideal gas equation.
Answer:
For the given reaction, K_P = \(\frac{\left(P_{N H_{3}}\right)^{2}}{\left(P_{N_{2}}\right)\left(P_{H_{2}}\right)^{3}}\)

[Note: The above question is modified to apply appropriate textual context, i. e., to indicate that students need to use ideal gas equation to derive expressions for P_N2, P_H2 and P_NH3]

Question 19.
For a chemical equilibrium reaction
H_2(g) + I_2(g) ⇌ 2HI_(g),
write an expression for K_P (and relate it to K_C).
Answer:

Question 20.
Write the relationship between K_C and K_P for the following equilibria:

Answer:

Question 21.
Write the expressions for K_C and K_P and the relationship between them for the equilibrium reaction,
2A_(g) + B_(g) ⇌ 3C_(g) + 2D_(g)
Answer:

Question 22.
Explain in short homogeneous equilibrium and heterogeneous equilibrium.
Answer:
i. In a homogeneous equilibrium, the reactants and products are in the same phase.
e.g. Dissociation of HI:
2HI_(g) ⇌ H_2(g) + I_2(g)
ii. In a heterogeneous equilibrium, the reactants and products exist in different phases, e.g. Formation of NH₄Cl:
NH_3(g) + HCl_(g) ⇌ NH₄Cl_(s)

Question 23.
The unit of K_C is different for different reactions. Explain this statement with suitable examples.
Answer:
Unit of equilibrium constant:
i. The unit of equilibrium constant depends upon the expression of K_C which is different for different equilibria. Therefore, the unit of K_C is also different for different reactions.
ii. Consider the following equilibrium reaction:

iii. Consider the following equilibrium reaction:

Question 24.
Write the equilibrium constant expression for the decomposition of baking soda. Deduce the unit of K_C from the above expression.
Answer:

[Note: Considering gaseous reactants and products, in this reaction, Δn = 2 – 0 = 2
∴ Units of K_C = (mol dm^-3)^Δn
= (mol dm^-3)²
= mol² dm^-6
Thus, the units of the above reaction is mol² dm^-6.]

Question 25.
What are the characteristics of equilibrium constant?
Answer:
Characteristics of equilibrium constant:

• The value of equilibrium constant is independent of initial concentrations of either the reactants or products.
• Equilibrium constant is temperature dependent. Hence, K_C and K_P change with change in temperature.
• Equilibrium constant has a characteristic value for a particular reversible reaction represented by a balanced equation at a given temperature.
• Higher value of K_C or K_P means more concentration of products is formed and the equilibrium point is more towards right hand side and vice versa.

Question 26.
Explain how equilibrium constant helps in predicting the direction of the reaction.
Answer:
Prediction of the direction of the reaction:
i. For the reaction, aA + bB ⇌ cC + dD,
The equilibrium constant (K_C) is given as:
K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
where, all the concentrations are equilibrium concentrations.
ii. When the reaction is not necessarily at equilibrium, the concentration ratio is called Q_C i.e.,
Q_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
iii. By comparing Q_C with K_C for a reaction under given conditions, we can decide whether the forward or the reverse reaction should occur to establish the equilibrium.
a. Q_C < K_C: The reaction will proceed from left to right, in forward direction, generating more product to attain the equilibrium.
b. Q_C > K_C: The reaction will proceed from right to left, removing product to attain the equilibrium.
c. Q_C = K_C: The reaction is at equilibrium and no net reaction occurs.

[Note: The prediction of the direction of the reaction on the basis of Q_C and K_C values makes no comment on the time required for attaining the equilibrium.]

Question 27.
Explain how K_C can be used to know the extent of the reaction?
Answer:
Extent of the reaction: The equilibrium constant expression indicates that the magnitude of K_C is:
i. directly proportional to the concentrations of the products.
ii. inversely proportional to the concentrations of the reactants.
a. Value of K_C is very high (K_C > 10³):
At equilibrium, there is a high proportion of products compared to reactants.
Forward reaction is favoured.
Reaction is in favour of products and nearly goes to completion.

b. Value of K_C is very low (K_C < 10^-3):
At equilibrium, only a small fraction of the reactants is converted into products.
Reverse reaction is favoured.
Reaction hardly proceeds towards the products.

c. Value of K_C is in the range of 10^-3 to 10³:
Appreciable concentrations of both reactants and products are present at equilibrium.

Question 28.
For the following reactions, write K_C expressions and predict direction of the reactions based on the magnitude of their equilibrium constants.
i. 2H_2(g) + O_2(g) ⇌ 2H2O_(g), K_C = 2.4 × 10⁴⁷ at 500 K
ii. 2H₂O_(g) ⇌ 2H_2(g) + O_2(g), K_C = 4.2 × 10^-48 at 500 K
Answer:
i. a. K_C expression:
K_C = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}{\left[\left[\mathrm{H}_{2(\mathrm{~g})}\right]\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}\)
b. For the reaction, K_C = 2.4 × 10⁴⁷ at 500 K
If the value of K_C >>> 10³, forward reaction is favoured.
Hence, the given reaction will proceed in the forward direction and will nearly go to completion.

ii. a. K_C expression:
K_C = \(\frac{\left[\mathrm{H}_{2(\mathrm{~g})}\right]^{2}\left[\mathrm{O}_{2(\mathrm{~g})}\right]}{\left[\mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\right]^{2}}\)
b. For the reaction, K_C = 4.2 × 10^-48 at 500 K
If the value of K_C <<< 10^-3, reverse reaction is favoured.
Hence, the given reaction will proceed in the backward direction and will nearly go to completion.

Question 29.
Describe how equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Answer:
An equilibrium constant can be used to calculate the composition of an equilibrium mixture.
Consider an equilibrium reaction, A_(aq) + B_(aq) ⇌ C_(aq) + D_(aq)
The equilibrium constant is 4.0 at a certain temperature.
Let the initial amount of A and B be 2.0 mol in ‘V’ litres. Let x mol be the equilibrium amount of C.
Hence, we can construct a table as shown below:

The expression for equilibrium constant can be written as:

Substituting the value of equilibrium concentration, we get

Therefore, equilibrium concentrations are 0.67 mol of A, 0.67 mol of B, 1.33 mol of C and 1.33 mol of D in V litres.

Question 30.
Explain the link between chemical equilibrium and chemical kinetics:
Answer:
Equilibrium constant (K_C) is related to rate or velocity constants of forward reaction (k_f) and reverse reaction (k_r) as:
K_C = \(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{r}}}\)
This equation can be used to determine the composition of the reaction mixture

[Note: The equilibrium refers to the relative amounts of reactants and products and thus a shift in equilibrium in a particular direction will imply the reaction in that direction will be favoured.]

Question 31.
Equal concentrations of hydrogen and iodine are mixed together in a closed container at 700 K and allowed to come to equilibrium. If the concentration of HI at equilibrium is 0.85 mol dm^-3, what are the equilibrium concentrations of H₂ and I₂ if K_C = 54 at this temperature?
Solution:
Given: [HI_(g)] = 0.85 mol dm^-3
K_C = 54 at 700 K
Equilibrium concentrations of H₂ and I₂
Formula: K_C = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
Balanced chemical reaction: 2HI_(g)

Equilibrium concentration of I_2(g) = Equilibrium concentration of H_2(g)

Ans: Equilibrium concentrations of H₂ and I₂ are equal to 0.12 mol dm^-3.

Question 32.
Calculate Kc at 500 K for the reaction,
2HI_(g) ⇌ H_2(g) + I_2(g) if the equilibrium concentrations are [HI] = 0.5 M, [H₂] = 0.08 M and [I₂] = 0.062 M.
Solution:
Given: T = 500 K,
At equilibrium, [HI] = 0.5 M, [H₂] = 0.08 M, [I₂] = 0.062 M.
To find: Equilibrium constant K_C
Formula: K_C = \(\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)
Calculation: The above equilibrium reaction is given as 2HI_(g) ⇌ H_2(g) + I_2(g)
The expression of K_C is

Ans: K_C at 500 K for the given reaction is 0.0198.

Question 33.
Calculate K_C and K_P for the reaction at 295 K, N₂O₄ ⇌ 2NO_2(g) if the equilibrium concentrations are [N₂O₄] = 0.75 M and [NO₂] = 0.062 M, R = 0.08206 L atm K^-1 mol^-1.
Solution:
Given: R = 0.08206 L atm K^-1 mol^-1, T = 295 K
At equilibrium , [N₂O₄] = 0.75 M, [NO₂] = 0.062 M
To find: Equilibrium constants, K_P and K_C
Formulae: i. K_C = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{A}]^{\mathrm{a}}[\mathrm{B}]^{\mathrm{b}}}\)
ii. K_P = K_C (RT)^Δn
Calculation : The equilibrium reaction is given as N₂O_4(g) ⇌ 2NO_2(g)
The expression of K_C is

K_P is related to K_C by expression: K_P = K_C (RT)^Δn
where, Δn = numbers of moles of gaseous products – number of moles of gaseous reactants
= 2 – 1 = 1
∴ K_P = K_C(RT)¹
∴ K_P = 5.13 × 10^-3 × 0.08206 × 295
∴ K_P= 123.9 × 10^-3 = 0.124
Ans: K_C and K_P for the reaction at 295 K are 5.13 × 10^-3 and 0.124 respectively.

Question 34.
The equilibrium constant K_C for the reaction of hydrogen with iodine is 54.0 at 700 K.

K_C = 54.0 at 700 K
If k_f is the rate constant for the formation of HI and k_r is the rate constant for the decomposition of HI, deduce whether k_r is larger or smaller than k_r.
ii. If the value of k_r at 700 K is 1.16 × 10^-3, what is the value of k_f ?
Solution:
Given: i. K_C = 54.0 at 700 K
ii. k_r = 1.16 × 10^-3 at 700 K
To find: i. Whether k_f is larger or smaller than k_r.
ii. Value of k_f.

Question 35.
Given the equilibrium reaction, H₂O_(g) + CH_4(g) ⇌ CO_(g) + 3H_2(g)
Using Le Chatelier’s principle, predict how concentration of CO will change when the equilibrium is disturbed by
i. adding CH₄
ii. adding H₂
iii. removing H₂O
iv. removing H₂
Answer:
i. Adding CH₄: Adding CH₄ will favour the forward reaction and the yield of CO and H₂ will increase.
ii. Adding H₂: Adding H₂ will favour the reverse reaction and the yield of CO and H₂ will decrease.
iii. Removing H₂O: Removing H₂O will favour the reverse reaction and the yield of CO and H₂ will decrease.
iv. Removing H₂: Removing H₂ will favour the forward reaction and the yield of CO and H₂ will increase.

Question 36.
By using Le Chatelier’s principle, explain the effect of change in pressure (due to volume change) on the composition of equilibrium mixture.
Answer:
Change in pressure:
i. The change in pressure has no effect on the concentrations of solids and liquids. However, it appreciably affects the concentrations of gases.
From the ideal gas equation, PV = nRT or P = \(\frac{\mathrm{n}}{\mathrm{V}}\)RT
∴ P ∝ \(\frac{\mathrm{n}}{\mathrm{V}}\)
where, the ratio n/V is an expression for the concentration of the gas in mol dm^-3.
ii. According to Le Chatelier’s principle at constant temperature, when pressure is increased, the equilibrium will shift in a direction in which the number of molecules decreases and when the pressure is decreased the equilibrium will shift in a direction in which the number of molecules increases.

[Note: For a reaction in which decrease in volume takes place, the reaction will be favoured by increasing pressure and for a reaction in which increase in volume takes place, the reaction will be favoured with lowering pressure, temperature being constant.]

Question 37.
An equilibrium mixture of dinitrogen tetroxide (colourless gas) and nitrogen dioxide (brown gas) is set up in a sealed flask at a particular temperature. Observe the effect of change of pressure on the gaseous equilibrium and complete the following table:

Answer:

Question 38.
By using Le Chatelier’s principle, explain the effect of change in pressure for the following equilibrium:
H_2(g) + I_2(g) ⇌ 2HI_(g)
Answer:
As there is the same number of molecules of gas on both sides, change of pressure has no effect on the equilibrium.

Question 39.
Explain the effect of change in pressure due to volume change of the following equilibria:
i. 2NO_(g) + Cl_2(g) ⇌ 2NOCl_(g)
ii. 2NO_(g) ⇌ N_2(g) + O_2(g)
Answer:
i. 2NO_(g) + Cl_2(g) ⇌ 2NOCl_(g)
In the forward reaction, the number of molecules decreases (3 to 2) and in the reverse reaction the number of molecules increases (2 to 3).
a. Effect of increase in pressure: According to Le Chatelier’s principle, when pressure is increased the forward reaction is favoured as the number of molecules decreases. Thus, when the pressure of the equilibrium system is increased at constant temperature by reducing the volume, the yield of NOCl increases.
b. Effect of decrease in pressure: When the pressure is decreased the equilibrium will shift from right to left. Therefore, the yield of NOCl will decrease.

ii. 2NO_(g) ⇌ N_2(g) + O_2(g)
As both reactants and products have equal numbers of moles (or molecules), there is no effect of change in pressure (due to volume change) on the composition of the equilibrium mixture.

Question 40.
Explain the effect of change in temperature on the value of K_C.
Answer:

• The value of equilibrium constant is unaffected if temperature remains constant.
• However, a change in temperature alters the value of equilibrium constant.
• In a reversible reaction, one of the reactions is exothermic (heat is released) and the other is endothermic (heat is absorbed).
• The value of equilibrium constant for an exothermic reaction decreases with increase in the temperature and that of endothermic reaction increases with the increase in temperature.

Question 41.
Explain the effect of change in temperature on the following equilibria:
CO_(g) + 2H_2(g) ⇌ CH₃OH_(g) ; ΔH = – 90 kJ
Answer:
i. The forward reaction is exothermic and reverse reaction is endothermic. According to Le Chatelier’s principle, when the temperature of the equilibrium mixture increases, the equilibrium shifts from right to left in endothermic direction. Therefore, the yield of CH₃OH decreases at high temperature.

ii. When the temperature decreases, the forward exothermic reaction is favoured. Therefore, the yield of CH₃OH increases at low temperature.
Thus, the decomposition of CH₃OH into CO and H₂ is favoured with increase in temperature, whereas formation of CH₃OH is favoured with decrease in temperature.

Question 42.
By using Le Chatelier’s principle, explain the effect of addition of a catalyst on the composition of equilibrium mixture.
Answer:

• When a catalyst is added to the equilibrium mixture, the rates of forward and reverse reactions increases to the same extent. Hence, the position of equilibrium remains unaffected.
• A catalyst does not change the composition of equilibrium mixture. The equilibrium concentrations of reactants and products remain same and catalyst does not shift the equilibrium in favour of either reactants or products.
• The value of equilibrium constant is also not affected by the presence of a catalyst.

[Note: A catalyst does not appear in the balanced chemical equation and in the equilibrium constant expression.]

Question 43.
Consider an esterification reaction:

What will happen if H+ ions are added to the reaction mixture?
Answer:
H⁺ ions act as catalyst in the esterification reaction. Hence, the addition of H⁺ ions reduces the time for the completion of reaction.

Question 44.
Complete the following table that shows the shifts in the equilibrium position for the reaction:
N₂O_4(g) + Heat ⇌ 2NO_2(g)

Answer:

Question 45.
Summarize effects of following four factors on the position of equilibrium and value of K_C:
i. Concentration
ii. Pressure
iii. Temperature
iv. Catalyst
Answer:

Effect of	Position of equilibrium	Value of KC
Concentration	Changes	No change
Pressure	Changes if reaction involves change in number of gas molecules	No change
Temperature	Change	Change
Catalyst	No change	No change

Question 46.
State TRUE or FALSE. Correct the false statement.
i. The value of equilibrium constant depends on temperature.
ii. If Q_C < K_CC, the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. Any change in the pressure of a gaseous reaction mixture at equilibrium, changes the value of K_C.
iv. In a reversible reaction, the reverse reaction has an energy change that is equal and opposite to that of the forward reaction.
Answer:
i. True
ii. False
If Q_C > K_C the reaction will proceed from right to left consuming more product to attain equilibrium.
iii. False
Any change in the pressure of a gaseous reaction mixture at equilibrium, does not change the value of K_C.
iv. True

Question 47.
Draw the flowchart showing the manufacture of NH₃ by Haber process.
Answer:

Question 48.
Explain in short: The Haber process.
Answer:
Haber process:

• The Haber process is the process of synthesis of ammonia gas by reacting together hydrogen gas and nitrogen gas in a particular stoichiometric ratio by volumes and at selected optimum temperature and pressure.
• The chemical reaction is: \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \stackrel{\text { Catalyst }}{\rightleftharpoons} 2 \mathrm{NH}_{3(\mathrm{~g})}+\text { Heat }\)
  The reaction proceeds with a decrease in number of moles (Δn = -2) and the forward reaction is exothermic.
• Iron (containing a small quantity of molybdenum) is used as catalyst.
• The optimum temperature is about 773 K and the optimum pressure is about 250 atm.

Question 49.
Consider the reaction P_(g) + Q_(g) ⇌ PQ_(g). Diagram ‘X’ represents the reaction at equilibrium.

i. If each molecule (sphere) represents a partial pressure of 1 atm, calculate the value of K_P.
ii. Predict the change in equilibrium, when the volume is increased by 50 percentage.
Answer:
i. For the given equilibrium mixture:

Chemical species	P	Q	PQ
Partial pressure	4	6	7

K_P = \(\frac{\mathrm{p}_{\mathrm{PQ}}}{\mathrm{p}_{\mathrm{p}} \times \mathrm{p}_{\mathrm{Q}}}=\frac{7}{4 \times 6}\) = 0.29
ii. Increasing the volume will shift the equilibrium position to the side with higher number of gaseous moles. In the given reaction, the equilibrium will shift to the left (toward reactant) resulting in an increase in the concentration of P and Q accompanied by a corresponding decrease in concentration of PQ.

Multiple Choice Questions

1. Which of the following is expression of K_C for
2NH_3(g) ⇌ N_2(g) + 3H_2(g)?

Answer:
(A) \(\frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}{\left[\mathrm{NH}_{3}\right]^{2}}\)

2. For the system 3A + 2B ⇌ C, the expression for equilibrium constant is …………..

Answer:
(D) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

3. For the reaction C_(s) + CO_2(g) ⇌ 2CO_(g) the partial pressure of CO₂ and CO are 4 and 8 atm, respectively, then K_P for the reaction is ……………
(A) 16 atm
(B) 2 atm
(C) 5 atm
(D) 4 atm
Answer:
(A) 16 atm

4. The equilibrium constant value for the reaction:
2H_2(g) + O_2(g) ⇌ 2H₂O_(g) is 2.4 × 10⁴⁷ at 500 K. What is the value of equilibrium constant for the reaction:
2H₂O_(g) ⇌ 2H_2(g) + O_2(g) ?
(A) 0.41 × 10^-46
(B) 0.41 × 10⁴⁷
(C) 0.41 × 10^-48
(D) 0.41 × 10^-47
Answer:
(D) 0.41 × 10^-47

5. For the reaction CO_(g) + Cl_2(g) ⇌ COCl_2(g), K_P/K_C is equal to ……………
(A) \(\frac{1}{\mathrm{RT}}\)
(B) RT
(C) \(\sqrt{\mathrm{RT}}\)
(D) 1.0
Answer:
(A) \(\frac{1}{\mathrm{RT}}\)

6. For which of the following reaction, K_P = K_C?
(A) PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)
(B) N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
(C) H_2(g) + I_2(g) ⇌ 2HI_(g)
(D) 2NO_2(g) ⇌ N₂O_4(g)
Answer:
(C) H_2(g) + I_2(g) ⇌ 2HI_(g)

7. For the equilibrium reaction
2NO_2(g) ⇌ N₂O_4(g) + 60.0 kJ, the increase in temperature ……………..
(A) favours the formation of N₂O₄
(B) favours the decomposition of N₂O₄
(C) does not affect the equilibrium
(D) stops the reaction
Answer:
(B) favours the decomposition of N₂O₄

8. The following reaction occurs in the blast furnace where iron ore is reduced to iron metal:
3Fe₂O_3(s) + 3CO_(g) ⇌ 2Fe_(l) + 3CO_2(g)
Using the Le Chatelier’s principle, predict which one of the following will NOT disturb the equilibrium?
(A) Removal of CO
(B) Removal of CO₂
(C) Addition of CO₂
(D) Addition of Fe₂O₃
Answer:
(D) Addition of Fe₂O₃

9. The reaction A + B ⇌ C + D + heat, has reached equilibrium. The reaction may be made to proceed forward by
(A) adding more C
(B) adding more D
(C) decreasing the temperature
(D) increasing the temperature
Answer:
(C) decreasing the temperature

10. Identify the CORRECT statement.
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.
(B) The value of equilibrium constant decreases in presence of a catalyst.
(C) Catalyst affect the position of the equilibrium.
(D) Catalyst changes the equilibrium composition of a reaction mixture.
Answer:
(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.

11. The equilibrium constant for the reaction:
N_2(g) + O_2(g) ⇌ 2NO_(g) is 4 × 10^-4 at 2000 K. In presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant in presence of catalyst of 2000 K is …………..
(A) 40 × 10^-4
(B) 4 × 10^-2
(C) 4 × 10^-3
(D) 4 × 10^-4
Answer:
(D) 4 × 10^-4

12. The rate of formation of NH₃ can be increased by using catalyst …………….
(A) Fe + Co
(B) Mo + Fr
(C) Fe + Mo
(D) Fe + Mg
Answer:
(C) Fe + Mo

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 Chemistry Important Questions Chapter 13 Nuclear Chemistry and Radioactivity Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 13 Nuclear Chemistry and Radioactivity

Question 1.
Explain the term nuclear chemistry. Give few examples of nuclear reactions.
Answer:
Nuclear chemistry is a branch of physical chemistry and it deals with the study of reactions involving changes in atomic nuclei. This branch started with the discovery of natural radioactivity by physicist Antoine Henri Becquerel.

Examples of nuclear reactions are as follows:

• Radioactive decay
• Artificial transmutation
• Nuclear fission
• Nuclear fusion

Question 2.
Write a short note on similarity between the solar system and structure of atom.
Answer:
Solar system: It consists of the Sun and planets in which Sun is at the centre of solar system and planets move around it under the force of gravity.

Atomic system: It consists of tiny central core called as nucleus at the centre of atom around which electrons are present. Like in solar system, electrostatic attractions hold subatomic particles in a structure of atom. The nucleus consists of protons and neutrons.

Question 3.
Answer the following.
i. Give the symbolic representation for calcium, (no. of protons = 20, mass number = 40)
ii. Calculate the number of neutrons for calcium.
Answer:
i. \({ }_{20}^{40} \mathrm{Ca}\), in which Z = 20 and A = 40.
ii. Number of neutrons: It can be calculated from formula (A = Z + N).
For calcium, N = A – Z = 40 – 20 = 20
Nucleus of the calcium atom contains 20 neutrons.

Question 4.
Explain the term nucleons with examples.
Answer:
The term nucleon refers to the sum of protons (p) and neutrons (n) present in atom, e.g. Number of nucleons present in \({ }_{20}^{40} \mathrm{Ca}\) are 40 (i.e., 20 protons and 20 neutrons). Number of nucleons present in \({ }_{11}^{23} \mathrm{Na}\) are 23 (i.e., 11 protons and 12 neutrons).

Question 5.
Define: Nuclide
Answer:
The nucleus of a specific isotope is called as nuclide.

Question 6.
Atom as a whole is electrically neutral. Justify.
Answer:

• The magnitude of electronic charge (e) on the nucleus is +Ze and that of outer sphere is -Ze. Number of protons and number of electrons are always equal in an atom.
• As a result of this, the charges get nullified, therefore, the atom as a whole is electrically neutral.

Question 7.
Define:
i. Isotopes
ii. Isobars
Answer:
i. Isotopes: Nuclides which contain same number of protons but different number of neutrons in their nuclei are called as isotopes. e.g. \({ }_{11}^{22} \mathrm{Na}\), \({ }_{11}^{23} \mathrm{Na}\) and \({ }_{11}^{24} \mathrm{Na}\)
ii. Isobars: Nuclides (of different element) which have same mass number but have different number of protons and neutrons in their nuclei are called as isobars.
OR
The atoms of different elements having the same mass number but different atomic numbers are called isobars.
e.g. \({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{N}\)

Question 8.
Define mirror nuclei and isotones.
Answer:

• Isobars in which the number of protons and neutrons differ by 1 unit and are interchanged are called as mirror nuclei.
• Isotones are defined as nuclides having the same number of neutrons but different number of protons and hence, different mass numbers.

Question 9.
Name the following.
i. Nuclides in which number of protons and neutrons differ by 1 and are interchanged.
ii. Nuclides having the same number of neutrons but different number of protons.
iii. Nuclides with the same mass number which differ in energy states.
Answer:
i. Mirror nuclei
ii. Isotones
iii. Nuclear isomers

Question 10.
Explain the term nuclear isomers?
Answer:

• The nuclides with the same number of protons (Z) and neutrons (N) or the same mass number (A) which differ in energy states are called nuclear isomers.
• In this, the isomer of higher energy is said to be in the metastable state which is represented by writing “m” after the mass number.
  e.g. Nuclear isomers of cobalt can be represented as, ^60mCo and ⁶⁰Co.

Question 11.
State true or false. Correct the false statement.
i. The number of nucleons in C-12 atom is 6.
ii. N-13 and C-13 are mirror nuclei.
iii. Nuclear isomers have same number of protons and neutrons.
Answer:
i. False,
The number of nucleons in C-12 atom is 12.
ii. True
iii. True

Question 12.
Give classification of nuclides on the basis of nuclear stability.
Answer:
Nuclides can be classified into stable and unstable/radioactive nuclides on the basis of nuclear stability.

• Stable nuclides: In this type of nuclides, the number of electrons and the location of nuclei may change in outer sphere but the number of protons and neutrons remain unchanged.
• Radioactive (unstable) nuclides: These nuclides undergo spontaneous change forming new nuclides.

Question 13.

	Number of protons (Z)	Number of neutrons (N)	Number of such nuclides
i.	Even	Even	165
ii.	Even	Odd	55

What conclusion can be drawn from the above given data?
Answer:

• Number of nuclides with even ‘Z’ and even ‘N’ are higher in number as compared to nuclides with even ‘Z’ and odd ‘N’
• Nuclides with even number of ‘Z’ and odd number of ‘N’ are about 1/3rd of nuclides where both ‘Z’ and ‘N’ are even.
• Nuclides with even number of protons (Z) and even number of neutrons (N) are most stable. These nuclides tend to fonn proton-proton and neutron-neutron pairs. This impart stability to the nucleus.

Question 14.
Write a note on naturally occurring nuclides with either odd number of protons or odd number of neutrons.
Answer:
i. The number of stable nuclides with either Z or N odd is about one third of nuclides where both are even.
ii. These nuclides are less stable than those having even number of protons and neutrons.
iii. In these nuclides one nucleon has no partner and therefore, these nuclides are less stable.
iv. Further the number of nuclides with odd A are nearly the same, irrespective of Z or N is odd. This indicates that protons and neutrons behave similarly in the respect of stability.
v. Following table gives the estimate of such nuclides occurring in nature.

	Number of protons (Z)	Number of neutrons (N)	Number of such nuclides
i.	Even	Odd	55
ii.	Odd	Even	50

Question 15.
State true or false. Correct the false statements.
i. The nuclides with even Z and even N constitute 85% of earth crust.
ii. Nuclides with either ‘Z’ or ‘N’ odd are more stable than those having even number of both ‘Z’ and ‘N’
iii. The number of nuclides with odd number of ‘Z’ and odd number of ‘N’ are only four.
Answer:
i. True
ii. False
Nuclides with either ‘Z’ or ‘N’ odd are less stable than nuclides having even number of both ‘Z’ and ‘N’.
iii. True

Question 16.
Heavier nuclides require greater number of neutrons (than protons) to attain stability. Justify.
Answer:

• The heavier nuclides with the increasing number of protons lead to large coulombic repulsions.
• Increased number of neutrons will separate the protons within the nuclei, which will impart stability. Thus, in order to attain stability heavier nuclide need more number of neutrons.

Question 17.
Consider the graph of neutron (N) plotted against proton number (Z). How will you identify radioactive nuclides from the graph?
Answer:
Nuclides which fall outside the belt or stability zone are radioactive nuclides.

Question 18.
Write a note Magic numbers.
Answer:
Magic numbers: The nuclei with 2, 8, 20, 28, 50, 82 and 126 neutrons or protons are particularly stable and abundant in nature. These numbers are known as magic numbers.
e.g. Lead (\({ }_{82}^{208} \mathrm{~Pb}\)) has two magic numbers, 82 protons and 126 neutrons.

Question 19.
What is the order of distance between two protons present in the nucleus?
Answer:
The order of distance between two protons present in the nucleus is typically of order of 10^-15 m.

Question 20.
Which factor is responsible for nuclear stability?
Answer:
Nuclear forces of attractions exist within nuclei. These are attractions between proton-proton (p-p), neutron-neutron(n-n) and proton-neutron(p-n). They constitute or give rise to nuclear potential which is responsible for nuclear stability.

Question 21.
Write short notes on: nuclear potential.
Answer:

• Nuclear potential is the attraction between p-p, n-n and p-n.
• These attractive forces are independent of the charge on nucleons or attraction between p-p, n-n and p-n are equal.
• These attractive forces operate over short range within the nucleus.
• Nuclear potential is responsible for the nuclear stability.

Question 22.
State true or false. Correct the false statement.
i. The nuclear forces of attractions are dependent on the charge on the nucleons.
ii. The actual mass of an atom is observed to be more than sum of the masses of its constituents.
Answer:
i. False
The nuclear forces of attractions are independent of the charge on the nucleons.
ii. False
The actual mass of an atom is observed to be less than sum of the masses of its constituents.

Question 23.
Define: Nuclear binding energy
Answer:
An energy equivalent to the mass lost is released during the formation of nucleus. This is called the nuclear binding energy.
OR
The energy requiredfor holding the nucleons together within the nucleus of an atom is called as the nuclear binding energy.

Question 24.
Explain the term: mass defect.
Answer:
During the formation of nucleus, certain mass is lost. This phenomenon is known as mass defect (Δm).
The exact mass of nucleus is slightly less than sum of the exact masses of the constituent nucleons. This difference is called as mass defect. It is represented by symbol Δm.
Formulae: Δm = calculated mass – observed mass

Question 25.
Explain the relation between nuclear mass and energy? Also give the energy released in the conversion of one atomic mass unit into energy.
Answer:
i. The nuclear mass is expressed in atomic mass unit (u) which is exactly 1/12^th of the mass of ¹²C atom. Thus, u = 1/12^th mass of C-12 atom = 1.66 × 10^-2 kg.
ii. The conversion of mass into energy is established through Einstein’s equation, E = mc².
Where m is the mass of matter converted into energy (E) and velocity of light (c).
iii. The energy released in the conversion of one u mass into energy is given by:
E = mc² = (1.66 × 10^-27kg) × (3 × 10⁸ m s^-1)²

Question 26.
Derive the expression for nuclear binding energy for a nuclide.
Answer:
Expression for nuclear binding energy:
i. Consider a nuclide \({ }_{z}^{A} X\) that contains Z protons and (A – Z) neutrons. Suppose the mass of the nuclide is m. The mass of proton is m_p and that of neutron is mn.

ii. Total mass = (A – Z)m_n + Zm_p + Zm_e …..(1)
Δm = [(A – Z)m_n + Zm_p + Zm_e] – m
= [(A – Z)m_n + Z(m_p + m_e] – m
= [(A – Z)m_n + Zm_H] – m …..(2)
Where (m_p + m_e) = m_H = mass of H atom.
Thus, (Δm) = [Zm_p + (A – Z)m_n] – m
Where Z = atomic number
A = mass number
(A – Z) = neutron number
m_p and m_n = masses of proton and neutron, respectively
m = mass of nuclide

iii. The mass defect, Δm is related to binding energy of nucleus by Einstein’s equation,
ΔE = Δm × c²
Where, ΔE = Binding energy, Δm = mass defect.
iv. Nuclear energy is measured in million electro volt (MeV).
v. The total binding energy is then given by,
B.E. = Δm (u) × 931.4
Where 1.00 u = 931.4 MeV
B.E. = 931.4 [Zm_H + (A – Z)m_n – m] ……(3)
Total binding energy of nucleus containing A nucleons is the B.E.
vi. The binding energy per nucleon is then given by,
\(\bar{B}\) = B.E./A

Question 27.
Calculate the mean binding energy per nucleon for the formation of \({ }_{8}^{16} \mathrm{O}\) nucleus. The mass of oxygen atom is 15.994 u. The masses of H atom and neutron are 1.0078 u and 1.0087 u, respectively.
Solution:
Given: m_H = 1.0078 u
m_n= 1.0087 u
m= 15.994 u
Z = 8, A= 16
To find: Mean binding energy per nucleon (\(\bar{B}\))
Formulae: i. Δm = Zm_H + (A – Z)m_n – m
ii. B.E. = Δm × 931.4 MeV
iii. \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
Calculation: i. The mass defect, Δm = Zm_H + (A – Z)m_n – m
Δm = 8 × 1.0078 u + 8 × 1.0087 u – 15.994 u = 0.138 u
ii. Total binding energy, B.E. (MeV) = Δm (amu) × 931.4
Hence, B.E. = 0.138 × 931.4 = 128.533 MeV
iii. Binding energy per nucleon, \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
Hence, \(\bar{B}\) = \(\frac{128.533}{16}\) = 8.033 MeV/nucleon
Ans: Binding energy per nucleon for the formation of \({ }_{8}^{16} \mathrm{O}\) nucleus = 8.033 MeV/nucleon

Question 28.
Calculate the binding energy per nucleon for the formation of \({ }_{2}^{4} \mathrm{He}\) nucleus. Mass of \({ }_{2}^{4} \mathrm{He}\) atom = 4.0026 u.
Solution:
Given: m = 4.0026 u
Z = 2, A = 4
To find: Binding energy per nucleon (\(\bar{B}\))
Formulae: i. Δm = Zm_H + (A – Z)m_n – m
ii. B.E. = Δm × 931.4 MeV
iii. \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
The mass defect, Δm = [Zm_H + (A – Z)m_n] – m
Δm = [(2 × 1.0078) + (2 × 1.0087)] – 4.0026 = 0.0304 u
Total binding energy, B.E. (MeV) = Δm (amu) × 931.4
= 0.0304 × 931.4
= 28.315 MeV
iii. B.E. per nucleon, \(\overline{\mathrm{B}}=\frac{\mathrm{B} . \mathrm{E} .}{\mathrm{A}}\)
\(\bar{B}\) = \(\frac{28.315}{4}\) = 7.079 Mey/nucleon
Ans: Binding energy per nucleon for formation of \({ }_{2}^{4} \mathrm{He}\) nucleus = 7.079 MeV/nucleon

Question 29.
Define radioactivity and give examples of two radioactive elements.
Answer:
Radioactivity is a phenomenon in which the nuclei spontaneously emit a nuclear particle and gamma radiation transforming to a different nuclide. e.g. Uranium and radium
[Note: Radioactivity is the phenomenon related to the nucleus.]

Question 30.
What is the criteria for an element to be known as radioactive element?
Answer:

• An element is considered to be radioactive if the nuclei of its atoms are unstable.
• That is, when element undergoes nuclear changes (i.e., emission of nuclear particles and gamma radiation), it is said to be radioactive.

Question 31.
What are the different types of radiations emitted by radioactive element?
Answer:
The radiations emitted by radioactive elements are as follows:

• Alpha (α) radiations
• Beta (β) radiations
• Gamma (γ) radiations

Question 32.
Write the unit of rate of decay.
Answer:
The rate of decay is expressed in the form of disintegrations per second (dps).

Question 33.
Derive the equation λ = \(\frac{\left(-\frac{\mathbf{d} \mathbf{N}}{\mathbf{d} \mathbf{t}}\right)}{\mathbf{N}}\) and write what does λ denotes.
Answer:
The rate of decay of a radioelement at any instant is proportional to the number of nuclei (atoms) present at that instant. It can be represented as,
\(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N} \quad \text { or }-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\lambda \mathrm{N}\) …….(i)
Where, \(-\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\) = Rate of decay at any time, t
λ = Decay constant
N = Number of nuclei (atoms) present at time, t
From equation (i),

Decay constant (λ) is the fraction of nuclei decaying in unit time.
OR
It is the ratio of the amount of substance disintegrated per unit time to the amount of substance present at that time.

Question 34.
Derive the expression for decay constant.
Answer:
Decay constant (λ) is the fraction of nuclei decaying in unit time.
Thus,
λ = \(-\frac{d N}{d t} \times \frac{1}{N}\) …(i)
Rearranging equation (i) we get,
\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}\) = -λ dt
On integrating above equation, we get
∫\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}\) = -∫ λ dt …(ii)
On performing the integration, we get lnN = -λt + C ……(iii)
where C is the constant of integration whose value is obtained as follows:
Let N₀ be the number of nuclei present at some arbitrary zero time. At time t, the number of nuclei is N. So, at t = 0, N = N₀, substituting in equation (iii), we get
lnN₀ = C
With this value of C, equation (iii) becomes
lnN = -λt + lnN₀
or λt = lnN₀ – InN = ln \(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) ……(iv)
Hence, λ = \(\frac{1}{t} \ln \frac{N_{0}}{N}\) …….(v)
Converting natural logarithm (ln) to logarithm to the base 10, equation (v) becomes
λ = \(\frac{2.303}{t} \log _{10} \frac{N_{0}}{N}\) ………(vi)
The equation (iv) can be expressed as ln \(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) = -λt. Taking antilog of both sides, we get
\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}} \text { or } \mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}\) …….(vii)
The equation (vi) and equation (vii) give the decay constant.

Question 35.
Write a note on half-life of a radioelement.
Answer:

• Half-life of a radioelement (t_1/2): It is the time needed for a given number of nuclei (atoms) of radioelement to decay exactly to half of its initial value.
• Each radio isotope has its own half-life.
• Half-life of a radioelement can be expressed in seconds, minutes, hours, days or years.
• Mathematical expression for half-life of a radioelement can be given as,
  \(t_{1 / 2}=\frac{0.693}{\lambda}\)

Question 36.
Complete the following statements based on the given graph.
i. As decay progresses, the number of radioactive atoms will ……….. with time.
ii. As decay progresses, the rate of decay will …………..
iii. Rate of radioactive decay at any instant is ………… to the number of atoms of the radioactive element present at that instant.
Answer:
i. decrease
ii. decrease
iii. proportional

Question 37.
²¹⁸Po decays initially at a rate of 816 dps. The rate falls to 408 dps after 24 min. Calculate the decay constant.
Solution:

Question 38.
After how many seconds will the concentration of radioactive element X will be halved, if the decay constant is 1.155 × 10^-3 s^-1?
Solution:

Ans: Concentration of radioactive element (X) will be halved in 600 s.

Question 39.
⁴¹Ar decays initially at a rate of 575 Bq. The rate falls to 358 dps after 75 minutes. What is the half-life of ⁴¹Ar?
Solution:


Ans: The half-life of Ar is 109.7 min.

Question 40.
The half-life of ³²P is 14.26 d. What percentage of ³²P sample will remain after 40 d?
Solution:
Given: t_1/2 = 14.26 d,
N₀ = 100,
t = 40 d
To find: Percentage of ³²P sample remaining after 40 d

Question 41.
The half-life of ³⁴Cl is 1.53 s. How long does it take for 99.9 % of sample of ³⁴Cl to decay?
Solution:
Given: t_1/2 = = 1.53 s,
N₀ = 100,
N = 100 – 99.9 = 0.1,
To find: Time (t)

Question 42.
The half-life of ²⁰⁹Po is 102 y. How much of 1 mg sample of polonium decays in 62 y?
Solution:
Given: t_1/2 = 102y,
t = 62 y,
N₀ = 1 mg
To find: Amount of polonium that decayed in 62 y

Taking antilog of both sides we get,
\(\frac{\mathrm{N}_{0}}{\mathrm{~N}}\) = antilog (0.1829) = 1.524
N = \(\frac{\mathrm{N}_{0}}{1.524}=\frac{1 \mathrm{mg}}{1.524}\) = 0.656 mg
N is the amount that remains after 62 y.
Hence, the amount decayed in 62 y = 1 mg – 0.656 mg = 0.344 mg
Ans: The amount decayed in 62 y is 0.344 mg

Question 43.
What will be the approximate time taken for 90 % decay of ¹⁷⁴Ir in terms of its half-life?
Solution:
Given: N₀ = 100
N = 100 – 90 = 10
To find: Time (t)

Ans: Thus, the approximate time required for 90 % decay of ¹⁷⁴Ir in terms of its half-life is 3.3t_1/2.

Question 44.
A radioactive decay of element X (Z = 35) is 30 % complete in 2 hours. Calculate its half-life period.
Solution:
Given: t = 2 hrs,
N₀ = 100
N= 100 – 30 = 70
To find: t_1/2

Question 45.
What are the different modes by which radio elements decay?
Answer:
There are 3 modes by which radio elements decay: α-decay, β-decay and γ-emission.

Question 46.
What is α-decay?
Answer:
Radioactive isotope/radioelement when undergoes decay by the emission of α-particle from the nuclei then the process involved is referred to as α-decay.

Question 47.
Give equation for radium-222 when it undergoes decay by emission of an α-particle.
\({ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{X}+{ }_{2}^{4} \mathrm{He}\)
Answer:
\({ }_{88}^{226} \mathrm{Ra} \longrightarrow{ }_{86}^{222} \mathrm{X}+{ }_{2}^{4} \mathrm{He}\)
Thus, atomic number of element ‘X’ will be 86 and atomic mass number will be 222.

Question 48.
Identify the mode of decay and state whether following equation is CORRECT or NOT. Justify.
\({ }_{92}^{238} \mathbf{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathbf{H e}\)
Ans:
i. It involves α-decay process.
ii. As uranium undergoes decay by emission of an α-particle (i.e., \({ }_{2}^{4} \mathrm{He}\)), daughter nuclei (in this case thorium) ‘will observe the decrease in atomic number by 2 units and decrease in atomic mass number by 4 units.
Hence, the given equation is correct.

Question 49.
If radioactive element ‘X’ undergoes α-emission then what will be the position of daughter nuclei in the periodic table with respect to element ‘X’.
Answer:
If radioactive element ‘X’ undergoes α-emission, then corresponding daughter nuclei formed will occupy two places to the left of the periodic table with respect to element ‘X’.

Question 50.
What is β^– – decay? Also explain the changes that occur in the parent nuclei due to β-emission with one example.
Answer:
β^– – decay: The emission of negatively charged stream of β^– particles from the nucleus is called β^– – decay.
i. β^– – Particles are electrons with a charge and mass of an electron, mass being negligible as compared to the nuclei.
ii. When a nucleus decays by emitting a high-speed electron called a beta particle (β), a new nucleus is formed with the same mass number as the original nucleus and with an atomic number that is one unit greater than the parent nuclei.
General equation:

Note: The mass number A does not change, the atomic number changes when a nuclei undergoes β-decay. e.g. Neptunium-238 decays to form plutonium-238:

Question 51.
Mention the atomic number and atomic mass number of the parent radioelement ‘X’ in the following case if parent nuclei undergo β-emission.
i. \(\mathrm{X} \longrightarrow{ }_{94}^{238} \mathrm{Pu}\)
ii. \(\mathrm{X} \longrightarrow{ }_{95}^{241} \mathrm{Am}\)
Answer:

Question 52.
How many α and β-particles are emitted in the following?
\({ }_{93}^{237} \mathrm{~Np} \longrightarrow{ }_{83}^{209} \mathrm{Bi}\)
Answer:
The emission of one α-particle decreases the mass number by 4 whereas the emission of β particles has no effect on mass number.
Net decrease in mass number = 237 – 209 = 28. This decrease is only due to α- particles. Hence, number of α- particles emitted = \(\frac {28}{4}\) = 7
Now, the emission of one α-particle decreases the atomic number by 2 and one β-particle emission increases it by 1.
The net decrease in atomic number = 93 – 83 = 10
The emission of 7 α-particles causes decrease in atomic number by 14. However, the actual decrease is only 10. It means atomic number increases by 4. This increase is due to emission of 4 β-particles.
Thus, 7 α and 4 β- particles are emitted.

Question 53.
Explain the process of γ-decay in detail with a suitable example.
Answer:
γ-decay:
i. γ-Radiation is always accompanied with α and β^– decay processes.
ii. During γ-radiation, the daughter nucleus is left in energetically excited state which decays to the ground state of product with emission of γ-rays.
For example, \({ }_{92}^{238} \mathrm{U} \longrightarrow{ }_{90}^{234} \mathrm{Th}+{ }_{2}^{4} \mathrm{He}+\gamma\)
iii. \({ }_{92}^{238} \mathrm{U}\) emits α-particles of two different energies, 4.147 MeV (23%) and 4.195 MeV (77%).
iv. When α-particles of energy 4.147 MeV are emitted, ²³⁴Th is left in an excited state which de-excites to the ground state with emission of γ-ray photons with energy 0.0048 MeV.

Question 54.
Half-life of ²⁰⁹Po is 102 y. How many α-particles are emitted in 1 s from 2 mg sample of Po?
Solution:
Given: t_1/2 = 102 y,
t = 1 s,
Amount of sample = 2 mg
To find: Number of α-particles emitted

Question 55.
Nuclear transmutation is a spontaneous or non-spontaneous process?
Answer:
Nuclear transmutation is a non-spontaneous (man-made) process.

Question 56.
What is nuclear transmutation?
Answer:
Nuclear transmutation:

• It is the process of transformation of a stable nucleus into another nucleus which can be stable or unstable.
• It can occur by the radioactive decay of a nucleus or the reaction of a nucleus with another particle.

Question 57.
Differentiate between chemical reactions and nuclear reactions.
Answer:
Chemical reactions:

• Rearrangement of atoms by breaking and forming of chemical bonds.
• Different isotopes of an element have same behaviour.
• Only outer shell electrons take part in the chemical reaction.
• The chemical reaction is accompanied by relatively small amounts of energy.
  e.g. chemical combustion of 1.0 g methane releases only 56 kJ energy.
• The rates of reaction are influenced by the temperature, pressure, concentration and catalyst.

Nuclear reactions:

• Elements or isotopes of one element are converted into another element in a nuclear reaction.
• Isotopes of an element behave differently.
• In addition to electrons, protons, neutrons, other elementary particles may be involved.
• The nuclear reaction is accompanied by a large amount of energy change, e.g. The nuclear transformation of 1 g of Uranium – 235 release 8.2 × 10⁷ kJ
• The rate of nuclear reactions is unaffected by temperature, pressure and catalyst.

Question 58.
What will happen when a nucleus of J’B is bombarded with α-particle? Identify the process involved.
Answer:
i. When a stable nucleus of \({ }_{5}^{10} \mathrm{~B}\) is is bombarded with α-particle, it transforms into \({ }_{7}^{13} \mathrm{~N}\), which is radioactive and spontaneously emits positrons to produces \({ }_{6}^{13} \mathrm{C}\).
This can be represented as,

ii. The process involved is known as induced radioactivity or artificial radioactivity.

Question 59.
Define: Nuclear fission
Answer:
Nuclear fission is defined as a process which involves splitting of the heavy nucleus of an atom into two nearly equal fragments accompanied by release of the large amount of energy.

Question 60.
Nuclear fission of ²³⁵U is a chain process. Justify.
Answer:

• Nuclear fission of ²³⁵U occurs when nucleus absorbs neutron. When a uranium nucleus absorbs neutron, it breaks into two lighter fragments and releases energy (heat), more neutrons, and other radiation.
• When one uranium 235 nucleus undergoes fission, three neutrons are emitted.
• These neutrons emitted in fission cause more fission of the uranium nuclei which yield more neutrons. These neutrons again bring forth fission producing further neutrons.
• The process continues indefinitely leading to chain reaction which continues even after the removal of bombarding neutrons.

Question 61.
Explain the term: Nuclear fusion and give one example.
Answer:
Nuclear fusion: In this process, the lighter nuclei combine (fuse) together and form a heavy nucleus which is accompanied by an enormous amount of energy.
e. g. The energy received by earth from the sun is due to the nuclear fusion reactions.

Question 62.
Which will produce more energy: Nuclear fission or fusion?
Answer:
Nuclear fusion will produce relatively more energy per given mass of fuel.

Question 63.
What is the range of temperature required to carry out nuclear fusion reaction?
Answer:
Nuclear fusion reaction requires extremely high temperature typically of the order of 108 K.

Question 64.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission:

• It is the process in which a heavy nucleus splits up into two lighter nuclei of nearly equal masses.
• About 200 MeV of energy is available per fission in case of \({ }_{92}^{235} \mathrm{U}\).
• The products of nuclear fission are, in general, radioactive.

Nuclear fusion:

• It is the process in which two lighter nuclei combine together to form a heavy nucleus.
• Energy available per fusion is much less but the energy per unit mass of material is much greater than that for fission of heavy nuclei.
• The products of fusion are, in general, non-radioactive.

Question 65.
Estimate the energy released in the fusion reaction.
\({ }_{1}^{2} \mathbf{H}+{ }_{2}^{3} \mathbf{H e} \longrightarrow{ }_{2}^{4} \mathbf{H e}+{ }_{1}^{1} \mathbf{H}\)
(Given atomic masses: ²H = 2.0141 u. ³He = 3.0160 u, ⁴He = 4.0026 u, ¹H = 1.0078 u)
Answer:

Question 66.
Explain the term: Radiocarbon dating in detail.
Answer:
Radiocarbon dating: The technique is used to find the age of historic and archaeological organic samples such as old wood samples and animal or human fossils.
Radioisotope used for carbon dating is ¹⁴C.
i. Radioactive ¹⁴C is formed in the upper atmosphere by bombardment of neutrons from cosmic ray on ¹⁴N.
\({ }_{7}^{14} \mathrm{~N}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{6}^{14} \mathrm{C}+{ }_{1}^{1} \mathrm{H}\)
ii. ¹⁴C combines with atmospheric oxygen to form ¹⁴CO₂ which mixes with ordinary ¹²CO₂.
iii. This carbon dioxide is absorbed by plants during photosynthesis.
iv. Animals eat plants which have absorbed a carbon dioxide (¹⁴CO₂ + ¹²CO₂). Hence, ¹⁴C becomes a part of plant and animal bodies.
v. As long as the plant is alive, the ratio ¹⁴C/¹²C remains constant.
vi. When the plant dies, photosynthesis will not occur and the ratio ¹⁴C/¹²C decreases with the decay of radioactive ¹⁴C which has a half-life 5730 years.
vii. The decay process of ¹⁴C is given below:
\({ }_{6}^{14} \mathrm{C} \longrightarrow{ }_{7}^{14} \mathrm{~N}+{ }_{-1}^{0} \mathrm{e}\)
viii. The activity (N) of given wood sample and that of fresh sample of live plant (N0) is measured, where, N0 denotes the activity of the given sample at the time of death.
ix. The age of the given wood sample, can be determined by applying following Formulae:

Question 67.
What is nuclear power?
Answer:
Nuclear power is the electricity generated from the fission of uranium and plutonium.

Question 68.
Nuclear power is a clean source of energy. Justify.
Answer:
Nuclear power offers huge environmental benefits in producing electricity because,

• it releases zero carbon dioxide.
• it releases zero sulphur and nitrogen oxides.
• these are atmospheric pollutants which pollute the air.

Thus, nuclear power is a clean source of energy.

Question 69.
How much energy will be produced by fission of 1 gram of ²³⁵U?
Answer:
Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy.

Question 70.
Nuclear fission is an alternative energy source. Explain.
Answer:

• Fission of 1 gram of uranium-235 produces about 24,000 kW/h of energy.
• This is the same amount of energy produced by burning 3 tons of coal or 12 barrels of oil, or nearly 5000 m³ of natural gas.
• The sources like coal, oil, natural gas are depleting very fast.
• Also, the costs of petrol and other products from petroleum industry is increasing.
• Thus, we need to depend on the nuclear fission as an alternative source of energy for electricity.

Question 71.
Label the follow ing diagram of simplified nuclear reactor.

Answer:

Question 72.
Explain in brief: Nuclear reactor
Answer:
Nuclear reactor: Nuclear reactor is a device for using atomic energy in controlled manner for peaceful purposes. During nuclear fission energy is released. The released energy can be utilized to generate electricity in a nuclear reactor.

Working of a nuclear reactor:

• In a nuclear reactor, U²³⁵ or U²³⁹, a fissionable material is stacked with heavy water (D₂O deuterium oxide) or graphite called moderator.
• The neutrons produced in the fission pass through the moderator and lose a part of their energy. The slow neutrons produced during the process are captured which initiate new fission.
• Cadmium rods are inserted in the moderator as they have ability to absorb neutrons. This controls the rate of chain reaction.
• The energy released during the reaction appears as heat and removed by circulating a liquid (coolant). The coolant which has absorbed excess of heat from the reactor is passed over a heat exchanger for producing steam.
• Steam is then passed through the turbines to produce electricity. Thus, the atomic energy produced with the use of fission reaction can be controlled in the nuclear reactor.
• This process can be explored for peaceful purpose such as conversion of atomic energy into electrical energy which can be used for civilian purposes, ships, submarines, etc.

Note: Schematic diagram of nuclear power plant:

Question 73.
Why cadmium rods are used in nuclear reactor?
Answer:
Cadmium rods are inserted in the moderator as they have ability to absorb neutrons which help to control the rate of chain reaction.

Question 74.
Why short-lived isotopes are used for diagnostic purposes?
Answer:
For diagnostic purpose, short-lived isotopes are used in order to limit the exposure time to radiation. Note: Diagnostic Radioisotopes are listed below:

Question 75.
Give one application of therapeutic radioisotopes.
Answer:
Therapeutic radioisotopes are used to destroy abnormal cell growth in the body, e.g. cancerous cells.
Note: Therapeutic Radioisotopes are listed below:

Question 76.
Give example of isotopes used in following.
i. Isotope used in the treatment of leukaemia.
ii. Isotopes used in the preservation of agricultural products by irradiation.
Answer:
i. Isotope of phosphorus, \({ }_{15}^{35} \mathrm{P}\).
ii. ⁶⁰Co or ¹³⁷Cs

Question 77.
At which places has BARC Mumbai set up irradiation plants for preservation of agricultural produce?
Answer:
Bhabha Atomic Research Centre (BARC) Mumbai has set up irradiation plants for preservation of agricultural produce such as mangoes, onion and potatoes at Vashi (Navi Mumbai) and Lasalgaon (Nashik).

Question 78.
Why radiotracer technique is used in chemistry?
Answer:
Radiotracer technique is used to trace the path/mechanism followed by a reaction in the system.

Question 79.
The half-life for radioactive decay of an element X is 140 days. Complete the following flow chart showing decay of 1 g of X.

Answer:

Shortcut method:
Amount of the element X left after n half-lives is given as [X] = \(\frac{[\mathrm{X}]_{0}}{2^{n}}\)
e.g. \(\frac{1}{2^{4}} \mathrm{~g}=\frac{1}{16} \mathrm{~g}\)

Question 80.
A sample of 35S complete its 10% decay in 20 min, then calculate the time required to complete decay by 19%.
Answer:
When decay is 10 % complete, if N₀ = 100 , then N = 100 – 10 = 90 and t = 20 minutes
When decay is 19 % complete, N = 100 – 19 = 81
Substituting these values in formula we get,

Multiple Choice Questions

1. Radius of the nucleus is related to the mass number A by ………….
(A) R = R₀A^1/2
(B) R = R₀A
(C) R = R₀A²
(D) R = R₀A^1/3
Answer:
(D) R = R₀A^1/3

2. Which of the following nuclides has the magic number of both protons and neutrons?
(A) \({ }_{50}^{115} \mathrm{Sn}\)
(B) \({ }_{81}^{206} \mathrm{Pb}\)
(C) \({ }_{82}^{208} \mathrm{Pb}\)
(D) \({ }_{50}^{118} \mathrm{Pb}\)
Answer:
(C) \({ }_{82}^{208} \mathrm{Pb}\)

3. The probability of decay of a radioactive element depends on …………..
i. the age of nucleus
ii. the presence of catalyst
iii. pressure
iv. temperature
(A) only i. and iv
(B) all of these
(C) only ii. And iii.
(D) none of these
Answer:
(D) none of these

4. The decay constant for 67Ga is 7.0 × 10^-4 s^-1. If initial concentration of is 0.07 g, what is the half-life of ⁶⁷Ga?
(A) 990 s
(B) 79.2 s
(C) 12375 s
(D) 10.10 × 10^-4 s
Answer:
(A) 990 s

5. The half-life of radioactive element X having decay constant of 1.7 × 10^-5 s^-1 is …………
(A) 21.5 h
(B) 19.7 h
(C) 11.3 h
(D) 2.8 h
Answer:
(C) 11.3 h

6. A radioactive decay of element X (Z = 90) is 30 % complete in 30 minutes. It has a half-life period of ……………
(A) 24.3 min
(B) 58.3 min
(C) 102.3 min
(D) 120.3 min
Answer:
(B) 58.3 min

7. The half-life of radium is 1600 years. The fraction of a sample of radium that would remain after 6400 year is ……….
(A) \(\frac {1}{2}\)
(B) \(\frac {1}{4}\)
(C) \(\frac {1}{8}\)
(D) \(\frac {1}{16}\)
Answer:
(D) \(\frac {1}{16}\)

8. The half-life of an element is 5 d. How much time is required for the decay of 7/8^th of the sample?
(A) 5 d
(B) 10 d
(C) 15 d
(D) 35/8 d
Answer:
(C) 15 d

9. The composition of an α-particle can be expressed as ……………….
(A) 1p + 1n
(B) 1p + 2n
(C) 2p + 1n
(D) 2p + 2n
Answer:
(D) 2p + 2n

10. If a radioactive nuclide of group 15 element undergoes β-particle emission, the daughter element will be found in ………………..
(A) 16 group
(B) 14 group
(C) 13 group
(D) same group
Answer:
(A) 16 group

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 9 Elements of Group 13, 14 and 15 Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 9 Elements of Group 13, 14 and 15

Question 1.
What are p-block elements?
Answer:

• Elements in which the differentiating electron (the last filling electron) enters the outermost p orbital are p-block elements.
• Since maximum six electrons can be accommodated in p-subshell i.e., three p-orbitals, the p-block contains six groups numbered from 13 to 18 in the modem periodic table.
• The p-block elements show greater variation in the properties than s-block elements.

Question 2.
Write the names of the elements present in groups 13, 14 and 15.
Answer:

Group	Name of family	Name of the elements
13	Boron family	Boron (5B), aluminium (13Al), gallium (31Ga), indium (49In), thallium (81Tl)
14	Carbon family	Carbon (6C), silicon (14Si), germanium (32Ge), tin (50Sn), lead (82Pb)
15	Nitrogen family	Nitrogen (7N), phosphorus (15P), arsenic (33AS), antimony (51Sb), bismuth (83Bi)

Question 3.
i. Write the general outer electronic configuration of the elements of group 13, group 14 and group 15.
ii. By how many electrons do their outer electronic configurations differ from their nearest inert gas?
Answer:
i.

Group	General outer electronic configuration
13	ns2 np1
14	ns2 np2
15	ns2 np3

ii. The outer electronic configurations of the elements group 13, group 14 and group 15 differ from their nearest inert gas by 5, 4 and 3 electrons, respectively.

Question 4.
In which form do the elements of groups 13,14 and 15 occur in nature?
Answer:

• The elements of groups 13, 14 and 15 do not occur in free monoatomic state and are found as compounds with other elements.
• They also occur in the form of polyatomic molecules (such as N₂, P₄, C₆₀) or polyatomic covalent arrays (such as graphite, diamond).

Question 5.
Write condensed electronic configurations of the following elements.
₁₃Al, ₄₉In, ₁₄Si, ₅₀Sn, ₁₅P, ₃₃As
Answer:
Condensed electronic configurations of
i. ₁₃Al: [Ne]3s² 3p¹
ii. ₄₉In: [Kr]4d¹⁰5s²5p¹
iii. ₁₄Si: [Ne]3s²3p²
iv. ₅₀Sn: [Kr]4d¹⁰5s²5p²
v. ₁₅P: [Ne]3s²3p³
vi. ₃₃As: [Ar]3d¹⁰4s²4p³

Note: Condensed electronic configurations of elements of groups 13, 14 and 15 are given in the table below.

Question 6.
Name the following.
i. A metalloid present in group 13.
ii. A group 13 element which is the third most abundant element in the earth’s crust.
Answer:
i. Boron
ii. Aluminium

Question 7.
Why boron is classified as a metalloid?
Answer:
Boron is glossy and hard solid like metals but a poor conductor of electricity like nonmetals. Since it exhibits properties of both metals and nonmetals, boron is classified as a metalloid.

Question 8.
Describe the variation in the electronegativity of group 13 elements.
Answer:

• In group 13, on moving down the group, the electronegativity decreases from B to Al.
• However, there is a marginal increase in the electronegativity from Al to Tl.
• This trend is a result of the irregularities observed in atomic size of elements.

Question 9.
Atomic numbers of the group 13 elements are in the order B < Al < Ga < In < Tl. Arrange these elements in increasing order of ionic radii of M³⁺.
Answer:

• The given elements are in an increasing order of their atomic number.
• The general outer electronic configuration of group 13 elements is ns²np¹.
• M³⁺ ion is formed by the removal of three electrons from the outermost shell ‘n’.
• In the M³⁺ ions, the ‘n-1’ shell becomes the outermost shell. Size of the ‘n-1’ shell increases down the group.

Therefore, the ionic radii of M³⁺ ion increases down the group in the following order:
B³⁺ < Al³⁺ < Ga³⁺ < In³⁺ < Tl³⁺

Question 10.
Why the atomic radius of Gallium is less than that of aluminium?
Answer:

• Atomic radius of the elements increases down the group due to addition of new shells.
• Electronic configuration of Al is [Ne]3s²3p¹ while that of Ga is [Ar]3d¹⁰4s²4p¹.
• As Al does not have d-electrons, it offers an exception to this trend.
• As we go from Al down to Ga the nuclear charge increases by 18 units. Out of the 18 electrons added, 10 electrons are in the inner 3d subshell of Ga. These d-electrons offer poor shielding effect.
• Therefore, the effect of attraction due to increased nuclear charge is experienced prominently by the outer electrons of Ga and thus, its atomic radius becomes smaller than that of Al.

Hence, the atomic radius of gallium is less than that of aluminium.

Question 11.
The values of the first ionization enthalpy of Al, Si and P are 577, 786 and 1012 kJ mol^-1 respectively. Explain the observed trend.
Answer:

• The trend shows increasing first ionization enthalpy from Al to Si to P.
• Al, Si and P belong to the third period in the periodic table and hence, they have same valence shell.
• As we move across a period from left to right, the nuclear charge increases. Due to this, electrons in the valence shell are held more tightly by the nucleus as we go from Al to Si to P.
• Therefore, more energy is required to remove an electron from its outermost shell.

Hence, the value of first ionization enthalpy increases from Al to Si to P.

Note: Atomic and physical properties of group 13 elements.

Question 12.
Name metal(s), nonmetal(s) and metalloid(s) of group 14.
Answer:
i. Metal: Tin, lead
ii. Nonmetal: Carbon
iii. Metalloid: Silicon, germanium

Question 13.
Explain the variation in the following properties of the group 14 elements,
i. Atomic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic radii (Covalent radii):

• In the periodic table as we move down the group 14 from C to Pb, the atomic radii increases due to the addition of new shell at each succeeding element.
• However, the increase is comparatively less after silicon due to poor shielding by inner d- and f-electrons in the atoms.

ii. Ionization enthalpy:

• Due to increased effective nuclear charge, group 14 elements have higher value of ionization enthalpy than corresponding group 13 elements.
• In the periodic table, as we move down the group 14 from C to Sn, the ionization enthalpy decreases.
• From Si to Sn, the ionization enthalpy decreases slightly.
• However, from Sn to Pb, the ionization enthalpy increases slightly. It is due to the poor shielding effect of intervening d and f orbitals and increase in the size of the atoms.

iii. Electronegativity:

• Due to small atomic size, group 14 elements are slightly more electronegative than the corresponding group 13 elements.
• As we move down the group 14 from C to Si in the periodic table, the electronegativity decreases.
• The electronegativity values for elements from Si to Pb are almost the same.
• Among group 14 elements, carbon is the most electronegative with electronegativity of 2.5.

Question 14.
Explain why there is a phenomenal decrease in ionization enthalpy from carbon to silicon.
Answer:

• Carbon is the first element of group 14 and thus, it has the smallest atomic size.
• The ionization enthalpy of carbon (1086 kJ mol^-1) is very high due to its small atomic size (77 pm) and high electronegativity (2.5).
• However, the ionization enthalpy of silicon (786 kJ mol^-1) decreases phenomenally due to the increase in its atomic size (118 pm) and low electronegativity (1.8).

Note: Atomic and physical properties of group 14 elements.

Question 15.
Which type of elements are present in group 15? Mention their physical state.
Answer:

• Group 15 includes all the three traditional types of elements i.e., metals, nonmetals and metalloids.
• Nitrogen is a gas whereas the remaining group 15 elements are solids.
• The gaseous nitrogen and brittle phosphorus are nonmetals.
• Arsenic and antimony are metalloids while bismuth is moderately reactive metal.

Question 16.
Explain the trends in physical properties of group 15 elements.
i. Atomic and ionic radii
ii. Ionization enthalpy
iii. Electronegativity
Answer:
i. Atomic and ionic radii:

• Atomic size increases down the group with increasing atomic number.
• The effective nuclear charge in case of group 15 elements is larger than that of group 14 elements. Due to the increased effective nuclear charge, electrons are strongly attracted by the nucleus. Thus, the atomic and ionic radii of group 15 elements are smaller than the atomic and ionic radii of the corresponding group 14 elements.
• On moving down the group, number of shells increases which leads to increased shielding effect and as a result atomic radii and ionic radii increases.

ii. Ionization enthalpy:

• Due to extra stability of half-filled p-orbitals and relatively smaller size of group 15 elements, ionization enthalpy of group 15 elements is much greater than that of the group 14 elements in the corresponding periods.
• On moving down the group, increase in atomic size and screening effect overcome the effective nuclear charge and thus, ionization enthalpy decreases.

iii. Electronegativity:

• Due to smaller size and greater effective nuclear charge of atoms, group 15 elements have higher electronegativity values than group 14 elements.
• On moving down the group, electronegativity values decreases due to increase in the size of the atoms and shielding effect.
• Nitrogen is the most electronegative element among group 15 elements. However, there is not much of a difference between the electronegativity values of other elements of group 15.

Note: Atomic and physical properties of group 15 elements.

Question 17.
Write a note on the oxidation state of p-block elements with respect to groups 13, 14 and 15 elements.
Answer:

• Oxidation state is the primary chemical property of all elements.
• The highest oxidation state exhibited by the p-block elements is equal to the total number of valence electrons i.e., the sum of s- and p-electrons present in the valence shell. This is sometimes called the group oxidation state.
• In boron, carbon and nitrogen families, the group oxidation state is the most stable oxidation state for the lighter elements.
• Besides, the elements of groups 13, 14 and 15 exhibit other oxidation states which are lower than the group oxidation state by two units.
• The lower oxidation states become increasingly stable as we move down to heavier elements in the groups.

Note: Group oxidation states and common oxidation states with examples for groups 13, 14 and 15.

Question 18.
What are general oxidation states of group 13 elements? Explain.
Answer:

• The general oxidation states of group 13 are +1 and +3.
• The group 13 elements have the outermost electronic configuration ns² np¹.
• If only np¹ electron takes part in bonding, the oxidation state is +1 and if all the three electrons i.e., ns² np¹ take part in bonding, the oxidation state is +3. Hence, the expected oxidation states are +1 and +3.

Question 19.
Give reason: The increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.
Answer:

• The increased stability of the oxidation state lowered by 2 units than the group oxidation state in heavier p-block elements is due to inert pair effect.
• In these elements, the two s-electrons are involved less readily in chemical reactions.
• This is because, in heavier p-block elements, the s-electrons of valence shell experience poor shielding than valence p-electrons due to ten inner d-electrons.

Hence, the increased stability of the oxidation state is lowered by 2 units than the group oxidation state in heavier p-block elements.

Question 20.
Why Tl¹⁺ ion is more stable than Tl³⁺?
Answer:

• Tl is a heavy element which belongs to group 13 of the p-block.
• The common oxidation state for this group is +3.
• In p-block, the lower oxidation state is more stable for heavier elements due to inert pair effect.

Hence, Tl¹⁺ ion is more stable than Tl³⁺ ion.

Question 21.
How can you explain the higher stability of BCl₃ as compared to TlCl₃?
Answer:

• Boron is a light element in group 13 and has outermost electronic configuration 2s² 2p¹ whereas thallium is a heavy element in group 13 and has outermost electronic configuration 6s² 6p¹.
• Because of its small ionic radius, boron forms stable compounds in +3 oxidation state.
• Thallium has a large atomic size and due to the inert pair effect forms more stable compounds with lower oxidation state +1 than compounds with +3 oxidation state.

Therefore, BCl₃ has higher stability than TlCl₃.

Question 22.
State the oxidation state for the following:
i. The group oxidation state of group 14 elements.
ii. The stable oxidation state for lead.
iii. Oxidation state of carbon in CH₄.
Answer:
i. +4
ii. +2
iii. -4

Question 23.
GeCl4 is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄. Explain.
Answer:

• Elements Ge and Pb belong to 4th and 6th period in the group 14.
• The group oxidation state of group 14 elements is +4.
• However, the stability of other oxidation state which is lower by 2 units i.e., +2, increases down the group due to inert pair effect.
• Therefore, the stability of the oxidation state +4 is more in Ge than in Pb while the stability of the oxidation state +2 is more in Pb than in Ge.

Hence, GeCl₄ is more stable than GeCl₂ while PbCl₂ is more stable than PbCl₄.

Question 24.
Name the elements of group 14 which are generally occur in +2 oxidation state.
Answer:
The elements of group 14 that are generally occur in +2 oxidation state are tin (Sn) and lead (Pb).

Question 25.
Discuss the nature of bonding in compounds of group 13, 14 and 15 elements.
Answer:

• The lighter elements in groups 13, 14 and 15 have small atomic radii and high ionization enthalpy values. They form covalent bonds with other atoms by overlapping of valence shell orbitals.
• As we move down the group, the value of ionization enthalpy decreases. The atomic radius increases since the valence shell orbitals are more diffused.
• The heavier elements in these groups tend to form ionic bonds. The first member of these groups belongs to second period and do not have d orbitals and hence, B, C and N cannot expand their octet.
• The subsequent elements in the group possess vacant d orbital in their valence shell, which can expand their octet forming a variety of compounds.

Question 26.
Explain the reactivity of groups 13, 14 and 15 elements towards air.
Answer:
i. Group 13 elements:
a. On heating with air or oxygen, group 13 elements form oxide of the type E₂O₃.
\(4 \mathrm{E}_{(\mathrm{s})}+3 \mathrm{O}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{E}_{2} \mathrm{O}_{3(\mathrm{~s})}\) (where, E = B, Al, Ga, In, Tl)

b. At high temperature, group 13 elements also react with nitrogen present in the air to form corresponding nitrides.
\(2 \mathrm{E}_{(\mathrm{s})}+\mathrm{N}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{EN}_{(\mathrm{s})}\) (where, E = B, Al, Ga, In, Tl)

ii. Group 14 elements: The elements of group 14 on heating in air or oxygen form oxide of the type EO and EO₂ in accordance with the stable oxidation state and availability of oxygen.

iii. Group 15 elements: The elements of group 15 on heating in air or oxygen forms two types of oxide i.e., E₂O₃ and E₂O₅.

Due to increase in metallic character down the groups 13, 14 and 15, the nature of their oxides gradually varies from acidic through amphoteric to basic.
[Note: The temperature required for the reaction of nitrogen with oxygen is very high. This is produced by striking an electric arc.]

Question 27.
Classify the following oxides into acidic, basic or amphoteric.
B₂O₃, Ga₂O₃, Tl₂O₃, In₂O₃, Al₂O₃
Answer:

Acidic oxide	B2O3
Basic oxides	In2O3, Tl2O3
Amphoteric oxides	Al2O3, Ga2O3

Question 28.
Match the following.

	Column A		Column B
i.	N2O5	a.	Amphoteric
ii.	Bi2O3	b.	Acidic
iii.	Sb2O3	c.	Basic

Answer:
i – b,
ii – c,
iii – a

Note: Nature of stable oxides of groups 13, 14 and 15 elements

Question 29.
State TRUE or FALSE. Correct the false statement.
i. Sb is more stable in +3 oxidation state.
ii. Oxides of the type E₂O₅ are formed by group 15 elements.
iii. As₄O₆ is an acidic oxide.
Answer:
i. True
ii. True
iii. False
As₄O₆ is an amphoteric oxide.

Question 30.
What happens when the elements of groups 13, 14 and 15 react with water?
Answer:
i. Most of the elements of groups 13, 14 and 15 are unaffected by water.
ii. Aluminium reacts with water on heating and forms hydroxide while tin reacts with steam to form oxide.

iii. Lead is unaffected by water due to the formation of a protective film of oxide.

Question 31.
Why is phosphorus stored under water?
Answer:
Phosphorus is highly reactive and hence, it is stored under water to prevent its reaction with air as it catches fire on being exposed to air.

Question 32.
Explain the reactivity of group 13 elements towards halogens.
Answer:
i. All the elements of group 13 react directly with halogens to form trihalides (EX3).
2E_(S) + 3X_2(g) → 2EX_3(s) (where, E = B, Al, Ga, In and X = F, Cl, Br, I)
ii. Thallium is an exception as it forms monohalides (TlX).

Question 33.
Describe the reactivity of group 14 elements with halogens.
Answer:

• All the elements of group 14 (except carbon) react directly with halogens to form tetrahalides (EX₄).
• The heavy elements Ge and Pb form dihalides as well.
• Stability of dihalides increases down the group due to inert pair effect.
• The ionic character of halides also increases steadily down the group.

Question 34.
Discuss the reactivity of group 15 elements with halogens.
Answer:

• Elements of the group 15 reacts with halogens to form two series of halides i.e., trihalides (EX₃) and pentahalides (EX₅).
• The pentahalides possess more covalent character due to availability of vacant d orbitals of the valence shell for bonding.
• Nitrogen being second period element, does not have d orbitals in its valence shell, and therefore, does not form pentahalides.
• Trihalides of the group 15 elements are predominantly covalent except BiF₃. The only stable trihalide of nitrogen is NF₃.

Question 35.
Nitrogen does not form pentahalides. Give reason.
Answer:

• The electronic configuration of 7N is 1s² 2s² 2p³. It has 3 unpaired electrons which can form 3 covalent bonds, thus forming NX₃ molecule.
• Valence shell of nitrogen (n = 2) contains only s and p orbitals.
• Thus, due to the absence of d orbitals in the valence shell, nitrogen cannot expand its octet, therefore, it cannot form compounds like NCl₅ and NF₅.

Hence, nitrogen does not form pentahalides.

Question 36.
Nitrogen does not form NCl₅ or NF₅ but phosphorus can. Explain.
Answer:

• The electronic configuration of 7N is 1s2 2s2 2p3 while that of 15P is 1s² 2s² 2p⁶ 3s² 3p³.
• As phosphorus contains d orbitals, it can expand its octet to form MX₃ as well MX₅ compounds.
• However, due to absence of d orbitals, nitrogen cannot form MX₃ or MX₅.

Hence, Nitrogen does not form NCl₅ or NF₅ but phosphoms can form compounds like PCl₅ or PF₅.

Question 37.
Define catenation.
Answer:
The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.

Question 38.
Explain catenation of group 14 elements.
Answer:
i. The property of self-linking of atoms of an element by covalent bonds to form chains and rings is called catenation.
ii. The strength of the element-element bond determines the tendency of an element to form a chain.
iii. Among the elements of group 14, the bond strength is maximum for C-C bond (348 kJ mol^-1). Hence, carbon has maximum tendency for catenation.

Bond	Bond strength (Bond enthalpy kJ mol-1)
C-C	348
Si-Si	297
Ge-Ge	260
Sn-Sn	240

iv. From the values of bond enthalpy, it can be concluded that the tendency to form chains is maximum for carbon and much lesser for silicon. Germanium has still lesser tendency and tin has hardly any tendency for catenation. Lead does not show catenation.
Therefore, the order of catenation of group 14 elements is C >> Si > Ge = Sn.

Question 39.
State TRUE or FALSE. Correct the false statement.
i. Among the group 14 elements, Ge does not show the property of catenation.
Answer:
i. False
Among the group 14 elements, Pb (lead) does not shows the property of catenation.

Question 40.
Define allotropy.
Answer:
When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy.

Question 41.
i. What are allotropes?
ii. Name various allotropes of carbon.
Answer:
i. When a solid element exists in different crystalline forms with different physical properties such as colour, density, melting point, etc. the phenomenon is called allotropy and the individual crystalline forms are called allotropes.
ii. Diamond, graphite, fiillerenes, graphene and carbon nanotubes are various allotropes of carbon.

Question 42.
Explain the structure of diamond.
Answer:
Structure of diamond:

• In diamond, each carbon atom undergoes sp3 hybridization and is linked to four other carbon atoms in tetrahedral manner.
• The C – C bond length is 154 pm.
• The tetrahedra are linked together forming a three-dimensional network structure involving strong C-C single bonds which makes diamond the hardest natural substance.

Question 43.
Write physical properties of diamond. Also, state its uses.
Answer:
i. Physical properties

• Diamond is the hardest natural substance.
• It has abnormally high melting point (3930 °C).
• It is a bad conductor of electricity.

ii. Uses: Diamond is used

• for cutting glass and in drilling tools.
• for making dies for drawing thin wire from metal.
• for making jewellery.

Question 44.
Describe the structure of graphite.
Answer:

• Graphite is composed of layers of two-dimensional sheets of carbon atoms.
• Each sheet is made up of hexagonal net of sp² carbons bonded to three neighbours forming three bonds.
• The fourth electron in the unhybrid p-orbital of each carbon is shared by all carbon atoms resulting in a π bond. These it electrons are delocalized over the whole layer.
• The C – C bond length in graphite is 141.5 pm.
• The individual layers are held by weak van der Waals forces and separated by 335 pm.
• Graphite is soft and slippery and is thermodynamically most stable allotrope of carbon.

Question 45.
Diamond is very hard whereas graphite is soft. Explain.
Answer:

• Diamond has three-dimensional network of sp³ hybridized carbon atoms joined by extended covalent bonds which are difficult to break. Therefore, diamond is hard.
• Graphite has two-dimensional sheet like structure, like layers of hexagonal rings formed from sp² hybridized carbon atoms. These layers are held by weak van der Waals forces, which can be broken easily. Therefore, graphite is soft and slippery.

Hence, diamond is very hard whereas graphite is soft.

Question 46.
i. What are fullerenes?
ii. How are they prepared?
Answer:
i. Fullerenes are allotropes of carbon in which carbon molecules are linked by a definite numbers of carbon atoms, for example as in C₆₀.
ii. Fullerenes are produced when an electric arc is struck between the graphite electrodes in an inert atmosphere of argon or helium. The soot formed contains significant amount of C₆₀ fullerene and smaller amounts of other fullerenes C₃₂, C₅₀, C₇₀ and C₈₄.

Question 47.
Discuss the structure and properties of fullerene (C6o).
Answer:

• C₆₀ has a shape like soccer ball and called Buckminsterfullerene or bucky ball.
• It contains 20 hexagonal and 12 pentagonal fused rings of carbon.
• The C₆₀ fullerene structure exhibit separations between the neighbouring carbons as 143.5 pm and 138.3 pm.
• Fullerenes are covalent and soluble in organic solvents.
• Fullerene C₆₀ reacts with group 1 metals forming solids such as K₃C₆₀.
• The compound K₃C₆₀ behaves as a superconductor below 18 K, which means that its carries electric current with zero resistance.

Question 48.
Explain the structure of carbon nanotubes.
Answer:

• Carbon nanotubes are cylindrical in shape consisting of rolled-up graphite sheet.
• Nanotubes can be single-walled (SWNTs) with a diameter of less than 1 nm or multi-walled (MWNTs) with diameter reaching more than 100 nm.
• Their lengths range from several micrometres to millimetres.

Question 49.
Describe the properties of carbon nanotubes.
Answer:

• Carbon nanotubes are robust. They can be bent, and when released, they will spring back to the original shape.
• Carbon nanotubes have high electrical or heat conductivities and highest strength-to-weight ratio for any known material to date.
• The researchers of NASA are combining carbon nanotubes with other materials into composites that can be used to build lightweight spacecraft.

Question 50.
What is graphene?
Answer:

• Isolated layer of graphite is called graphene.
• Graphene sheet is a two dimensional solid.
• It has unique electronic properties.

Question 51.
Explain the structure of various allotropes of phosphorus.
Answer:
Phosphorus is found in different allotropic forms. White and red phosphorus are important allotropes of phosphorus.
i. White (yellow) phosphorus:

1. White (yellow) phosphorus consists of discrete tetrahedral P₄ molecules.
2. The P – P – P bond angle is 60°.
3. White phosphorus is less stable and hence more reactive, because of angular strain in the P₄ molecule.

ii. Red phosphorus:

• Red phosphorus consists of chains of P₄ linked together by covalent bonds.
• Thus, it is polymeric in nature.

Question 52.
Enlist properties of
i. white phosphorus.
ii. red phosphorus.
Answer:
i. Properties of white phosphorus:

• It is translucent white waxy solid.
• It glows in the dark (chemiluminescence).
• It is insoluble in water but dissolves in boiling NaOH solution.
• It is poisonous.

ii. Properties of red phosphorus:

• It is stable and less reactive.
• It is odourless and possess iron grey lustre.
• It does not glow in the dark.
• It is insoluble in water.
• It is nonpoisonous.

Question 53.
How is red phosphorus prepared?
Answer:
Red phosphorus is prepared by heating white phosphorus at 573 K in an inert atmosphere.

Question 54.
State whether the following statement is TRUE or FALSE. Correct if false.
i. Covalent molecules have irregular shape described with the help of bond lengths and bond angles.
ii. It is difficult to understand the reactivity of covalent inorganic compounds from their structures.
iii. Inorganic molecules are often represented by molecular formulae indicating their elemental composition.
Answer:
i. False
Covalent molecules have definite shape described with the help of bond lengths and bond angles.
ii. False
The reactivity of covalent inorganic compounds is better understood from their structures.
iii. True

Question 55.
Describe structure of the following molecules.
i. Boron trichloride
ii. Aluminium chloride
iii. Orthoboric acid
Answer:
i. Structure of boron trichloride (BCl₃) molecule:

• Boron trichloride (BCl₃) is a covalent compound.
• In BCl₃ molecule, boron atom is sp² hybridized having one vacant unhybridized p orbital.
• B in BCl₃ has incomplete octet.
• BCl₃ is a nonpolar trigonal planar molecule.
• Each Cl – B – Cl bond angle is 120°.

ii. Structure of aluminium chloride (AlCl₃) molecule:

• Aluminium atom in aluminium chloride is sp² hybridized, with one vacant unhybrid p-orbital.
• Aluminium chloride exists as the dimer (Al₂Cl₆) formed by overlap of vacant 3d orbital of Al with a lone pair of electrons of Cl.

iii. Structure of orthoboric or boric acid (H₃BO₃) molecule:

• Orthoboric acid has central boron atom bound to three -OH groups.
• The solid orthoboric acid has layered crystal structure in which trigonal planar B(OH)₃ units are joined together by hydrogen bonds.

Question 56.
Which are the different crystalline forms of silica?
Answer:
Quartz, cristobalite and tridymite are the different crystalline forms of silica.
[Note: These crystalline forms are inter-convertible at a suitable temperature.]

Question 57.
Explain the structure of silicon dioxide.
Answer:

• Silicon dioxide (SiO₂), is also known as silica.
• It is a covalent three-dimensional network solid.
• In SiO₂, each silicon atom is covalently bound in tetrahedral manner to four oxygen atoms.
• The crystal contains eight membered rings having alternate silicon and oxygen atoms.

Question 58.
Discuss the nature and structure of the following compounds.
i. Nitric acid
ii. Phosphoric acid
Answer:
i. Nitric acid:

• Nitric acid (HNO₃) is a strong, oxidizing mineral acid.
• The central nitrogen atom is sp² hybridized.
• HNO₃ exhibits resonance phenomenon.
• Figure (a) represents resonating structures of HNO₃ while figure (b) represents resonance hybrid of HNO₃.

ii. Phosphoric acid (Orthophosphoric acid):

• Phosphorus forms number of oxyacids. Orthophosphoric acid (H₃PO₄) is a strong nontoxic mineral acid.
• It contains three ionizable acidic hydrogens.
• The central phosphorus atom is tetrahedral.

Question 59.
Give the molecular formula of crystalline borax.
Answer:
The crystalline borax has formula Na₂B₄O₇.10H₂O or Na₂[B₄O₅(OH)₄].8H₂O.

Question 60.
How is borax obtained from its mineral colemanite?
Answer:
Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

Question 61.
Why is the aqueous solution of borax alkaline?
Answer:
On hydrolysis, borax forms a strong base (NaOH) and a weak acid (H₃BO₃). The presence of the strong base makes borax solution alkaline.

Question 62.
What happens when borax is heated strongly?
Answer:
Borax is a white crystalline solid. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

Question 63.
Explain borax bead test.
Answer:
i. Borax bead test is used to detect coloured transition metal ions.
ii. On heating, borax loses water molecules and swells up. On further heating, it melts and forms a transparent liquid, which solidifies into a glass like material known as borax bead.

iii. The borax bead consists of sodium metaborate and boric anhydride, which reacts with metals salts to form coloured bead.
e.g. When borax is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured Co(BO₂)₂ bead is formed.

Question 64.
Write the uses of borax.
Answer:
Borax is used

• to manufacture optical and hard borosilicate glasses.
• as a flux for soldering and welding.
• as a mild antiseptic in the preparation of medical soaps.
• in qualitative analysis for borax bead test.
• as a brightener in washing powder.

Question 65.
How are silicones prepared? Write their properties.
Answer:
i. Preparation of silicones:
a. Alkyl or aryl substituted silicon chlorides having general formula R_nSiCl_(4-n) (R = alkyl or aryl group) are used as the starting materials for manufacture of silicones.
b. When methyl chloride reacts with silicon in the presence of copper catalyst at a temperature 573 K, various types of methyl substituted chlorosilane of formula MeSiCl₃, Me₂SiCl₂, Me₃SiCl with small amounts of Me₄Si are formed.

c. Hydrolysis of dimethyldichlorosilane, (CH₃)₂SiCl₂ followed by condensation polymerisation yields straight chain silicone polymers.

d. The chain length of polymer can be controlled by adding (CH₃)₃SiCl at the end.

ii. Properties:

• Silicones are water repellent.
• They have high thermal stability.
• They are good electrical insulators.
• They are resistant to oxidation and chemicals.

Question 66.
Explain the preparation of ammonia from nitrogeneous organic matter.
Answer:
Ammonia is formed by the decomposition of nitrogeneous organic matter such as urea. It is therefore, present naturally in small quantities in air and soil.

Question 67.
Describe laboratory method for preparation of ammonia.
Answer:
Ammonia is prepared on laboratory scale by decomposition of the ammonium salts with calcium hydroxide or caustic soda.

Question 68.
How is ammonia manufactured by Haber process?
Answer:

• On the large scale, ammonia is prepared by direct combination of dinitrogen and dihydrogen by Haber process.
• In this process, dinitrogen reacts with dihydrogen under high pressure of 200 × 10⁵ Pa (200 atm) and temperature around 700 K to produce ammonia.
  N_2(g) + 2H_2(g) ⇌ 2NH_3(g); Δ_fH° = -46.1 kJ mol^-1
• Iron oxide with trace amounts of K₂O and Al₂O₃ is used as catalyst in Haber process.
• High pressure favours the formation of ammonia as equilibrium is attained rapidly under these conditions.

Question 69.
State the physical properties of ammonia.
Answer:

• Ammonia is a colourless gas with pungent odour.
• It has freezing point of 198.4 K and boiling point of 239.7 K.
• It is highly soluble in water.

Question 70.
What is liquor ammonia?
Answer:
The concentrated aqueous solution of ammonia (NH₃) is called liquor ammonia.

Question 71.
Give reason: Ammonia has higher melting and boiling points.
Answer:

• In solid and liquid state, NH₃ molecules get associated together through hydrogen bonding.
• As a result, extra amount of energy is required to break such intermolecular hydrogen bonds. Hence, ammonia has higher melting and boiling points.

Question 72.
Why is ammonia basic in aqueous solution?
Answer:
i. As ammonia is highly soluble in water, it readily forms OH^– ions in its aqueous solution.
\(\mathrm{NH}_{3(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(a q)}^{-}\)
ii. Thus, due to the formation of OH^– ions, aqueous solution of ammonia is basic in nature.

Question 73.
How does the aqueous solution of ammonia react with the following salt solutions?
i. ZnSO₄
ii. FeCl₃
Answer:
Aqueous solution of ammonia precipitates out as hydroxides (or hydrated oxides) of metals solutions.

Question 74.
Write applications of ammonia.
Answer:
Ammonia is used in

• manufacture of fertilizers such as urea, diammonium phosphate, ammonium nitrate, ammonium sulphate etc.
• manufacture of some inorganic compounds like nitric acid.
• refrigerant (liq. ammonia).
• laboratory reagent in qualitative and quantitative analysis (aq. solution of ammonia).

Question 75.
Give reactions involved in the formation of Nessler’s reagent.
Answer:

Question 76.
How does ammonia react with Nessler’s reagent?
Answer:
Ammonia react with Nessler’s reagent (an alkaline solution of K₂HgI₄) to form a brown precipitate (Millon’s base).

Question 77.
Complete and write the balanced chemical equations for:
i. Ca₂B₆O₁₁ + Na₂CO₃ →
ii. CoO + B₂O₃ →
iii. AgCl + NH₃ →
iv. ZnSO₄ + 2NH₄OH →
v. a. 2KI + HgCl₂ →
b. 2KI + HgI₂ →
Answer:

Question 78.
Naina was preparing a compound in the laboratory. She added compound ‘A’ to (CaOH)₂ solution. As a result of this, a compound ‘B’ was obtained which had a pungent smell. On adding Nessler’s reagent to the compound ‘B’, a brown precipitate of compound ‘C’ was obtained.
Write the chemical reactions involved and identify ‘A’, ‘B’ and ‘C’.
Answer:
i. When ammonium chloride is mixed with (CaOH)₂ solution, ammonia is formed which has a pungent odour.

ii. Ammonia react with Nessler’ s reagent (an alkaline solution of K₂Hgl₄) to form a brown precipitate (Millon’ s base).

Multiple Choice Questions

1. The electronic configuration of boron family is ……………
(A) ns² np²
(B) ns² np⁵
(C) ns² np⁶
(D) ns² np¹
Answer:
(D) ns² np¹

2. ………… has noble gas core plus 14 f-electrons and 10 d-electrons.
(A) Gallium
(B) Indium
(C) Thallium
(D) Boron
Answer:
(C) Thallium

3. The group 15 element having inner electronic configuration as of argon is …………..
(A) Phosphorus (Z = 15)
(B) Antimony (Z = 51)
(C) Arsenic (Z = 33)
(D) Nitrogen (Z = 7)
Answer:
(C) Arsenic (Z = 33)

4. Which of the following is NOT a metalloid?
(A) B
(B) Sn
(C) Ge
(D) Sb
Answer:
(B) Sn

5. Among the group 13 elements, melting point is highest for …………..
(A) B
(B) Al
(C) Ga
(D) In
Answer:
(A) B

6. On moving down the group 14, the ionization enthalpy
(A) increases slightly from Si to Sn and decreases slightly from Sn to Pb
(B) increases throughout uniformly from Si to Pb
(C) decreases throughout uniformly from Si to Pb
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb
Answer:
(D) decreases slightly from Si to Sn and increases slightly from Sn to Pb

7. ………… is the most electronegative element of group 14.
(A) Carbon
(B) Silicon
(C) Germanium
(D) Tin
Answer:
(A) Carbon

8. The stability of +3 oxidation state in aqueous solution is in order ……………
(A) Al > Ga > In > Tl
(B) Tl > In > Ga > Al
(C) Al > Tl > Ga > In
(D) Tl > Al > Ga > In
Answer:
(A) Al > Ga > In > Tl

9. Group oxidation state of group 15 elements is ……………
(A) +4
(B) +1
(C) +3
(D) +5
Answer:
(D) +5

10. …………. cannot expand its octet due to absence of d orbital in its valence shell.
(A) Ga
(B) C
(C) As
(D) Ge
Answer:
(B) C

11. Which one of the following statements about boron and aluminium is INCORRECT?
(A) Both exhibit oxidation state of +3.
(B) Both form oxides of the formula M₂O₃.
(C) Both form trihalides, MX₃.
(D) Both form amphoteric oxides.
Answer:
(D) Both form amphoteric oxides.

12. Which of the following is basic oxide?
(A) Bi₂O₃
(B) CO₂
(C) B₂O₃
(D) SiO₂
Answer:
(A) Bi₂O₃

13. The reaction of Al with H₂O produces ……………
(A) Al₂O₃
(B) AlH₃
(C) Al(OH)₃
(D) Al₂H₆
Answer:
(C) Al(OH)₃

14. Which of the following is a stable halide of nitrogen?
(A) NF₃
(B) NCl₅
(C) NF₅
(D) NBr₅
Answer:
(A) NF₃

15. Catenation is the ability of …………..
(A) atoms to form strong bonds with similar atoms
(B) elements to form giant molecules
(C) an element to form multiple bonds
(D) an element to form long chains of identical atoms
Answer:
(D) an element to form long chains of identical atoms

16. Among the group 13 elements, the property of allotropy is shown by ………………
(A) indium
(B) aluminium
(C) thallium
(D) boron
Answer:
(D) boron

17. Thermodynamically stable allotrope of carbon is …………..
(A) diamond
(B) graphite
(C) buckyball
(D) all of these
Answer:
(B) graphite

18. White phosphorus contains discrete …………… molecules.
(A) P₅
(B) P₄
(C) P₆
(D) P₅₂
Answer:
(B) P₄

19. In white phosphorus, the P-P-P bond angle is ……………
(A) 60°
(B) 90°
(C) 109.5
(D) 120°
Answer:
(A) 60°

20. 3c-2e bonds are present in ………………
(A) NH₃
(B) B₂H₆
(C) H₃BO₃
(D) SiCl₄
Answer:
(B) B₂H₆

21. Which of the following is borax?
(A) Na₂B₄O₇.4H₂O
(B) Na₂B₄O₇.10H₂O
(C) H₃BO₃
(D) NaBO₂
Answer:
(B) Na₂B₄O₇.10H₂O

22. In Borax bead test, the coloured ions give characteristic coloured beads due to formation of …………….
(A) metal borates
(B) metal metaborates
(C) metal phosphates
(D) metal tetraborates
Answer:
(B) metal metaborates

23. The catalyst used in Haber process contains …………..
(A) nickel
(B) palladium
(C) iron
(D) platinum
Answer:
(C) iron

24. Which of the following is used as refrigerant?
(A) Nessler’s reagent
(B) Liq. ammonia
(C) Borax
(D) Diborane
Answer:
(B) Liq. ammonia