Class 12 – Page 25 – Maharashtra Board Solutions

12th Biology Chapter 6 Exercise Plant Water Relation Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Prasanna

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Plant Water Relation Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 6 Exercise Solutions

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
Passive absorption is the chief method of absorption (98%).
There is no expenditure of energy in passive absorption.
Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
It occurs during daytime when there is active transpiration.
Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
Active absorption may be osmotic or non- osmotic type.
For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
The movement is against the gravity and described as ascent of sap.
The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

The loss of water in the form of vapour is called transpiration.
Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
Opening and closing of stomata is controlled by turgidity of guard cells.
When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

Removal of excess of water
Helps in passive absorption of water and minerals
Helps in ascent of sap – transpiration pull
Maintains turgor of cells
Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

Root pressure theory is proposed by J. Pristley.
For translocation of water, activity of living cells of root is responsible.
Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
Root pressure is an osmotic phenomenon.

Limitation of this theory:

Not applicable to tall plants above 20 metres.
Even in absence of root pressure ascent of sap is noticed.
In actively transpiring plants, root pressure is not developed.
In taller gymnosperms, root pressure is zero.
Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

Capillarity theory of water translocation is proposed by Bohem.
Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
Xylem vessels and tracheids are tubular elements having their lumen.
In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

The loss of water in the form of water vapour is called transpiration.
About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
The process is necessary evil because water which is important for plant is lost in the process.
At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
Root hair cell becomes flaccid and ready to absorb soil water.
Water is passed on similarly in inner cortical cells.
Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Osmotic Pressure (O.R)	Turgor Pressure (T.P.)
1. The pressure exerted due to osmosis is called osmotic pressure.	1. The pressure exerted by turgid cell sap on cell membrane and cell wall, is called turgor pressure.
2. It is pressure caused by water when it moves by osmosis.	2. It is pressure caused by content of cell (cell sap).
3. It is generated by the osmotic flow of water through a semipermeable membrane.	3. It is maintained by osmosis.

Question 11.
How are the minerals absorbed by the plants ?
Answer:

Soil is the chief source of minerals for the plants.
Minerals get dissolved in the soil water.
Minerals are absorbed by the plants in the ionic form mainly through roots.
Absorption of minerals is independent of water.
Absorbed minerals are pulled upwards along with xylem sap.
Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:

Water from soil is absorbed by plants with the help of root hairs.
Root hairs are present in zone of absorption.
Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
Root hairs are nothing but cytoplasmic extensions of epiblema cell.
Root hairs are long tube like structures of about 1 to 10 mm.
They are colourless, unbranched and very delicate structures.
A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
Water molecules get adsorbed on cell wall of root hair (imbibition).
They enter the root hair cell by diffusion through cell wall which is freely permeable.
By process of osmosis they enter through plasma membrane which is semipermeable.
The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
The pathway of water is by apoplast and symplast.
When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
Ascent of sap occurs through lumen of xylem elements.
Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
When guard cells are flaccid that results in closure of stomata.
According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

Plants absorb minerals from the soil with their root system.
MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
Ions get accumulated in the root hair against the concentration gradient.
These ions pass into cortical cells and finally reach xylem of roots.
Along with the water these minerals are carried to other parts of plant.

12th Std Biology Questions And Answers:

12th Chemistry Chapter 13 Exercise Amine Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amine Textbook Exercise Questions and Answers.

Amine Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 13 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 13 Exercise Solutions

1. Choose the most correct option.

Question i.
The hybridisation of nitrogen in primary amine is ………………………. .
a. sp
b. sp²
c. sp³
d. sp³d
Answer:
c. sp³

Question ii.
Isobutylamine is an example of ………………………. .
a. 2° amine
b. 3° amine
c. 1° amine
d. quaternary ammonium salt.
Answer:
a. 2° amine

Question iii.
Which one of the following compounds has the highest boiling point?
a. n-Butylamine
b. sec-Butylamine
c. isobutylamine
d. tert-Butylamine
Answer:
a. n-Butylamine

Question iv.
Which of the following has the highest basic strength?
a. Trimethylamine
b. Methylamine
c. Ammonia
d. Dimethylamine
Answer:
d. Dimethylamine

Question v.
Which type of amine does produce N₂ when treated with HNO₂?
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both primary and secondary amines
Answer:
a. Primary amine

Question vi.
Carbylamine test is given by
a. Primary amine
b. Secondary amine
c. Tertiary amine
d. Both secondary and tertiary amines
Answer:
a. Primary amine

Question vii.
Which one of the following compounds does not react with acetyl chloride?
a. CH₃-CH₂-NH₂
b. (CH₃-CH₂)₂NH
c. (CH₃-CH₂)₃N
d. C₆H₅-NH₂Answer:
Answer:
c. (CH₃ – CH₂)₃N

Question viii.
Which of the following compounds will dissolve in aqueous NaOH after undergoing reaction with Hinsberg reagent?
a. Ethylamine
b. Triethylamine
c. Trimethylamine
d. Diethylamine
Answer:
a. Ethyl amine

Question ix.
Identify ‘B’ in the following reactions

Answer:
d. CH₃-CH₂-OH

Question x.
Which one of the following compounds contains azo linkage?
a. Hydrazine
b. p-Hydroxyazobenzene
c. N-Nitrosodiethylamine
d. Ethylenediamine
Answer:
b. p-Hydroxyazobenzene

2. Answer in one sentence.

Question i.
Write reaction of p-toluenesulfonyl chloride with diethylamine.
Answer:

Question ii.
How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine?
Answer:
2 moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine.

Question iii.
Which amide does produce ethanamine by Hofmann bromamide degradation reaction?
Answer:
Propanamide (CH₃ – CH₂ – CONH₂) produces ethanamine by Hofmann bromamide degradation reaction.

Question iv.
Write the order of basicity of aliphatic alkylamine in gaseous phase.
Answer:
The order of basicity of aliphatic alkyl amines in the gaseous follows the order : tertiary amine > secondary amine > primary amine > NH₃.

Question v.
Why are primary aliphatic amines stronger bases than ammonia?
Answer:
The alkyl group tends to increase the electron density on the nitrogen atom. As a result, amines can donate the lone f pair of electrons on nitrogen more easily than ammonia. Hence, aliphatic amines are stronger bases than ammonia.

Question vi.
Predict the product of the following reaction. Nitrobenzene Sn/Conc. HCl?
Answer:
The product is aniline/

Question vii.
Write the IUPAC name of benzylamine.
Answer:
The IUPAC name is Phenylmethanamine.

Question viii.
Arrange the following amines in an increasing order of boiling points. n-propylamine, ethylmethyl amine, trimethylamine.
Answer:
Amines in an increasing order of boiling points : trimethyl amine, ethyl methyl amine, n-propyl amine

Question ix.
Write the balanced chemical equations for the action of dil H₂SO₄ on diethylamine.
Answer:

Question x.
Arrange the following amines in the increasing order of their _pK_b values. Aniline, Cyclohexylamine, 4-Nitroaniline
Answer:
Cyclohexyl amine (_pK_A 3.34), aniline (_pK_A 9.13) 4-nitroaniline (_pK_A 12.99)

3. Answer the following

Question i.
Identify A and B in the following reactions.

Answer:

Question ii.
Explain the basic nature of amines with suitable example.
Answer:
The basic strength of amines is expressed in terms of K_b or _pK_b value. According to Lowry-Bron-sted theory the basic nature of amines is explained by the following equilibrium equation.

In this equilibrium amine accepts H⁺, hence an amine is a Lowry-Bronsted base.

According to Lewis theory, the species which donates a pair of electrons is called a base.

The nitrogen atom in amiqes has a lone pair of electrons, which can be donated to suitable acceptor like proton H⁺.

The aqueous solutions of amines are basic in nature due to release of free OH^– ions in solutions. Hence amines are Lewis bases. There exists an equilibrium in their aqueous solutions as follows :

R – NH₂ + H₂O ⇌ RNH₃ + OH^–

Since OH^– is a stronger base, equilibrium shifts towards left-hand side giving less concentration of OH^–.

Here, K_b value is smaller and _pK_b value is larger.

Hence amines are weak bases.

Question iii.
What is diazotisation? Write diazotisation reaction of aniline.
Answer:
Aryl amines react with nitrous acid in cold condition (273 – 278 K) forms arene diazonium salts. The conversion of primary aromatic amine into diazonium salts is called diazotisation.

Diazotisation of aniline :

Question iv.
Write reaction to convert acetic acid into methylamine.
Answer:

Question v.
Write a short note on coupling reactions.
Answer:

Reactions involving retention of diazo group : (Coupling reactions) :

Question vi.
Explain Gabriel phthalimide synthesis.
Answer:
Phthalimide is reacted with alcoholic KOH to form potassium phthalimide. Further potassium phthalimide is treated with an ethyl iodide. The product N-ethylphthalimide is hydrolysed with aq NaOH to form ethyl amine. This reaction is known Gabriel phthalimide synthesis.

Question vii.
Explain carbylamine reaction with suitable examples.
Answer:
Aliphatic or aromatic primary amines on heating with chloroform and alcoholic potassium hydroxide solution form carbyl amines or isocyanides with extremely unpleasant smell. This reaction is a test for primary amines.

Secondary and tertiary amines do not give this test.

Question viii.
Write reaction to convert
(i) methanamine into ethanamine
(ii) Aniline into p-bromoaniline.
Answer:
(1) Methanamine into ethanamine

(2) Aniline into p-bromo aniline

Question ix.
Complete the following reactions :
a. C₆H₅N₂Cl + C₂H₅OH →
b. C₆H₅NH₂ + Br₂(aq) → ?
Answer:
(a)

(b)

Question x.
Explain Ammonolysis of alkyl halides.
Answer:
When an alkyl halide is heated with alcoholic ammonia in a sealed tube under pressure at 373 K, a mixture of primary, secondary, tertiary amines and a quaternary ammonium salt is obtained. In this reaction, breaking of C – X bond by ammonia is called ammonolysis of alkyl halides. The reaction is also known as alkylation. For example, when methyl bromide is heated with alcoholic ammonia at 373 K, it gives a mixture of methylamine (a primary amine), dimethylamine (a secondary amine), trimethyl amine (a tertiary amine) and tetramethylam- monium bromide (a quaternary ammonium salt).

The order of reactivity of alkyl halides with ammonia is R – I > R – Br > R – Cl.

Question xi.
Write reaction to convert ethylamine into methylamine.
Answer:

4. Answer the following.

Question i.
Write the IUPAC names of the following amines :

Answer:

Question ii.
What are amines? How are they classified?
Answer:
Amines are classified on the basis of the number of hydrogen atoms of ammonia that are replaced by alkyl group. Amines are classified as primary (1°), secondary (2°) and tertiary (3°).

(1) Primary amines (1° amines) : The amines in which only one hydrogen atom of ammonia is replaced by an alkyl group or aryl group are called primary (1°) amines.

Examples :
(i) CH₃ – NH₂ methylamine
(ii) CH₃ – CH₂ – NH₂ ethylamine

(2) Secondary amines (2° amines) : The amines in which two hydrogen atoms of ammonia are replaced by two, same or different alkyl or aryl groups are called secondary (2°) amines.

Examples :
(i) C₂H₅ – NH – CH₃ ethylmethylamine
(ii) CH₃ – NH – CH₃ dimethylamine

(3) Tertiary amines (3° amines) : The amines in which all the three hydrogen atoms of ammonia are replaced by three same or different alkyl or aryl groups are called tertiary (3°) amines.

Examples :

Secondary and tertiary amines are further classified as (1) Simple or symmetrical amines (2) Mixed or unsymmetrical amines.

(i) Simple or symmetrical amines : In simple amines same alkyl groups are attached to the nitrogen e.g.

(ii) Mixed or unsymmetrical amines : In mixed amines different alkyl groups are attached to the nitrogen.

Question iii.
Write IUPAC names of the following amines.

Answer:

Question iv.
Write reactions to prepare ethanamine from
a. Acetonitrile
b. Nitroethane
c. Propionamide
Answer:
a. Ethanamine from acetonitrile :

b. Ethanamine from nitroethane :

c. Ethanamine from Propionamide :

Question v.
What is the action of acetic anhydride on ethylamine, diethylamine and triethylamine?
Answer:
Acetylation of amines : The reaction in which the H atom attached to nitrogen in amine is replaced by acetyl group is called acetylation of amines.

(1) Ethylamine on reaction with acetic anhydride forms monoacetyl derivative, N-acetylethylamine.

(2) Diethylamine (a secondary amine) on reaction with acetic anhydride forms a monoacetyl derivative, N-acetyldiethyl amine (or N,N-diethyl acetamide).

(3) Triethylamine does not react with acetic anhydride as it does not have any H atom attached nitrogen atom of amin e

Question vii.
Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg’s reagent?
Answer:
This reaction is useful for the distinction of primary, secondary and tertiary amines.

(i) Primary amine (like ethyl amine) is treated with Hinsberg’s reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

(iii) Tertiary amine like triethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH, however it dissolves in dil. HCl to give a clear solution due to formation of ammonium salt.

Question viii.
Write reactions to bring about the following conversions :
a. Aniline into p-nitroaniline
b. Aniline into sulphanilic acid?
Answer:
(1) Aniline into p-nitroaniline

(2) Aniline into sulphanilic acid.

Activity :

Prepare a chart of azodyes, colours and its application.
Prepare a list of names and structures of N-containing ingredients of diet.

12th Chemistry Digest Chapter 13 Amines Intext Questions and Answers

Use your brain power! (Textbook Page No 282)

Question 1.
Classify the following amines as simple/mixed; 1°, 2°, 3° and aliphatic or aromatic. (C₂H₅)₂NH, (CH₃)₃N, C₂H₅ – NH – CH₃,

Answer:

(A) Common Names : Rules

According to common naming system, the amines are named as alkylamines.
The common name of a primary amine is obtained by writing the name of the alkyl group followed by the word ‘amine’.
Example : CH₃ – NH₂ : methyl-amine
The simple {symmetrical) secondary and tertiary amines are written by adding prefix ‘di- (forpresence of two alkyl groups) and ‘tri’- (for presence of three alkyl groups) respectively to the name of alkyl groups.
Examples: (i) CH₃– NH – CH₃ dimethylamine, (ii) (C₂H₅)₃ N triethylamine
The mixed (or unsymmetrical) secondary and tertiary amines are given names by writing the names of alkyl groups in alphabetical order, followed by the word ‘amine’.
Example : CH₃ – CH₂ – NH – CH₃ ethyhnethylamine

(B) IUPAC names : Rules

According to IUPAC system of nomenclature of amines, aliphatic amines are named as alkanarnines.
The name of the amine is obtained by replacing the suffix ‘e’ from parent alkane’s name by ‘amine’.
The position of the amino group is indicated by the lowest possible locant.
Example :
In case of secondary and tertiary amines, the largest alkyl group is considered to be the parent alkane and other alkyl groups are written as N-substituents.
Example : ClH₅NH – CH₃ N – Methylethanamine
A complete name of amine is written as one word.

Try this….. (Textbook Page No 283)

Question 1.
Draw possible structures of all the isomers of C₄H₁₁N. Write their common as well as IUPAC names.
Answer:

Use your brain power! (Textbook Page No 283)

Question 1.
Write chemical equations for

(i) reaction of alc. NH with C₂H₅I.
Answer:

(ii) Amonolysis of benzyl chloride followed by the reaction with 2 moles of CH3I.
Answer:

(2) Ammonolysis of alkyl halides is not suitable method to prepare primary amines.
Answer:
In the laboratory, ammonolysis of alkyl halides is not a suitable method to prepare primary amines as it gives a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. (Refer to the reaction in answer to Question 16). The separation of primary amine becomes difficult.

Problem 13.1 : (Textbook Page No 285)

Question 1.
Write reaction to convert methyl bromide into ethyl amine? Also, comment on the number of carbon atoms in the starting compound and the product.
Solution :
Methyl bromide can be converted into ethyl amine in two stage reaction sequence as shown below.

The starting compound methyl bromide contains one carbon atom while the product ethylamine contains two carbon atoms. A reaction in which number of carbons increases involves a step up reaction. The overall conversion of methyl bromide into ethyl amine is a step up conversion.

Use your brain power! (Textbook Page No 285)

Identify ‘A’ and ‘B’ in the following conversions.

Answer:

Use your brain power! (Textbook Page No 286)

Question 1.
Write the chemical equations for the following conversions :
(1) Methyl chloride to ethylamine.
(2) Benzamide to aniline.
(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine.
(4) Benzamide to benzylamine.
Answer:
(1) Methyl chloride to ethylamine

(2) Benzamide to aniline

(3) 1, 4-Dichlorobutane to hexane-1, 6-diamine

(4) Benzamide to benzylamine

Use your brain power! (Textbook Page No 287)

Question 1.
Arrange the following :
(1) In decreasing order of the boiling point C₂H₅ – OH, C₂H₅ – NH₂, (CH₃)₂ NH
(2) In increasing order of solubility in water: C₂H₅– NH₂, C₃H₇– NH₂, C₆H₅– NH₂
Answer:
(1) Decreasing order of the boiling point : C₂H₅ — OH, C₂H₅ — NH₂, (CH₃)₂ NH
(2) Increasing order of solubility in water : C₆H₅NH₂, C₃H₇ — NH₂, C₂H₅ — NH₂

Use your brain power! (Textbook Page No 288)

Question 1.
Refer to pK_b values and answer which compound from the following pairs is the stronger base?
(1) CH₃ – NH₂ and (CH₃)₂ NH
(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(3) NH₃ and (CH₃)₂ CH – NH₂
Answer:
(1) CH₃ -NH₂ and (CH₃)₂ NH
(CH₃)₂ NH is a stronger base

(2) (C₂H₅)₂ NH and (C₂H₅)₃ N
(C₂H₅)₂ NH is a stronger base

(3) NH₃ and (CH₃)₂ CHNH₂(CH₃)₂ CHNH₂ is a stronger base

Use your brain power! (Textbook Page No 290)

Question 1.
Arrange the following amines in decreasing order of their basic strength :
NH₃, CH₃ – NH₂, (CH₃)₂NH, C₆H₅NH₂
Answer:
Decreasing order of basic strength :
(CH₃)₂NH, CH₃ -NH₂, NH₃, C₆H₅NH₂

Use your brain power! (Textbook Page No 291)

Question 1.

Answer:

Use your brain power! (Textbook Page No 291)

Question 1.
Complete the following reaction :

Answer:

Use your brain power! (Textbook Page No 292)

Answer:

Use your brain power! (Textbook Page No 292)

Question 1.
Write the carbylamine reaction by using aniline as starting material.
Answer:

Can you tell? (Textbook Page No 292)

(1) What is the formula of nitrous acid ?
(2) Can nitrous acid be stored in bottle ?
Answer:
(1) Formula of nitrous acid : H – O – N = O
(2) Nitrous acid cannot be stored in bottle.

Use your brain power! (Textbook Page No 294)

Question 1.
How will you distinguish between methyl amine, dimethylamine and trimethylamine by Hinsberg’s test?
Answer:
(1) Methyl amine (primary amine) reacts with benzene sulphonyl chloride to form N-methylbenzene sulphona- mide

(2) Dimethyl amine reacts with benzene sulphonyl chloride to give N, N – dimethylbenzene sulphonamide.

(3) Trimethyl amine does not react with benzene sulphonyl chloride and remains insoluble in KOH

Problem 13.1 : (Textbook Page No 295)

Question 1.
Write the scheme for preparation of p-bromoaniline from aniline. Justify your answer.
Solution :
NH₂ – group in aniline is highly ring activating and o – /p – directing due to involvement of the lone pair of electrons on ‘N’ in resonance with the ring. As a result, on reaction with Br₂ it gives 2,4,6-tribromoaniline. To get a monobromo product, it is necessary to decrease the ring activating effect of – NH₂ group. This is done by acetylation of aniline. The lone pair of ‘N’ in acetanilide is also involved in resonance in the acetyl group. To that extent, ring activation decreases.

Hence, acetanilide on bromination gives a monobromo product p-bromoacetanilide. After monobromination the original – NH₂ group is regenerated. The protection of – NH₂ group in the form of acetyl group is removed by acid catalyzed hydrolysis to get p-bromoaniline, as shown in the following scheme.

Use your brain power! (Textbook Page No 296)

Question 1.
(1) Can aniline react with a Lewis acid?
(2) Why aniline does not undergo Frledel – Craft’s reaction using aluminium chloride?
Answer:
(1) Aniline reacts with a Lewis acid, forms salt.
(2) Aniline does not undergo Friedcl-Crafr’s reaction (alkylation and acetylation) due to salt formation with aluminium chloride (Lewis acid), which is used as catalyst. Due to this, nitrogen of anime acquires + ve charge and hence acts as strong deactivating effect on the ring and makes it difficult for electrophilic attack.

Can you tell? (Textbook Page No 294)

(1) Do tertiary amines have ‘H’ bonded to ‘N?
(2) Why do tertiary amines not react with benzene sulfonyl chloride?
Answer:
(1) Tertiary amines do not have ‘H’ bonded to ‘N’.
(2) Tertiary amine does not undergo reaction with benzene sulphonyl chloride as it does not have any H atom attached to nitrogen atom of amine.

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 13 Amines Textbook Exercise Questions and Answers.

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 7 Exercise Elements of Groups 16, 17 and 18 Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 7 Elements of Groups 16, 17 and 18 Textbook Exercise Questions and Answers.

Elements of Groups 16, 17 and 18 Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 7 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 7 Exercise Solutions

1. Select appropriate answers for the following.

Question i.
Which of the following has the highest electron gain enthalpy?
A. Fluorine
B. Chlorine
C. Bromine
D. Iodine
Answer:
B. Chlorine

Question ii.
Hydrides of group 16 are weakly acidic. The correct order of acidity is
A. H₂O > H₂S > H₂Se > H₂Te
B. H₂Te > H₂O > H₂S > H₂Se
C. H₂Te > H₂Se > H₂S > H₂O
D. H₂Te > H₂Se > H₂O > H₂S
Answer:
C. H₂Te > H₂Se > H₂S > H₂O

Question iii.
Which of the following element does not show oxidation state of +4 ?
A. O
B. S
C. Se
D. Te
Answer:
A. O

Question iv.
HI acid when heated with conc. H₂SO₄ forms
A. HIO₃
B. KIO₃
C. I₂
D. KI
Answer:
C. I₂

Question v.
Ozone layer is depleted by
A. NO
B. NO₂
C. NO₃
D. N₂O₅Answer:
A. NO

Question vi.
Which of the following occurs in liquid state at room temperature?
A. HIO₃
B. HBr
C. HCl
D. HF
Answer:
D. HF

Question vii.
In pyrosulfurous acid oxidation state of sulfur is
A. Only +2
B. Only +4
C. +2 and +6
D. Only +6
Answer:
B. Only + 4

Question viii.
Stability of interhalogen compounds follows the order
A. BrF > IBr > ICl > ClF > BrCl
B. IBr > BeF > ICl > ClF > BrCl
C. ClF > ICl > IBr > BrCl > BrF
D. ICl > ClF > BrCl > IBr > BrF
Answer:
C. ClF > ICl > IBr > BrCl > BrF

Question ix.
BrCl reacts with water to form
A. HBr
B. Br₂ + Cl₂
C. HOBr
D. HOBr + HCl
Answer:
D. HOBr + HCl

Question x.
Chlorine reacts with excess of fluorine to form.
A. ClF
B. ClF₃
C. ClF₂
D. Cl₂F₃Answer:
B. ClF₃

Question xi.
In interhalogen compounds, which of the following halogens is never the central atom.
A. I
B. Cl
C. Br
D. F
Answer:
D. F

Question xii.
Which of the following has one lone pair of electrons?
A. IF₃
B. ICl
C. IF₅
D. ClF₃Answer:
C. IF₅

Question xiii.
In which of the following pairs, molecules are paired with their correct shapes?
A. [I₃] : bent
B. BrF₅ : trigonal bipyramid
C. ClF₃ : trigonal planar
D. [BrF₄] : square planar
Answer:
A. [I₃] : bent

Question xiv.
Among the known interhalogen compounds, the maximum number of atoms is
A. 3
B. 6
C. 7
D. 8
Answer:
D. 8

2. Answer the following.

Question i.
Write the order of the thermal stability of the hydrides of group 16 elements.
Answer:
The thermal stability of the hydrides of group 16 elements decreases in the order of H₂O > H₂S > H₂Se > H₂Te.

Question ii.
What is the oxidation state of Te in TeO₂?
Answer:
The oxidation state of Te in TeO₂ is + 4.

Question iii.
Name two gases which deplete ozone layer.
Answer:
Nitrogen oxide (NO) released from exhaust systems of car or supersonic jet aeroplanes and chlorofluorocarbons (Freons) used in aerosol sprays and refrigerators deplete ozone layer.

Question iv.
Give two uses of ClO₂Answer:
(i) ClO₂ is used as a bleaching agent for paper pulp and textiles.
(ii) It is also used in water treatment.

Question v.
What is the action of bromine on magnesium metal?
Answer:
Bromine reacts instantly with magnesium metal to give magnesium bromide.

Question vi.
Write the names of allotropic forms of selenium.
Answer:
Selenium has two allotropic forms as follows :
(i) Red (non-metallic) form
(ii) Grey (metallic) form

Question vii.
What is the oxidation state of S in H₂SO₄.
Answer:
The oxidation state of S in H₂SO₄ is + 6.

Question viii.
The pKa values of HCl is -7.0 and that of HI is -10.0. Which is the stronger acid?
Answer:
For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7.
For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. Hence HCl dissoClates more than HI.
Therefore HCl is a stronger acid than HI.

Question ix.
Give one example showing reducing property of ozone.
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O

For example :
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g)→ PbSO_(s) + 4O_2(g)
(ii) Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question x.
Write the reaction of conc. H₂SO₄ with sugar.
Answer:
Concentrated sulphuric acid when added to sugar, it is dehydrated giving carbon.

The carbon that is left behind is called sugar charcoal and the process is called char.

Question xi.
Give two uses of chlorine.
Answer:
Chlorine is used for :

for sterilization of drinking water.
bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
in the extraction of metals like gold and platinum.
in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄-S-C₂H₄-Cl), phosgene (COCl₂) used in warfare.
in the manufacture of tear gas (CCl₃NO₂).

Question xii.
Complete the following.
1. ICl₃ + H₂O …….. + …….. + ICl
2. I₂ + KClO₃ ……. + KIO₂
3. BrCl + H₂O ……. + HCl
4. Cl₂ + ClF₃ ……..
5. H₂C = CH₂ + ICl …….
6. XeF₄ + SiO₂ ……. + SiF₄
7. XeF₆ + 6H₂O …….. + HF
8. XeOF₄ + H₂O ……. + HF
Answer:
1. 2ICI₃ + 3H₂O → 5HCl + HlO₃ + ICl
2. I₂ + KCIO₃ → ICl + KIO₃
3. BrCl + H₂O → HOBr + HCl
4. Cl₂ + C1F₃ → 3ClF
5. CH₂ = CH₂ + ICl → CH₂I – CH₂Cl
6. 2XeF₆ + SiO₂→ 2XeOF₄ + SiF₄7. XeF₆ + 3H₂O → XeO₃ + 6HF
8. XeOF₄ + H₂O→ XeO₂F₂ + 2HF

Question xiii.
Match the following
A – B
XeOF₂ – Xenon trioxydifluoride
XeO₂F₂ – Xenon monooxydifluoride
XeO₃F₂ – Xenon dioxytetrafluoride
XeO₂F₄ – Xenon dioxydifluoride
Answer:
XeOF₂ – Xenon monooxydifluoride
XeO₂F₂ – Xenon dioxydifluoride
XeO₃F₂ – Xenon trioxydifluoride
XeO₂F₄– Xenon dioxytetrafluoride

Question xiv.
What is the oxidation state of xenon in the following compounds?
XeOF₄, XeO₃, XeF₅, XeF₄, XeF₂.
Answer:

Compound	Oxidation state of Xe
XeOF₄	+ 6
XeO₃	+ 6
XeF₆	+ 6
XeF₄	+ 4
XeF₂	+ 2

3. Answer the following.

Question i.
The first ionisation enthalpies of S, Cl and Ar are 1000, 1256 and 1520 kJ/mol^-1, respectively. Explain the observed trend.
Answer:
(i) The atomic number increases as, ₁₆S < ₁₇Cl < ₁₈Ar₁.
(ii) Due to decrease in atomic size and increase in effective nuclear charge, Cl binds valence electrons strongly.
(iii) Hence ionisation enthalpy of Cl (1256 kJ mol^-1) is higher than that of S(1000 kJ mol^-1)
(iv) Ar has electronic configuration 3s²3p⁶. Since all electrons are paired and the octet is complete, it has the highest ionisation enthalpy, (1520 kJ mol^-1)

Question ii.
“Acidic character of hydrides of group 16 elements increases from H₂O to H₂Te” Explain.
Answer:
(i) The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te. This is because the bond dissociation enthalpy of the H-E bond decreases down the group.
(ii) Thus, the acidic character increases from H₂O to H₂Te.

Question iii.
How is dioxygen prepared in laboratory from KClO₃?
Answer:
By heating chlorates, nitrates and permanganates.
Potassium chlorate in the presence of manganese dioxide on heating decomposes to form potassium chloride and oxygen.

Question iv.
What happens when
a. Lead sulfide reacts with ozone (O₃).
b. Nitric oxide reacts with ozone.
Answer:
(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄) changing the oxidation state of S from – 2 to +6.
PbS_(s) + 4O_3(g) → PbSO_(s) + 4O_2(g)

(ii) Ozone oxidises nitrogen oxide to nitrogen dioxide.
NO_(g) + O_3(g) → NO_2(g) + O_2(g)

Question v.
Give two chemical reactions to explain oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question vi.
Discuss the structure of sulfur dioxide.
Answer:
(i) SO₂ molecule has a bent V shaped structure with S-O-S bond angle 119.5° and bond dissoClation enthalpy is 297 kJ mol^-1.
(ii) Sulphur in SO₂ is sp² hybridised forming three hybrid orbitals. Due to lone pair electrons, bond angle is reduced from 120° to 119.5°.
(iii) In SO₂, each oxygen atom is bonded to sulphur by σ and a π bond.
(iv) a bond between S and O are formed by sp²-p overlapping.
(v) One of π bonds is formed by pπ – pπ overlapping while other n bond is formed by pπ – dπ overlap.
(vi) Due to resonance both the bonds are identical having observed bond length 143 pm due to resonance,

Question vii.
Fluorine shows only -1 oxidation state while other halogens show -1, +1, +3, +5 and +7 oxidation states. Explain.
Answer:

Halogens have outer electronic configuration ns² np⁵.
Halogens have tendency to gain or share one electron to attain the stable configuration of nearest inert element with configuration ns²np⁶.
Hence they are monovalent and show oxidation state – 1.
Since fluorine does not have vacant d-orbital, it shows only one oxidation state of – 1 while all other halogens show variable oxidation states from – 1 to +7.
These oxidation states are, – 1, +1, + 3, +5 and + 7. Cl and Br also show oxidation states + 4 and + 6 in their oxides and oxyaClds.

Question viii.
What is the action of chlorine on the following
a. Fe
b. Excess of NH₃Answer:
(a) Chlorine reacts with Fe to give ferric chloride.
2Fe + 3Cl₂ → 2FeCl₃

(b) Chlorine reacts with the excess of ammonia to form ammonium chloride, NH₄Cl and nitrogen.

Question ix.
How is hydrogen chloride prepared from sodium chloride?
Answer:

In the laboratory, hydrogen chloride, HCl is prepared by heating a mixture of NaCl and concentrated H₂SO₄.
Hydrogen chloride gas, is dried by passing it through a dehydrating agent like concentrated H₂SO₄ and then collected by upward displacement of air.

Question x.
Draw structures of XeF₆, XeO₃, XeOF₄, XeF₂.
Answer:

Question xi.
What are interhalogen compounds? Give two examples.
Answer:
Interhalogen compounds : Compounds formed by the combination of atoms of two different halogens are called interhalogen compounds. In an interhalogen compound, of the two halogen atoms, one atom is more electropositive than the other. The interhalogen compound is regarded as the halide of the more electropositive halogen.
For example ClF, BrF₃, ICl

Question xii.
What is the action of hydrochloric acid on the following?
a. NH₃
b. Na₂CO₃Answer:
a. Hydrochloric acid reacts with ammonia to give white fumes of ammonium chloride.
NH₃ + HCl → NH₄Cl

b. Hydrochloric acid reacts with sodium carbonate to give sodium chloride, water with the liberation of carbon dioxide gas.
Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Question xiii.
Give two uses of HCl.
Answer:
Hydrogen chloride (OR hydrochloric acid) is used :

in the manufacture of chlorine and ammonium chloride,
to manufacture glucose from com, starch
to manufacture dye
in mediClne and galvanising
as an important reagent in the laboratory
to extract glue from bones and for the purification of bone black.
for dissolving metals, Fe + 2HCl_(aq) → FeCl₂ + H_2(g)

Question xiv.
Write the names and structural formulae of oxoacids of chlorine.
Answer:

Question xv.
What happens when
a. Cl₂ reacts with F₂ in equal volume at 437 K.
b. Br₂ reacts with excess of F₂.
Answer:
(a) Cl₂ reacts with F₂ in equal volumes at 437 K to give chlorine monofluoride ClF.

(b) Br₂ reacts with excess of F₂ to give bromine trifluoride BF₃.

Question xvi.
How are xenon fluorides XeF₂, XeF₄ and XeF₆ obtained ? Give suitable reactions.
Answer:
Xenon fluorides are generally prepared by the direct reaction of xenon and fluorine in different ratios and under appropriate experimental conditions, such as temperature, in the presence of an electric discharge and by a photochemical reaction.
(i) Preparation of XeF₂ :

(ii) Preparation of XeF₄ :

(iii) Preparation of XeF₆:

Question xvii.
How are XeO₃ and XeOF₄ prepared?
Answer:
Preparation of XeO₃ : Xenon trioxide (XeO₃) is prepared by the hydrolysis of XeF₄ or XeF₆.

By hydrolysis of XeF₄ :
3XeF₄ + 6H₂0 → 2Xe + XeO₃ + 12 HF + \(1 \frac{1}{2} \mathrm{O}_{2}\)
By hydrolysis of XeF₆ :
XeF₆ + 3H₂O → XeO₃ + 6HF
Preparation of XeOF₄ :
Xenon oxytetrafluoride (XeOF₄) is prepared by the partial hydrolysis of XeF₆.
XeF₆ + H₂O → XeOF₄ + 2HF

Question xviii.
Give two uses of neon and argon.
Answer:
Uses of neon (Ne) :

Neon is used in the production of neon discharge lamps and signs by filling Ne in glass discharge tubes.
Neon signs are visible from a long distance and also have high penetrating power in mist or fog.
A mixture of neon and helium is used in voltage stabilizers and current rectifiers.
Neon is also used in the production of lasers and fluorescent tubes.

Uses of argon (Ar) :

Argon is used to fill fluorescent tubes and radio valves.
It is used to provide inert atmosphere for welding and production of steel.
It is used along with neon in neon sign lamps to obtain different colours.
A mixture of 85% Ar and 15% N2 is used in electric bulbs to enhance the life of the filament.

Question xix.
Describe the structure of Ozone. Give two uses of ozone.
Answer:
(A)

Ozone has molecular formula O₃.
The lewis dot and dash structures for O₃ are :
Infrared and electron diffraction spectra show that O₃ molecule is angular with 0-0-0 bond angle 117°.
Both 0-0 bonds are identical having bond length 128 pm which is intermediate between single and double bonds.
This is explained by considering resonating structures and resonance hybrid.

(B) Uses of Ozone :

Ozone sterilises drinking water by oxidising germs and bacteria present in it.
It is used as a bleaching agent for ivory, oils, starch, wax and delicate fabrics like silk.
Ozone is used to purify the air in crowded places like Clnema halls, railways, tunnels, etc.
In industry, ozone is used in the manufacture of synthetic camphor, potassium permanganate, etc.

Question xx.
Explain the trend in following atomic properties of group 16 elements.
i. Atomic radii
ii. Ionisation enthalpy
iii. Electronegativity.
Answer:
(1) Atomic and ionic radii :

As compared to group 15 elements, the atomic and ionic radii of group 16 elements are smaller due to higher nuclear charge.
The atomic and ionic radii increase down the group from oxygen to polonium. This is due to the addition of a new shell at each successive elements on moving down the group. The atomic radii increases in the order O < S < Se < Te < Po

(2) Ionisation enthalpy :

The ionisation enthalpy of group 16 elements has quite high values.
Ionisation enthalpy decreases down the group from oxygen to polonium. This is due to the increase in atomic volume down the group.
The first ionisation enthalpy of the lighter elements of group 16 (O, S, Se) have lower values than those of group 15 elements in the corresponding periods. This is due to difference in their electronic configurations.

Group 15 : (valence shell) ns² np_x¹ np_y¹ np_z¹
Group 16 : (valence shell) ns² np_x² np_y¹ np_z¹

Group 15 elements have extra stability of half-filled and more symmetrical orbitals, while group 16 elements acquire extra stability by losing one of paired electrons from npx- orbital forming half-filled p-orbitals.

Hence group 16 elements have lower first ionisation enthalpy than group 15 elements.

(3) Electronegativity :

The electronegativity values of group 16 elements have higher values than corresponding group 15 elements in the same periods.
Oxygen is the second most electronegative elements after fluorine. (O = 3.5, F = 4)
On moving down the group electronegativity decreases from oxygen to polonium.
On moving down the group atomic size increases, hence nuclear attraction decreases, therefore electro-negativity decreases.

Elements	O	S	Se	Te	Po
Electronegativity	3.5	2.44	2.48	2.01	1.76

4. Answer the following.

Question i.
Distinguish between rhombic sulfur and monoclinic sulfur.
Answer:

Rhombic sulphur	Monoclinic sulphur
1. It is pale yellow.	1. It is bright yellow.
2. Orthorhombic crystals	2. Needle-shaped monoclinic crystals
3. Melting point, 385.8 K	3. Melting point, 393 K
4. Density, 2.069 g/cm³	4. Density: 1.989 g/cm³
5. Insoluble in water, but soluble in CS₂	5. Soluble in CS₂
6. It is stable below 369 K and transforms to α-sulphur above this temperature.	6. It is stable above 369 K and transforms into β-sulphur below this temperature.
7. It exists as S₈ molecules with a structure of a puckered ring.	7. It exists as S₈ molecules with a structure of a puckered ring.
8. It is obtained by the evaporation of roll sulphur in CS₂	8. It is prepared by melting rhombic sulphur and cooling it till a crust is formed. Two holes are pierced in the crust and the remaining liquid is poured to obtain needle-shaped crystals of monoclinic sulphur (β-sulphur).

Question ii.
Give two reactions showing oxidizing property of concentrated H₂SO₄.
Answer:
Hot and concentrated H₂SO₄ acts as an oxidising agent, since it gives nascent oxygen on heating.

Question iii.
How is SO₂ prepared in the laboratory from sodium sulfite? Give two physical properties of SO₂.
Answer:
(A) Laboratory method (From sulphite) :

Sodium sulphite on treating with dilute H₂SO₄ forms SO₂.
Na₂SO₃ + H₂SO_4(aq) → Na₂SO₄ + H₂O₍₁₎ + SO_2(g)
Sodium sulphite, Na₂SO₃ on reaction with dilute hydrochloric acid solution forms SO₂.
Na₂SO_3(aq) + 2HCl_(aq) → 2NaCl_9aq0 + H₂O₍₁₎ + SO_2(g)

(B) Physical properties of SO₂

It is a colourless gas with a pungent smell.
It is highly soluble in water and forms sulphurous acid, H₂SO₃SO_2(g) + H₂O₍₁₎ → H₂SO_3(aq)
It is poisonous in nature.
At room temperature, it liquefies at 2 atmospheres. It has boiling point 263K.

Question iv.
Describe the manufacturing of H₂SO₄ by contact process.
Answer:
Contact process of the manufacture of sulphuric acid involves following steps :

(1) Preparation of SO₂ : Sulphur or pyrite ores like iron pyrites, FeS₂ on burning in excess of air, form SO₂.

(2) Oxidation of SO₂ to SO₃ : SO₂ is oxidised to SO₃ in the presence of a heterogeneous catalyst V₂O₅ and atmospheric oxygen. This oxidation reaction is reversible.

To avoid the poisoning of a costly catalyst, it is necessary to make SO₂ free from the impurities like dust, moisture, As₂O₃ poison, etc.

The forward reaction is exothermic and favoured by increase in pressure. The reaction is carried out at high pressure (2 bar) and 720 K temperature. The reacting gases, SO₂ and O₂ are taken in the ratio 2:3.

(3) Dissolution of SO₃ : SO₃ obtained from catalytic converter is absorbed in 98%. H₂SO₄ to obtain H₂S₂O₇, oleum or fuming sulphuric acid.

Flow diagram for the manufacture of sulphuric acid

Question 7.1 (Textbook Page No 141)

12th Chemistry Digest Chapter 7 Elements of Groups 16, 17 and 18 Intext Questions and Answers

Question 1.
Elements of group 16 generally show lower values of first ionisation enthalpy compared to the elements of corresponding period of group 15. Why?
Answer:
Group 15 elements have extra stable, half filled p-orbitals with electronic configuration (ns²np³). Therefore more amount of energy is required to remove an electron compared to that of the partially filled orbitals (ns²np⁴) of group 16 elements of the corresponding period.

Question 7.2 (Textbook Page No 141)

Question 1.
The values of first ionisation enthalpy of S and Cl are 1000 and 1256 kJ mol^-1, respectively. Explain the observed trend.
Answer :
The elements S and Cl belong to second period of the periodic table.
Across a period effective nuclear charge increases and atomic size decreases with increase in atomic number. Therefore the energy required for the removal of electron from the valence shell (I.E.) increases in the order S < Cl.

Question 7.4 (Textbook Page No 141)

Question 1.
Fluorine has less negative electron gain affinity than chlorine. Why?
Answer :
The size of fluorine atom is smaller than chlorine atom. As a result, there are strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and therefore, the incoming electron does not experience much attraction. Thus fluorine has less negative electron gain affinity than chlorine.

Try this… (Textbook Page No 140)

Question 1.
Explain the trend in the following properties of group 17 elements.

(1) Atomic size,
(2) Ionisation enthalpy,
(3) Electronegativity,
(4) Electron gain enthalpy.
Answer:
(1) Atomic size :

Atomic and ionic radii increase down the group as atomic number increases due to the addition of new electronic valence shell to each succeeding element.
The atomic radii increase in the order F < Cl < Br < 1
Halogens possess the smallest atomic and ionic radii in their respective periods since the effective nuclear charge experienced by valence electrons in halogen atoms is the highest.

(2) Ionisation enthalpy :

The ionisation enthalpies of halogens are very high due to their small size and large nuclear attraction.
The ionisation ethalpies decrease down the group since the atomic size increases.
The ionisation enthalpy decreases in the order F > Cl > Br > I.
Among halogens fluorine has the highest ionisation enthalpy due to its smallest size.

Element	F	Cl	Br	I
Ionisation enthalpy kJ/mol	1680	1256	1142	1008

(3) Electronegativity :

Halogens have the highest values for electronegativity due to their small atomic radii and high effective nuclear charge.
Each halogen is the most electronegative element of its period.
Fluorine has the highest electronegativity as compared to any element in the periodic table.
The electronegativity decreases as,
F > Cl > Br > I
4.0 3.2 3.0 2.7 (electronegativity)

(4) Electron gain enthalpy (Δ_egH) :

The halogens have the highest negative values for electron gain enthalpy.
Electron gain enthalpies of halogens are negative indicating release of energy.
Halogens liberate maximum heat by gain of electron as compared to other elements.
Since halogens have outer valence electronic configuration, ns² np⁵, they have strong tendency to accept an electron to complete an octet and acquire electronic configuration of the nearest inert elements.
In case of fluorine due to small size of 2 p-orbitals and high electron density, F has less negative electron gain enthalpy than Cl.
F_(g) + e^– → F^–_(g) Δ_egH = – 333 klmol^-1
Cl_(g) + e^– → Cl^–_(g) Δ_egH = – 349 kJ mol^-1
The variation in electron gain enthalpy is in the order of, Cl > F > Br > I.

Question 2.
Oxygen has less negative electron gain enthalpy than sulphur. Why?
Answer:

Oxygen has a smaller atomic size than sulphur.
It is more electronegative than sulphur.
It has a larger electron density.
Due to high electron density, oxygen does not accept the incoming electron easily and therefore has less electron gain enthalpy than sulphur.

Question 7.3 (Textbook Page No 141)

Question 1.
Why is there a large difference between the melting and boiling points of oxygen and sulphur?
Answer :
Oxygen exists as diatomic molecule (O₂) whereas sulphur exists as polyatomic molecule (S₈). The van der Waals forces of attraction between O₂ molecules are relatively weak owing to their much smaller size. The large van der Waals attractive forces in the S₈ molecules are due to large molecular size. Therefore oxygen has low m.p. and b.p. as compared to sulphur.

Question 7.5 (Textbook Page No 141)

Question 1.
Bond dissoClation enthalpy of F₂ (158.8 kj mol^-1) is lower than that of Cl₂ (242.6 kj mol^-1) Why?
Answer :
Fluorine has small atomic size than chlorine. The lone pairs on each F atom in F₂ molecule are so close together that they strongly repel each other, and make the F – F bond weak. Thus, it requires less amount of energy to break the F – F bond. In Cl₂ molecule the lone pairs on each Cl atom are at a larger distance and the repulsion is less.

Thus Cl – Cl bond is comparatively stronger. Therefore bond dissoClation enthalpy of F₂ is lower than that of Cl₂.

Question 7.6 (Textbook Page No 142)

Question 1.
Noble gases have very low melting and boiling points. Why?
Answer :
Noble gases are monoatomic, the only type of interatomic interactions which exist between them are weak van der Waals forces. Therefore, they can be liquefied at very low temperatures and have very low melting or boiling points.

Can you tell? (Textbook Page No 142)

Question 1.
The first member of the a group usually differs in properties from the rest of the members of the group. Why?
Answer:
The first member of a group usually differs in properties from the rest of the members of the group for the following reasons :

Its small size
High electronegativity
Absence of vacant d-orbitals in its valence shell.

Use your brain power! (Textbook Page No 142)

Question 1.
Oxygen forms only OF₂ with fluorine while sulphur forms SF₆. Explain. Why?
Answer:

Oxygen combines with the most electronegative element fluorine to form OF₂ and exhibits positive oxidation state (+ 2). Since, oxygen does not have vacant J-orbitals it cannot exhibit higher oxidation states.
Sulphur has vacant d-orbitals and hence can exhibit + 6 oxidation state to form SF₆.

Question 2.
Which of the following possesses hydrogen bonding? H₂S, H₂O, H₂Se, H₂Te
Answer:

Oxygen being more electronegative, is capable of forming hydrogen bonding in the compound H₂O.
The other elements S, Se and Te of Group 16, being less electronegative do not form hydrogen bonds.
Thus, hydrogen bonding is not present in the other hydrides H₂S, H₂Se and H₂Te.

Question 3.
Show hydrogen bonding in the above molecule with the help of a diagram.
Answer:

Try this….. (Textbook Page No 143)

Question 1.
Complete the following tables :

Answer:

Can you tell? (Textbook Page No 146)

Question 1.
What is allotropy?
Answer:
The property of some elements to exist in two or more different forms in the same physical state is called allotropy.

Question 2.
What is the difference between allotropy and polymorphism?
Answer:

Allotropy is the existence of an element in more than one physical form. It means that under different conditions of temperature and pressure an element can exist in more than one physical forms.
Coal, graphite and diamond etc., are different allotropic forms of carbon.
Polymorphism is the existence of a substance in more than one crystalline form.
It means that under different conditions of temperature and pressure, a substance can form more than one type of crystal. For example, mercuric iodide exists in the orthorhombic and trigonal form.

Question 7.7 (Textbook Page No 146)

Which form of sulphur shows paramagnetic behaviour?
Answer :
In the vapour state, sulphur partly exists as S₂ molecule, which has two unpaired electrons in the antibonding π^* orbitals like O₂. Hence it exhibits paramagnetism.

Try this….. (Textbook Page No 149)

Question 1.
Why water in a fish pot needs to be changed from time to time?
Answer:
A fish pot is an artificial ecosystem and the fish in it are selective and maintained in a restricted environment.

In a fish pot, the unwanted food and waste generated by the fish mix with the water and remain untreated due to lack of decomposers.

Accumulation of waste material will decrease the levels of dissolved oxygen in the water pot.

Hence, it is necessary to change the water from time to time.

Question 7.8 (Textbook Page No 149)

Dioxygen is paramagnetic in spite of having an even number of electrons. Explain.
Answer :
Dioxygen is a covalently bonded molecule.
The paramagnetic behaviour of O₂ can be explained with the help of molecular orbital theory.
Electronic configuration O₂
KK σ(2s)² σ(2s)² σ^*(2p_z)² π(2p_x)² π(2p_x)² π(2p_y)²π^*(2p_x)¹ π^*(2p_y)¹. Presence of two unpaired electrons in antibonding orbitals explains paramagnetic nature of dioxygen.

Question 7.9 (Textbook Page No 150)

High concentration of ozone can be dangerously explosive. Explain.
Answer :
Thermal stability : Ozone is thermodynamically unstable than oxygen and decomposes into O₂. The decomposition is exothermic and results in the liberation of heat (ΔH is – ve) and an increase in entropy (ΔS is positive). This results in large negative Gibbs energy change (ΔG). Therefore high concentration of ozone can be dangerously explosive. Eq O₃ → O₂ + O

Try this…… (Textbook Page No 151)

(a) Ozone is used as a bleaching agent. Explain.
Answer:

Ozone due to its oxidising property can act as a bleaching agent. O_3(g) → O_2(g) + O
It bleaches coloured matter. coloured matter + O → colourless matter
Ozone bleaches in the absence of moisture, so it is also known as dry bleach.
Ozone can bleach ivory and delicate fabrics like silk.

(b) Why does ozone act as a powerful oxidising agent?
Answer:
Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent. O_3(g) → O_2(g) + O
For example :

It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).
pbS_(s) + 4O_3(g)→ PbSO_(s) + 4O_2(g)
Potassium iodide, KI is oxidised to iodine, I₂ in the solution.
2KI_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

Question 7.10 : (Textbook Page No 154)

What is the action of concentrated H₂SO₄ on (a) HBr (b) HI
Answer :
Concentrated sulphuric acid oxidises hydrobromic acid to bromine.

2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
It oxidises hydroiodic acid to iodine.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O

Try this….. (Textbook Page No 156)

Question 1.
Give the reasons for the bleaching action of chlorine.
Answer:

Chlorine acts as a powerful bleaching agent due to its oxidising nature.
In moist conditions or in the presence of water it forms unstable hypochlorous acid, HOCl which decomposes giving nascent oxygen which oxidises the vegetable colouring matter of green leaves, flowers, litmus, indigo, etc.
Cl₂ + H₂O → HCl + HOCl
HOCl → HCl + [O]
Vegetable coloured matter + [O] → colourless matter.

Question 2.
Name two gases used in war.
Answer:
Phosgene : COCl₂
Mustard gas: Cl – CH₂ – CH₂ – S – CH₂ – CH₂ – Cl

Use your brain power! (Textbook Page No 157)

Question 1.
Chlorine and fluorine combine to form interhalogen compounds. The halide ion will be of chlorine or fluorine?
Answer:
Among the- two halogens, chlorine is more electropositive than fluorine (Electronegativity values: F = 4.0, Cl = 3.2)

The interhalogen compound is regarded as the halide of the more electropositive halogen. Hence, the interhalogen compound is the fluoride of chlorine, i.e. chlorine monofluoride, CiF.

Question 2.
Why does fluorine combine with other halogens to form maximum number of fluorides?
Answer:
Since fluorine is the most electronegative element and has the smallest atomic radius compared to other halogen fluorine forms maximum number of fluorides.

Use your brain power! (Textbook Page No 158)

Question 1.
What will be the names of the following compounds: ICl, BrF?
Answer:
ICl : Iodine monochloride
BrF : Bromine monofluoride

Question 2.
Which halogen (X) will have maximum number of other halogen (X ) attached?
Answer:
The halogen Iodine (I) will have the maximum number of other halogens attached.

Question 3.
Which halogen has tendency to form more interhalogen compounds?
Answer:
The halogen fluorine (F) has the maximum tendency to form more interhalogen compounds as it has a small size and more electronegativity.

Question 4.
Which will be more reactive?
(a) ClF₃ or ClF,
(b) BrF₅ or BrF
Answer:
ClF₃ is more reactive than ClF, while BrF₅ is more reactive than BrF. Both ClF₃ and BrF₅ are unstable compared to ClF and BrF respectively due to steric hindrance hence are more reactive.

Question 5.
Complete the table :

Formula	Name
ClF	Chlorine monofluoride
ClF₃	…………………………………
…………………………………	Chlorine pentachloride
BrF	…………………………………
…………………………………	Bromine pentafluoride
ICl	…………………………………
ICl₃	…………………………………

Answer:

Formula	Name
ClF	Chlorine monofluoride
ClF₃	Chlorine trifluoride
CIF₅	Chlorine pentafluoride
BrF	Bromine monofluoride
BrF₅	Bromine pentafluoride
ICl	Iodine monochloride
ICl₃	Iodine trichloride

Use your brain power! (Textbook Page No 159)

Question 1.
In the special reaction for ICl, identify the oxidant and the reductant? Denote oxidation states of the species.
Answer:

Potassium chlorate, KClO₃ is the oxidising agent or oxidant and iodine is the reducing agent or reductant.

Use your brain power! (Textbook Page No 162)

Question 1.
What are missing entries?

Formula	Name
XeOF₂…………… XeO₃F₂XeO₂F₄	Xenon monooxyfluoride Xenon dioxydifluoride …………………………………….. ……………………………………..

Answer:

Formula	Name
XeOF₂XeO₂F₂XeO₃F₂XeO₂F₄	Xenon monooxydifluoride Xenon dioxydifluoride Xenon trioxydifluoride Xenon dioxytetrafluoride

12th Std Chemistry Questions And Answers:

12th Chemistry Chapter 14 Exercise Biomolecules Solutions Maharashtra Board

January 8, 2025January 7, 2025 by Bhagya

Balbharti Maharashtra State Board 12th Chemistry Textbook Solutions Chapter 14 Biomolecules Textbook Exercise Questions and Answers.

Biomolecules Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 14 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 14 Exercise Solutions

1. Select the most correct choice.

Question i.
CH₂OH-CO-(CHOH)₄-CH₂OH is an example of
a. Aldohexose
b. Aldoheptose
c. Ketotetrose
d. Ketoheptose
Answer:
(d) Ketoheptose

Question ii.
Open chain formula of glucose does not contain
a. Formyl group
b. Anomeric hydroxyl group
c. Primary hydroxyl group
d. Secondary hydroxyl group
Answer:
(b) Anomeric hydroxyl group

Question iii.
Which of the following does not apply to CH₂NH₂ – COOH
a. Neutral amino acid
b. L – amino acid
c. Exists as zwitterion
d. Natural amino acid
Answer:
(d) Natural amino acid

Question iv.
Tryptophan is called essential amino acid because
a. It contains an aromatic nucleus.
b. It is present in all the human proteins.
c. It cannot be synthesized by the human body.
d. It is an essential constituent of enzymes.
Answer:
(c) It cannot be synthesised by human body.

Question v.
A disulfide link gives rise to the following structure of protein.
a. Primary
b. Secondary
c. Tertiary
d. Quaternary
Answer:
(c) Tertiary

Question vi.
RNA has
a. A – U base pairing
b. P – S – P – S backbone
c. double helix
d. G – C base pairing
Answer:
(a) A – U base pairing

2. Give scientific reasons :

Question i.
The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
Answer:
(1) In disaccharide sucrose, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a nonreducing sugar. As there is no free aldehyde group, it does not reduce Tollen’s reagent to metallic silver. Hence, sucrose gives negative Tollen’s test.

(2) While the disaccharide maltose is a reducing sugar because a free aldehyde group can be produced at C₁ of second sugar molecule. It is a reducing sugar. It reduces Tollen’s reagent to shining silver mirror. Hence, Maltose gives positive Tollen’s test.

Question ii.
On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
Answer:
On complete hydrolysis DNA yields 2-deoxy-D-ribose, adenine, thymine, guanine, cystosine and phosphoric acid. Since adenine always forms two hydrogen bonds with thymine, the hydrolysis of DNA gives equimolar quantities of adenine and thymine.

Question iii.
α – Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
Answer:

α-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, proton transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called zwitter ion.

Question iv.
Hydrolysis of sucrose is called inversion.
Answer:

Sucrose is dextro rotatory. On hydrolysis it gives equimolar mixture of D – ( + ) glucose and D – ( -) fructose. Since the laevorotation of fructose (- 92.4°) is more than dextrorotation of glucose ( + 52.7°), the hydrolysis product has net laevorotation. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro ( + ) to laevo (-) and the product is called as invert sugar and so the hydrolysis of sucrose is called inversion.

Question v.
On boiling, egg albumin becomes opaque white.
Answer:
Upon boiling the egg, denaturation αcurs. During denaturation, secondary and tertiary structures are destroyed, but primary structure remains intact. Egg contains soluble globular proteins, which forms insoluble fibrous proteins (opque) on boiling egg.

3. Answer the following

Question i.
Some of the following statements apply to DNA only, some to RNA only and some to both. Lable them accordingly.
a. The polynucleotide is double stranded. ( …………… )
b. The polynucleotide contains uracil. ( …………… )
c. The polynucleotide contains D-ribose ( …………… ).
d. The polynucleotide contains Guanine ( …………… ).
Answer:
(1) The polynucleotide is double stranded. (DNA)
(2) The polynucleotide contains uracil. (RNA)
(3) The polynucleotide contain D-ribose (RNA)
(4) Thc polynucleotide contains Guanine (DNA, RNA)

Question ii.
Write the sequence of the complementary strand for the following segments of a DNA molecule.
a. 5′ – CGTTTAAG – 3′
b. 5′ – CCGGTTAATACGGC – 3′
Answer:
(1) DNA molecule : 5′ – CGTTTAAG – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

(2) DNA molecule : 5′ – CCGGTTAATACGGC – 3′
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Question iii.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Answer:
(1) Dipeptide formed from alanine and glycine

(2) Dipeptide formed from alanine and tyrosine

(3) Dipeptide formed from glycine and tyrosine.

Question iv.
Give two pieces of evidence for the presence of the formyl group in glucose.
Answer:
(1) Glucose reacts with hydroxyl amine in an aqueous solution to form glucose oxime. This indicates the presence of – CHO (formyl group) in glucose.

(2) Glucose on oxidation with mild oxidising agent like bromine water gives gluconic acid which shows carbonyl group in glucose is aldehyde (formyl group) group.

4. Draw a neat diagram for the following:

Question i.
Haworth formula of glucopyranose
Answer:

Question ii.
Zwitter ion
Answer:

Question iii.
Haworth formula of maltose
Answer:

Question iv.
Secondary structure of the protein

Answer:
The structure of proteins can be studied at four different levels i.e. primary, secondary, tertiary and quaternary levels. Each level is more complex than the previous one.
(1) Primary structure of proteins :
(a) Representation by structural formula

(b) Representation with amino acid symbols

Primary structure of proteins is the sequence of constituent a-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residue creates different protein molecule. Primary structure of proteins is represented by writing the three letter symbols of amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes. According to the convention, the N-terminal amino acid residue as written at the left end and the C-terminal amino acid residue at the right end.

(2) Secondary structure of proteins : The three-dimensional arrangement of lαalized regions of a long polypeptide chain is called the secondary structure of protein. Hydrogen bonding between N-H proton of one amide linkage and C = O oxygen of another gives rise to the secondary structure. There are two different types of secondary structures i.e. α-helix and β-pleated sheet.

α-Helix : In a-helix structure, a polypeptide chain gets coiled by twisting into a right handed or clαkwise spiral known as a-helixn. The characteristic features of α-helical structure of protein are :

(1) Each turn of the helix has 3.6 amino acids.
(2) A C = O group of one amino acid is hydrogen bonded to N – H group of the fourth amino acid along the chain.
(3) Hydrogen bonds are parallel to the axis of helix while R groups extend outward from the helix core.
Myosin in muscle and a-keratin in hair are proteins with almost entire a-helical secondary structure.

β-Pleated sheet : In secondary structure, when two or more polypeptide chains (strands) line up side-by-side is called β-pleated sheets. The β-picate sheet structure of protein consists of extended strands of polypeptide chains held together by intermolecular hydrogen bonding. The characteristics of β-pleated sheet structure are :

The C = O and N – H bonds lie in the planes of the sheet.
Hydrogen bonding occurs between the N – H and C = O groups of nearby amino acid residues in the neighbouring chains.
The R groups are oriented above and below the plane of the sheet.

The β-pleated sheet arrangement is favoured by amino acids with small R groups.

(3) Tertiary structure of proteins :

The three-dimensional shape acquired by the entire polypeptide chain of a protein is called its tertiary structure. The structure is stabilized and has attractive interaction with the aqueous environment of the cell due to the folding of the chain in a particular manner. Tertiary structure gives rise to two major molecular shapes i.e. globular and fibrous proteins. The main forces which stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like -COO^–, \(\mathrm{NH}_{3}^{+}\) in the side chain) and also London dispersion forces. Finally, disulfide bonds formed by oxidation of nearby – SH groups (in cysteine residues) are the covalent bonds which stabilize the tertiary structure.

(4) Quaternary structure of proteins The two or more polypeptide chains with folded tertiary structures forms complex protein. The spatial arrangements of these polypeptide chains with respect to each other is known as quaternary structure. Each individual polypeptide chain is called a subunit of the overall protein. For example: Haemoglobin consists of four subunits called haeme held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.

Question v.
AMP
Answer:

Question vi.
dAMP
Answer:

Question vii.
One purine base from nucleic acid
Answer:

Question viii.
Enzyme catalysis
Answer:

Activity :

Draw the structure of a segment of DNA comprising at least ten nucleotides on a big chart paper.
Make a model of DNA double stranded structure as group activity.

12th Chemistry Digest Chapter 14 Biomolecules Intext Questions and Answers

Try ….. this (Textbook Page No 298)

Question 1.
Observe the following structural formulae carefully and answer the questions.

(1) How many OH groups are present in glucose, fructose and ribose respectively?
(2) Which other functional groups are present in these three compounds?
Answer:
(1) Glucose contains five hydroxyl (- OH) groups.
Fructose contains five hydroxyl ( – OH) groups.
Ribose contains four hydroxyl ( – OH) groups.

(2) Glucose contains aldehyde ( – CHO) as other functional group.
Fructose contains ketonic group as other functional group.
Ribose contains aldehyde ( – CHO) as other functional group.

Use your brain power! (Textbook Page No 299)

Question 1.
Give IUPAC names to the following monosaccharides.

Answer:
(1) Aldotriose
(2) Aldopentose
(3) Ketoheptose

Problem 14.1 : (Textbook Page No 300)

Question 1.
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. Molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (- OH) groups must be present in the original compound?
Solution :
In acetylation reaction H atom of an (- OH) group is replaced by an acetyl group (- COH₃).

This results in an increase in molecular mass by [(12 + 16 + 12 + 3 x 1) – 1] that has, 42 u. In the given alcohol, increase in molecular mass = 174 u – 90 u = 84 u
∴ Number of – OH groups \(=\frac{84 \mathrm{u}}{42 \mathrm{u}}=2\)

Use your brain power! (Textbook Page No 301)

(1) Write structural formula of glucose showing all the bonds in the molecule.
(2) Number all the carbons in the molecules giving number 1 to the ( – CHO) carbon.
(3) Mark the chiral carbons in the molecule with asterisk (*).
(4) How many chiral carbons are present in glucose?
Answer:
Refer structural formula of glucose for (1) (2) and (3).

(4) There are 4 chiral carbon atoms present in glucose.

Use your brain power! (Textbook Page No 306)

Question 1.
(1) Is galactose an aldohexose or a ketohexose?
(2) Which carbon in galactose has different configuration compared to glucose?
(3) Draw Haworth formulae of α-D-galactose and β-D-galactose.
(4) Which disaccharides among sucrose, maltose and lactose is/are expected to give positive Fehling test?
(5) What are the expected products of hydrolysis of lactose?
Answer:

Galactose is an aldohexose.
Fourth carbon in galactose has different configuration compared to glucose.
Maltose and lactose are expected to give positive Fehling solution test.
The expected products of hydrolysis of lactose are D – ( +) glucose and D – ( +) galactose.

Can you think? (Textbook Page No 307)

Question 1.
When you chew plain bread, chapati or bhaakari for long time, it tastes sweet. What could be the reason?
Answer:
When chapati, bread or bhakari are chewed for long time the pulp mixes with saliva and carbohydrate component in them diseminates and gives the sweet taste.

Use your brain power! (Textbook Page No 309)

Question 1.
Tryptophan and histidine have the structures (I) and (II) respectively. Classify them into neutral? acidic/basic &amino acids and justify your answer. (Hint: Consider învolvement of lone pair in resonance).

Answer:

In tryptophan, nitrogen atom present in cyclic structure cannot donate pair of electrons as it is stabilized by resonance. The other amino group and carboxylic group present in the side chain neutralize each other. Tryptophan has equal number of amino and carboxylic groups. Hence, tryptophan is a neutral amino acid.

In histidine, amino groups are more in number than carboxyl groups therefore histidine ¡s basic in nature.

Can you think? (Textbook Page No 309)

Question 1.
Compare the molecular masses of the following compounds and explain the observed melting points.

Answer:
Above compounds have same molecular masses but they have different melting points, a-amino acids have higher melting points compared to the corresponding amines or carboxylic acids of comparable masses. This property is due to the presence of both carboxylic group (acidic) and amino group (basic) in the molecule. In aqueous solution, protons transfer from acidic group to amino (basic) group of amino acid forms a salt, which is a dipolar ion called – Zwitter ion.

Use your brain power! (Textbook Page No 310)

Question 1.
(1) Write the structural formula of dipeptide formed by combination of carboxyl group of alanine and amino group of glycine.
(2) Name the resulting dipeptide.
(3) Is this dipeptide same as glycyalanine or its structural isomer?
Answer:

(2) ala-glycine. OR ala-gly
(3) It is a structural isomer.

Question 54.
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.

Problem 14.3 : (Textbook Page No 311)

Question 1.
Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin. Gly-Tyr-Gly-Ala-Phe-Val
Solution :
In the given hexapeptide hydroylsis by chymotripsin can take place at two points, namely, Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in required range will be :

Problem 14.4 : (Textbook Page No 311)

Question 1.
Write down the structures of amino acids constituting the following peptide.

Solution :
The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one H₂O molecule across the amide bond as follows :

Use your brain power! (Textbook Page No 313)

A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these giving rise to tertiary level structure of protein.

Answer:
Tertiary level structure from amino residues.

Can you tell? (Textbook Page No 313)

Question 1.
What is the physical change observed when (a) egg is boiled, (b) milk gets curdled on adding lemon juice?
Answer:
(a) When egg is boiled, coagulation of eggwhite (insoluble fibrous proteins) takes place. This is a common example of denaturation.
(b) When lemon juice is added to milk, it gets curdled due to the formation of lactic acid. This is another example of denaturation.

Can you tell? (Textbook Page No 315)

Question 1.
What is the single term that answers all the following questions?
(1) What decides whether you are blue eyed or brown eyed?
(2) Why does wheat grain germinate to produce wheat plant and not rice plant?
(3) Which acid molecules are present in nuclei of living cells?
Answer:
(1) Nucleic acid (DNA)
(2) Nucleic acid (DNA)
(3) Nucleic acid (DNA + RNA)

Use your brain power! (Textbook Page No 317)

Question 1.
Draw structural formulae of nucleosides formed from the following sugars and bases.
(1) D – ribose and guanine
(2) D – 2 – deoxyribose and thymine
Answer:
(1) D-ribose and guanine

(2) D – 2 – deoxyribose and thymine

Problem 14.5 (Textbook Page No 318)

Queston 1.
Draw a schematic representaion of trinucicotide segment ACT of a DNA molecule.
Solution :
In DNA molecule sugar is deoxyribose. The base ‘A’ in the given segment is at 5 end while the base T at the 3’ end. I-fence the schematic representation of the given segment of DNA is

Problem 14.6 : (Textbook Page No 320)

Question 1.
Write the sequence of the complementary strand of the following portion of a DNA molecule : 5 -ACGTAC-3
Solution :
The complementary strand runs in opposite direction from the 3′ end to the 5′ end. It has the base sequence decided by complementary base pairs A – T and C – G.

Problem 14.2 : (Text Page No 303)

Question 1.
Assign D/L configuration to the following monosaccharides.

Solution :
D/L configuration is assigned to Fischer projection formula of monosaccharide on the basis of the lowest chiral carbon.

Threose has two chiral carbons C-2 and C-3. The given Fischer projection formula of threose has OH groups at the lowest C -3 chiral carbon on the right side.
∴ It is D-threose.

Ribose has three chiral carbons C – 2, C – 3 and C -4.
The given Fischer projection formula of ribose has – OH group at the lowest C -4 chiral carbon on the left side.
∴ It is L-ribose

12th Std Chemistry Questions And Answers:

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: