Class 12 – Page 33 – Maharashtra Board Solutions

12th Chemistry Chapter 16 Exercise Green Chemistry and Nanochemistry Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 16 Green Chemistry and Nanochemistry Textbook Exercise Questions and Answers.

Green Chemistry and Nanochemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 16 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 16 Exercise Solutions

1. Choose the most correct option.

Question i.
The development that meets the needs of the present without compromising the ability of future generations to meet their own need is known as
a. Continuous development
b. Sustainable development
c. True development
d. Irrational development
Answer:
b. Sustainable development

Question ii.
Which of the following is ϒ-isomer of BHC?
a. DDT
b. lindane
c. Chloroform
d. Chlorobenzene
Answer:
b. lindane

Question iii.
The prefix ‘nano’ comes from
a. French word meaning billion
b. Greek word meaning dwarf
c. Spanish word meaning particle
d. Latin word meaning invisible
Answer:
(b) Greek word meaning dwarf

Question iv.
Which of the following information is given by FTIR technique?
a. Absorption of functional groups
b. Particle size
c. Confirmation of formation of nanoparticles
d. Crystal structure
Answer:
(a) Absorption of functional groups

Question v.
The concept of green chemistry was coined by
a. Born Haber
b. Nario Taniguchi
c. Richard Feynman
d. Paul T. Anastas
Answer:
(d) Paul T. Anastas

2. Answer the following

Question i.
Write the formula to calculate % atom economy.
Answer:

Question ii.
Name the ϒ-isomer of BHC.
Answer:
Lindane

Question iii.
Ridhima wants to detect structure of surface of materials. Name the technique she has to use.
Answer:
Scanning electron microscopy (SEM)

Question iv.
Which nanomaterial is used for tyres of car to increase the life of tyres?
Answer:
Carbon black

Question v.
Name the scientist who discovered scanning tunneling microscope (STM) in 1980.
Answer:
Gerd Binning and Heinrich Rohrer. (Nobel prize 1986)

Question vi.
1 nm = …..m?
Answer:
1 nm = 10⁹ m

3. Answer the following

Question i.
Define
(i) Green chemistry
(ii) sustainable development.
Answer:
(i) Green chemistry : Green chemistry is the use of chemistry for pollution prevention and it designs the use of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.

(ii) Sustainable development : Sustainable development is the development that meets the needs of the present, without compromising the ability of future generations to meet their own needs.

Question ii.
Explain the role of green chemistry.
Answer:
When the waste and pollution that society generates exceeds the Earth’s natural capacity for dealing with it, the green chemistry approach plays an important role.

To reduce or eliminate the use or generation of hazardous substances in the design, manufacture and use of chemical products by promoting innovative chemical technologies.
Capital expenditure required for prevention of pollution is controlled by the use of green chemistry.
Since green chemistry incorporates and promotes pollution prevention practices in the manufacturing process of chemicals it helps industrial ecology.
Green chemistry helps to protect the presence of ozone in the stratosphere. Ozone layer is essential for the survival of life on the earth.
Global warming (Greenhouse effect) is controlled by green chemistry. At present it is the beginning of the green revolution.
It is an exciting time with the new challenges for chemist involved with the discovery, manufacturing and use of chemicals. Green chemistry helps us to save environment and save earth, which is important for our future.

Question iii.
Give the full form (long form) of the names for the following instruments.
a. XRD
b. TEM.
c. STM
d. FTIR
e. SEM
Answer:
a. XRD-X-ray diffraction
b. TEM-Tunneling Electron Microscope
c. STM – Scanning Tunneling Microscope
d. FTIR-Fourier Transform Infrared Spectroscope
e. SEM-Scanning Electron Microscope

Question iv.
Define the following terms :
a. Nanoscience
b. Nanotechnology
c. Nanomaterial
d. Nanochemistry
Answer:
a. Nanoscience : The study of phenomena and manipulation of materials at atomic, molecular and macromolecular scales where properties differ significantly from those at a larger scale is called nanoscience.

b. Nanotechnology : The design, characterization, production and application of structures, device and system by controlling shape and size at nanometer scale is called nanotechnology.

c. Nanomaterial : A material having structural components with at least one dimension in the nanometer scale that is 1 -100 nm is called the nanomaterial. Nanomaterials are larger than single atoms but smaller than bacteria and cells.

d. Nanochemistry : It is the combination of chemistry and nanoscience. It deals with designing and synthesis of materials of nanoscale with different size and shape, structure and composition and their organization into functional architectures.

Question v.
How nanotechnology plays an important role in water purification techniques?
Answer:

Water purification is an important issue as 1.1 billion people do not have access to improved water supply. Water contains water bom pathogens like viruses, bacteria.
Silver nanoparticles are highly effective bacterial disinfectant to remove E. Coli from water. Hence, filter materials coated with silver nanoparticles is used to clean water.
Silver nanoparticles (AgNps) is a cost effective alternative technology (for e.g. water purifier).

Question vi.
Which nanomaterial is used in sunscreen lotion? Write its use.
Answer:
Zinc oxide (ZnO) and Titanium dioxide (TiO₂) nanoparticles are used sunscreen lotions. The chemicals protect the skin against harmful u.v (ultraviolet) rays by absorbing or reflecting the light and prevent the skin from damage.

Question vii.
How will you illustrate the use of safer solvent and auxiliaries?
Answer:

Use of safer solvents and auxiliaries – is a principle of green chemistry it states that safer solvent like water, supercritical CO₂ should be used in place of volatile halogenated organic solvents, like CH₂CI₂, CHCI₃, CCI₄ for chemical synthesis and other purposes.
Solvents dissolve solutes and form solutions, they facilitate many reactions. Water is a safer benign solvent while solvents like dichloromethane (CH₂CI₂), chloroform (CHCI₃) etc are hazardous.
Use of toxic solvents affect millions of workers every year and have implications for consumers and the environment. A large amount of waste is created by their use and they also have huge environmental and health impacts.
Finding safer solvents or designing processes which are solvent free is the best way to improve the process and the product.

Question viii.
Define catalyst. Give two examples.
Answer:
A substance which speeds up the rate of a reaction without itself being changed chemically in the reaction is called a catalyst. It helps to increase selectivity, minimise waste and reduce reaction time and energy demands. For example : Hydrogenation of oil the catalyst used are platinum or palladium, Raney nickel.

4. Answer the following

Question i.
Explain any three principles of green chemistry.
Answer:

Environment protection is the prime concern which has lead to the need for designing chemicals that degrade and can be discarded easily. These chemicals and their degradation products should be non-toxic, non-bioaccumulative or should not be environmentally persistent.
This principle aims at waste product being automatically degradable to clean the environment. Thus the preference for biodegradable polymers and pesticides.
To make the separation and segregation easier for the consumer an international plastic recycle mark is printed on larger items.
There is a dire need to develop improvised analytical methods to allow for real time, in process monitoring and control prior to the formation of hazardous substances.
It is very much important for the chemical industries and nuclear reactors to develop or modify analytical
methodologies so that continuous monitoring of the manufacturing and processing unit is possible.
It is needed to develop chemical processes that are safer and minimize the risk of accidents. It is important to select chemical substances used in a chemical reaction in such a way that they can minimize the occurrence of chemical accidents, explosions, fire and emissions.
For example : Chemical process that works with the gaseous substances can lead to relatively higher possibilities of accidents including explosion as compared to the system working with nonvolatile liquid and solid substances.

Question ii.
Explain atom economy with suitable example.
Answer:
(1) Atom economy is a measure of the amount of atoms from the starting material that are present in the final product at the end of a chemical process. Good atom economy means most of the atoms of the reactants are incorporated in the desired products. Only small amount of waste is produced, hence lesser problem of waste disposal.

(2) The atom economy of a process can be calculated using the following formula.

The atom economy of the above reacijon is less than 50% and waste produced is higher.

Question iii.
How will you illustrate the principle, minimization of steps?
Answer:
(1) The technique of protecting or blocking group is commonly used in organic synthesis. Finally on completion of reaction deprotection of the group is required. This leads to unnecessary increase in the number of steps and decreased atom economy.

(2) The green chemistry principle aims to develop processes to avoid necessary steps i.e. (minimization of steps). When biocatalyst is used very often there is no need for protection of selective group. For example, conversion of m-hydroxyl benzaldehyde to m-hydroxybenzoic acid.

Question iv.
What do you mean by sol and gel? Describe the sol-gel method of preparation for nanoparticles.
Answer:
(1) Sol : Sols are dispersions of colloidal particles in a liquid. Colloids are solid particles with diameter of 1-100 nm.

(2) Gel : A gel is interconnected rigid network with pores of submicrometer dimensions and polymeric chains whose average length is greater than a micrometer.

(3) Sol-gel Process : A sol-gel process is an inorganic polymerisation reaction. It is generally carried out at room temperature, it includes four steps : Hydrolysis, polycondensation, drying and thermal decomposition. This method is widely used to prepare oxide materials.

The reactions involved in the sol-gel process are as follows :
MOR + H₂O → MOH + ROH (hydrolysis)
metal alkoxide
MOH + ROM → M-O-M + ROH (condensation)

Formation of different stable solution of the alkoxide or solvated metal precursor.
Gelation involves the formation of an oxide or alcohol-bridged network (gel) by a polycondensation reaction.
Aging of the gel means during that period gel transforms into a solid mass.
Drying of the gel involves removal of water and other volatile liquids from the gel network.
Dehydration is achieved when the material is heated at temperatures up to 800°C.

Question v.
Which flower is an example of self-cleaning?
Answer:

Lotus is an example of self cleansing.
Nanostructures on the lotus plant leaves are super hydrophobic, they repel water which carries dirt as it rolls off.
Thus though lotus plant (Nelumbonucifera) grows in muddy water, its leaves always appear clean.

Activity :
Collect information about the application of nanochemistry in cosmetics and pharmaceuticals

12th Chemistry Digest Chapter 16 Green Chemistry and Nanochemistry Intext Questions and Answers

Do you know? (Textbook page 343)

Question 1.
Does plastic packaging impact the food they wrap ?
Answer:
Phthalates leach into food through packaging so you should avoid microwaving food or drinks in plastic and not use plastic cling wrap and store your food in glass container whenever possible. Try to avoid prepackaging, processed food so that you will reduce exposure to the harmful effects of plastic.

Used Catalyst (Textbook page 342)

Question 18.
Complete the chart:

Reaction	Name of Catalyst used
1. Hydrogenation of oil (Hardening)	…………………………………
2. Haber’s process of manufacture of ammonia	…………………………………
3. Manufacture of HDPE polymer	…………………………………
4. Manufacture of H₂S0₄ by contact process	…………………………………
5. Fischer-Tropsch process (synthesis of gasoline)	…………………………………

Answer:

Reaction	Name of Catalyst used
1. Hydrogenation of oil (Hardening)	Nickel (Ni)
2. Haber’s process of manufacture of ammonia	Iron
3. Manufacture of HDPE polymer	Zeigler-Natta catalyst
4. Manufacture of H₂S0₄ by contact process	Vanadium oxide (V₂0₅)
5. Fischer-Tropsch process (synthesis of gasoline)	Cobalt-based or Iron based

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 10 Exercise Halogen Derivatives Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 10 Halogen Derivatives Textbook Exercise Questions and Answers.

Halogen Derivatives Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 10 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 10 Exercise Solutions

1. Choose the most correct option.

Question i.
The correct order of increasing reactivity of C-X bond towards nucleophile in the following compounds is

a. I < II < III < IV
b. II < I < III < IV
c. III < IV < II < I
d. IV < III < I < II
Answer:
(d) IV < III < I < II

Question ii.

The major product of the above reaction is,

Answer:
(c)

Question iii.
Which of the following is likely to undergo racemization during alkaline hydrolysis?

Answer:
(a) Only I

Question iv.
The best method for preparation of alkyl fluorides is
a. Finkelstein reaction
b. Swartz reaction
c. Free radical fluorination
d. Sandmeyer’s reaction
Answer:
b. Swartz reaction

Question v.
Identify the chiral molecule from the following.
a. 1-Bromobutane
b. 1,1- Dibromobutane
c. 2,3- Dibromobutane
d. 2-Bromobutane
Answer:
(d) 2-Bromobutane

Question vi.
An alkyl chloride on Wurtz reaction gives 2,2,5,5-tetramethylhexane. The same alkyl chloride on reduction with zinc-copper couple in alchol give hydrocarbon with molecular formula C₅H₁₂. What is the structure of alkyl chloride

Answer:
(a)

Question vii.
Butanenitrile may be prepared by heating
a. propanol with KCN
b. butanol with KCN
c. n-butyl chloride with KCN
d. n-propyl chloride with KCN
Answer:
(d) n-propyl chloride with KCN

Question viii.
Choose the compound from the following that will react fastest by S_N1 mechanism.
a. 1-iodobutane
b. 1-iodopropane
c. 2-iodo-2 methylbutane
d. 2-iodo-3-methylbutane
Answer:
(c) 2-iodo-2 methylbutane

Question ix.

The product ‘B’ in the above reaction sequence is,

Answer:
(d)

Question x.
Which of the following is used as source of dichlorocarbene
a. tetrachloromethane
b. chloroform
c. iodoform
d. DDT
Answer:
(b) chloroform

2. Do as directed.

Question i.
Write IUPAC name of the following compounds

Answer:

Question ii.
Write structure and IUPAC name of the major product in each of the following reaction.

Answer:
Structure and IUPAC name

Question iii.
Identify chiral molecule/s from the following.

Answer:
Chiral molecule

Question iv.
Which one compound from the following pairs would undergo S_N2 faster from the?

Answer:
(1) Sincey is a primary halide it undergoes S_N2 reaction faster than .
(2) Since iodine is a better leaving group than chloride, 1-iodo propane (CH₃CH₂CH₂I) undergoes S_N2 reaction faster than l-chloropropane (CH₃CH₂CH₂CI).

Question v.
Complete the following reactions giving major product.

Answer:

Answer:

Question vi.
Name the reagent used to bring about the following conversions.
a. Bromoethane to ethoxyethane
b. 1-Chloropropane to 1 nitropropane
c. Ethyl bromide to ethyl isocyanide
d. Chlorobenzene to biphenyl
Answer:

Question vii.
Arrange the following in the increase order of boiling points
a. 1-Bromopropane
b. 2- Bromopropane
c. 1- Bromobutane
d. 1-Bromo-2-methylpropane
Answer:
l-Bromo-2-methylpropane, 2-Bromopropane, 1-Bromopropane, 1-Bromo butane

Question viii.
Match the pairs.

Answer:

3. Give reasons

Question i.
Haloarenes are less reactive than haloalkanes.
Answer:
Haloarenes (Aryl halides) are less reactive than (alkyl halides) haloalkanes due to the following reasons :

(1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with 7r-electrons of the benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.

Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult and are less reactive. On the other hand, in alkyl halides, carbon is attached to chlorine by a single bond and it can be easily broken.

(2) Aryl halides are stabilized by resonance but alkyl halides are not. Hence, the energy of activation for the displacement of halogen from aryl halides is much greater than that of alkyl halides.

(3) Different hybridization state of carbon atom in C-X bond :
(i) In alkyl halides, the carbon of C-X bond is sp³-hybridized with less 5-character and greater bond length of 178 pm, which requires less energy to break the C-X bond.

(ii) In aryl halides, the carbon of C-X bond is sp³-hybridized with more 5-character and shorter bond length which requires more energy to break C-X bond. Therefore, aryl halides are less reactive than alkyl halides.

(iii) Polarity of the C-X bond : In aryl halide C-X bond is less polar than in alkyl halides. Because sp³-hybrid carbon of C-X bond has less tendency to release electrons to the halogen than a sp³-hybrid carbon in alkyl halides. Thus halogen atom in aryl halides cannot be easily displaced by nucleophile.

(2) Aryl halides are extremely less reactive towards nucleophilic substitution reactions.
Answer:
Aryl halides are extremely less reactive towards nucleophilic substitution reaction due to the following reasons : (1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with 7r-electrons of the benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.

Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult and are less reactive towards nucleophilic substitution.

(2) Sp² hybrid state of C : Different hybridization state of carbon atom in C-X bond : In aryl halides, the carbon of C-X bond is sp²-hybridized with more 5-character and shorter bond length of 169 pm which requires more energy to break C-X bond. It is difficult to break a shorter bond than a longer bond, in alkyl chloride (bond length 178 pm) therefore, aryl halides are less reactive towards nucleophilic substitution reaction.

(3) Instability of phenyl cation : In aryl halides, the phenyl cation formed due to self ionisation will not be stabilized by resonance which rules out possibility of S_N1 mechanism. Also backside attack of nucleophile is blocked by the aromatic ring which rules out S_N2 mechanism. Thus cations are not formed and hence aryl halides do not undergo nucleophilic substitution reaction easily.

(4) As any halides are electron rich molecules due to the presence of re-bond, they repel electron rich nucleophilic, attack. Hence, aryl halides are less reactive towards nucleophilic substitution reactions. However, the presence of electron withdrawing groups at o/p position activates the halogen of aryl halides towards substitution.

(3) Aryl halides undergo electrophilic substitution reactions slowly.
Answer:
Aryl halides undergo electrophilic substitution reactions slowly and it can be explained as follows :

(1) Inductive effect : The strongly electronegative halogen atom withdraws the electrons from carbon, atom of the ring, hence aryl halides show reactivity towards electrophilic attack.

(2) Resonance effect : The resonating structures of aryl halides show increase in electron density at ortho and para position, hence it is o, p directing.

The inductive effect and resonance effect compete with each other. The inductive effect is stronger than resonance effect. The reactivity of aryl halides is controlled by stronger inductive effect and o, p orientation is controlled by weaker resonating effect.

The attack of electrophile (Y) on haloarenes at ortho and para positions are more stable due to formation of chloronium ion. The chloronium ion formed is comparatively more stable than other hybrid structures of carbonium ion.

(4) Reactions involving Grignard reagent must be carried out under anhydrous condition.
Answer:
(1) Grignard reagent (R Mg X) is an organometallic compound. The carbon-magnesium bond is highly polar and magnesium halogen bond is in ionic in nature. Grignard reagent is highly reactive.

(2) The reactions of Grignard reagent are carried out in dry conditions because traces of moisture may spoil the reaction and Grignard reagent reacts with water to produce alkane. Hence, reactions involving Grignard reagent must be carried out under anhydrous condition.

(5) Alkyl halides are generally not prepared by free radical halogenation of alkane.
Answer:
(1) Free radical halogenation of alkane gives a mixture of all different possible Monohaloalkanes as well as polyhalogen alkanes.
(2) In this method, by changing the quantity of halogen the desired product can be made to predominate over the other
products. Hence, alkyl halides are generally not prepared by free radical halogenation of alkane.

Question ii.
Alkyl halides though polar are immiscible with water.
Answer:
(1) In alkyl halide, the halogen atom is more electronegative than carbon atom, the C – X bond is polar.
(2) Though alkyl halide is polar, it is insoluble in water because alkyl halide is not able to form hydrogen bonds with water. Attraction between alkyl halide molecule is stronger than attraction between alkyl halide and water.

(2) C-F bond in CH₃F is the strongest bond and C-I bond in CH₃I is the weakest bond. Explain.
Answer:
(1) Methyl fluoride (CH₃F) is highly polar molecule and has the shortest C-F bond length (139 pm) and the strongest C-F bond due to greater overlap of orbitals of the same principal quantum number i.e., overlap of 2sp³ orbital of carbon with 2pz orbital of fluorine.
(2) Methyl iodide (CH₃I) is much less polar and has the longest (C-I) bond length (214 pm) and the weakest C-I bond due to poor overlap of 2sp³ orbital carbon with 5pz orbital of iodine i.e., 2sp³ orbital of carbon cannot penetrate into larger p-orbitals.

(3) The boiling point of alkyl iodide is higher than that of alkyl fluoride.
Answer:
For a given alkyl group, the boiling point increases with increasing atomic mass of the halogen, because magnitude of van der Waals force increases with increase in size and mass of halogen. Therefore, boiling point of alkyl iodide is higher than that of alkyl fluoride.

(4) The boiling point of isopropyl bromide is lower than that of it-propyl bromide.
Answer:
For isomeric alkyl halides (isopropyl bromide and n-propyl bromide), the boiling point decreases as the branching increases, surface area decreases on branching and van der Waals forces decrease, therefore, the boiling point of isopropyl bromide is lower than that of n-propyl bromide.

(5) p-Dichlorobenzene has mp. higher than those of o-and rn-isomers.
Answer:
p-Dichlorobenzene has higher melting point than those of o-and m-isomers. This is because of its symmetrical structure which can easily fits in crystal lattice. As a result intermolecular forces of attraction are stronger and therefore greater energy is required to overcome its lattice energy.

Question iii.
Reactions involving Grignard reagent must be carried out under anhydrous conditions.

Question iv.
Alkyl halides are generally not prepared by free radical halogenation of alkanes.
Answer:
(1) Direct fluorination of alkanes is highly exothermic, explosive and invariably leads to polyfluorination and decomposition of the alkanes. It is difficult to control the reaction.
(2) Direct iodination of alkanes is highly reversible and difficult to carry out.
(3) In direct chlorination and bromination, the reaction is not selective. It can lead to different isomeric monohalogenated alkanes (alkyl halides) as well as polyhalogenated alkanes.
Hence, halogenation of alkanes is not a good method of preparation of alkyl halides.

4. Distinguish between – S_N1 and S_N2 mechanism of substitution reaction ?
Answer:

5. Explain – Optical isomerism in 2-chlorobutane.
Answer:
(1) 2-Chlorobutane contains an asymmetric. carbon atom (the starred carbon atom) which is attached to four different groups, i.e., ethyl (-CH₂ – CH₃), methyl (CH₃), chloro (Cl) and hydrogen (H) groups.

(2) Two different arrangements of these groups around the carbon atom are possible as shown in the figure. Hence, it exists as a pair of enanti¬omers. The two enantiomers are mirror images of each other and are not superimposable.

(3) One of the enantiomers will rotate the plane of plane-polarized light to the left hand side and is called the laevorotatory isomer (/-isomer). The other enantiomer will rotate the plane of plane-polarized light to the right hand side and is called the dextrorotatory isomer (d-isomer).

(4) Equimolar mixture of the d- and the 1-isomers is optically inactive and is called the racemic mixture or the racemate (dl-mixture). The optical inactivity of the racemic mixture is due to external compensation.

6. Convert the following.

Question i.
Propene to propan-1-ol
Answer:

Question ii.
Benzyl alcohol to benzyl cyanide
Answer:

Question iii.
Ethanol to propane nitrile
Answer:

Question iv.
But-1-ene to n-butyl iodide
Answer:

Question v.
2-Chloropropane to propan-1-ol
Answer:

Question vi.
tert-Butyl bromide to isobutyl bromide
Answer:

Question vii.
Aniline to chlorobenzene
Answer:

Question viii.
Propene to 1-nitropropane
Answer:

7. Answer the following

Question i.
HCl is added to a hydrocarbon ‘A’ (C₄H₈) to give a compound ‘B’ which on hydrolysis with aqueous alkali forms tertiary alcohol ‘C’ (C₄H10O). Identify ‘A’ , ‘B’ and ‘C’.
Answer:

Question ii.
Complete the following reaction sequences by writing the structural formulae of the organic compounds ‘A’, ‘B’ and ‘C’.

Answer:

Question iii.
Observe the following and answer the questions given below.

a. Name the type of halogen derivative
b. Comment on the bond length of C-X bond in it
c. Can react by S_N1 mechanism? Justify your answer.
Answer:
a. Vinyl halide
b. C – X bond length shorter in vinyl halide than alkyl halide. Vinyl halide has partial double bond character due to resonance.

In vinyl halide, carbon is sp hybridised. The bond is shorter and stronger and the molecule is more stable.

c. Yes, It reacts by S_N1 mechanism. S_N1 mechanism involves formation of carbocation intermediate. The vinylic carbocation intermediate formed is resonance stabilized, hence S_N1 mechanism is favoured.

Activity :
1. Collect detailed information about Freons and their uses.
2. Collect information about DDT as a persistent pesticide.
Reference books
i. Organic chemistry by Morrison, Boyd, Bhattacharjee, 7^th edition, Pearson
ii. Organic chemistry by Finar, Vol 1, 6^th edition, Pearson

12th Chemistry Digest Chapter 9 Halogen Derivatives Intext Questions and Answers

Use your brain power….. (Textbook page 212)

Question 1.
Write IUPAC names of the following:

Answer:

Question 10.1 : (Textbook page 213)

How will you obtain 1.bromo.1-methylcyclohexane from alkene? Write possible structures of alkene and the reaction involved.

Answer:

Use your brain power ….. (Textbook page 213)

Question 1.
Rewrite the following reaction by filling the blanks:

Answer:

Question 10.2 : (Textbook page 216)

Arrange the following compounds in order of increasing boiling points : bromoform, chloromethane, dibromomethane, bromomethane.
Answer:
The comparative boiling points of halogen derivatives are mainly related with van der Waals forces of attraction which depend upon the molecular size. In the present case all the compounds contain only one carbon. Thus the molecular size depends upon the size of halogen and number of halogen atoms present.

Thus increasing order of boiling point is, CH₃CI < CH₃Br < CH₂Br₂ < CHBr₃

Try this ….. (Textbook page 2016)

Question 1.
(1) Make a three-dimensional model of 2-chlorobutane.
(2) Make another model which is a mirror image of the first model.
(3) Try to superimpose the two models on each other.
(4) Do they superimpose on each other exactly ?
(5) Comment on whether the two models are identical or not.
Answer:
(1) (2) and (3)

(4) Two models are non-superimposable mir ror images of each other called enantiomers.

(5) Two enantiomers are identical. Theyhave the same physical properties (such as melting points, boiling points, densities refractive index). They also have identical chemical properties. The magnitude of their optical rotation is equal but the sign of optical rotation is opposite.

Try this ….. (Textbook page 219)

Question 1.
1. Draw structares of enantiomers of lactic acid using Fischer projection formulae.
2. Draw structures of enantiomers of 2-bromobutane using wedge formula.
Answer:
(1)

(2) Wedge formula : 2-brornobutane

Can you tell? (Textbook page 220)

Question 1.
Alkyl halides, when treated with alcoholic solution of silver nitrite, give nitroalkanes whereas with sodium nitrite they give alkyl nitrites. Explain.
Answer:
Nitrite ion is an ambident nucleophile, which can attack through ‘O’ or ‘N’.

Both nitrogen and oxygen are capable of donating electron pair. C – N bond, being stronger than N – O bond, attack occurs through C atom from alkyl halide forming nitroalkane.

However, sodium nitrite (NaNO₂) is an ionic compound and oxygen is free to donate pair of electrons. Hence, attack occurs through oxygen resulting in the formation of alkyl nitrite.

Use your brain power! (Textbook page 222)

Question 1.
Draw the Fischer projection formulae of two products obtained when compound (A) reacts with OHe by S_N1 mechanis.

Answer:

Question 2.
Draw the Fischer projection formula of the product formed when compound (B) reacts with OHΘ by S_N2 mechanism.

Answer:

Question 10.4 : (Textbook page 223)

Allylic and benzylic halides show high reactivity towards the S_N1 mechanism than other primary alkyl halides. Explain.
Answer:
In allylic and benzylic halide, the carbocation formed undergoes stabilization through the resonance. Hence, allylic and benzylic halides show high reactivity towards the S_N1 reaction. The resonating structures are

Resonance stabilization of allylic carbocation

Resonance stabilization of benzylic carbocation

Question 10.5 : (Textbook page 224)

Which of the following two compounds would react faster by SN2 mechanism and Why?

Answer :
In S_N2 mechanism, a pentacoordinate T.S. is involved. The order of reactivity of alkyl halides towards S_N2 mechanism is.
Primary > Secondary > Tertiary, (due to increasing crowding in T.S. from primary to tertiary halides.
1- Chlorobutane being primary halide will react faster by S_N2 mechanism, than the secondary halide 2- chlorobutane.)

Can you tell? (Textbook page 227)

Question 1.
Conversion of chlorobenzene to phenol by aqueous sodium hydroxide requires a high temperature of about 623K and high pressure. Explain.

Answer:
Due to the partial double bond character in chlorobenzene, the bond cleavage in chlorobenzene is difficult and is less reactive. Hence, during the conversion of chlorobenzene to phenol by a question NaOH requires high temperature & high pressure.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 9 Exercise Coordination Compounds Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

Coordination Compounds Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 9 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 9 Exercise Solutions

1. Choose the most correct option.

Question i.
The oxidation state of cobalt ion in the complex [Co(NH₃)₅Br]SO₄ is ……………………….
a. + 2
b. + 3
c. + 1
d. + 4
Answer:
(b) + 3

Question ii.
IUPAC name of the complex [Pt(en)₂(SCN)₂]²⁺ is ………………………
a. bis (ethylenediamine dithiocyanatoplatinum (IV) ion
b. bis (ethylenediamine) dithiocyantoplatinate (IV) ion
c. dicyanatobis (ethylenediamine) platinate IV ion
d. bis (ethylenediammine)dithiocynato platinate (IV) ion
Answer:
(a) bis(ethylenediamine dithiocyanatoplatinum (IV) ion

Question iii.
Formula for the compound sodium hexacynoferrate (III) is
a. [NaFe(CN)₆]
b. Na₂[Fe(CN)₆]
c. Na[Fe(CN)₆]
d. Na₃[Fe(CN)₆]
Answer:
(d) Na₃[Fe(CN)₆]

Question iv.
Which of the following complexes exist as cis and trans isomers?
1. [Cr(NH₂)₂Cl₄]^⊕
2. [Co(NH₃)₅Br]^2⊕
3. [PtCl₂Br₂]^2⊕ (square planar)
4. [FeCl₂(NCS)₂]^2⊕ (tetrahedral)
a. 1 and 3
b. 2 and 3
c. 1 and 3
d. 4 only
Answer:
(a) 1 and 3

Question v.
Which of the following complexes are chiral?
1. [Co(en)₂Cl₂]^⊕
2. [Pt(en)Cl₂]
3. [Cr(C₂O₄)₃]^3⊕
4. [Co(NH₃)4CI₂]^⊕
a. 1 and 3
b. 2 and 3
c. 1 and 4
d. 2 and 4
Answer:
(a) 1 and 3

Question vi.
On the basis of CFT predict the number of unpaired electrons in [CrF₆]³^⊕.
a. 1
b. 2
c. 3
d. 4
Answer:
(c) 3

Question vii.
When an excess of AgNO₃ is added to the complex one mole of AgCl is precipitated. The formula of the complex is ……………..
a. [CoCl2(NH₃)₄]Cl
b. [CoCl(NH₃)₄] Cl₂
c. [CoCl₃(NH₃)₃]
d. [Co(NH₃)₄]Cl₃Answer:
(a) [COCI₃(NH₃)₄]CI

Question viii.
The sum of coordination number and oxidation number of M in [M(en)₂C₂O₄]Cl is
a. 6
b. 7
c. 9
d. 8
Answer:
(c) 9

2. Answer the following in one or two sentences.

Question i.
Write the formula for tetraammineplatinum (II) chloride.
Answer:
Formula of tetraamineplatinum(II) chloride : [Pt(NH₃)₄]CI₂

Table 9.1 : IUPAC names of anionic and neutral ligands

Table 9.2: IUPAC names of anionic complexes

Metal	Name
A1 Cr Cu Co Au(Gold) Fe Pb Mn Mo Ni Zn Ag Sn	Aluminate Chromate Cuprate Cobaltate Aurate Ferrate Plumbate Manganate Molybdate Nickelate Zincate Argentate Stannate

Table 9.3 : IUPAC names of some complexes

Complex	IUPAC name
(i) Anionic complexes :
(a) [Ni(CN)J^2-(b) [Co(C₂0₄)₃]^3-(c) [Fe(CN)₆]^4-	Tetracyanonickelate(II) ion Trioxalatocobaltate(III) ion Hexacyanoferrate(II) ion
(ii) Compounds containing complex anions and metal cations :
(a) Na₃[Co(N0₂)₆] (b) K₃[A1(C₂0₄)₃] (c) Na₃[AIF₆]	Sodium hexanitrocobaltate(III) Potassium trioxalatoaluminate(III) Sodium hexafluoroaluminate(III)
(iii) Cationic complexes :
(a) [Cu(NH₃)₄]²⁺(b) [Fe(H₂0)₅(NCS)]²⁺(c) [Pt(en)₂(SCN)₂]²⁺	Tetraamminecopper(II) ion Pentaaquai sothiocyanatoiron(III) ionBis(ethylenediamine)dithiocyanatoplatinum(IV)
(iv) Compounds containing complex cation and anion :
(a) [PtBr₂(NH₃)₄]Br₂(b) [Co(NH₃)₅C0₃]CI (c) [Co(H₂0)(NH₃)₅]I₃	Tetraamminedibromoplatinum(IV) bromide, Pentaamminecarbonatocobalt(III) chloride, Pentaammineaquacobalt(III) iodide
(v) Neutral complexes :
(a) Co(N0₂)₃(NH₃)₃(b) Fe(CO)₅(c) Rh(NH₃) ₃(SCN) ₃	Triamminetrinitrocobalt(III) Pentacarbonyliron(0) Triamminetrithiocyanatorhodium(III)

Question ii.
Predict whether the [Cr(en)₂(H₂O)₂]³⁺ complex is chiral. Write structure of its enantiomer.
Answer:
(i) Complex is chiral.
(ii) The following are its enantiomers

Question iv.
Name the Lewis acids and bases in the complex [PtCl₂(NH₃)₂].
Answer:
Lewis acid : Pt²⁺
Lewis bases : Cl^– and NF₃

Question v.
What is the shape of a complex in which the coordination number of central metal ion is 4?
Answer:
A complex with the coordination number of central metal ion equal to 4 may be tetrahedral or square planar.

Question vi.
Is the complex [CoF₆] cationic or anionic if the oxidation state of cobalt ion is +3?
Answer:
In the complex, Co carries + 3 charge while 6F^– carry – 6 charge. Hence the net charge on the complex is – 3.
Therefore it is an anionic complex.

Question vii.
Consider the complexes [Cu(NH₃)₄][PtCl₄] and [Pt(NH₃)₄] [CuCl₄]. What type of isomerism these two complexes exhibit?
Answer:
Since in these two given complexes, there is an exchange of ligands between cationic and anionic constituents, they exhibit coordination isomerism.

Question viii.
Mention two applications of coordination compounds.
Answer:
(1) In biology : Several biologically important natural compounds are metal complexes which play an important role in number of processes occurring in plants and animals.

For example, chlorophyll in plants is a complex of Mg²⁺ ions, haemoglobin in blood is a complex of iron, vitamin B₁₂ is a complex of cobalt.

(2) In medicine : The complexes are used on a large scale in medicine. Many medicines in the complex form are more stable, more effective and can be assimilated easily.

For example, platinum complex [Pt(NH₃)₂CI₂] known as cisplatin is effectively used in cancer treatment. EDTA is used to treat poisoning by heavy metals like lead.

(3) To estimate hardness of water :

The hardness of water is due to the presence Mg²⁺ and Ca²⁺ ion in water.
The strong field ligand EDTA forms stable complexes with Mg²⁺ and Ca²⁺. Hence these ions can be removed by adding EDTA to hard water.

Similarly these ions can be selectively estimated due to the difference in their stability constants.

(4) Electroplating : This involves deposition of a metal on the other metal. For smooth plating, it is necessary to supply continuously the metal ions in small amounts.
For this purpose, a solution of a coordination compound is used which dissociates to a very less extent. For example, for uniform and thin plating of silver and gold, the complexes K[Ag(CN)₂] and K[Au(CN)₂] are used.

3. Answer in brief.

Question i.
What are bidentate ligands? Give one example.
Answer:
Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H₂N – (CH₂)₂– NH₂.

Question ii.
What is the coordination number and oxidation state of metal ion in the complex [Pt(NH₃)Cl₅]²?
Answer:
Coordination number = 6
Oxidation state of Pt = +4.

Question iii.
What is the difference between a double salt and a complex? Give an example.
Answer:

Double salt	Coordination compound (complex)
(1) Double salts exist only in the solid state and dissociate into their constituent ions in the aqueous solutions.	(1) Coordination compounds exist in the solid-state as well as in the aqueous or non-aqueous solutions.
(2) Double salts lose their identity in the solution.	(2) They do not lose their identity completely.
(3) The properties of double salts are same as those of their constituents.	(3) The properties of coordination compounds are different from their constituents.
(4) Metal ions in the double salts show their normal valence.	(4) Metal ions in the coordination compounds show two valences namely primary valence and secondary valence satisfied by anions or neutral molecules called ligands.
(5) For example in K₂SO₄. K₂SO₄. A1₂(SO₄)₃. 24H₂O. The ions K⁺, Al^{3 +} and SO₄ show their properties.	(5) In K₄[Fe(CN)₆], ions K⁺ and [Fe(CN)₆]⁴‘~ ions show their properties.

Question iv.
Classify the following complexes as homoleptic and heteroleptic
[Cu(NH₃)₄]SO₄, [Cu(en)₂(H₂O)Cl]^3⊕, [Fe(H₂O)₅(NCS)]^2⊕, tetraammine zinc (II) nitrate.
Answer:
Homoleptic complex :
(a) [Cu(NH₃)₄]SO₄
(d) Tetraaminezinc (II) nitrate : [Zn(NH₃)₄](NO₃)₂

Heteroleptic Complex :
(b) [Cu(en)₂(H₂O)CI]²⁺
(c) [Fe(H₂O)₅(NCS)]²⁺

Question v.
Write formulae of the following complexes
a. Potassium ammine-tri chloroplatinate (II)
b. Dicyanoaurate (I) ion
Answer:
(a) Potassium amminetrichloroplatinate(II) K[Pt(NH₃)CI₃]
(b) Dicyanoaurate (I) ion [AU(CN)₂]^–

Question vi.
What are ionization isomers? Give an example.
Answer:
Ionisation isomers : The coordination compounds having same molecular composition but differ in the compositions of coordination (or inner) sphere and outer sphere and produce different ions on ionisation in the solution are called ionisation isomers. For example, Pentaamminesulphatocobalt (III) bromide [Co(NH₃)₅SO₄] Br, Pentaamminebromocobalt(III) sulphate [Co(NH₃)₅Br] SO₄.

Question vii.
What are the high-spin and low-spin complexes?
Answer:
(1) High spin complex (HS) :

The complex which has greater iwmher of unpaired electrons and hence a higher value of resultant spin and magnetic moment is called high spin (or spin free) or IlS complex.
It is formed with weak field ligands and the complexes have lower values for crystal field splitting energy (CFSE). Δ₀
The paramagnetism of HS complex is larger.

(2) Low spin complex (LS) :

The complex which has the Icasi number of unpaired electrons or all electrons paired and hence the lowest
(or no) resultant spin or magnetic moment is called low spin (or spin paired) or LS complex.
It is formed with strong tickl ligands and the complexes have higher values of crystal field splitting energy (Δ₀).
Low spin complex is diamagnetic or has low paramagnetism.

Table 9.5 : d-orbitai diagrams fir high spin and low spin complexes

(Only the electronic configurations c4 to d1 render the high spin and low spin complexes)

Question viii.
[CoCl₄]^2⊕ is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.
Answer:
₂₇Co [Ar] 3d⁷4s²
Oxidation state of Co = +2 Co²⁺ [Ar] 3d⁷ 4s°
Since CI^– is a weak ligand, there is no pairing of electrons. Since C.N. is 4, there is sp³ hybridisation.

Question ix.
What are strong field and weak field ligands? Give one example of each.
Answer:
The ligands are then classified as (a) strong field and (b) weak field ligands. Strong field ligands are those in which donor atoms are C,N or P. Thus CN^–, NC^–, CO, HN₃, EDTA, en (ethylenediammine) are considered to be strong ligands. They cause larger splitting of d orbitals and pairing of electrons is favoured. These ligands tend to form low spin complexes. Weak field ligands are those in which donor atoms are halogens, oxygen or sulphur.

For example, F^–, CI^–, Br^–, I^–, SCN^–, C₂O₄^2-. In case of these ligands the A0 parameter is smaller compared to the energy required for the pairing of electrons, which is called as electron pairing energy. The ligands then can be arranged in order of their increasing field strength as
I^– < Br^– < CI^– < S^2- < F^– < OH^– < C₂O₄^2- < H₂O < NCS^– < EDTA < NH₃ < en < CN^– < CO.

Question x.
With the help of a crystal field energy-level diagram explain why the complex [Cr(en)₃]^3⊕ is coloured?
Answer:

Since (en) is a strong field ligand there is pairing of electrons. The electrons occupy the t_2g orbitals of lower energy. It has one unpaired electron. Due to d-d transition, it is coloured.

4. Answer the following questions.

Question i.
Give valence bond description for the hybrid orbitals are used by the metal? State the number of unpaired electrons.
Answer:
Since CI^– is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = 2
Type of hybridisation = sp³

Geometry of complex ion = Tetrahedral
The complex ion is paramagnetic.

Question ii.
Draw a qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment. Predict the number of unpaired electrons in the complex [Fe(CN)₆]^4⊕. Is the complex diamagnetic or paramagnetic? Is it coloured? Explain.
Answer:
(A) r-orbital splitting in the octahedral environment :

(B) [Fe (CN)₆]^4- is an octahedral complex.
(C) Since CN^– is a strong ligand, there is pairing of electrons and the complex is diamagnetic.
(D) The complex exists as lemon yellow crystals.
(In the complex all electrons in t_2g are paired and requires high radiation energy for excitation.)

Question iii.
Draw isomers in each of the following
a. [Pt(NH₃)₂ClNO₂]
b. [Ru(NH₃)4Cl₂]
c. [Cr(en₂)Br₂]^⊕Answer:
(a) [Pt(NH₃)₂CINO₂]

(b) [RU(NH₃)₄CI₂]

(c) [Cr(en₂)Br₂]⁺

Question iv.
Draw geometric isomers and enantiomers of the following complexes.
a. [Pt(en)₃]^4⊕
b. [Pt(en)₂ClBr]^2⊕Answer:
The complex [Pt(en)₃]⁴⁺ has two optical isomers.

Question v.
What are ligands? What are their types? Give one example of each type.
Answer:
Ligands : The neutral molecules or negatively charged anions (or rarely positive ions) which are bonded by coordinate bonds to the central metal atom or metal ion in a coordination compound are called ligands or donor groups. For example in [Cu(CN)₄]^2-, four CN^– ions are ligands coordinated to central metal ion Cu²⁺. Ligands can be classified on the basis of number of electron donor atoms in the ligand i.e. denticity.

(1) Monodentate or unidentate ligand : A ligand molecule or an ion which has only one donor atom with a lone pair of electrons forming only one coordinate bond with metal atom or ion in the complex is called monodentate or unidentate ligand. For example NH₃, Cl^–, OH^–, H₂O, etc.

(2) Polydentate or multidentate ligand : A ligand molecule or an ion which has two or more donor atoms with the lone pairs of electrons forming two or more coordinate bonds with the central metal atom or ion in the complex is called polydentate or multidentate ligand. For example, ethylene diamine, H₂N – (CH₂)₂– NH₂.
According to the number of donor atoms they are classified as follows :

Bidentate ligand : This ligand has two donor atoms in the molecule or ion. For example, ethylenediamine, H₂N – (CH₂)₂– NH₂.
Tridentate ligand : This ligand molecule has three donor atoms or three sites of attachment.
E.g. Diethelene triamine, H₂N – CH₂– CH₂– NH – CH₂– CH₂– NH₂. This has three N donor atoms.
Tetradentate (or quadridentate) ligand : This ligand molecule has four donor atoms.
Eg. Triethylene tetraamine which has four N donor atoms.
Hexdentate ligand : This ligand molecule has six donor atoms. E.g. Ethylenediamine tetracetato.

(3) Ambidentate ligand : A ligand molecule or an ion which has two or more donor atoms, however in the formation of a complex, only one donor atom is attached to the metal atom or an ion is called ambidentate ligand. For example, NO^–₂ which has two donor atoms N and O forming a coordinate bond, M ← ONO (nitrito) or M ← NO₂ (nitro).

(4) Bridging ligand : A monodentate ligand having more than one lone pairs of electrons, hence can attach to two or more metal atoms or ions and hence acts as a bridge between different metal atoms is called bridging ligand. For example : OH^–, F^–, SO₄^-2, etc.

Question vi.
What are cationic, anionic and neutral complexes? Give one example of each.
Answer:
(1) Cationic sphere complexes : A positively charged coordination sphere or a coordination compound having a positively charged coordination sphere is called cationic sphere complex.

For example : [Zn(NH₃)₄]²⁺ and [Co(NH₃)₅CI] SO₄ are cationic complexes. The latter has coordination sphere [Co(NH₃)₅CI]²⁺, the anion SO₄²⁺ makes it electrically neutral.

(2) Anionic sphere complexes : A negatively charged coordination sphere or a coordination compound having negatively charged coordination sphere is called anionic sphere complex. For example, [Ni(CN)₄]²⁺ and K₃[Fe(CN)₆] have anionic coordination sphere; [Fe(CN)₆]^3- and three K+ ions make the latter electrically neutral.

(3) Neutral sphere complexes : A neutral coordination complex does not possess cationic or anionic sphere.

[Pt(NH₃)₂CI₂] or [Ni(CO)₄] are neither cation nor anion but are neutral sphere complexes.

Question vii.
How stability of the coordination compounds can be explained in terms of equilibrium constants?
Answer:
Stability of the coordination compounds : The stability of coordination compounds can be explained on the basis of their stability constants. The stability of coordination compounds depends on metal-ligand interactions. In the complex, metal serves as electron-pair acceptor (Lewis acid) while the ligand as Lewis base (since it is electron
donor). The metal-ligand interaction can be realized as the Lewis acid-Lewis base interaction. Stronger the interaction greater is stability of the complex.

Consider the equilibrium for the metal-ligand interaction :
M^a+ + nL^x- ⇌ [ML_n]^a+(-nx)
where a, x, [a + ( – nx)] denote the charge on the metal, ligand and the complex, respectively. Now, the equilibrium constant K is given by

Stability of the complex can be explained in terms of K. Higher the value of K larger is the thermodynamic stability of the complex hence K is called stability constant, and denoted by Kstah. The equilibria for the complex formation with the corresponding K values are given below.

Ag⁺ + 2CN^– ⇌ [Ag(CN)2]^– K = 5.5 x 10¹⁸
Cu²⁺ + 4CN^– ⇌ [CU(CN)4]^2- K = 2.0 x 10²⁷
Co³⁺ + 6NH₃ ⇌ [CO(NH3)6]³⁺ K = 5.0 x 10³³

From the above data, the stability of the complexes is [Co(NH₃)₆]³⁺ > [Cu(CN)₄]^2- > [Ag(CN)₂]^–.

Question viii.
Name the factors governing the equilibrium constants of the coordination compounds.
Answer:
The equilibrium constant of the complex depends on the following factors :

(a) Charge to size ratio of the metal ion : Higher the ratio greater is the stability. For the divalent metal ion complexes their stability shows the trend : Cu²⁺ > Ni²⁺ > Co²⁺ > Fe²⁺ > Mn²⁺ > Cd²⁺. The above stability order is called the Irving-William order. In the above list both Cu and Cd have the charge + 2, however, the ionic radius of Cu2 + is 69 pm and that of Cd2 + is 97 pm. The charge to size ratio of Cu²⁺ is greater than that of Cd²⁺. Therefore the Cu²⁺ forms stable complexes than Cd²⁺.

(b) Nature of the ligand : A second factor that governs stability of the complexes is related to how easily the ligand can donate its lone pair of electrons to the central metal ion that is, the basicity of the ligand. The ligands those are stronger bases tend to form more stable complexes.

Activity :
1. The reaction of chromium metal with H 2SO4 in the absence of air gives blue solution of chromium ion.
Cr(s) + 2H^⊕(aq) → Cr^2⊕(aq) + H₂(s)
Cr^2⊕ forms octahedral complex with H₂O ligands.
a. Write formula of the complex
b. Describe bonding in the complex using CFT and VBT.
Draw crystal field splitting and valence bond orbital diagrams.

2. Reaction of complex [Co(NH₃)₃(NO₂)₃ with HCl gives a complex [Co(NH₃)₃H₂OCl₂]^⊕ in which two chloride ligands are trans to one another.
a. Draw possible stereoisomers of starting material
b. Assuming that NH₃ groups remain in place, which of two starting isomers would give the observed product?

12th Chemistry Digest Chapter 9 Coordination Compounds Intext Questions and Answers

Use your brain power ……. (Textbook page 192)

Question 1.
Draw Lewis structures of the following ligands and identify the donor atom in them :
NH₃, H₂O.
Answer:

Try this ………. (Textbook page 193)

Question 1.
Can you write ionisation of [Ni (NH₃)₆] CI₂?
Answer:
[Ni (NH₃)₆] CI₂ → [Ni(NH₃)₆]²⁺ + 2CI^–

Question 2.
Identify coordination sphere and counter ions.
Answer:
Coordination sphere : [Ni(NH₃)₆]²⁺
Counter ions : CI^–

Can you tell ? (Textbook page 193)

Question 1.
A complex is made of Co (III) and consists of four NH₃ molecules and two CI^– ions as ligands. What is the charge number and formula of complexion?
Answer:
The complex ion has formula, [Co(NH₃)₄CI₂]⁺.
The charge number is + 1.

Use vour brain power ……………… (Textbook page 193)

Question 1.
Coordination number used in coordination of compounds is somewhat different than that used in solid state. Explain.
Answer:

In a coordination compound the coordination number is the number of donor atoms of ligands directly attached to metal atom or ion.
In a solid state, the number of closest constituent atoms or ions in contact with a particular atom in the crystal lattice is called coordination number.
In a coordination compound, coordination number depends upon nature of metal atom or ion, and its electronic configuration.
In a solid state, the coordination number depends upon the crystalline structure of the unit cell.

Can you tell? ………………. (Textbook page 194)

Question 1.
What is the coordination number of
(a) Co in [CoCl₂(en)₂]⁺ = 6
(b) Ir in [Ir(C₂O₄)₂Cl₂]³⁺ and
(c) Pt in [Pt(NO₂)₂(NH₃)₂] ?
Answer:
(a) Coordination number of Co in [CoCl₂(en)₂]⁺ = 6
(b) Coordination number of Ir in [Ir(C₂O₄)₂Cl₂]³⁺ = 6
(c) Coordination number of Pt in [Pt(NO₂)₂(NH₃)₂] = 4

Use your brain power ……… (Textbook page 195)

Question 1.
Classify the complexes as homoleptic and heteroleptic:
(a) [Co (NH₃)₅CI]SO₄,
(b) [CO(ONO)(NH₃)₅]CI₂,
(c) [CoCl(NH₃)(en)₂]²⁺ and
(d) [Cu(C2O₄)₃]^3-
Answer:
Homoleptic Complexes : (d) [Cu(C₂O₄)₃]^3-
Heteroleptic Complexes : (a) [CO(NH₃)₅CI]SO₄
(b) [CO(ONO)(NH₃)₅]CI₂,
(C) [CoCl(NH₃)(en)₂]²⁺

Use your brain power ……… (Textbook page 195)

Question 1.
Classify the complexes as cationic, anionic or Cr(H₂O)₂(C₂O₄)_2^3-, PtCI₂(en)₂ and Cr(CO)₆.
Answer:
Cationic complexes : [CO(NH₃)₆]CI₂
Anionic complexes : Na4[Fe(CN)₆], [Cr(H₂O)₂ (C₂O₄)2]^3-
Neutral complexes : Cr(CO)₆, Pt CI₂(en)₂

Try this ……. (Textbook page 197)

Question 1.
Write the representation of the following :
(i) Tricarbonatocobaltate(III) ion.
(ii) Sodium hexacyanoferrate(III).
(iii) Potassium hexacyafioferrate(II).
(iv) Aquachlorobis(ethylenediamine)cobalt(III).
(v) Tetraaquadichlorochromium(III) chloride.
(vi) Diamminedichloroplatinum(II).
Answer:
(i) [Co(C0₃)₃]^3-
(ii) Na₃[Fe(CN)₆]
(iii) K₄[Fe(CN)₆]
(iv) Co(en)₂(H₂O)(Cl)
(v) [Cr(H₂O)₄CI₂]CI
(vi) Pt(NH₃)₂CI₂

Try this …… (Textbook page 196)

Question 1.
Find out the EAN of
(a) [Zn(NH₃)₄]²⁺
(b) [Fe(CN)₆]⁴⁺
Answer:
(a) For the complex ion, [Zn(NH₃)₄]²⁺ :
Atomic number of Zn = Z = 30
Charge on metal ion = + 2
∴ Number of electrons lost by Zn atom = X = 2 Total number of electrons donated by 4NH^₂3
ligands = Y = 2 x 4 = 8
EAN = Z – X + Y
= 30 – 2 + 8
= 36

(Note : This is atomic number of the nearest inert element 3₆Kr.)

(b) For the complex ion, [Fe(CN)J^4-:
For Fe, Z = 26 (Atomic number)
X = 2 (Due to + 2 charge on Fe)
Y = 12 (Due to 6 CN^– ligands)
∴ EAN = Z – X + Y
= 26 – 2 + 12
= 36

Use your brain power …… (Textbook page 197)

Question 1.
Do the following complexes follow the EAN rule
(a) Cr(CO)₄,
(b) Ni(CO)₄,
(c) Mn(CO)₅,
(d) Fe(CO)₅?
Answer:
(a) Cr(CO)₄ : EAN = Z – X + Y
(b) Ni(C0)4 : EAN = Z – X + Y
= 24 – 0 + 8
= 28 – 0 + 8
= 32
= 36
(c) Mn(CO)₅ : EAN = Z – X + Y
= 25 – 0 + 10
= 35

(d) Fe(CO)5 : EAN = Z – X + Y
= 26 – 0 + 10
= 36

Conclusion :
(a) Cr(CO)₄ and (c) Mn(CO)₅ do not follow EAN Rule.

Try this ….. (Textbook page 199)

Question 1.
Draw structures of ci,c and trans isomers of [Fe(NH₃)₂(CN)₄]
Answer:

Remember ….. (Textbook page 199)

Our hands are non-superimposable mirror images. When you hold your left hand up to a mirror the image looks like right hand.

Try this ….. (Textbook page 199)

Question 1.
Draw enantiomers of [Cr(OX)₂]³ where OX = C₂O4 :
Answer:

Question 2.
Draw (A) enantiomers and (B) cis and trans isomers of [Cr(H₂O)₂(OX)₂] :
Answer:
(A) Enantiomers :

(B) as and trans isomers :

Can you tell ? ….. (Textbook page 200)

Question 1.
Can you write IUPAC names of isomers (I) [Co(NH₃)₅SO₄]Br and (II) [Co(NH₃)₅Br]SO₄?
Answer:

Question 2.
Write linkage isomers of [Fe(H₂O)₅SCN]⁺. Write their IUPAC names.
Answer:

Use your brain power …..(Textbook page 201)

Question 1.
The stability constant K of the [Ag(CN)₂]^– is 5.5 x 10 while that for the corresponding [Ag(NH₃)₂]⁺ is 1.6 x 10⁷. Explain why [Ag(CN)₂]^2- is more stable.
Answer:
Stability constant of [Ag(CN)₂]^2- is larger than that of [Ag(NH₃)₂]⁺ and hence [Ag(CN)₂]^2- is more stable. Also, CN is a stronger ligand than NH₃.

Remember …… (Textbook Page 202)

Question 1.
Complete the missing entries.

(Note : The missing entries are underlined.)

Table 9.3: Type of hybridisation and geometry of a complex

Try this ….. (Textbook page 204)

Question 1.
Based on the VBT predict structure and magnetic behaviour of the [Ni(NH₃)₆]
Answer:
₂₈Ni [Ar] 3d⁸ 4s²
Ni³⁺ [Ar] 3d⁷ 4s°
Hybridisation : sp³d²
Geometry : Octahedral
Magnetic property : Paramagnetic

Try this …… (Textbook page 202)

Question 1.
Give VBT description of bonding in each of following complexes. Predict their magnetic behaviour.
(a) [ZnCI₄]²⁺
(b) [CO(H₂O)₆]^2- (high spin)
(c) [Pt(CN)₄]^2- (square planar)
(d) [CoCI₄]^2- (tetrahedral)
(e) [Cr(NH₃)₆]³⁺

Try this ……. (Textbook page 206)

Question 1.
Sketch qualitatively crystal field d orbital energy level diagrams for each of the following complexes :
(a) [Ni(en)₃]²⁺ (b) [Mn(CN)₆]^3- (c) [Fe(H₂O)₆]²⁺
Predict whether each of the complexes is diamagnetic or paramagnetic.
Answer:
(a) The complex ion, [Ni(en)₃]²⁺ is octahedral.
₂₈Ni [Ar] 3d⁸ 4s²
Ni²⁺ [Ar] 3d⁸ 4s°.

Since en is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t_2g, orbitals
Magnetic moment = \(\mu=\sqrt{n(n+2)}\)
\(=\sqrt{2(2+2)}=2.83 \mathrm{~B} . \mathrm{M} .\)

The complex ion is paramagnetic.

(b) The complex ion [Mn(CN)₆]^3- is octahedral.
₂₅Mn [Ar] 3d⁵ 4s²
Mn³⁺ [Ar] 3d⁴ 4s°

Since CN^– is a strong ligand there is pairing of electrons.
Number of unpaired electrons = n = 2 in t_2g, orbitals
Magnetic moment = \(\mu=\sqrt{n(n+2)}\)
\(=\sqrt{2(2+2)}=2.83 \mathrm{~B} . \mathrm{M}\).

The complex ion is paramagnetic.

Since H₂O is a weak ligand, there is no pairing of electrons.
Number of unpaired electrons = n = 4 in t_2g and e_g orbitals.
Magnetic moment
\(\begin{aligned}
=\mu &=\sqrt{n(n+2)} \\
&=\sqrt{4(4+2)} \\
&=4.90 \mathrm{~B} . \mathrm{M} .
\end{aligned}\)
The complex ion is paramagnetic.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 3 Exercise Ionic Equilibria Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 3 Ionic Equilibria Textbook Exercise Questions and Answers.

Ionic Equilibria Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 3 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 3 Exercise Solutions

1. Choose the most correct answer:

Question i.
The pH of 10^-8 M of HCl is
(a) 8
(b) 7
(c) less than 7
(d) greater than 7
Answer:
(c) less than 7

Question ii.
Which of the following solution will have pH value equal to 1.0?
(a) 50 mL of 0.1M HCl + 50mL of 0.1 M NaOH
(b) 60 mL of 0.1M HCl + 40mL of 0.1 M NaOH
(c) 20 mL of 0.1M HCl + 80mL of 0.1 M NaOH
(d) 75 mL of 0.2M HCl + 25mL of 0.2 M NaOH
Answer:
(d) 75 mL of 0.2M HCl + 25mL of 0.2 M NaOH

Question iii.
Which of the following is a buffer solution ?
(a) CH₃COONa + NaCl in water
(b) CH₃COOH + HCl in water
(c) CH₃COOH + CH₃COONa in water
(d) HCl + NH₄Cl in water
Answer:
(c) CH₃COOH + CH₃COONa in water

Question iv.
The solubility product of a sparingly soluble salt AX is 5.2 x 10^-13. Its solubility in mol dm^-3 is
(a) 7.2 × 10^-7
(b) 1.35 × 10^-4
(c) 7.2 × 10^-8
(d) 13.5 × 10^-8
Answer:
(a) 7.2 × 10^-7

Question v.
Blood in human body is highly buffered at pH of
(a) 7.4
(b) 7.0
(c) 6.9
(d) 8.1
Answer:
(a) 7.4

Question vi.
The conjugate base of [Zn(H₂O)₄]²⁺ is
(a) [Zn(H₂O)₄]²⁺ NH3
(b) [Zn(H₂O)₃]²⁺
(c) [Zn(H₂O)₃OH]⁺
(d) [Zn(H₂O)H]³⁺
Answer:
(c) [Zn(H₂O)₃OH]+

Question vii.
For pH > 7 the hydronium ion concentration would be
(a) 10^-7 M
(b) < 10^-7 M
(c) > 10^-7 M
(d) ≥ 10^-7 M
Answer:
(b) < 10^-7 M

2. Answer the following in one sentence :

Question i.
Why cations are Lewis acids ?
Answer:
Since cations are deficient of electrons they accept a pair of electrons, hence they are Lewis acids.

Question ii.
Why is KCl solution neutral to litmus?
Answer:

Since KCl is a salt of strong base KOH and strong acid HCl, it does not undergo hydrolysis in its aqueous solution.
Due to strong acid and strong base, concentrations [H₃O⁺] = [OH^–] and the solution is neutral.

Question iii.
How are basic buffer solutions prepared?
Answer:

Basic buffer solution is prepared by mixing aqueous solutions of a weak base like NH₄OH and its salt of a strong acid like NH₄Cl.
A weak base is selected according to the required pH or pOH of the solution and dissociation constant of the weak base.

Question iv.
Dissociation constant of acetic acid is 1.8 × 10^-5. Calculate percent dissociation of acetic acid in 0.01 M solution.
Answer:
Given : K_a = 1.8 x 10^-5; C = 0.01 M
Percent dissociation = ?

∴ Percent dissociation = α × 100
= 4.242 × 10^-2 × 10²
= 4.242%
Percent dissociation = 4.242%

Question v.
Write one property of a buffer solution.
Answer:
Properties (or advantages) of a buffer solution :

The pH of a buffer solution is maintained appreciably constant.
By addition of a small amount of an acid or a base pH does not change.
On dilution with water, pH of the solution doesn’t change.

Question vi.
The pH of a solution is 6.06. Calculate its H⁺ ion concentration.

Question vii.
Calculate the pH of 0.01 M sulphuric acid.
Answer:
Given : C = 0.01 M H₂SO₄, pH = ?
\(\mathrm{H}_{2} \mathrm{SO}_{4(\mathrm{aq})} \longrightarrow 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
∴ [H₃O⁺] = 2 × 0.01 = 0.02 M
PH = -log₁₀ [H₃O⁺]
= -log₁₀ 0.02
= –\((\overline{2} .3010)\)
= 2 – 0.3010
= 1.6990
pH = 1.6990.

Question viii.
The dissociation of H₂S is suppressed in the presence of HCl. Name the phenomenon.
Answer:
The weak dibasic acid H₂S is dissociated as follows :

When HCl is added, it increases the concentration of common ion H₃O⁺.
\(\mathrm{HCl}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightarrow \mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence by Le Chaterlier’s principle, the equilibrium is shifted from right to left, suppressing the dissociation of weak electrolyte H₂S.

Question ix.
Why is it necessary to add H₂SO₄ while preparing the solution of CuSO₄?
Answer:
CuSO₄ is a salt of strong acid H₂SO₄ and weak base Cu(OH)₂. CuSO₄ in aqueous solution undergoes hydrolysis and forms a precipitate of Cu(OH)₂ and solution becomes turbid.
CuSO₄ + 2H₂O ⇌ CU(OH)₂↓ + H₂SO₄
OR
CuSO₄ + 4H₂O ⇌ Cu(OH)₂ + 2H₃O⁺ + \(\mathrm{SO}_{4}^{2-}\)
When H₂SO₄ is added, the hydrolysis equilibrium is shifted to left hand side and Cu(OH)₂ dissolves giving clear solution.

Question x.
Classify the following buffers into different types :
a. CH₃COOH + CH₃COONa
b. NH₄OH + NH₄Cl
c. Sodium benzoate + benzoic acid
d. Cu(OH)₂ + CuCl₂
Answer:
(a) Acidic buffer (CH₃COOH + CH₃COONa)
(b) Basic buffer (NH₄OH + NH₄Cl)
(c) Acidic buffer (Sodium benzoate + benzoic acid)
(d) Basic buffer (Cu(OH)₂ + CuCl₂)
[Note : Cu(OH)₂ being insoluble is not used to prepare a buffer solution.]

3. Answer the following in brief :

Question i.
What are acids and bases according to Arrhenius theory ?
Answer:
According to Arrhenius theory :
Acid : It is a substance which contains hydrogen and on dissolving in water produces hydrogen ions (H⁺) E.g. HCl
\(\mathrm{HCl}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)

Base : It is a substance which contains OH group and on dissolving in water produces hydroxyl ions (OH^–). E.g. NaOH
\(\mathrm{NaOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)

Question ii.
What is meant by conjugate acid-base pair?
Answer:
Conjugate acid-base pair : A pair of an acid and a base differing by a proton is called a conjugate acid-base pair.

Question iii.
Label the conjugate acid-base pair in the following reactions
a. HCl + H₂O ⇌ H₃O⁺ + Cl^–
b. \(\mathrm{CO}_{3}^{2-}\) + H₂O ⇌ OH^– + \(\mathrm{HCO}_{3}^{-}\)
Answer:

Question iv.
Write a reaction in which water acts as a base.
Answer:

Since water accepts a proton, it acts as a base.

Question v.
Ammonia serves as a Lewis base whereas AlCl₃ is Lewis acid. Explain.
Answer:

Since ammonia molecule, NH3 has a lone pair of electrons to donate it acts as a Lewis base.
AlCl₃ is a molecule with incomplete octet hence it is electron deficient and acts as a Lewis acid.

Question vi.
Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid.
Answer:
Given : C = 0.1 M; Dissociation = 5%, K_a=2 Percent dissociation

Dissociation constant of acid = K_a = 2.63 × 10^-4

Question vii.
Derive the relation pH + pOH = 14.
Answer:
The ionic product of water, Kw is given by,
Kw = [H₃O⁺] × [OH^–]
At 298 K, K_w = 1 × 10^-14
∴ pK_w = -log₁₀K_w = log₁₀ 1 x 10^-14 = 14
∵ [H₃O⁺] × [OH^–] = 1 × 10^-14
Taking logarithm to base 10 of both sides,
log₁₀ [H₃O⁺] + log₁₀ [OH^–] = log₁₀ 1 x 10^-14
Multiplying both the sides by -1,
-log₁₀ [H₃O⁺] -log₁₀ [OH^–] = -log₁₀ 1 x 10^-14
∵ pH = -log₁₀ [H₃O⁺]; pOH = -log₁₀ [OH^–];
pKw = – log₁₀ K_w
∴ pH + pOH = pK_w
OR pH + pOH =14

Question viii.
Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
Answer:
(A) (i) Sodium carbonate is a salt of weak acid and strong base.
(ii) In aqueous solution it undergoes hydrolysis.

(iii) Strong base dissociates completely while weak acid dissociates partially since [OH^–] > [H₃O⁺], the solution is basic.

(B) (i) Ammonium chloride is a salt of strong acid and weak base.
(ii) In aqueous solution it undergoes hydrolysis

(iii) Since [H⁺] or [H₃O⁺ ] > [OH^–] the solution is acidic.

Question ix.
pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.
Answer:
Given : pH = 3.2; C = 0.02 M; K_a = ?
pH = -log₁₀ [H⁺]
∴ [H⁺] = Antilog – pH
= Antilog – 3.2
= Antilog \(\overline{4} .8\)
= 6.31 × 10^-4M

K_a = cα²
= 0.02 × (0.0315)²
= 1.984 × 10^-5
Dissociation constant = K_a = 1.984 × 10^-5

Question x.
In NaOH solution [OH^–] is 2.87 × 10^-4. Calculate the pH of solution.
Answer:
Given : [OH^–] = 2.87 × 10^-4 M, pH = ?
pOH = -log₁₀ [OH^–]
= -log₁₀ 2.87 × 10^-4
= –\((\overline{4} .4579)\)
= (4 – 0.4579)
= 3.5421
∵ pH + pOH = 14
∴ pH = 14 – pOH = 14 – 3.5421 = 10.4579
pH = 10.4579.

4. Answer the following :

Question i.
Define degree of dissociation. Derive Ostwald’s dilution law for the CH₃COOH.
Answer:
(A) Degree of dissociation :
It is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium.
It is denoted by a and represented by,
α = \(\frac{\text { number of moles dissociated }}{\text { total number of moles of an electrolyte }}\)
Or α = \(\frac{\text { Per cent dissociation }}{100}\)
∴ Per cent dissociation = α × 100

(B) Consider V dm³ of a solution containing one mole of CH₃COOH. Then the concentration of acid is, C = \(\frac{1}{V}\) mol dm³. Let α be the degree of dissociation

This is Ostwald’s dilution law.

Question ii.
Define pH and pOH. Derive relationship between pH and pOH.
Answer:
(1) pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H⁺ is known as the pH of a solution.
pH = -log₁₀[H⁺]

(2) pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH^– is known as the pOH of a solution.
pOH = -log₁₀[OH^–]

Relationship between pH and pOH:
The ionic product of water, Kw is given by,
K_w = [H₃O⁺] × [OH^–]
At 298 K, K_w = 1 × 10^-14
∴ pK_w = -log₁₀K_w = log₁₀ 1 x 10^-14 = 14
∵ [H₃O⁺] × [OH^–] = 1 × 10^-14
Taking logarithm to base 10 of both sides,
log₁₀ [H₃O⁺] + log₁₀ [OH^–] = log₁₀ 1 x 10^-14
Multiplying both the sides by -1,
-log₁₀ [H₃O⁺] – log₁₀ [OH^–] = -log₁₀ 1 x 10^-14
∵ pH = -log₁₀ [H₃O⁺]; pOH = -log₁₀ [OH^–];
pK_w = – log₁₀ K_w
∴ pH + pOH = pK_w
OR pH + pOH =14

Question iii.
What is meant by hydrolysis ? A solution of CH₃COONH₄ is neutral. why ?
Answer:
Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.

A salt of weak acid and weak base for which K_a = K_b:
Consider hydrolysis of CH₃COONH₄.

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^–] and the solution reacts neutral after hydrolysis.

Question iv.
Dissociation of HCN is suppressed by the addition of HCl. Explain.
Answer:
The weak acid HCN is dissociated as follows :
\(\mathrm{HCN}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{CN}_{(\mathrm{aq})}^{-}\)
The dissociation constant K_a is represented as,
K_a = \(\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}\)
When HCl is added, it increases the concentration of H₃O⁺, hence in order to keep the ratio constant, then by Le Chatelier’s principle, the equilibrium is shifted from right to left, suppressing the dissociation of HCN.

Question v.
Derive the relationship between degree of dissociation and dissociation constant in weak electrolytes.
Answer:
Expression of Ostwald’s dilution law in the case of a weak electrolyte : Consider the dissociation of a weak electrolyte BA. Let V dm³ of a solution contain one mole of the electrolyte. Then the concentration of a solution is, C = \(\frac{1}{V}\)mol dm^-3. Let α be the degree of dissociation of the electrolyte.

Applying the law of mass action to this dissociation equilibrium, we have,

As the electrolyte is weak, α is very small as compared to unity, ∴ (1 – α) ≈ 1.

This is the expression of Ostwald’s dilution law. Thus, the degree of dissociation of a weak electrolyte is directly proportional to the square root of the volume of the solution containing 1 mole of an electrolyte.

Question vi.
Sulfides of cation of group II are precipitated in acidic solution (H₂S + HCl) whereas sulfides of cations of group IIIB are precipitated in ammoniacal solution of H₂S. Comment on the relative values of solubility product of sulfides of these.
Answer:
(1) In qualitative analysis, the cations of group II are precipitated as sulphides, namely HgS, CuS, PbS, etc., while cations of group IIIB are precipitated as sulphides, namely, CoS, NiS, ZnS.

(2) The sulphides of group II have extremely low solubility product (K_sp) about 10^-29 to 10^-53 while the sulphides of group IIIB have slightly higher K_sp values about 10^-20 to 10^-23.

(3) In group II, sulphides are precipitated by adding H₂S in acidic solution while in IIIB group they are precipitated in a basic solution like ammonical solution.

(4) In acidic medium due to common ion H⁺, H₂S is dissociated to very less extent but gives sufficient S^2- ion to exceed solubility product of group II sulphides of cations and precipitate them.
\(\mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} ; \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{aq})} \rightleftharpoons 2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{S}_{(\mathrm{aq})}^{2-}\)

(5) In basic medium, H⁺ from H₂S are removed by OH^– in solution, or by NH₄OH, increasing the dissociation of H₂S and concentration of S^2-, so that IP > K_sp.

(6) Therefore group II cations are precipitated in an acidic medium while cations of group IIIB are precipitated in ammonical solution.

Question vii.
Solubility of a sparingly soluble salt get affected in presence of a soluble salt having one common ion. Explain.
Answer:
Consider the solubility equilibrium of a sparingly soluble salt, AgCl.
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
The solubility product, K_sp is given by,
K_sp = [Ag⁺] × [Cl^–]
Consider addition of a strong electrolyte AgNO₃ with a common ion Ag⁺.
\(\mathrm{AgNO}_{3(\mathrm{aq})} \longrightarrow \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{3(\mathrm{aq})}^{-}\)
The concentration Ag⁺ in the solution is increased, hence by Le Chatelier’s principle the equilibrium of AgCl is shifted to left hand side since the value of K_sp is constant.
Thus in the presence of a common ion, the solubility of a sparingly soluble salt is suppressed.

Question viii.
The pH of rain water collected in a certain region of Maharashtra on particular day was 5.1. Calculate the H₃O⁺ ion concentration of the rain water and its percent dissociation.
Answer:
Given : pH = 5.1, [H₃O⁺] = ?
PH = -log₁₀ [H₃O⁺]
∴ log₁₀ [H₃O⁺] = -pH
∴ [H₃O⁺] = Antilog – pH
= Antilog – 5.1
= Antilog \(\overline{6} .9\)
= 7.943 × 10^-6 M
(H₃O⁺ in rainwater is due to dissolved gases, CO₂, SO₂, etc. forming acids which dissociate giving H₃O⁺ and acidity to rainwater.)
[H₃O⁺] = 7.943 × 10^-4M

Question ix.
Explain the relation between ionic product and solubility product to predict whether a precipitate will form when two solutions are mixed?
Answer:
If ionic product and solubility product are indicated by IP and K_sp respectively then,
(I) When IP = K_sp, the solution is saturated.
(II) When IP > K_sp, the solution is supersaturated and hence precipitation will occur, when two solutions are mixed.
(Ill) When IP < K_sp, the solution is unsaturated and precipitation will not occur, when two solutions are mixed.

12th Chemistry Digest Chapter 3 Ionic Equilibria Intext Questions and Answers

Use your brain power (Textbook Page No. 47)

Question 1.
Which of the following is a strong electrolyte ?
HF, AgCl, CuSO₄, CH₃COONH₄, H₃PO₄.
Answer:
CH₃COONH₄ is a strong electrolyte since in aqueous solution it dissociates completely. Sparingly soluble salts AgCl, CuSO₄ are also strong electrolytes.

Use your brain power (Textbook Page No. 49)

Question 1.
All Bronsted bases are also Lewis bases, but all Bronsted acids are not Lewis acids. Explain.
Answer:
NH₃ is a Bronsted base since it can accept a proton while it is also a Lewis base since it has a lone pair of electrons to donate.

(2) HCl is a Bronsted acid since it can donate a proton but it is not a Lewis acid since it can’t accept a pair of electrons.

Use your brain power (Textbook Page No. 53)

Question 1.
Suppose that pH of monobasic and dibasic acid is the same. Does this mean that the molar concentrations of both acids are identical ?
Answer:
Even if monobasic acid and dibasic acid give same pH, their molar concentrations are different. One mole of monobasic acid like HCl gives 1 mol of H⁺ while one mole of dibasic acid gives 2 mol of H⁺ in solution. Hence the concentration of dibasic acid will be half of the concentration of monobasic acid. For example, for same pH. [Monobasic acid] = [Dibasic acid]/2

Question 2.
How does pH of pure water vary with temperature ? Explain.
Answer:
Since the increase in temperature, increases the dissociation of water, its pH decreases.

Can you tell ? (Textbook Page No. 54)

Question 1.
Why (i) an aqueous solution of NH₄Cl is acidic.
(ii) while that of HCOOK basic ?
Answer:
(i) (i) Ammonium chloride is a salt of strong acid and weak base.
(ii) In aqueous solution it undergoes hydrolysis

(iii) Since [H⁺] or [H₃O⁺] > [OH^–] the solution is acidic.

(ii) HCOOK is a salt of weak acid HCOOH and strong base KOH. In aqueous solution it undergoes hydrolysis giving weak acid and strong base KOH which dissociates completely,

∴ [OH^–] > [H₃O⁺], and the solution reacts basic.

Can you think ? (Textbook Page No. 56)

Question 1.
Home made jams and jellies without any added chemical preservative additives spoil in a few days whereas commercial jams and jellies have a long shelf life. Explain. What role does added sodium benzoate play ?
Answer:
Sodium benzoate added to jams and jellies in commercial products maintains the pH constant and acts as a preservative. Hence jams and jellies are not spoiled for a very long time unlike homemade products.

Can you tell ? (Textbook Page No. 56)

Question 1.
It is enough to add a few mL of a buffer solution to maintain its pH. Which property of buffer is used here ?
Answer:
The important property of reserve acidity and reserve basicity of a buffer solution is used to maintain constant pH. Weak acid or weak base along with ions (cations or anions) from salt react with excess of added acid (H⁺) or base [OH^–] and maintain pH constant.

Use your brain power (Textbook Page No. 59)

Question 1.
What is the relationship between molar solubility and solubility product for salts given below : (i) Ag₂CrO₄ (ii) Ca₃(PO₄)₂ (iii) Cr(OH)₃.
Answer:

Can you tell ? (Textbook Page No. 60)

Question 1.
How is the ionization of NH₄OH suppressed by addition of NH₄Cl to the solution of NH₄OH ?
Answer:
Ionisation of NH₄OH is represented as follows :
\(\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
It has ionisation constant,
K_b = \(\frac{\left[\mathrm{NH}^{4+}\right] \times\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{4} \mathrm{OH}\right]}\)
K_b has constant value at constant temperature. When strong electrolyte NH₄Cl is added to its solution, it dissociates completely.
\(\mathrm{NH}_{4} \mathrm{Cl}_{(\mathrm{aq})} \longrightarrow \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Due to common ion \(\mathrm{NH}_{4}^{+}\), by Le Chatelier’s principle, the equilibrium is shifted from right to left, suppressing the ionisation of NH₄OH.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 2 Exercise Solutions Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 2 Solutions Textbook Exercise Questions and Answers.

Solutions Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 2 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 2 Exercise Solutions

1. Choose the most correct answer.

Question i.
The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
Answer:
(d) 12 mm Hg

Question ii.
The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
Answer:
(c) osmotic pressure

Question iii.
In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass per cent
Answer:
(a) molarity

Question iv.
Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1 m
(b) 1 M
(c) 1 mass%
(d) 1 mole fraction of solute
Answer:
(a) 1 m

Question v.
Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
Answer:
(a) nature of solvent

Question vi.
Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
Answer:
(b) boiling point of solvent is lower than that of solution

Question vii.
A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer:
(c) 4.92 atm

Question viii.
The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
(a) 5.41%
(b) 3.54%
(c) 4.53%
(d) 53.4%
Answer:
(a) 5.41%

Question ix.
Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
Answer:
(d) directly proportional to the mole fraction of the solvent

Question x.
Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature
Answer:
(a) boiling point of water involved in cooking is increased

Question xi.
Henry’s law constant for a gas CH₃Br is 0.159 mol dm^-3 atm at 250°C. What is the solubility of CH₃Br in water at 25 °C and a partial pressure of 0.164 atm?
(a) 0.0159 mol L^-1
(b) 0.164 mol L^-1
(c) 0.026 M
(d) 0.042 M
Answer:
(c) 0.026 M

Question xii.
Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
Answer:
(c) they are isotonic solutions

2. Answer the following in one or two sentences

Question i.
What is osmotic pressure ?
Answer:
(1) Definition : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.

Osmosis and osmotic pressure

(2) Explanation : Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened. It is filled with the experimental solution and immersed in a solvent like water. As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel. This increases the hydrostatic pressure. At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Question ii.
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why ?
Answer:

While calculating osmotic pressure by equation, π = CRT, the concentration is expressed in molarity but not in molality.
This is because the measurements of osmotic pressure are made at a certain constant temperature.
Molarity depends upon temperature but molality is independent of temperature.
Hence in osmotic pressure measurements, concentration is expressed in molarity.

Question iii.
Write the equation relating boiling point elevation to the concentration of solution.
Answer:
The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Hence, if ΔT_b is the elevation in the boiling point of a solution of molal concentration m then,
ΔT_b ∝ m
∴ ΔT_b = K_b m
where K_b is a proportionality constant.
If m = 1 molal,
ΔT_b = K_b
K_b is called the ebullioscopic constant or molal elevation constant. K_b is characteristic of the solvent.

Question iv.
A 0.1 m solution of K₂SO₄ in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_f for water is 1.86 K kg mol^-1?
Answer:
Given : m = 0.1 m, ΔT_f = 0 – (-0.43) = 0.43 °C
K_f = 1.86 K kg mol^-1, i = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{0.43}{1.86 \times 0.1}\) = 2.312
van’t Hoff factor = i = 2.312

Question v.
What is van’t Hoff factor?
Answer:
Definition of the van’t Hoff factor, i : It is defined as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.

The van’t Hoff factor can be represented as,

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

When the solute neither undergoes dissociation or association in the solution, then, i = 1
When the solute undergoes dissociation in the solution, then, i > 1
When the solute undergoes association in the solution, then i < 1

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

Question vi.
How is van’t Hoff factor related to degree of ionization?
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question vii.
Which of the following solutions will have higher freezing point depression and why ?
a. 0.1 m NaCl b. 0.05 m Al₂(SO₄)₃
Answer:
(1) Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
(2) More the number of particles in the solution, higher is the depression in freezing point.
(3) The number of particles (ions) from electrolytes are,

(4) Therefore Al₂(SO₄)₃ solution will have higher freezing point depression.

Question viii.
State Raoult’s law for a solution containing a nonvolatile solute.
Answer:
Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

Question ix.
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm³ of water? Why?
Answer:

The boiling point of water (or any liquid) depends on its vapour pressure.
Higher the vapour pressure, lower is the boiling point.
When 1 mole of volatile methyl alcohol is added to 1 dm³ of water, its vapour pressure is increased decreasing the boiling point of water.

Question x.
Which of the four colligative properties is most often used for molecular mass determination? Why?
Answer:

Since osmotic pressure has large values, it can be measured more precisely.
The osmotic pressure can be measured at a suitable constant temperature.
The molecular masses can be measured more accurately.
Therefore, it is more useful to determine molecular masses of expensive substances by osmotic pressure.

3. Answer the following.

Question i.
How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀ and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b = T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀ – P, where P₀ and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b ∝ (P₀ – P) or ΔT_b ∝ ΔP]

Question ii.
What are isotonic and hypertonic solutions?
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_A and n_B moles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A = \(\frac{n_{\mathrm{A}}}{V}\) (in mol dm^-3) and
C_B = \(\frac{n_{\mathrm{B}}}{V}\) (in mol dm^-3)

If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A = C_ART and π_B = C_BRT, where π_A and π_B are their osmotic pressures.
For the isotonic solutions,
π_A = π_B
∴ C_A = C_B
∴ \(\frac{n_{\mathrm{A}}}{V}=\frac{n_{\mathrm{B}}}{V}\)
∴ n_A = n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation : Consider two solutions of substances A and B having osmotic pressures π_A and π_B. If π_B is greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_A and C_B are their concentrations, then C_B > C_A. Hence, for equal volume of the solutions, n_B > n_A.

Question iii.
A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
Answer:

When a solvent and a solution containing a non-volatile solute are separated by a semipermeable membrane, there arises a flow of solvent molecules from solvent to solution as well as from solution to solvent.
Due to higher vapour pressure of solvent than solution, the rate of flow of solvent molecules from solvent to solution is higher.
As more and more solvent passes into solution due to osmosis, the solvent content increases, and the rate of backward flow increases.
At a certain stage an equilibrium is reached where both the opposing rates become equal attaining an equilibrium.

Question iv.
The osmotic pressure of CaCl₂ and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate van’t Hoff factor for CaCl₂.
Answer:
Given : π_Cacl2 = 0.605 atm;
π_Urea = 0.245 atm
For urea solution, van’t Hoff factor, i = 1
π_Cacl2 = i × (CRT)_Cacl2
π_Urea = (CRT)_Urea

van’t Hoff factor = i = 2.47

Question v.
Explain reverse osmosis.
Answer:
Reverse osmosis : The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like seawater into pure water through a semipermeable membrane is called reverse osmosis.

The osmotic pressure of seawater is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure water from seawater enters the other side of pure water.

Purification of seawater by reverse osmosis

For this purpose of suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.
This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Question vi.
How molar mass of a solute is determined by osmotic pressure measurement?
Answer:
Consider V dm³ (litres) of a solution containing W₂ mass of a solute of molar mass M₂ at a temperature T.
Number of moles of solute, n₂ = \(\frac{W_{2}}{M_{2}}\)
The osmotic pressure π is given by,
π = \(\frac{W_{2} R T}{M_{2} V}\)
∴ M₂ = \(\frac{W_{2} R T}{\pi V}\)
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.
Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances.

Question vii.
Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
Answer:
Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.

If P₀ is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀ – P) is the lowering of the vapour pressure.

Question viii.
Using Raoult’s law, how will you show that ∆P = \(\boldsymbol{P}_{1}^{0}\)x₂ ? Where x₂ is the mole fraction of solute in the solution and \(\boldsymbol{P}_{1}^{0}\) vapour pressure of pure solvent.
Answer:
If x₁ and x₂ are the mole fractions of solvent and solute respectively, then
x₁ + x₂
By Raoult’s law,
P = x₁ × P₀
where P⁰ is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.

Question ix.
While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
Answer:

Boiling point elevation and freezing point depression involve temperature changes, (ΔT_b and ΔT_f).
Since molarity depends on temperature but molality is independent of temperature we use molality and not molarity in considering boiling point elevation and freezing point depression.

Question 4.
Derive the relationship between degree of dissociation of an electrolyte and van’t Hoff factor.
Answer:
Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y+ + yB^x-
For 1 mole of electrolyte : 1 – α, xα, yα and
For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 – xα + yα – α]
= m[1 + α(x + y – 1)]
The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = \(\frac{i-1}{n-1}\) ……..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question 5.
What is effect of temperature on solubility of solids in water? Give examples.
Answer:
The solubility of a solid solute depends upon temperature.

Variation of solubilities of some ionic solids with temperature

Generally rise in temperature increases the solubility. This is due to expansion of holes or empty spaces in the liquid solvent. Generally 10 °C rise in temperature, increases the solubility of solids two fold.
Dissolution process may be endothermic or exothermic.
The solubility of the substances like NaBr, NaCl, KCl, etc. changes slightly with the increase in temperature.
The solubility of the salts like NaNO₃, KNO₃, KBr, etc. increases appreciably with the increase in temperature.
The solubility of Na₂SO₄ first increases and after 30 °C decreases with the increase in temperature.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional crystallisation.

Question 6.
Obtain the relationship between freezing point depression of a solution containing nonvolatile nonelectrolyte and its molar mass.
Answer:
The freezing point depression, ΔT_f of a solution is directly proportional to molality (m) of the solution.
∴ ΔT_f ∝ m
∴ ΔT_f = K_f m
where Kf is a molal depression constant.
The molality of a solution is given by,

If W₁ grams of a solvent contain W₂ grams of a solute of the molar mass M₂, then the molality m of the solution is given by,

If the weights are expressed in kg then,
ΔT_f = K_f × \(\frac{W_{2}}{W_{1} M_{2}}\)
The unit of K_f is K kg mol^-1
Hence, from the measurement of the depression in the freezing point of the solution, the molar mass of the substance can be determined.

Question 7.
Explain with diagram the boiling point elevation in terms of vapour pressure lowering.
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³ Nm^-2).

Question 8.
Fish generally needs O₂ concentration in water at least 3.8 mg/L for survival. What partial pressure of O₂ above the water is needed for the survival of fish? Given the solubility of O₂ in water at 0 °C and 1 atm partial pressure is 2.2 × 10^-3 mol/L (0.054 atm)
Answer:
Given : Required concentration of O₂
= 3.8 mg/L
= \(\frac{3.8 \times 10^{-3}}{32} \mathrm{~mol} \mathrm{~L}^{-1}\)
Solubility of O₂ = 2.2 × 10^-3 mol L^-1
P = 1 atm
Partial pressure of O₂ needed = Po₂ = ?

Pressure needed = Po₂ = 0.05397 atm.

Question 9.
The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of solution containing 2.8 g urea in 50 g of water? (16.17 mm Hg)
Answer:
Given : Vapour pressure of pure solvent (water) = P0
= 17 mm Hg
Weight of solvent = W₁ = 50 g
Weight of solute (urea) = 2.8 g
Molecular weight of a solvent = M₁ = 18
Molecular weight of a solute (urea) = M₂
= 60 g mol^-1
\(\frac{P_{0}-P}{P_{0}}=\frac{W_{2} \times M_{1}}{W_{1} \times M_{2}}\)
∴ \(\frac{17-P}{17}=\frac{2.8 \times 18}{50 \times 60}\) = 0.0168
∴ 17 – P = 17 × 0.0168
17 – P = 0.2856
∴ P= 17 – 0.2856
= 16.7144 mm Hg
Vapour pressure of solution = 16.7144 mm Hg

Question 10.
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) has freezing point of 271K. Calculate the freezing point of 5% aqueous glucose solution.
Answer:
Given : W₂ = 5 g cane sugar; W₁ = 100 – 5 = 95 g
M₂ = 342 g mol^-1; T_f₁ = 271 K;
ΔT_f₁ = 273 – 271 = 2 K; T_f = ?
W₂ = 5 g glucose, W’₁ = 100 – 5 = 95 g,
M’₂ = 180 g mol-1, ΔT_f₂ = ?

= 12.996 K kg mol^-1
≅ 13 K kg mol^-1

∴ Freezing point of solution = T_f
= 273 – 3.801 = 269.2 K
Freezing point of solution = 269.2 K.

Question 11.
A solution of citric acid C₆H₈O₇ in 50 g of acetic acid has a boiling point elevation of 1.76 K. If K_b for acetic acid is 3.07 K kg mol^-1, what is the molality of solution?
Answer:
Given : W₁ = 50 g acetic acid
ΔT_b = 1.76 K
K_b = 3.07 K kg mol^-1
m = ?
ΔT_b = K_b × m
∴ m = \(\frac{\Delta T_{\mathrm{b}}}{K_{\mathrm{b}}}\)
= \(\frac{1.76}{3.07}\)
= 0.5733 m
Molality of solution = 0.5733 m

Question 12.
An aqueous solution of a certain organic compound has a density of 1.063 gmL^-1, an osmotic pressure of 12.16 atm at 25°C and a freezing point of -1.03°C. What is the molar mass of the compound? (334 g/mol)

Question 13.
A mixture of benzene and toluene contains 30% by mass of toluene. At 30°C, vapour pressure of pure toluene is 36.7 mm Hg and that of pure benzene is 118.2 mm Hg. Assuming that the two liquids form ideal solutions, calculate the total pressure and partial pressure of each constituent above the solution at 30°C.
Answer:
Given : 30% by mass of toluene (T) and 70% by mass of benzene (B).
W_T = 30 g; W_B = 70g
\(P_{\mathrm{T}}^{0}\) = 36.7 mm Hg; \(P_{\mathrm{B}}^{0}\) = 118.2 mm Hg
M_T = 92 g mol^-1; MB = 78 g mol^-1
P_T = ?, P_B = ?, P_soln = ?

Total number of moles = n_Total = n_T + n_B
= 0.326 + 0.8974
= 1.2234 mol

Mole fractions :
x_T = \(\frac{n_{\mathrm{T}}}{n_{\text {Total }}}=\frac{0.326}{1.2234}\) = 0.2665
x_B = 1 – 0.2665 = 0.7335
Psoln = x_T + \(P_{\mathrm{T}}^{0}\) + x_B × \(P_{\mathrm{B}}^{0}\)
= 0.2665 × 36.7 + 0.7335 × 118.2
= 9.780 + 86.7
= 96.48 mm Hg

Partial pressures :
P_T = x_T × P_soln
= 0.2665 × 96.48
= 25.71 mm Hg
P_B = x_B × P_soln
= 0.7335 × 96.48
= 70.77 mm Hg
Total pressure P_soln = 96.48 mm Hg
Partial pressures : P_Toluene = 25.71 mm Hg
P_Benzene = 70.77 mm Hg

Question 14.
At 25 °C a 0.1 molal solution of CH₃COOH is 1.35% dissociated in an aqueous solution. Calculate freezing point and osmotic pressure of the solution assuming molality and molarity to be identical.
Answer:
Given : T = 273 + 25 = 298 K
C = 0.1 m ≅ 0.1 M; K_f = 1.86 K kg mol^-1
Per cent dissociation = 1.35
Freezing point = t_f = ?
π = ?
α = \(\frac{1.35}{100}\) = 0.0135

i = 1 – α + α + α = 1 + α = 1 + 0.0135 = 1.0135
(i) ΔT_f = i × K_f × m
= 1.0135 × 1.86 × 0.1
= 0.1885 °C
∴ Freezing point of solution = 0 – 0.1885
= – 0.1885 °C

(ii) n = iCRT
= 1.035 × 0.1 × 0.08206 × 298
= 2.53 atm

(i) Freezing point of solution = – 0.1885 °C
(ii) Osmotic pressure = π = 2.53 atm

Question 15.
A 0.15 m aqueous solution of KCl freezes at -0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume volume of solution equal to that of water.
Answer:
Given : c = 0.15 m KCl ≅ 0.15 M KCl
ΔT_f = 0 – T_f = 0 – (-0.510) = 0.510 °C
T = 273 K; K_f = 1.86 K kg mol^-1
i = ?; π = ?
ΔT_f = i × K_f × m
∴ i = \(\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}\)
= \(\frac{0.510}{1.86 \times 0.15}\)
= 1.828
π = iCRT
= 1.828 × 0.15 × 0.08206 × 273
= 6.143 atm
i = 1.828, Osmotic pressure = π = 6.143 atm

12th Chemistry Digest Chapter 2 Solutions Intext Questions and Answers

Can you tell ? (Textbook Page No. 29)

Question 1.
Why naphthalene dissolves in benzene but not in water ?
Answer:
Since naphthalene is a covalent nonpolar substance, it is soluble in a nonpolar solvent like benzene but insoluble in polar solvent like water.

Question 2.
Anhydrous sodium sulphate dissolves in water with the evolution of heat. What is the effect of temperature on its solubility ?
Answer:
Since the dissolution of anhydrous sodium sulphate in water is an exothermic process due to evolution of heat, according to Le Chatelier’s principle its solubility decreases with the increase in temperature.

(Textbook Page No. 42)

Question 1.
If 1.25 m sucrose solution has ΔT_f of 2.32 °C, what will be the expected value of ΔT_f for 1.25 m CaCl₂ solution?
Answer:
Sucrose being nonelectrolyte, it has i = 1 but for CaCl₂,
(CaCl₂ → Ca²⁺ + 2Cl^–) the value of i = 3.
Hence
ΔT_f = i × 2.32
= 3 × 2.32
= 6.92 °C
∴ ΔT_f = 6.92 °C.

(Textbook Page No. 44)

Question 1.
Which of the following solutions will have maximum boiling point elevation and which have minimum freezing point depression assuming the complete dissociation? (a) 0.1m KCl (b) 0.05 m NaCl (c) 1 m AlPO₄ (d) 0.1 m MgSO₄.
Solution :
Boiling point elevation and freezing point depression are colligative properties that depend on number of particles in solution. The solution having more number of particles will have large boiling point elevation and that having less number of particles would show minimum freezing point depression.

AlPO₄ solution contains highest moles and hence highest number particles and in turn, the maximum ΔT_b. NaCl solution has minimum moles and particles, it has minimum ΔT_f.

Question 2.
Arrange the following solutions in order of increasing osmotic pressure. Assume complete ionization. (a) 0.5 m Li₂SO₄ (b) 0.5 m KCl (c) 0.5 m Al₂(SO₄)₃ (d) 0.1 m BaCl₂.
Answer:
Consider 1 dm³ of each solution.

Osmotic pressure being a colligative property, it depends on number of particles in the solution.
Therefore, increasing order of osmotic pressure is,

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 1 Exercise Solid State Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 1 Solid State Textbook Exercise Questions and Answers.

Solid State Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 1 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 1 Exercise Solutions

1. Choose the most correct answer.

Question i.
Molecular solids are
(a) crystalline solids
(b) amorphous solids
(c) ionic solids
(d) metallic solids
Answer:
(b) amorphous solids

Question ii.
Which of the following is an n-type semiconductor?
(a) Pure Si
(b) Si-doped with As
(c) Si-doped with Ga
(d) Ge doped with In
Answer:
(b) Si-doped with As

Question iii.
In Frenkel defect
(a) electrical neutrality of the substance is changed.
(b) density of the substance is changed.
(c) both cation and anion are missing
(d) overall electrical neutrality is preserved
Answer:
(d) overall electrical neutrality is preserved

Question iv.
In crystal lattice formed by bcc unit cell the void volume is
(a) 68%
(b) 74%
(c) 32%
(d) 26%
Answer:
(c) 32%

Question v.
The coordination number of atoms in bcc crystal lattice is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question vi.
Which of the following is not correct ?
(a) Four spheres are involved in the formation of tetrahedral void.
(b) The centres of spheres in octahedral voids are at the a pices of a regular tetrahedron.
(c) If the number of atoms is N the number of octahedral voids is 2N.
(d) If the number of atoms is N/2, the number of tetrahedral voids is N.
Answer:
(c) If the number of atoms is N the number of octahedral voids is 2N.

Question vii.
A compound forms hcp structure. Number of octahedral and tetrahedral voids in 0.5 mole of substance is respectively
(a) 3.011 × 10²³, 6.022 × 10²³
(b) 6.022 × 10²³, 3.011 × 10²³
(c) 4.011 × 10²³, 2.011 × 10²³
(d) 6.011 × 10²³, 12.022 × 10²³
Answer:
(a) 3.011 × 10²³, 6.022 × 10²³

Question viii.
Pb has fcc structure with edge length of unit cell 495 pm. Radius of Pb atom is
(a) 205 pm
(b) 185 pm
(c) 260 pm
(d) 175 pm
Answer:
(d) 175 pm

2. Answer the following in one or two sentences.

Question i.
What are the types of particles in each of the four main classes of crystalline solids?
Answer:
The smallest constituents or particles of various solids are atoms, ions or molecules.

Question ii.
Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use ?
Answer:
fcc has the most efficient packing of particles while scc has the least efficient packing.

Question iii.
The following pictures show population of bands for materials having different electrical properties. Classify them as insulator, semiconductor or a metal.

Answer:
Picture A represents metal conductor,
Picture B represents insulator,
Picture C represents semiconductor.

Question iv.
What is a unit cell?
Answer:

Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice) which when repeated in different directions produces the crystalline solid (lattice).
The crystal is considered to consist of an infinite number of unit cells.
The unit cell possesses all the characteristics of the crystalline solid.

Question v.
How does electrical conductivity of a semiconductor change with temperature ? Why?
Answer:

Since the energy difference between valence band and conduction band in semiconductor is not large, the electrons from valence band can be promoted to conduction by heating.
Hence electrical conductivity of a semiconductor increases with temperature.

Question vi.
The picture represents bands of MOs for Si. Label valence band, conduction band and band gap.

Answer:

Question vii.
A solid is hard, brittle and electrically non-conductor. Its melt conducts electricity. What type of solid is it?
Answer:
A solid crystalline electrolyte like NaCl is hard, brittle and electrically nonconductor. But its melt conducts electricity.

Question viii.
Mention two properties that are common to both hep and ccp lattices.
Answer:
In hcp and ccp crystal lattices coordination number is 12 and packing efficiency is 74%.

Question ix.
Sketch a tetrahedral void.
Answer:

Question x.
What are ferromagnetic substances?
Answer:

The substances which possess unpaired electrons and high paramagnetic character and when placed in a magnetic field are strongly attracted and show permanent magnetic moment even when the external magnetic field is removed are said to be ferromagnetic. They can be permanently magnetised.
In the solid state, the metal ions of ferromagnetic substance are grouped together into small regions called domains, where each domain acts as a tiny magnet.

For example : Fe, Co, Gd, Ni, CrO₂, etc.

3. Answer the following in brief.

Question i.
What are valence band and conduction band?
Answer:
There are two types of bands of molecular orbitals as follows :

Valence band : The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
Conduction band : Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct, heat and electricity.

Question ii.
Distinguish between ionic solids and molecular solids.
Answer:

Type/ Property	Ionic solids	Molecular solids
1. Particles of unit cell	Cations and anions	Monoatomic or polyatomic molecules
2. Interparticle forces	Electrostatic	London, dipole-dipole forces and/or hydrogen bonds
3. Hardness	Hard and brittle	Soft
4. Melting points	High 600 °C to 3000 °C	Low (-272 °C to 400 °C)
5. Thermal and electrical conductivity	Poor electrical conductors in solid state. Good conductors when melted or dissolved in water.	Poor conductor of heat and electricity
6. Examples	NaCl, CaF₂	ice, benzoic acid

Question iii.
Calculate the number of atoms in fcc unit cell.
Answer:
Number of atoms in face-centred cubic (fcc) unit cell :
In this unit cell, there are 8 atoms at 8 corners and 6 atoms at 6 face centres. Each corner contributes 1/8th atom to the unit cell, hence due to 8 corners,
Number of atoms = \(\frac {1}{8}\) × 8 = 1.
Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres,
Number of atoms = \(\frac {1}{2}\) × 6 = 3.
∴ Total number of atoms present in fee unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.

Face centered unit cell

Question iv.
How are the spheres arranged in first layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer ?
Answer:
(i) Stacking of square close packed layers :

Stacking of square close packed layers

In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type.

Each sphere is in contact with six surrounded spheres, hence the coordination number of each sphere is six.

(ii) Stacking of two hexagonal close packed layers :
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner.

In this the spheres of second layer are placed in the depression of the first layer.
In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently.

Two layers of closed packed spheres

In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers.

The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.

Question v.
Calculate the packing efficiency of metal crystal that has simple cubic structure.
Answer:
Step 1 : Radius of sphere : In simple cubic lattice, the atoms (spheres) are present at eight corners and in contact along the edge in the unit cell.
If ‘a’ is the edge length of the unit cell and ‘r’ is the radius of the atom, then
a = 2r or r = a/2

scc structure

Step 2 : Volume of sphere :
Volume of one particle = \(\frac{4 \pi}{3}\) × r³
= \(\frac{4 \pi}{3}\) × (a/2)³ = \(\frac{\pi a^{3}}{6}\)

Step 3 : Total volume of particles : Since the unit cell contains one particle. Volume occupied by one particle in unit cell = \(\frac{\pi a^{3}}{6}\)

Step 4 : Packing efficiency :
Packing efficiency

∴ Packing efficiency = 52.36%
Percentage of void space = 100 – 52.36
= 47.64%

Question vi.
What are paramagnetic substances? Give examples.
Answer:
(1) The magnetic properties of a substance arise due to the presence of electrons.
(2) An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and magnetic properties.
(3) If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They experience a net force of attraction when placed in the magnetic field. This phenomenon is called paramagnetism and the substance is said to be paramagnetic.
For example, O₂, Cu²⁺, Fe³⁺ , Cr³⁺ , NO, etc.

Question vii.
What are the consequences of Schottky defect?
Answer:
Consequences of Schottky defect :

Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.
As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same.
This defect arises in ionic crystals like NaCl, AgBr, KCl, etc.

Question viii.
Cesium chloride crystallizes in cubic unit cell with Cl^– ions at the corners and Cs⁺ ion in the centre of the cube. How many CsCl molecules are there in the unit cell ?
Answer:
Number of Cs⁺ ion at body centre = 1
Number of Cl^– ions due to 8 comers = 8 × \(\frac {1}{8}\) = 1
Hence unit cell contains 1 CsCl molecule.

Question ix.
Cu crystallizes in fee unit cell with edge length of 495 pm. What is the radius of Cu atom ?
Answer:
Given : a = 495 pm
Radius, r = ?
For fee structure,
radius = r = \(\frac{a}{2 \sqrt{2}}=\frac{495}{2 \times \sqrt{2}}\) = 175 cm.
Radius of Cu atom = 175 pm

Question x.
Obtain the relationship between density of a substance and the edge length of unit cell.
Answer:
(1) Consider a cubic unit cell of edge length ‘a’.
The volume of unit cell = a³

(2) If there are ‘n’ particles per unit cell and the mass of particle is ‘m’, then
Mass of unit cell = m × n.

(3) If the density of the unit cell of the substance is p then
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
ρ = \(\frac{m \times n}{a^{3}}\)

Question 4.
The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.
Answer:
Given : Crystal structure of iridium = fcc
Molar mass of iridium = 192.2 gmol^-1
Density = ρ = 22.4 gcm^-3
Radius of iridium = ?
In fcc structure, there are 8 Ir atoms at 8 comers and 6 Ir atoms at 6 face centres.
∴ Total number of Ir atoms = \(\frac {1}{8}\) × 8 + \(\frac {1}{2}\) × 6
= 1 + 3
= 4
Mass of Ir atom = \(\frac{192.2}{6.022 \times 10^{23}}\)
= 31.92 × 10^-23 g
∴ Mass of 4 Ir atoms = 4 × 31.92 × 10^-23
= 1.277 × 10^-21 g
∴ Mass of unit cell = 1.277 × 10^-21 g
Density of unit cell = \(\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}\)
22.4 = \(\frac{1.277 \times 10^{-21}}{a^{3}}\)
∴ a³ = \(\frac{1.277 \times 10^{-21}}{22.4}\)
= 57 × 10^-24 cm³
∴ a = (57 × 10^-24)³ = 3.848 × 10^-8 cm
If r is the radius of iridium atom, then for fcc structure,
r = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{3.848 \times 10^{-8}}{2 \times 1.414}\)
= 1.36 × 10^-8 cm
= 136 pm
Radius of iridium atom = 136 pm

Question 5.
Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is the radius of Al atom ? How many unit cells are there in 1.00 cm³ of Al ?
Answer:
Given : Structure of Al
= Cubic close packed structure
= ccp structure
Edge length of unit cell = a = 353.6 pm
= 3.536 × 10^-8 cm
r = ?
Number of unit cells in 1.00 cm³ of Al = ?
Radius of Al atom = r = \(\frac{a}{2 \sqrt{2}}=\frac{353.6}{2 \sqrt{2}}\)
= \(\frac{353.6}{2 \times 1.414}\) = 125 pm
Volume of one unit cell = a³ = (3.536 × 10^-8)³
= 4.421 × 10^-23 cm³
Number of unit cells = \(\frac{1.00}{4.421 \times 10^{-23}}\)
= 2.26 × 10²²
Radius of Al atom = 125 pm
Number of unit cells = 2.26 × 10²²

Question 6.
In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of tetrahedral voids. What is the formula of the compound?
Answer:
In the given hcp lattice, Y atoms are present at 12 corners and 2 face centres.
∴ Number of Y atoms = \(\frac {1}{2}\) × 12 + 2 × \(\frac {1}{2}\) = 3
There are 6 tetrahedral voids, the number of X atoms = \(\frac {1}{3}\) × 6 = 2
∴ Formula of the compound is X₂Y₃.

Question 7.
How are tetrahedral and octahedral voids formed?
Answer:
Tetrahedral void : The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void.

The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

Octahedral void : The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.

Question 8.
Third layer of spheres is added to second layer so as to form hcp or ccp structure. What is the difference between the addition of third layer to form these hexagonal close-packed structures?
Answer:

In the formation of hexagonal closed-packed (hcp) structure, the first one dimensional row shows depressions between neighbouring atoms.
When a second row is arranged so that spheres fit in these depressions then a staggered arrangement is obtained. If the first row is A then the second row is B.
When third row is placed in staggered manner in contact with second row then A type arrangement is obtained.
Similarly, the spheres in fourth row can be arranged as B type layer. This results in ABAB… type setting of the layers. This gives hexagonal close packing (hcp) structure.

Hexagonal close packing (hcp)

Question 9.
An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell ? (fcc or ccp)
Answer:
Given : Molar mass = M = 27 g mol^-1
Nature of crystal = cubic unit cell
Edge length = a = 405 pm = 4.05 × 10^-8 cm
Density = ρ = 2.7 g cm^-3
Nature of unit cell = ?

= 3.997
≅ 4
Hence the nature of unit cell = face-centred cubic unit cell
Radius of Al atom = 125 pm
The nature of cubic unit cell is fcc.

Question 10.
An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (1.16 × 10²⁴, 2.32 × 10²⁴)

Question 11.
Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.
Answer:
Conductor:

A substance which conducts heat and electricity to a greater extent is called conductor.
In this, conduction bands and valence bands overlap or are very closely spaced.
There is no energy difference or very less energy difference between valence bands and conduction bands.
There are free electrons in the conduction bands.
The conductance decreases with the increase in temperature.
E.g., Metals, alloys.
The conducting properties can’t be improved by adding third substance.

Insulator:

A substance which cannot conduct heat and electricity under any conditions is called insulator.
In this, conduction bands and valence bands are far apart.
The energy difference between conduction bands and valence bands is very large.
There are no free electrons in the conduction bands and electrons can’t be excited from valence bands to conduction bands due to large energy difference.
No effect of temperature on conducting properties.
E.g., Wood, rubber, plastics.
No effect of addition of any substance.

Semiconductor:

A substance that has poor electrical conductance at low temperature but higher conductance at higher temperature is called semiconductor.
In this, conduction bands and valence bands are spaced closely.
The energy difference between conduction bands and valence bands is small.
The electrons can be easily excited from valence bands to conduction bands by heating.
Conductance increases with the increase in temperature.
E.g., Si, Ge
By doping, conducting properties improve. E.g. n-type, p-type semiconductors.

Question 12.
What are n-type semiconductors? Why is the conductivity of doped n-type semiconductor higher than that of pure semiconductor ? Explain with diagram.
Answer:
n-type semiconductor:

n-type semiconductor contains increased number of electrons in the conduction band.
When Si semiconductor is doped with 15^th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms.
P has five valence electrons, out of which four are involved in covalent bonding with neighboring Si atoms while one electrons remains free and delocalised.
These free electrons increase the electrical conductivity of the semiconductor.
The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

P atom occupying regular site of Si atom

Question 13.
Explain with diagram. Frenkel defect. What are the conditions for its formation? What is its effect on density and electrical neutrality of the crystal?
Answer:

Frenkel defect : This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points.
Cations have smaller size than anions, hence generally cations occupy the interstitial sites.
This creates a vacancy defect at its original position and interstitial defect at new position.
Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect :

This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
The ionic compounds must have ions with low coordination number.

Consequences of Frenkel defect :

Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
The crystal remains electrically neutral.
This defect is observed in ZnS, AgCl, AgBr, Agl, CaF₂, etc.

Question 14.
What is an impurity defect? What are its types? Explain the formation of vacancies through aliovalent impurity with example.
Answer:
Impurity defect : This defect arises when foreign atoms, that is, atoms different from the host atoms are present in the crystal lattice.

There are two types of impurity defects namely

Substitutional defects and
Interstitial defects.

(1) Substitutional defects : These defects arises when foreign atoms occupy the lattice sites in place of host atoms, due to their displacements.
Examples : Solid solutions of metals (alloys). For example. Brass in which host atoms are of Cu which are replaced by impurity of Zn atoms. In this Zn atoms occupy regular sites while Cu atoms occupy substituted sites.

Brass

Vacancy through aliovalent impurity :
By addition of impurities of aliovalent ions :

Vacancy through aliovalent ion

When aliovalent ion like Sr²⁺ in small amount is added by additing SrCl₂ to NaCl during its crystallisation, each Sr²⁺ion (oxidation state 2+) removes 2 Na⁺ ions from their lattice points, to maintain electrical neutrality. Hence one of vacant lattice site is occupied by Sr²⁺ ion while other site remains vacant.

Interstitial impurity defect :

Stainless steel

A defect in solid in which the impurity atoms occupy interstitial vacant spaces of lattice structure is called interstitial impurity defect.

For example, in steel, normal lattice sites are occupied by Fe atoms but interstitial spaces are occupied by carbon atoms.

12th Chemistry Digest Chapter 1 Solid State Intext Questions and Answers

Try this… (Textbook Page No. 1)

Observe the above figure carefully. The two types of circles in this figure represent two types of constituent particles of a solid.

Question 1.
Will you call the arrangement of particles in this solid regular or irregular ?
Answer:
The arrangement of particles in this solid is regular.

Question 2.
Is the arrangement of constituent particles in directions \(\overrightarrow{\mathbf{A B}}\), \(\overrightarrow{\mathbf{C D}}\) and \(\overrightarrow{\mathbf{E F}}\) same or different?
Answer:
\(\overrightarrow{\mathbf{A B}}\) represents arrangement of identical particles of one type.
\(\overrightarrow{\mathbf{C D}}\) represents arrangement of identical particles of another type.
\(\overrightarrow{\mathbf{E F}}\) represents regular arrangement of two different particles in alternate positions.

Use your brain power ! (Textbook Page No. 2)

Question 1.
Identify the arrangements A and B as crystalline or amorphous.

Answer:
Arrangement in image A indicates the substance is crystalline.
Arrangement in image B indicates the substance is amorphous.

Try this… (Textbook Page No. 3)

Question 1.
Graphite is a covalent solid yet soft and good conductor of electricity. Explain.
Answer:

Each carbon atom in graphite is sp² hybridised and covalently bonded to other three sp2 hybridised carbon atoms forming σ bonds and the fourth electron in 2p_z orbital of each carbon atom is used in the formation of a π bond. This results in the formation of hexagonal rings in two dimensions.
In graphite, the layers consisting of hexagonal carbon network are held together by weak van der Waal’s forces imparting softness.
The electrons in π bonds in the ring are delocalised and free to move in the delocalised molecular orbitals giving good electrical conductance.

Use your brain power ! (Textbook Page No. 13)

Question 1.
Which of the three lattices scc, bcc and fcc has the most efficient packing of particles ? Which one has the least efficient packing ?
Answer:
fcc has the most efficient packing of particles while see has the least efficient packing.

Can you think ? (Textbook Page No. 20)

Question 1.
When ZnO is heated it turns yellow and returns back to original white colour on cooling. What could be the reason ?
Answer:
When colourless ZnO is strongly heated, the metal atoms are deposited on crystal surface and anions O^2- migrate to the surface producing vacancies at anion lattice points.

These anions combine with Zn atoms forming ZnO and release electrons.
Zn + O^2- → ZnO + 2e^–

These released electrons diffuse into the crystal and occupy vacant sites of anions and produce F- centres. Due to these colour centres, ZnO turns yellow.

Can you tell ? (Textbook Page No. 23)

Let a small quantity of phosphorus be doped into pure silicon.

Question 1.
Will the resulting material contain the same number of total number of electrons as the original pure silicon ?
Answer:
Total number of electrons in doped silicon will be more than in original silicon.

Question 2.
Will the material be electrically neutral or charged?
Answer:
Material will be electrically neutral.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 8 Exercise Transition and Inner Transition Elements Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 8 Transition and Inner Transition Elements Textbook Exercise Questions and Answers.

Transition and Inner Transition Elements Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 8 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 8 Exercise Solutions

1. Choose the most correct option.

Question i.
Which one of the following is diamagnetic
a. Cr^3⊕
b. Fe^3⊕
c. Cu^2⊕
d. Sc^3⊕Answer:
d. Sc^3⊕

Question ii.
Most stable oxidation state of Titanium is
a. +2
b. +3
c. +4
d. +5
Answer:
c. +4

Question iii.
Components of Nichrome alloy are
a. Ni, Cr, Fe
b. Ni, Cr, Fe, C
c. Ni, Cr
d. Cu, Fe
Answer:
(c) Ni, Cr

Question iv.
Most stable oxidation state of Ruthenium is
a. +2
b. +4
c. +8
d. +6
Answer:
(b) +4

Question v.
Stable oxidation states for chromium are
a. +2, +3
b. +3, +4
c. +4, +5
d. +3, +6
Answer:
d. +3, +6

Question vi.
Electronic configuration of Cu and Cu⁺¹
a. 3d¹⁰, 4s⁰; 3d⁹, 4s⁰
b. 3d⁹, 4s¹; 3d⁹4s⁰
c. 3d¹⁰, 4s¹; 3d¹⁰, 4s⁰
d. 3d⁸, 4s¹; 3d¹⁰, 4s⁰Answer:
c. 3d¹⁰, 4s¹; 3d¹⁰, 4s°

Question vii.
Which of the following have d0s0 configuration
a. Sc^3⊕
b. Ti^4⊕
c. V^5⊕
d. all of the above
Answer:
d. All of the above

Question viii.
Magnetic moment of a metal complex is 5.9 B.M. Number of unpaired electrons in the complex is
a. 2
b. 3
c. 4
d. 5
Answer:
d. 5

Question ix.
In which of the following series all the elements are radioactive in nature
a. Lanthanoids
b. Actinoids
c. d-block elements
d. s-block elements
Answer:
b. Actinides

Question x.
Which of the following sets of ions contain only paramagnetic ions
a. Sm^3⊕, Ho^3⊕, Lu^3⊕
b. La^3⊕, Ce^3⊕, Sm^3⊕
c. La^3⊕, Eu^3⊕, Gd^3⊕
d. Ce^3⊕, Eu^3⊕, Yb^3⊕Answer:
d. Ce^3⊕, Eu^3⊕, Yb^3⊕

Question xi.
Which actinoid, other than uranium, occur in a significant amount naturally?
a. Thorium
b. Actinium
c. Protactinium
d. Plutonium
Answer:
a. Thorium

Question xii.
The flux added during extraction of Iron from hematite are its?
a. Silica
b. Calcium carbonate
c. Sodium carbonate
d. Alumina
Answer:
b. Calcium carbonate

2. Answer the following

Question i.
What is the oxidation state of Manganese in
\(\text { (i) } \mathrm{MnO}_{4}^{2-}(\mathrm{ii}) \mathrm{MnO}_{4}^{-} \text {? }\)
Answer:
Oxidation state of Manganese in
\((i) \mathrm{MnO}_{4}^{2-} is +6
(ii) \mathrm{MnO}_{4}^{-}is +7\)

*Question ii.
Give uses of KMnO₄

Question iii.
Why salts of Sc^3⊕, Ti^4⊕, V^5⊕ are colorless?
Answer:
(i) Sc³⁺ salts are colourless :

The electronic configuration of ₂₁Sc [Ar| 3d¹ 4s² and Sc³⁺ [Ar] d°.
Since there are no unpaired electrons in 3d subshell, d → d transition is not possible.
Therefore, Sc³⁺ ions do not absorb the radiations in the visible region. Hence salts of Sc³⁺ are colourless (or white).

(ii) Ti⁴⁺ salts are colourless :

The electronic configuration of ₂₂Ti [Ar] 3d²4s² and Ti⁴⁺ : [Ar] d°
Since there are no unpaired electrons in 3d subshell, d-*d transition is not possible.
Therefore, Ti³⁺ ions do not absorb the radiation in visible region. Hence salts of Ti³⁺ are colourless.

(iii) Vs⁵⁺ salts are eolourless :

The electronic configuration of ₂₃V : [Ar] 3d³4s² and V⁵⁺ : [Ar] 3d°
Since there are no unpaired electrons in 3d-subshell, d – d transition is not possible.
Therefore, V⁵⁺ ions do not absorb the radiations in the visible region. Hence, V⁵⁺ salts are colourless, a

Question iv.
Which steps are involved in the manufacture of potassium dichromate from chromite ore?
Answer:
Steps in the manufacture of potassium dichromate from chromite ore are :

Concentration of chromite ore.
Conversion of chromite ore into sodium chromate (Na₂CrO₄).
Conversion of Na₂CrO₄ into sodium dichromate (Na₂Cr₂O₇).
Conversion of Na₂Cr₂O₇ into K₂Cr₂O₇.

Question v.
Balance the following equation
(i) KMnO₄+ H₂C₂O4 + H₂SO₄ → MnSO₄ + K₂SO₄ + H₂O + O₂
(ii) K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂Answer:
(i) 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂(ii) Acidified potassium dichromate oxidises potassium iodide (KI) to iodine (I2). Potassium dichromate is reduced to chromic sulphate. Liberated iodine turns the solution brown K₂Cr₂O₇ + 6KI + 7H₂SO₄ → 4K₂SO₄ + Cr₂(SO₄)₃ + 7H₂O + 3I₂[Oxidation state of iodine increases from – 1 to zero]

Question vi.
What are the stable oxidation states of plutonium, cerium, manganese, Europium?
Answer:
Stable oxidation states :
Plutonium + 3 to + 7
Cerium + 3, + 4
Manganese + 2, + 4, + 6, + 7
Europium +2, +3

Question vii.
Write the electronic configuration of chromium and copper.
Answer:
Chromium (₂₄Cr) has electronic configuration,
₂₄Cr (Expected) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
(Observed) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

Explanation :

The energy difference between 3d- and 45-orbitals is very low.
d-orbitals being degenerate, they acquire more stability when they are half-filled (3d⁵).
Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cr giving more stable half-filled orbital. Hence, the configuration of Cr is [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².

Copper (₂₉CU) has electronic configuration,
₂₉Cu (Expected) : Is² 2s³ 2p⁶ 3s³ 3p⁶ 3d⁹ 4s²
(Observed) : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Explanation :

The energy difference between 3d- and 45-orbitals is very low.
d-orbitals being degenerate, they acquire more stability when they are completely filled.
Therefore, there arises a transfer of one electron from 45-orbital to 3d-orbital in Cu giving completely filled more stable d-orbital.

Hence, the configuration of Cu is [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s².

Question viii.
Why nobelium is the only actinoid with +2 oxidation state?
Answer:

Nobelium has the electronic configuration ₁₀₂NO : [Rn] 5f¹⁴6d°7s²
No²⁺ : [Rn] 5f¹⁴6d°
Since the 4f subshell is completely filled and 6d° empty, + 2 oxidation state is the stable oxidation state.
Other actinoids in + 2 oxidation state are not as stable due to incomplete 4f subshell.

Question ix.
Explain with the help of balanced chemical equation, why the solution of Ce(IV) is prepared in acidic medium.
Answer:
Ce⁴⁺ undergoes hydrolysis as, Ce⁴⁺+ 2H₂O → Ce(OH)₄ + 4H⁺.
Due to the presence of H⁺ in the solution, the solution is acidic.

Question x.
What is meant by ‘shielding of electrons’ in an atom?
Answer:
The inner shell electrons in an atom screen or shield the outermost electron from the nuclear attraction. This effect is called the shielding effect.

The magnitude of the shielding effect depends upon the number of inner electrons.

Question xi.
The atomic number of an element is 90. Is this element diamagnetic or paramagnetic?
Answer:
The 90th element thorium has an electronic configuration, [Rn] 6d²7s². Since it has 2 unpaired electrons it is paramagnetic.

3. Answer the following

Question i.
Explain the trends in atomic radii of d-block elements
Answer:

The atomic or ionic radii of 3-d series transition elements are smaller than those of representative elements, with the same oxidation states.
For the same oxidation state, there is an increase in nuclear charge and a gradual decrease in ionic radii of 3d-series elements is observed. Thus ionic radii of ions with oxidation state + 2 decreases with increase in atomic number.
There is slight increase is observed in Zn²⁺ ions. With the higher oxidation states, effective nuclear charge increases. Therefore ionic radii decrease with increase in oxidation state of the same element. For example, Fe²⁺ ion has ionic radius 77 pm whereas Fe³⁺ has 65 pm.

Question ii.
Name different zones in the Blast furnace. Write the reactions taking place in them.
Answer:
(i) Zone of combustion : The hot air oxidises coke to CO which is an exothermic reaction, due to which the temperature of furnace rises.
C + ^1/2O₂ → CO ΔH= – 220kJ
Some part of CO dissociates to give finely divided carbon and O₂.
2CO → 2C + O₂The hot gases with CO rise up in the furnace and heats the charge coming down. CO acts as a fuel as well as a reducing agent.

(ii) Zone of reduction : At about 900 °C, CO reduces Fe₂O₃ to spongy (or porous) iron.
Fe₂O₃ + 3CO → 2Fe + 3CO₂Carbon also reduces partially Fe₂0₃ to Fe.
Fe₂O₃ + 3C → 2Fe + 3CO

(iii) Zone of slag formation : At 1200 K limestone, CaCO₃ in the charge, decomposes and forms a basic flux CaO which further reacts at 1500 K with gangue (SiO₂, Al₂O₃) and forms a slag of CaSiO₃ and Ca₃AlO₃.
CaCO₃ + CaO + CO₂.
CaO + SiO₂→ CaSiO₃12CaO + 2Al₂O₃ → 4Ca₃AlO₃ + 3O₂

The slag is removed from the bottom of the furnace through an outlet.

(iv) Zone of fusion : The impurities in ore like MnO₂ and Ca₃(PO₄)₂ are reduced to Mn and P while SiO₂ is reduced in Si. The spongy iron moving down in the furnace melts in the fusion zone and dissolves the impurities like C, Si, Mn, phosphorus and sulphure. The molten iron collects at the bottom of furnace. The lighter slag floats on the molten iron and prevents its oxidation.

The molten iron is removed and cooled in moulds. It is called pig iron or cast iron. It contains about 4% carbon.

Question iii.
What are the differences between cast iron, wrought iron and steel?
Answer:

Cast iron	Wrought iron	Steel
(1) Hard and brittle (2) Contains 4% carbon. (3) Used for making pipes, manufacturing automotive parts, pots, pans, utensils	(1) Very soft (2) Contains less than 0.2% carbon. (3) Used for making pipes, bars for stay bolts, engine bolts and rivets.	(1) Neither too hard nor too soft. (2) Contains 0.2 to 2% carbon (3) Used in buildings infrastructure, tools, ships, automobiles, weapons etc.

Question iv.
Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?
Answer:
The electronic configuration of Fe^{2 +} and Fe³⁺ :
Fe²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶Fe³⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Due to loss of two electrons from the 4.v-orbital and one electron from the 3d-orbital, iron attains 3⁺ oxidation state. Since in Fe³⁺, the 3d-orbital is half-filled, it gets extra stability, hence Fe³⁺ is more stable than Fe²⁺.

Question v.
Give the similarities and differences in elements of 3d, 4d and 5d series.
Answer:
Similarity :

They are placed between .s-block and p-block of the periodic table.
All elements are metals showing metallic characters.
Some are paramagnetic.
Most of them give coloured compounds.
They have catalytic properties.
They form complexes.
They have variable oxidation states.

Differences :

In 4d and 5d series lanthanide and actinoid contraction is observed. In 3d series atomic size changes are less marked.
4d and 5d elements have high coordination numbers compared to 3d elements.
4d and 5d series have similar properties whereas 3d series have different properties.

Question vi.
Explain trends in ionisation enthalpies of d-block elements.
Answer:

The ionisation enthalpies of transition elements are quite high and lie between those of 5-block and p-block elements. This is because the nuclear charge and atomic radii of transition elements lie between those of 5-block and p-block elements.
As atomic number of transition elements increases along the period and along the group, first ionisation enthalpy increases even though the increase is not regular.
If IE_1; IE₂ and IE₃ are the first, second and third ionisation enthalpies of the transition elements, then IE₁ < IE₂ < IE₃.
In the transition elements, the added last differentiating electron enters into (n – 1) d-orbital and shields the valence electrons from the nuclear attraction. This gives rise to the screening effect of (n – 1) d-electrons.
Due to this screening effect of (n – 1) d electrons, the ionisation enthalpy increases slowly and the increase is not very regular.

Question vii.
What is meant by diamagnetic and paramagnetic metal? Give one example of diamagnetic and paramagnetic transition metal and lanthanoid metal.
Answer:

Paramagnetic substances : When a magnetic field is applied, substances which are attracted towards the applied magnetic field are called paramagnetic substances. Example : Ni²⁺, Pr⁴⁺
Diamagnetic substances : When a magnetic field is applied, substances which are repelled by the magnetic fields are called diamagnetic substances. Example : Zn²⁺, La³⁺
Ferromagnetic substances : When a magnetic field is applied, substances which are attracted very strongly are called ferromagnetic substances. These substances can be magnetised. For example, Fe, Co, Ni are ferromagnetic.

Question viii.
Why the ground-state electronic configurations of gadolinium and lawrencium are different than expected?

Question ix.
Write steps involved in the metallurgical process
Answer:
The various steps and principles involved in the extraction of pure metals from their ores are as follows.:

Concentration of ores in which impurities (gangue) are removed.
Conversion of ores into oxides or other reducible compounds of metals.
Reduction of ores to obtain crude metals.
Refining of metals giving pure metals.

Question x.
Cerium and Terbium behaves as good oxidising agents in +4 oxidation state. Explain.
Answer:

The most stable oxidation state of lanthanoids is +3.
Hence, Ce⁴⁺ (cerium) and Tb⁴⁺ (terbium) tend to get + 3 oxidation state which is more stable.
Since they get reduced by accepting electron, they are good oxidising agents in + 4 oxidation state.

Question xi.
Europium and Ytterbium behave as good reducing agents in +2 oxidation state explain.
Answer:

The most stable oxidation state of lanthanoids is + 3.
Hence, Eu²⁺ and Yb²⁺ tend to get + 3 oxidation states by losing one electron.
Since they get oxidised, they are good reducing agents in + 2 oxidation state.

Activity :
Make groups and each group prepare a PowerPoint presentation on the properties and applications of one element. You can use your imagination to create some innovative ways of presenting data.

You can use pictures, images, flow charts, etc. to make the presentation easier to understand. Don’t forget to cite the reference(s) from where data for the presentation is collected (including figures and charts). Have fun!

12th Chemistry Digest Chapter 8 Transition and Inner Transition Elements Intext Questions and Answers

Do you know? (Textbook Page No 165)

Question 1.
In which block of the modern periodic table are the transition and inner transition elements placed?
Answer:
The transition elements are placed in d-block and inner transition elements are placed in f-block of the modern periodic table.

Use your brain power! (Textbook Page No 167)

Question 1.
Fill in the blanks with correct outer electronic configurations.
Answer:
Answers are given in bold.

Try this….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Cr and Cu.
Answer:
₂₄Cr : [Ar] 3d⁵4s¹ ₃₀Cu : [Ar] 3d¹⁰4s¹

Can you tell? (Textbook Page No 168)

Question 1.
Which of the first transition series element shows the maximum number of oxidation states and why?
Answer:

₂₅Mn shows the maximum number of oxidation states, + 2 to +7.
₂₅Mn : [Ar] 3d⁵4s³
Mn has incompletely filled J-subshell.
Due to small difference in energy between 3d and 4s -orbitals, Mn can lose (or share) electrons from both the orbitals.
Hence Mn shows oxidation states from + 2 to +7.

Question 2.
Which elements in the 4d and 5d-series will show maximum number of oxidation states?
Answer:
In 4d-series maximum number of oxidation states are for Ruthenium Ru ( + 2, +3, + 4„ +6, +7, + 8). In 5d-series, maximum number of oxidation states are for Osmium, Os ( + 2 to + 8).

Try this ….. (Textbook Page No 168)

Question 1.
Write the electronic configuration of Mn⁶⁺, Mn⁴⁺, Fe⁴⁺, Co⁵⁺, Ni²⁺.
Answer:

Ions	Electronic configuration of valence shell
Mn⁶⁺	[Ar] 3d¹
Mn⁴⁺	[Ar] 3d³
Fe⁴⁺	[Ar] 3d⁴
Co⁵⁺	[Ar] 3d⁴
Ni²⁺	[Ar] 3d⁸

Try this ….. (Textbook Page No 171)

Question 1.
Pick up the paramagnetic species from the following : Cu¹⁺, Fe³⁺, Ni²⁺, Zn²⁺, Cd²⁺, Pd²⁺.
Answer:
The following ions are paramagnetic : Fe³⁺, Ni²⁺, Pd²⁺

Try this ….. (Textbook Page No 171)

Question 1.
What will be the magnetic moment of transition metal having 3 unpaired electrons?
(a) equal to 1.73 B.M.?
(b) less than 1.73 BM.
(c) more than 1.73 B.M.?
Answer:
By spin-only formula, \(\mu=\sqrt{n(n+2)}\) where n is number of unpaired electrons.
\(\mu=\sqrt{3(3+2)}=\sqrt{3(5)}=3.87 \mathrm{~B} . \mathrm{M}\)
Thus the value is more than 1.73 B.M.

Use your brain power! (Textbook Page No 171)

Question 1.
A metal ion from the first transition series has two unpaired electrons. Calculate the magnetic moment.
Answer:
\(\)\begin{aligned}
\mu &=\sqrt{n(n+2)} \\
&=\sqrt{2(2+2)} \\
&=\sqrt{8} \\
&=2.84 \text { B.M. }
\end{aligned}\(\)

Problem (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of divalent cation of a transition metal with atomic number 25.
Answer:
For element with atomic number 25. electronic configuration of its divalent cation will be : [Ar] 3d⁵.

Try this….. (Textbook Page No 172)

Question 1.
Calculate the spin-only magnetic moment of a divalent cation of element Slaving atomic number 27.
Answer:
Electronic configuration of divalent ion of an element with atomic number 27 : [Ar] 3d⁷;

Can you tell? (Textbook Page No 172)

Question 1.
Compounds of s and p-block elements are almost white. What could be the absorbed radiation? (uv or visible)?
Answer:
The white colour of a compound indicates the absorption of uv radiation.

Can you tell? (Textbook Page No 181)

Question 1.
Why f-block elements are called inner transition metals?
Answer:
f-block elements are called inner transition elements since f-orbital lies much inside the f-orbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

Question 2.
Are there an similarities between transition and inner transition metals?
Answer:
There are some properties similarity between transition and inner transition metals.

They are placed between s and p-block elements.
They are metals with filling of inner suhshells in their electronic configuration.
They show variable oxidation slates.
They show magnetism.
They form coloured compounds.
They have catalytic property.

Problem (Textbook Page No 184)

Question 1.
Which of the following will have highest fourth ionisation enthalpy, La⁴⁺, Gd⁴⁺, Lu⁴⁺.
Answer:
La : 4f°5d¹6s²Gd : 4f¹5d¹6s²Lu : 4f¹⁴5d¹6s²
Lu will have the highest fourth ionisation enthalpy since Lu³⁺ has the most stable configuration of 4f¹⁴.

Use your brain power! (Textbook Page No 185)

Question 1.
Do you think that lanthanoid complex would show magnetism?
Answer:
Lanthanoid complexes may show magnetism.

Question 2.
Can you calculate the spin only magnetic moment of lanthanoid complexes using the same formula that you used for transition metal complexes?
Answer:
You cannot calculate magnetic moment of lanthanoid complexes using spin only formula as you have to consider orbital momentum also.

Question 3.
Calculate the spin only magnetic moment of La³⁺. Compare the value with that given in the table.
Answer:
La³⁺ ion has no unpaired electron.

La³⁺ ion has zero value of magnetic moment same as given in the table.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 12 Exercise Aldehydes, Ketones and Carboxylic Acids Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Textbook Exercise Questions and Answers.

Aldehydes, Ketones and Carboxylic Acids Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 12 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 12 Exercise Solutions

1. Choose the most correct option.

Question i.
In the following resonating structures A and B, the number of unshared electrons in valence shell present on oxygen respectively are

Answer:
c. 4, 6

Question ii.
In the Wolf -Kishner reduction, alkyl aryl ketones are reduced to alkylbenzenes. During this change, ketones are first converted into
a. acids
b. alcohols
c. hydrazones
d. alkenes
Answer:
c. hydrazones

Question iii.
Aldol condensation is
a. electrophilic substitution reaction
b. nucleophilic substitution reaction
c. elimination reaction
d. addition – elimination reaction
Answer:
d. addition-elimination reaction

Question iv.
Which one of the following has the lowest acidity?

Answer:
(c)

Question v.
Diborane reduces
a. ester group
b. nitro group
c. halo group
d. acid group
Answer:
d. acid group

Question vi.
Benzaldehyde does NOT show positive test with
a. Schiff reagent
b. Tollens’ ragent
c. Sodium bisulphite solution
d. Fehling solution
Answer:
d. Fehling solution

2. Answer the following in one sentence

Question i.
What are aromatic ketones?
Answer:
The compounds in which group is attached to either two aryl groups or one aryl and one alkyl group are called aromatic ketones.

For example :

Question ii.
Is phenylacetic acid an aromatic carboxylic acid?
Answer:
Phenylacetic acid is not an aromatic carboxylic acid.

Question iii.
Write reaction showing conversion of ethanenitrile into ethanol.
Answer:

Question iv.
Predict the product of the following reaction:
\(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOCH}_{3} \frac{\mathrm{i} \cdot \mathrm{AlH}(\mathrm{i}-\mathrm{Bu})_{2}}{\text { ii. } \mathrm{H}_{3} \mathrm{O}^{\oplus}}{\longrightarrow} ?\)
Answer:

Question v.
Name the product obtained by reacting toluene with carbon monoxide and hydrogen chloride in presence of anhydrous aluminium chloride.
Answer:

Question vi.
Write reaction showing conversion of Benzonitrile into benzoic acid.
Answer:

Question vii.
Name the product obtained by the oxidation of 1,2,3,4-tetrahydronaphthalene with acidified potassium permanganate.
Answer:

Question viii.
What is formalin?
Answer:
The aqueous solution of formaldehyde (40%) is known as formalin.

Question ix.
Arrange the following compounds in the increasing order of their boiling points : Formaldehyde, ethane, methyl alcohol.
Answer:
Ethane, formaldehyde, methyl alcohol.

Question x.
Acetic acid is prepared from methyl magnesium bromide and dry ice in presence of dry ether. Name the compound which serves not only reagent but also as cooling agent in the reaction.
Answer:
The cooling agent used in the above reaction is dry ice (O = C = O).

3. Answer in brief.

Question i.
Observe the following equation of reaction of Tollens’ reagent with aldehyde. How do we know that a redox reaction has taken place. Explain.
\(\begin{array}{r}
\mathrm{R}-\mathrm{CHO}+2 \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}+\mathrm{OH}^{-} \stackrel{\Delta}{\longrightarrow} \\
\mathrm{R}-\mathrm{COO}^{-}+2 \mathrm{Ag} \downarrow+4 \mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}
\end{array}\)
Answer:
Tollen’s reagent oxidises acetaldehyde to acetic acid (carboxylate ion) and Ag in Tollen’s reagent complex are reduced to silver. In this reaction, oxidation and reduction takes place simultaneously hence, it is a redox reaction.

Question ii.
Formic acid is stronger than acetic acid. Explain.
Answer:

In acetic acid, the methyl group is an electron-donating group. The acetate ion formed gets destabilized due to the electron releasing effect of methyl group ( +1 effect) which is higher than that of H-atom in the corresponding formic acid. As a result, acetic acid dissociates to a lesser extent. Thus decreasing the acidity of acetic acid.

Formic acid having lower pK_a value than acetic acid. Hence, formic acid is a stronger acid than acetic acid.

Question iii.
What is the action of hydrazine on cyclopentanone in presence of —. KOH in ethylene glycol?
Answer:

Question iv.
Write reaction showing conversion of Acetaldehyde into acetaldehyde dimethyl acetal.
Answer:

Question v.
Aldehydes are more reactive toward nucleophilic addition reactions than ketones. Explain.
Answer:
The reactivity of aldehydes and keones is due to the polarity of carbonyl group which results in electrophilicity of carbon. The reactivity is further explained on the basis of electronic effect and steric effects.

(1) Influence of electronic effects: A ketone has two electron-donating aJ.kyl groups ( + I effect) bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity. In contrast. aldehydes have only electron-donating group bonded to carbonyl carbon. This shows aldehydes are more electrophilic than ketones.

(2) Steric effects : Two bulky alkyl groups in ketone come in the way of the incoming nucleophile. This is called steric hindrance to nucleophilic attack.

On the other hand. nucleophile can easily attack the carbonyl carbon in aldehyde because has one alkyl group and is less crowded or sterically less hindered.

Hence aldehydes are more reactive and can easily be attacked by nucleophiles.

Question vi.
Write reaction showing the action of the following reagent on propane nitrile
a. Dilute NaOH
b. Dilute HCl ?
Answer:
(1) Action of dil NaOH:

(2) Action of dil HCl:

Question vi.
Arrange the following carboxylic acids with increasing order of their acidic strength and justify your answer.

Answer:
The increasing order of acidity will be 1 <3 <2. Acidity depends on mainly two factors : (1) ease of proton release (2) stability of conjugate base formed. In example (3) the ether O exerts a I effect and is closer to COOH group than in 2 (1 effect diminishes). Also the conjugate base formed will be stabilized by the same – I effect by delocalization of charge.

4. Answer the following

Question i.
Write a note on
a. Cannizaro reaction
b. Stephen reaction.
Answer:
(1) The carbon atom adjacent to carbonyl carbon atom is called a-carbon atom (α – C) and the hydrogen atom attached to a-carbon atom is called α-hydrogen atom (α – H).

(2) The α-hydrogen of aldehydes and ketones is acidic in nature due to (i) the strong-I effect of carbonyl group (ii) resonance stabilization of the carbanion.

(3) Aldol condensation reaction is characteristic reaction of aldehydes and ketones containing active α-hydrogen atom.

(4) When aldehydes or ketones containing α – H atoms are warmed with a dilute base or dilute acid, two molecules of them undergo self condensation to give β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol) respectively. The reaction is known as Aldol addition Reaction.

(5) In aldol condensation, the product is formed by the nucleophilic addition of α-carbon atom of a second molecule which gets attached to carbonyl carbon atom of the first molecule and α-hydrogen atom of the second molecule gets attached to carbonyl oxygen atom of the first molecule forming (- OH) group to give β-hydroxy aldehyde or ketone.

(6) This is a reversible reaction, establishing an equilibrium favouring aldol formation to a greater extent than ketol formation.

(7) For aldehyde :

Acetaldol on heating undergoes subsequent elimination of water giving rise to α, β unsaturated aldehyde.

The overall reaction is called aldol condensation. It is a nucleophilic addition-elimination reaction. For ketone :

Diacetone alcohol on dehydration by heating forms α, β unsaturated ketone.

Question ii.
What is the action of the following reagents on toluene ?
a. Alkaline KMnO₄, dil. HCl and heat
b. CrO₂Cl₂ in CS₂
c. Acetyl chloride in presence of anhydrous AlCl₃.
Answer:
(1) Action of alkaline KMnO₄ : When toluene is heated with alkaline KMnO₄. (methyl group gets oxidised to earboxy lic group) benzoic acid is obtained

(2) Action of CrO₂Cl₂ in C₂:
Answer:
When Loluenc is ircated with soluion of chromyl chloride (CrO₂Cl₂) in Cs₂, brown chromium complex is obtained, which on acid hydrolysis gives benzaldehyde.

(3) Action of acetyl chloride in presence of anhyd. AlCl₃.
Answer:
When toluene is treated with acetyl chloride in presence of anhydrous AlCl₃ 4-methyl acetophenone is obtained.

Question iii.
Write the IUPAC names of the following structures :

Question iv.
Write reaction showing conversion of p- bromoisopropyl benzene into p-Isopropyl benzoic acid (3 steps).
Answer:

Question v.
Write reaction showing aldol condensation of cyclohexanone.
Answer:

Activity :
Draw and complete the following reaction scheme which starts with acetaldehyde. In each empty box, write the structural formula of the organic compound that would be formed.

12th Chemistry Digest Chapter 12 Aldehydes, Ketones and Carboxylic Acids Intext Questions and Answers

Use your brain power! (Textbook Page No 254)

Question 1.
Classify the following as aliphatic and aromatic aldehydes.

Answer:

Use your brain power! (Textbook Page No 255)

Question 1.
Classify the following as simple and mixed ketones. Benzoplienone, acetoneq hutanoneq acetophenone.

Compoun
Benzophenone	……………………………………………..
Acetone	……………………………………………..
Butanone	……………………………………………..
Acetophenone	……………………………………………..

Answer:

Compound
Benzophenone	Simple ketone
Acetone	Simple ketone
Butanone	Mixed ketone
Acetophenone	Mixed ketone

Use your brain power! (Textbook Page No 264)

Write IUPAC names for the following compounds.

Try this ….. (Textbook Page No 260)

Question 1.
Draw structures for the following :
(1) 2-Methylpentanal
(2) Hexan-2-one
Answer:

Can you tell? (Textbook Page No 260)

Question 1.
Which is the reagent that oxidizes primary alcohols to only aldehydes and does not oxidize aldehydes further into carboxylic acid ?
Answer:
The reagent that is used to make only aldehydes is-heated Cu at 573 K.

Use your brain power! (Textbook Page No 261)

Question 1.
Write the structure of the product formed on Rosenmund reduction of ethanoyl chloride and benzoyl chloride.
Answer:

Can you think? (Textbook Page No 261)

Question 1.
What is the alcohol formed when benzoyl chloride is reduced with pure palladium as the catalyst ?
Answer:

Use your brain power! (Textbook Page No 262)

Question 1.
Name the compounds which are used for the preparation of benzophenone by Friedel-Crafts acylation reaction. Draw their structures.
Answer:
The compounds which are used in preparation of benzophenone by Friedel – Crafts reaction are :

Use your brain power! (Textbook Page No 263)

Identify the reagents necessary to achieve each of the following transformations:

Answer:

Use your brain power! (TextBook Page No 264)

Predict the products (name and structure) in the following reactions.

Answer:

Problem 12.1 : (Textbook Page No 276)

Question 1.
Alcohols (R – OH), phenols (Ar – OH) and carboxylic acids (R – COOH) can undergo ionization of O – H bond to give away proton H⁺; yet they have different pK_a values, which are 16, 10 and 4.5 respectively. Explain.
Solution :
pK_a value is indicative of acid strength. Lower of pK_a value stronger the acid. Alcohols, phenols and carboxylic acids, all involve ionization of an O – H bond. But their different pK_a values indicate that their acid strengths are different. This is because the resulting conjugate bases are stabilized to different extents.

As the conjugate base of carboxylic acid is best stabilized, among the three, carboxylic acids are strongest and have the lowest pKa value. As conjugate base of alcohols is destabilized, alcohols are weakest acids and have highest pIQ value. As conjugate base of phenols is moderately stabilized, phenols are moderately acidic and have intermediate pBQ value.

Try this….. (Textbook Page No 277)

Question 1.
Compare the following two conjugate bases and answer.

(1) Indicate the inductive effects of CH₃ – group in (a) and Cl-group in (b) by putting arrowheads in the middle of appropriate covalent bonds.
(2) Which species is stabilized by inductive effect, (a) or (b) ?
(3) Which species is destabilized by inductive effect, (a) or (b) ?
Answer:

(2) The monochloroacetate ion formed gets stabilised due to electron-withdrawing of Cl atom (- I effect).
(3) The acetate ion formed gets destabilised due to electron releasing effect of methyl group

Use your brain power! (Textbook Page No 277)

Question 1.
(1) Compare the pK_a values and arrange the following in an increasing order of acid strength. CI₃CCOOH, ClCH₂COOH, CH₃COOH, Cl₂CHCOOH
(2) Draw structures of conjugate bases of monochloroacetic acid and dichloroacetic acid. Which one is more stabilized by – 1 effect?
Answer:

The dichloroacetate ion formed gets stabilised due electron-withdrawing effect of two chlorine atoms. (- 1 effect)

Try this….. (Textbook Page No 277)

Question 1.
Arrange the following acids in order of their decreasing acidity.

Answer:
Acidity in the decreasing order

Try this ….. (Textbook Page No 267)

Question 1.
Draw the structure of propanone and indicate its polarity.
Answer:

Can you tell? (Textbook Page No 268)

Question 1.
Simple hydrocarbons, ethers, ketones and alcohols do not get oxidized by Tollen’s reagent. Explain, Why?
Answer:
(1) Due to the presence of hydrogen atom in aldehyde group , an aldehyde is oxidised to carboxylic acid which is not possible in case of ethers, ketones, alcohols and hydrocarbons.
(2) In ketones, carbonyl atom is attached to C-atom, hence C – C bond in can’t be broken easily.
(3) H atom attached to carbonyl carbon can be oxidised to – OH group giving carboxylic group Therefore, aldehyde reduces Tollen’s reagent, whereas simple hydrocarbons, ethers, ketones and alcohols do not reduce Tollen’s reagent.

Use your brain power! (Textbook Page No 269)

Question 1.
Sodium bisulfite is sodium salt of sulfurous acid, write down its detailed bond structure.
Answer:

Bond structure of sodium bisulfite

Use your brain power! (Textbook Page No 270)

Predict the product of the following reaction:

Answer:

Use your brain power! (Textbook Page No 271)

Question 1.
Draw the structures of
(1) The semicarbazone of cyclohexanone
(2) The imine formed in the reaction between 2-methylhexanal and ethyl amine
(3) 2, 4-dinitrophenyl hydrazone of acetaldehyde.
Answer:
(1) The semi carbazone of cyclohexanone.

(2) The imine formed between 2-methyl hexanal and ethyl amine.

(3) 2, 4-dinitrophenylhydrazone of acetaldehyde.

Try this….. (Textbook Page No 272)

Question 1.
Write chemical reactions taking place when propan-2-ol is treated with iodine and sodium hydroxide.
Answer:

Question 2.
When acetaldehyde Is treated with dilute NaOH, the following reaction is observed.

(1) What are the functional groups in the product?
(2) Can another product be formed during the same reaction? (Deduce the answer by doing atomic audit of reactant and product)
(3) Is this an addition reaction or condensation reaction?
Answer:
(1) There arc two functiona’ groups in the product: -CRO and -OH
(2) No other product can be formed in the same reaction.
(3) This is an addition reaction.

Use your brain power! (Textbook Page No 273)

Question 1.
Observe the following reaction.
Answer:

Question 2.
Will this reaction give a mixture of products like a cross aldol reaction ?
Answer:
No, since benzaldehyde, does not have a-hydrogen atom, it will not undergo self aldol condensation.

Use your brain power! (Textbook Page No 274)

Question 1.
Can isobutyraldehyde undergo a Cannizzaro reaction? Explain.
Answer:
Since isobutyraldehydecontains a-carbon atom, it cannot undergo Cannizzaro reaction.

Can you tell? (Textbook Page No 279)

What is the term used for elimination of water molecule ?
Answer:
Dehydration.

Use your brain power! (Textbook Page No 278)

Question 1.
Fill in the blanks and rewrite the balanced equations.

Answer:

12th Std Chemistry Questions And Answers:

Maharashtra Board Class 12 BK & Accountancy Previous Year Question Papers with Solutions Pdf

October 28, 2024October 25, 2024 by sachin

Maharashtra Board Class 12th BK & Accountancy Previous Year Question Paper, BK & Accountancy Question Paper Class 12 Maharashtra State Board Pdf 2024, Maharashtra Board 12th BK & Accountancy Question Paper Pdf Download, BK & Accountancy Previous Year Question Paper Class 12 Maharashtra State Board Pdf, 12th BK & Accountancy Paper Pattern 2024 Maharashtra Board, Maharashtra Board Model Paper Class 12th BK & Accountancy Pdf with Solutions, BK & Accountancy Class 12 Maharashtra Board Question Paper.

Maharashtra State Board Class 12th BK & Accountancy Question Papers with Answers Pdf Download

Maharashtra Board Class 12 BK & Accountancy Previous Year Question Paper, Previous Year BK & Accountancy Question Paper Class 12 Maharashtra State Board Pdf, Class 12 Maharashtra Board BK & Accountancy Previous Year Question Paper, BK & Accountancy Paper Pattern Class 12 Maharashtra Board 2024, 12th BK & Accountancy Question Paper Maharashtra State Board Pdf with Answers, Class 12 BK & Accountancy Question Paper 2024 Maharashtra Board.

Maharashtra Board Class 12th Previous Year Question Paper

Maharashtra State Board 12th Commerce Book Keeping & Accountancy Solutions Digest | 12th BK Textbook Solutions

August 30, 2024August 29, 2024 by Bhagya

Maharashtra State Board Std 12th Commerce Book Keeping & Accountancy Textbook Solutions Digest

Book Keeping and Accountancy Class 12 Solutions | 12th BK Textbook Solutions