NCERT - Learn Chapter-wise Free NCERT Text books

NCERT Solutions for Class 12 Chemistry 2021

Students studying in CBSE class 12 searching for NCERT Solutions for Class 12 Chemistry 2021 can get the same here. This is the place where you will get the NCERT Solutions for Class 12 Chemistry 2021 in a manner that is simple and easy to comprehend and understand. Class 12 is one crucial stage in the life of a student and if you are a student who has opted for science stream and is struggling with the subject Chemistry we have come to make the subject a tad bit more friendly with the NCERT Solutions for Class 12 Chemistry 2021. NCERT Class 12 Chemistry 2021 has 16 Chapters and we have in the sections below covered all these 13 Chapters when giving the NCERT Solutions for Class 12 Chemistry 2021.

Chapters In NCERT Class 12 Chemistry

Chapter 1 - The Solid State

Chapter 2 - Solutions

Chapter 3 - Electrochemistry

Chapter 4 - Chemical Kinetics

Chapter 5 - Surface Chemistry

Chapter 6 - General Principles and Processes of Isolation of Elements

Chapter 7 - The P-block elements

Chapter 8 - The d and f block elements

Chapter 9 - Coordination compounds

Chapter 10 - Haloalkanes and Haloarenes

Chapter 11 - Alcohols, Phenols and Ethers

Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

Chapter 13 - Amines

Chapter 14 - Biomolecules

Chapter 15 - Polymers

Chapter 16 - Chemistry in Everyday life

Chapter Wise NCERT Solutions for Class 12 Chemistry 2021

Chapter 1 - The Solid State

1. Why do solids have a definite volume?

Answer :

At room temperature, intermolecular forces bring molecules so close that they cling to one another and occupy fixed positions. They oscillate about their mean position but they have fixed volume. So, due to the fixed position of particles and strong intermolecular force, solids have a definite volume.

2. Classify the following as amorphous or crystalline solids: Polyurethane

Answer :

Polyurethane is an amorphous solid.

3 Classify the following as amorphous or crystalline solids:

naphthalene

Answer :

Naphthalene is a crystalline solids.

4.. Classify the following as amorphous or crystalline solids: benzoic acid,

Answer :

Benzoic acid is a crystalline solid.

5. Classify the following as amorphous or crystalline solids:

potassium nitrate

Answer :

Potassium nitrate is a crystalline solid.

Chapter 2 - Solutions

1. Give an example of a solid solution in which the solute is a gas.

Answer:

Solution of hydrogen in palladium is such an example in which solute is a gas and solvent is solid.

2. Define the following terms:

Mole fraction

Answer :

Mole fraction is defined as the ratio of number of moles of a component and total number of moles in all components.

i.e.,

3. Define the following terms:

Molality

Answer :

It is defined as the number of moles of solute dissolved per kg (1000g) of solvent

i.e.,

It is independent of temperature.

4. Concentrated nitric acid used in laboratory work is

nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is

Answer :

According to given question, in 100 g of solution 68 g is nitric acid and rest is water.

So moles of 68 g HNO 3 :-

Density of solution is given to be 1.504.

So volume of 100 g solution becomes :-

Thus, molarity of nitric acid is :

5. Why do gases always tend to be less soluble in liquids as the temperature is raised?

Answer:

It is known that dissolution of gas in a liquid is an exothermic process. So, by Le Chatelier principle we know that equilibrium shifts backwards as we increase temperature in case of exothermic process. Thus gases always tend to be less soluble in liquids as the temperature is raised.

Chapter 3 - Electrochemistry

1. Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.

Answer :

The oxidising strength of elements increases as the standard electrode potential increases.

So all the elements having greater standard potential than iron can oxidise it to a higher state.

Few such elements are :- F 2 , Cl 2 , Br 2 , Ag +1 etc.

2. Why does the conductivity of a solution decrease with dilution?

Answer :

The conductivity of a solution depends upon the number of ions and the distance between them. In the process of dilution, we don't increase the number of ions in the solution instead we increase the distance between them. So the conductivity of the solution decreases due to dilution.

3. Suggest a way to determine the

value of water.

Answer :

We know :

If we draw a straight line between

and , its slope will be -A and the intercept on the y-axis will be .

In this way, we can obtain the value of limiting molar conductivity.

4. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Answer :

The two materials are methane and methanol that can be used as fuels in fuel cells.

5. Explain how rusting of iron is envisaged as setting up of a electrochemical cell.

Answer :

The chemistry of corrosion is quite complex but it can be understood by considering it as an electrochemical phenomenon. Consider a particular spot on an object where corrosion takes place. At here oxidation takes place and this spot behaves as the anode. The released electrons at anodic spot go through the metal and go to another spot on the metal and reduction of oxygen takes place in the presence of H+. This spot behaves as a cathode with the reaction. In this way this analogy is possible.

Chapter 4 - Chemical Kinetics

1. What will be the effect of temperature on rate constant ?

Answer :

The rate constant of the reaction is nearly doubled on rising in 10-degree temperature.

Arrhenius equation depicts the relation between temperature and rates constant.

A= Arrhenius factor

Ea = Activation energy

R = gas constant

T = temperature

2 Mention the factors that affect the rate of a chemical reaction.

Answer :

The following factors that affect the rate of reaction-

• the concentration of reactants

• temperature, and

• presence of catalyst

3. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled

Answer :

Let assume the concentration of reactant be x

So, rate of reaction,

Now, if the concentration of reactant is doubled then

. So the rate of reaction would be

Hence we can say that the rate of reaction increased by 4 times.

4. A reaction is first order in A and second order in B.

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Answer :

If the concentration of [A] and[B] is increased by 2 times, then

Therefore, the rate of reaction will increase 8 times.

5. A first order reaction takes 40 min for 30% decomposition. Calculate

Answer :

For the first-order reaction,

(30% already decomposed and remaining is 70%)

therefore half life = 0.693/k

=

= 77.7 (approx)

Chapter 5 - Surface Chemistry

1. Write any two characteristics of Chemisorption.

Answer :

Two characteristics of Chemisorption-

• The bond formation between the solid(adsorbent) and the gas molecules is by chemical bonds like a covalent bond or ionic bond.

• It is highly specific in nature and also irreversible.

2. Why is it essential to wash the precipitate with water before estimating it quantitatively?

Answer :

The precipitate is formed from reactions so it contains some chemical or unwanted substances traces (like impurities, catalyst and reactant). For quantitative estimation of the product, we need to remove these additional substances which get adsorbed onto the surface of the required product and it gives us the wrong information about the product of the reaction. So to avoid this problem we wash the precipitate with the water (most of the impurities are water soluble).

3. Why are powdered substances more effective absorbents than their crystalline forms?

Answer :

Since adsorption is a surface phenomenon. So, the extent of adsorption directly depends on the surface area. Thus, the powdered substances more effective absorbents than their crystalline forms due to their large surface area which can adsorb more gas as compared to the crystalline form.

4. What is the role of desorption in the process of catalysis?

Answer :

In catalytic reaction, the reactant attached on the surface of the catalyst and form complex and after some procedure, it becomes the product of the reaction. We need to detach it from the surface of the catalyst to get the product and catalyst separately and hereby desorption process, we can get the product.

5. What modification can you suggest in the Hardy Schulze law?

Answer :

According to Hardy-Schulze law that ‘the higher the valence of the flocculating ion added, the higher is its power to cause precipitation. Here he does not about the size of the ion, which polarises the other oppositely charged ion. Hence the modified law should be "the greater the polarising power of flocculating ion added, the greater is its power to cause precipitation."

Chapter 6 - General Principles and Processes of Isolation of Elements

1. Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO? What are those conditions ?

Answer :

Yes, temperature below 1623K, Mg can reduce

to Al and temperature above 1623K Aluminium can reduce to

2. What is the role of depressant in froth floatation process?

Answer :

Role of depressant in froth flotation method is to separate two sulphid ores by adjusting one ore to form froth. For example-

is used as a depressant it selectively prevents to coming to the froth but allows PbS with the froth. NaCN react with to form

Reaction-

3. Explain: (i) Zone refining

Answer :

ZONE REFINING- This method is based on that the impurities are more soluble in the molten state than n the solid state. In this process, a circular mobile heater surrounding the rod of impure metal is fixed at one end. Move the heater from one end to another end so that the impurities present in the rod also moves to the other end of the rod. Repeat the process several times again and again. Impurities get concentrated at separate ends of the rod. This end is cut off.

4. Explain: (ii)Column chromatography.

Answer :

Column chromatography-

The principle of chromatography is based on, the different component of the mixture is getting adsorbed to a different extent on an adsorbent. In this, there are two phases one is mobile and the other is immobile phase. In column chromatography, Al_{2}O_{3} column is generally used as a stationary phase. The mobile phase is may be a gas or liquid. The mobile phase is forced to move over the stationary phase. The adsorbed component is removed with the help of a suitable solvent.

5 Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?

Answer :

The following elements present in the anode mud in electrolytic refining of copper are selenium, silver, tellurium, gold, platinum and antimony. They are present because they are very less reactive and are not affected by the purification process. So, therefore they settle down below as anode mud.

Chapter 7 - The P-block elements

1. Why are pentahalides more covalent than trihalides ?

Answer :

Pentahalides are more covalent than trihalides. This is due to the fact that in pentahalides +5 oxidation state exists while in the case of trihalides +3 oxidation state exists. So, Higher the +ve O.S of the central atom more will be the polarising power and more will be the covalent character in the bond between the central atom and a halogen atom. Since elements in +5 oxidation state will have more polarising power than in +3 oxidation state, the covalent character of bonds is more in pentahalides.

2. Why is

the strongest reducing agent amongst all the hydrides of Group 15 elements ?

Answer :

We see that the stability of hydrides becomes lesser as we go from

to .this can be seen from dissociation enthalpy of their bond. due to that, the reducing character of the hydrides will be more. Ammonia is only a very mild reducing agent while is the strongest reducing agent amongst all of the hydrides.

3. Why is

a liquid and a gas ?

Answer :

has oxygen as the centre atom. oxygen is small in size as well as high electronegative when we compare it with sulpher.molecules of water are highly associated through hydrogen bonding which is not present in sulpher.molecules of are connected to each other through weak van der Wal force only. Hence a liquid and a gas.

4. Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe

Answer :

Since Platinum(Pt) is a noble metal .it will not react with oxygen directly

5. Mention three areas in which

plays an important role.

Answer :

1. Manufacture of fertilisers (e.g., ammonium sulphate, superphosphate) from

   
   .

2. Use is petroleum refining

3. Manufacture of pigments, paints and dyestuff intermediates and detergent industry.

Chapter 8 - The d and f block elements

1. Silver atom has completely filled d-orbitals

in its ground state. How can you say that it is a transition element?

Answer :

Silver atom(atomic no. = 47) has completely filled d-orbital in its ground state(4

). However, in +2 oxidation state, the electron of d-orbitals get removed. As a result, the d-orbital become incomplete( ) . Hence it is a transition element.

2. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Answer :

In 3d series of transition metals Manganese shows largest number of oxidation states because it has highest number of unpaired electrons in its

-orbitals.So, that by removing its all electrons we get different oxidation states.

Example-

etc.

3. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer :

Oxygen and fluorine are strong oxidising agents and both of their oxides and fluorides are highly electronegative in nature and also small in size. Because of these properties, they can oxidise the metal to its highest oxidation states.

4. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer :

Actinoid contraction is greater from element to element than lanthanoid contraction. The reason behind it is the poor shielding effect of 5 f (in actinoids) orbitals than 4 f orbitals( in lanthanoids). As a result, the effective nuclear charge experienced by valence electrons is more in actinoids than lanthanoids elements.

5. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Answer :

On moving along the lanthanoid series, the atomic number is gradually increased by one. It means the no. of electrons and protons of the atom is also increases by one. And because of it the effective nuclear charge increases (electrons are adding in the same shell, and the nuclear attraction overcomes the interelectronic repulsion due to adding of a proton). Also, with the increase in atomic number, the number of electrons in orbital also increases. Due to the poor shielding effect of the electrons, the effective nuclear charge experienced by an outer electron is increased, and also the attraction of the nucleus for the outermost electron is increased. As a result, there is a gradual decrease in the atomic size as an increase in atomic number. This is known as lanthanoid contraction.

Chapter 9 - Coordination compounds

1. Write the formulas for the following coordination compounds:

(i) Tetraamminediaquacobalt(III) chloride

Answer :

The chemical formula for the coordination compound Tetraamminediaquacobalt(III) chloride is

2. Write the formulas for the following coordination compounds:

(ii) Potassium tetracyanidonickelate(II)

Answer :

The formula for the coordination compound Potassium tetracyanidonickelate II is :

3. Write the IUPAC names of the following coordination compounds:

Answer :

The IUPAC name of the coordination compound

is :

Hexaamminecobalt(III) chloride

4. Write the IUPAC names of the following coordination compounds:

Answer :

The IUPAC name of the following coordination compound

is

Potassium trioxalatoferrate(III)

5. Write the IUPAC names of the following coordination compounds:

Answer :

The IUPAC names of the coordination compound

is

Potassium tetrachloridopalladate(II).

Chapter 10 - Haloalkanes and Haloarenes

1. Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer :

We don't use sulphuric acid because it acts as an oxidising agent and the required alkyl iodide is not produced. The reactions are given below :-

2KI + H 2 SO 4 → 2KHSO 4 + 2HI

2HI + H 2 SO 4 → I 2 + SO 2 + H 2 O

2. Arrange each set of compounds in order of increasing boiling points.

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.

Answer :

It is known that boiling point increases with increase in molecular mass when the alkyl group is the same.

So the order of increasing boiling point is Chloromethane < Bromomethane < Dibromomethane < Bromoform

3. Arrange each set of compounds in order of increasing boiling points.

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Answer :

In the given compounds the halide groups are the same. In these cases, the boiling point depends on the bulkiness of the alkyl group. The boiling point increases with an increase in the chain length. Also, the boiling point decreases with an increase in branching.

So the order is :- 1- Chlorobutane > 1- Chloropropane > Isopropyl Chloride

4. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

Answer :

(i) 2-Chloro-3-methylbutane. And it is a secondary alkyl halide.

5. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(ii)

Answer :

3-Chloro-4-methylhexane. And it is primary alkyl halide.

Chapter 11 - Alcohols, Phenols and Ethers

1. Classify the following as primary, secondary and tertiary alcohols:

Answer :

To classify we look at the OH bonded carbon.

Here, only 1 carbon is attached to it, hence it is primary alcohol.

2. Classify the following as primary, secondary and tertiary alcohols:

Answer :

To classify we look at the OH bonded carbon.

Here, only 1 carbon is attached to it, hence it is primary alcohol.

3. Identify allylic alcohols in the above examples.

Answer :

The alcohols (ii) and (vi) are allylic alcohols. Because -C=C-C-OH is the skeleton of allylic alcohol.

4. Give structures of the products you would expect when each of the following alcohol reacts with

(a)

with Butan-1-ol

Answer :

Primary alcohols do no react with Lucas’ reagent.

Hence no reaction.

5. Classify the isomers of alcohols in question 11.3 (i) as primary, secondary and tertiary alcohols.

Answer :

Primary Alcohol: Pentan-1-ol; 2-Methylbutan-1-ol; 3-Methylbutan-1-ol; 2,2-Dimethylpropan-1-ol

Secondary Alcohol: Pentan-2-ol; 3-Methylbutan-2-ol; Pentan-3-ol

Tertiary Alcohol: 2-Methylbutan-2-ol

Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

1. Write the structures of the following compounds.

Di-sec. butyl ketone

Answer :

The structure of the compound Di-sec. butyl ketone is shown here-

2. Write the structure of products of the following reactions;

(ii)

Answer :

Reaction of acyl chloride with dialkylcadmium (

), prepared from reaction of cadmium chloride and grignard reagents ,gives ketone

3. Arrange the following compounds in increasing order of their boiling points.

Answer :

Increasing order in their boiling points-

Alcohol has the highest boiling point due to more extensive intermolecular H-bonding. Aldehyde is more polar than ether so, ethanal has high BP than ethyl ether. And alkane has the lowest BP.

4. Give the IUPAC names of the following compounds:

Answer :

The IUPAC name of the compound

is-

3-phenyl propanoic acid

5. Give the IUPAC names of the following compounds:

Answer :

The IUPAC name of the compound

is -

3-methyl but-2-en-1-oic acid

Chapter 13 - Amines

1. How will you convert

Benzene into N, N-dimethylaniline

Answer :

Nitration of benzene gives nitrobenzene and after catalytic hydrogenation of nitrobenzene, it gives aniline. Aniline on reacting with two moles of chloromethane to form N, N-dimethylaniline.

2. Arrange the following in increasing order of their basic strength:

and

Answer :

Considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkylamines. order of basic strength in ethyl substituted amine is-

and order in benzene substituted ring-

Due to the -R effect of benzene

has less electron density than ammonia. So, the final order is -

3. Arrange the following in increasing order of their basic strength:

Answer :

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group the increasing order of basicity in ethyl as a substituted group is shown above.

4. Arrange the following in increasing order of their basic strength:

(iii)

Answer :

On considering the inductive effect, solvation effect and steric hindrance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines are -

5. Complete the following acid-base reactions and name the products:

Answer :

The above-given reaction is an acid-base reaction. so salt is formed.

Chapter 14 - Biomolecules

1. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?

Answer :

When pentaacetate of glucose is made to react with hydroxylamine , it does not react indicating the absence of free —CHO group.

In aqueous solution, the cyclic structure (have -OH group at C-1 ) should get converted to open chain structure which has an aldehyde group at C-1. It then should react with hydroxylamine and give glucose oxime. But such case is not observed. This suggests that in aqueous solution open chain structure doesn't exist and as a result, the aldehyde group is absent in pentaacetate of glucose.

2. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.

Answer :

Amino acids have both acidic and basic group(amine) which lead to the formation of zwitterions and make them dipolar compounds. Due to this property, they form strong molecular bonds themselves and also with water. This result high melting point and good solubility in water as compared to haloacids.

Halo acids doesn't exhibit the property of dipolar compounds. Only carboxyl group of haloacid is involved in H-bonding, not the halogen atom. That's why they have low melting points and less solubility than amino acids.

3. Where does the water present in the egg go after boiling the egg?

Answer :

Due to denaturation of proteins, globules unfold and helix gets uncoiled which changes its biological activity. In denaturation, secondary and tertiary structures are destroyed whereas primary structure remains the same.

Due to this process (denaturation of proteins) coagulation of egg take place while boiling. In egg, the globular protein changes into a rubber-like structure which is responsible for absorption of water.

4. What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?

Answer :

When a nucleotide from the DNA containing thymine is hydrolyzed, the products are thymine β-D-2-deoxyribose and phosphoric acid.

5. When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?

Answer :

When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained, this fact suggests that RNA is a single strand structure. Unlike DNA which is a double strand structure in which pairing of bases occurs (for e.g. adenine pairs with thymine). Thus, on hydrolysis, the amount of adenine produced will be the same as the amount produced by thymine. In RNA there is no relationship with quantities of bases, meaning bases don't occur in pairs or it is single strand structure.

Chapter 15 - Polymers

1. Classify the following as addition and condensation polymers: Terylene, Bakelite, Polythene, Teflon.

Answer :

• Addition Polymers are formed by the direct addition of repeated monomers. Example- polyethene and Teflon

• Condensation polymers are formed by condensation of two or more than two monomers by eliminating by-product like water and HCl. Example- terylene and bakelite

2. Arrange the following polymers in increasing order of their intermolecular forces.

Nylon 6,6, Buna-S, Polythene

Answer :

Increasing order in their intermolecular forces-

Buna-S(elastomers)<Polyethene<Nylon6, 6(fibres

• (Thermoplastics, intermediate forces between elastomers and fibres)

• ( strong H-bond or dipole-dipole interaction)

• elastomers weakest force of attraction

3. How do you explain the functionality of a monomer?

Answer :

The functionality of a monomer is the number of binding sites present in it. For example, for propene and ethene functionality is one but for adipic acid and 1,3- butadiene is two.

4. How do you explain the functionality of a monomer?

Answer :

The functionality of a monomer is the number of binding sites present in it. For example, for propene and ethene functionality is one but for adipic acid and 1,3- butadiene is two.

5. Why do elastomers possess elastic properties?

Answer :

In elastomers, the polymeric chains are held by weak intermolecular forces of attraction. These weak binding forces allow them to stretch and a few cross-links are there in between the chains, which helps them to retract after stretching or releasing forces. Due to this elastomers are elastic in nature.

Example- Buna-S, Buna-N and Neoprene etc.

Chapter 16 - Chemistry in Everyday life

1. Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why ?

Answer :

Sleeping pills contain drugs and when drugs are taken in doses higher than recommended may cause harmful effects on our body, it may affect the nervous system; also it may lead to person death. Hence, a doctor should always be consulted before taking any medicins.

2. With reference to which classification has the statement, “ranitidine is an antacid” been given?

Answer :

The given statement refers to the classification of the pharmacological effects of the drug. This is because drugs which are used to counteract(decrease the level of HCl) the effect of excess acid secretion in the stomach is known as an antacid. Ranitidine belongs to the class of histamine receptors blockers, where they block the activity of H_{2} receptors present in the stomach(cause of secretion of gastric acid).

3. Why do we require artificial sweetening agents ?

Answer :

Natural sweeteners like sucrose provide calories to the body. For a diabetic patient, we need artificial sweetener because it does not add any type of calories which affect the diabetic patient But in natural sweetener the amount of fat, carbohydrate, protein is high. Examples of artificial sweeteners are- Saccharin, aspartame etc.

4. Why do we need to classify drugs in different ways?

Answer :

We need to classify the drugs in different ways because of-

• Pharmacological effect

This classification provides doctors with the whole range of drugs available for the treatment of a certain type of problem

• Drug action

The action of drugs on a particular biochemical process

• Chemical structure

This classification provides a range of drugs sharing common structural features

• molecular targets

Drugs have some common structural features may have the same mechanism of action on the targets. Hence it is the most useful classification for medicinal chemists

5. Explain the term, target molecules or drug targets as used in medicinal chemistry.

Answer :

In medicinal chemistry, drug targets are the key molecules, which involved in certain metabolic pathways that result in specific diseases. Drugs mainly interact with biomolecules such as carbohydrates, lipids, proteins and nucleic acids. Drugs are the chemical agents, and they are designed to inhibit these target molecules by binding with the active sites of the key molecules.