# NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

## NCERT Solutions for Class 11 Chemistry Chapter 9 – Free PDF Download

*According to the latest update on the CBSE Syllabus 2022-23, this chapter has been removed.

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen includes all the questions provided in the textbook according to the CBSE syllabus, along with extra questions, worksheets, exemplar questions, MCQs and HOTS (High Order Thinking Skills). The NCERT Solutions consists of step-by-step answers to the questions along with detailed explanations to help students learn and understand concepts easily.

If you are looking for the most detailed, accurate and free NCERT Solutions for Class 11 Chemistry updated as per the latest CBSE Syllabus 2022-23, you have come to the right place. By clicking on the link below, students can download the NCERT Solutions for this chapter right away.

Hydrogen is the first element in the periodic table and the lightest element, with just one proton and one electron. Dihydrogen is the most abundant element in this universe. The NCERT Solutions for Class 11 Chemistry Chapter 9 – Hydrogen are designed to help the students attempt the CBSE Class 11 Chemistry first-term exam easily. These solutions are free, and students can either view them online on the website or download the PDF and practise them without any time constraints.

## NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

The Class 11 Chemistry NCERT Solutions for Chapter 9 includes all the questions provided in the NCERT textbook that is prescribed for Class 11 CBSE schools. BYJU’S NCERT Solutions for Class 11 Chemistry Chapter 9 – Hydrogen are designed to help the students attempt the term – I CBSE Class 11 Chemistry exam easily. The solution consists of structured answers to the questions along with detailed explanations to help students learn and understand concepts easily. So if you are looking for the most detailed, accurate and free Solutions for Class 11 NCERT Chemistry, you have come to the right place. The solutions are free, and you can either view them online on the website or download the PDF.

Class 11 Chemistry Chapter 9 Hydrogen is intended to make students accustomed to the properties, reactions and uses of hydrogen and other relative factors of hydrogen. Besides, hydrogen is an important element as it is the most common item that makes up more than 80% of the mass of the entire universe. Therefore, it is important to know all about this element. We are providing a set of NCERT Solutions for Class 11 Chemistry Chapter 9 to help students test their knowledge about the chapter, and also they can learn how to answer the questions accurately. Students can access the downloadable solutions for NCERT Class 11 Chemistry Chapter 9 PDF from anywhere and at any time convenient for them to learn about different topics effectively.

### Subtopics of Class 11 Chemistry Chapter 9 – Hydrogen

1. Position of Hydrogen in the Periodic Table
2. Dihydrogen, H2
1. Occurrence
2. Isotopes of Hydrogen
3. Preparation of Dihydrogen, H2
1. Laboratory Preparation of Dihydrogen
2. Commercial Production of Dihydrogen
4. Properties of Dihydrogen
1. Physical Properties
2. Chemical Properties
3. Uses of Dihydrogen
5. Hydrides
1. Ionic or Saline Hydrides
2. Covalent or Molecular Hydride
3. Metallic or Non-stoichiometric (or Interstitial ) Hydrides
6. Water Ex
1. Physical Properties of Water Ex
2. Structure of Water
3. Structure of Ice
4. Chemical Properties of Water
5. Hard and Soft Water
6. Temporary Hardness
7. Permanent Hardness
7. Hydrogen Peroxide (H2O2)
1. Preparation
2. Physical Properties
3. Structure
4. Chemical Properties
5. Storage
6. Uses
8. Heavy Water, D2O
9. Dihydrogen as a Fuel

### Class 11 Chemistry NCERT Solutions for Chapter 9 Hydrogen – Important Questions

Q 9.1

Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.

Ans:

The 1st element in the periodic table is hydrogen. Hydrogen exhibits dual behaviour because it has only 1 electron on its one ‘S’ shell, i.e. hydrogen resembles both halogens and alkali metals.

Electronic configuration of hydrogen = [1s1]

Hydrogen’s resemblance with alkali metals:

Hydrogen has 1 valence electron on its valence shell, like alkali metals.

[He] 2s 1 – Li

1s1 – H

[Ne] 3s1 – Na

Therefore, to form a uni-positive ion, it can lose one of its electrons.

To form halides, oxides and sulphides, it combines with electro -ve elements, which are the same as alkali metals.

Hydrogen’s resemblance with halogens:

Only 1 electron is required to complete their respective octets for both the halogen and hydrogen.

H: 1s 1

F: 1s 2 2s 2 2p 5

Cl: 1s 2 2s 2 2p 6 3s 2 3p 5

It forms several covalent compounds and diatomic molecules like halogens. Even though hydrogen has certain similarities to both halogen and alkali metals, it differs from them. Hydrogen won’t possess metallic characteristics; it possesses higher ionisation enthalpy and reacts less than halogens.

Due to these reasons, hydrogen can’t be replaced with the alkali metal of 1st group or with the halogens of 2nd group. Therefore, it is best to place hydrogen separately in the periodic table.

Q 9.2

Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?

Ans:

3 isotopes:

(i)  tritium   $^{3}_{1}H$ or T

(ii)  protium $^{3}_{1}H$

(iii) deuterium $^{2}_{1}H$ or D

Mass Ratio:

Tritium : Protium : deuterium = 1 : 2 : 3

Q 9.3

Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?

Ans:

The ionisation enthalpy of a hydrogen atom is higher. Therefore, it is harder to remove its electron. This results in its tendency to exist in the low monoatomic form. Instead of that, a covalent bond is formed by hydrogen with another hydrogen atom and exists as a diatomic molecule.

Q 9.4

How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?

Ans:

By the process of coal gasification, di hydrogen is produced as,

$C_{(s)}+H_{2}O_{(g)}\rightarrow CO_{(g)}+H_{2(g)}$ [C – Coal ]

Reaction with carbon monoxide with steam in the presence of a catalyst (iron chromate) results in an increase in the yield of dihydrogen.

$CO_{(g)}+H_{2}O_{(g)}\rightarrow CO_{2(g)}+H_{2(g)}$

The above reaction is known as the water-gas shift reaction. The carbon dioxide can be removed by scrubbing it with sodium arsenite solution.

Q 9.5

Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?

Ans:

The preparation of dihydrogen is by the electrolysis of acidified or alkaline water using platinum electrodes. Generally, 15 – 20% of an acid (H2 SO4) or a base (NaOH) is used.

At the cathode, the reduction of water occurs as:

$2H_{2}O+2e^{-}\rightarrow 2H_{2}+2OH^{-}$

At the anode, oxidation of OH– ions takes place as:

$2OH^{-}\rightarrow H_{2}O+\frac{1}{2}O_{2}+2e^{-}$

The net reaction is represented as:

$H_{2}O_{(1)}\rightarrow H_{2(g)}+\frac{1}{2}O_{2(g)}$

Due to the absence of ions, the electrical conductivity of pure water is too low. Hence, electrolysis of pure water takes place at a low rate. The rate of electrolysis increases if an electrolyte, such as a base or acid, is added to the process. The electrolyte is added, which makes the ions available in the process for the conduction of electricity and for electrolysis to take place.

Q 9.6

Complete the following reactions:

(i)  $H_{2(g)}+M_{m}O_{0(g)}\rightarrow$

(ii)  $CO_{(g)}+H_{2(g)}\rightarrow$

(iii)  $C_{3}H_{8(g)}+3H_{2}O_{(g)}\rightarrow$

(iv) $Zn_{(g)}+NaOH_{(aq)}\rightarrow$

Ans :

(i)

$H_{2(g)}+M_{m}O_{0(g)}\rightarrow$$mM_{(s)}+H_{2}O_{(l)}$

(ii)

$CO_{(g)}+H_{2(g)}\rightarrow$ $CH_{3}OH_{(l)}$

(iii)

$C_{3}H_{8(g)}+3H_{2}O_{(g)}\rightarrow$$3CO_{(g)}+7H_{2(g)}$

(iv)

$Zn_{(g)}+NaOH_{(aq)}\rightarrow$$Na_{2}ZnO_{2(aq)}+H{2(g)}$

Q 9.7

Discuss the consequences of the high enthalpy of the H–H bond in terms of the chemical reactivity of dihydrogen.

Ans :

The ionisation enthalpy of the H–H bond is higher (1312 kJ mol–1 ), which shows that hydrogen has a low tendency to form H+ ions. Its ionisation enthalpy value is comparable to that of halogens. Hence, it forms

→ a large number of covalent bonds

→ diatomic molecules (H2)

→ hydrides with an element

Hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals because ionisation enthalpy is very high.

Q 9.8

What do you understand by (i) electron-rich, (ii) electron-precise, and (iii) electron-deficient – compounds of hydrogen? Provide justification with suitable examples.

Ans:

Molecular hydride is classified on the basis of the presence of the bonds, and the total number of electrons in their Lewis structures are as follows:

1. Electron-deficient hydrides
2. Electron-precise hydrides
3. Electron-rich hydrides

1. An electron-deficient hydride has very less electrons, less than that required for representing its conventional Lewis structure.

For example, BH3, AlH3 etc.

They exist in diametric forms such as B2H6 and Al2H6 to make up for their deficiency.

In B2H6, there are 6 bonds in all, out of which only 4 bonds are regular and 2 centred-2 electron bonds. The remaining 2 bonds are 3 centred-2 electron bonds, i.e. 2 electrons are shared by 3 atoms.

2. An electron-precise hydride has a sufficient number of electrons to be represented to form a covalent bond.

For example, CH4, SiH4 etc.

In this compound, 4 regular bonds are formed where 2 electrons are shared by 2 atoms.

3. An electron-rich hydride compound contains excess valence electrons to form covalent bonds.

For example, NH3, PH3

There are 3 regular bonds in all, with a lone pair of electrons on the nitrogen atom.

Q 9.9

What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?

Ans:

Electron-deficient compounds of hydrogen do not have a sufficient number of electrons to form an octet.  Examples of group 13 hydrides, such as BH3 AlH3 etc., act as electron-deficient compounds.

They exist in polymeric forms such as B2H6 and Al2H6​ to overcome their deficiency.

These compounds act as Lewis acids. They form a complex by accepting electron pairs from Lewis bases.

B2H6 + 2NMe3 →2BH3.NMe3

​B2H6  + 2CO→2BH3. CO

Q 9.10

Do you expect the carbon hydrides of the type (Cn H2n+2 ) to act as a ‘Lewis’ base or acid? Justify.

Ans:

For carbon hydrides which belong to type (Cn H2n+2), the following hydrides are possible for

n = 1$\Rightarrow$CH4

n = 2 $\Rightarrow$C2 H6

n = 3$\Rightarrow$C3 H8

For a hydride to act as a Lewis acid, it should be electron deficient.

Lewis acid = electron accepting

Also, for a hydride to act as a Lewis base, it should be electron rich.

Lewis base = electron donating

Taking C2 H6 as an example, the total number of electrons is 14, and the total covalent bonds are 7. Hence, the bonds are regular 2e -2centered bonds.

Hence, hydride C2H6 has sufficient electrons to be represented by a conventional Lewis structure, it is an electron-precise hydride, having all atoms with octets. Thus, it can neither accept nor donate electrons to act as a Lewis base or Lewis acid.

Q 9.11

What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of hydride to be formed by alkali metals? Justify your answer.

Ans:

Non-stoichiometric hydrides are hydrogen-deficient compounds which are formed by the reaction of dihydrogen with d- block and f- block elements. These hydrides do not follow the law of constant composition.

For example, LaH2.87, YbH2.55, TiH 1.5 – 1.8  etc.

Alkali metals form stoichiometric hydrides, which are naturally ionic. Hydride ions have comparable sizes (208 pm) with alkali metal ions. This results in a strong binding force between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed.

Alkali metals will not form non-stoichiometric hydrides.

Q 9.12

How do you expect the metallic hydrides to be useful for hydrogen storage? Explain

Ans:

Metallic hydrides are hydrogen deficient. They don’t follow the law of constant composition.

It has been established that in the hydrides of Pd, Ac, Ni, and Ce, hydrogen occupies the interstitial position in lattices which allows further absorption of hydrogen on these metals.

Metals like Pt and Pd have the capacity to accommodate a large volume of hydrogen. Hence, metallic hydrides serve as a source of energy and are used for the storage of hydrogen.

Q 9.13

How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.

Ans:

The atomic hydrogen torch is also known as the oxyhydrogen torch. These atoms are produced through dihydrogen dissociation with the help of an electric arc which results in a huge amount of energy.

Energy released = 435.88 kJ mol-1

This energy is used in the generation of 4000 K temperature, which is used in the cutting and welding of metals.

Therefore, atomic hydrogen torches are used for this purpose, i.e., it allows to recombine on a particular surface to be welded for the generation of a particular temperature.

Q 9.14

Among NH3, H2O and HF, which would you expect to have the highest magnitude of hydrogen bonding and why?

Ans:

The extent of hydrogen bonding mainly depends on

(i) Electronegativity

(ii) Number of hydrogen atoms available for bonding.

Among oxygen, fluorine and nitrogen, the increasing order of their electro-negativities is N < O < F.

Therefore, the expected order of the extent of hydrogen bonding is HF > H2O > NH3.

But, the actual order is H2O > HF > NH3.

Even though fluorine is more electronegative than oxygen, the extent of hydrogen bonding is high in water.

There is a shortage of hydrogens in HF, whereas there are exactly the right numbers of hydrogens in water. As a result, only straight-chain bonding takes place.

On the other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.

The extent of hydrogen bonding is limited in the case of ammonia because nitrogen has only 1 lone pair. Therefore, it cannot satisfy all hydrogens.

Q 9.15

Saline hydrides are known to react with water violently, producing fire. Can CO2, a well-known fire extinguisher, be used in this case? Explain.

Ans:

Saline hydrides, i.e. LiH, NaH etc., react with water to form hydrogen gas and a base. The chemical equation to represent this reaction is

$MH_{(s)}+H_{2}O_{(aq)}\rightarrow MOH_{(aq)}+H_{2(g)}$

This reaction behaves violently, and also fire is produced from this.

Dioxygen weighs lighter than CO2. CO2 is commonly used as a fire extinguisher as it covers the fire like a blanket and inhibits the dioxygen supply, thereby dousing the fire.

It can be used in this scenario also – It weighs higher than dihydrogen and is effective in isolating the burning surface from dioxygen and dihydrogen.

Q 9.16

Arrange the following

(i) CaH2, BeH2 and TiH2 in order of increasing electrical conductance.

(ii) LiH, NaH and CsH in order of increasing ionic character.

(iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy.

(iv) NaH, MgH2 and H2O in order of increasing reducing properties.

Ans:

(i) The electrical conductance of a molecule mainly depends on its covalent or ionic nature. CaH2 is an ionic hydride which conducts electricity in the molten state. Titanium hydride, TiH2, is metallic in nature and conducts electricity at room temperature.  Covalent compounds do not conduct, whereas Ionic compounds conduct.  BeH2 is a covalent hydride. Hence, it does not conduct.

Hence, the increasing order of electrical conductance is as follows:

BeH2 < CaH2 < TiH2

(ii) The ionic character of a bond is dependent on the electro-negativities of the atoms involved. The higher the difference between the electro-negativities of atoms, the smaller the ionic character. Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase, as shown below.

LiH < NaH < CsH

(iii) The bond pair in the D–D bond is more strongly attracted by the nucleus than the bond pair in the H–H bond. This is because of the higher nuclear mass of D2. The stronger the attraction, the greater will be the bond strength and the higher the bond dissociation enthalpy.

Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the repulsive and attractive forces present in a molecule.

Hence, the bond dissociation enthalpy of D–D is higher than H–H. However, bond dissociation enthalpy is the minimum in the case of F–F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre.

Therefore, the increasing order of bond dissociation enthalpy is as follows:

F–F < H–H < D–D

(iv) Ionic hydrides are strong reducing agents.  NaH can easily donate its electrons. Hence, it is most reducing in nature. Both MgH2 and H2O are covalent hydrides. H2O is less reducing than MgH2 since the bond dissociation energy of H2O is higher than MgH2. Hence, the increasing order of the reducing property is H2O < MgH2 < NaH.

Q 9.17

Compare the structures of H2O and H2 O2.

Ans:

The water molecule will be displayed with a bond angle of 104 .5and has a bent form in the gaseous phase. The O-H bond length is 95.7 pm.

Structure :

Hydrogen peroxide has a non-planar structure both in the solid and gas phases.

The dihedral angle in the gas and solid phases is 90.2o and 111.5o.

Q 9.18

What do you understand by the term ’auto-protolysis’ of water? What is its significance?

Ans:

Auto-protolysis (self-ionisation) of water is a chemical reaction in which 2 water molecules react to produce a hydroxide ion (OH) and a hydronium ion (H3O+).

The reaction involved can be represented as:

$H_{2}O_{(l)}+H_{2}O_{(l)}\leftrightarrow H_{3}O^{+}_{(aq)}+OH^{-}_{(aq)}$

Auto-protolysis of water indicates its amphoteric nature, i.e., its ability to act as an acid as well as a base.

The acid-base reaction can be written as follows:

$H_{2}O_{(l)}+H_{2}O_{(l)}\leftrightarrow H_{3}O^{+}_{(aq)}+OH^{-}_{(aq)}$

Q 9.19

Consider the reaction of water with F2 and suggest, in terms of oxidation and reduction, which species are oxidised/reduced.

Ans:

The reaction between water and fluorine can be represented as:

$2F_{2(g)}+2H_{2}O_{(l)}\rightarrow 4H^{+}_{(aq)}+4F^{-}_{(aq)}+O_{2(g)}$

This is an example of a redox reaction

$2F_{2(g)}+2H_{2}O_{(l)}\rightarrow 4H^{+}_{(aq)}+4F^{-}_{(aq)}+O_{2(g)}$

Water is getting oxidised to oxygen, and fluorine is being reduced to fluoride ions.

The oxidation number of various species can be represented as:

$2F_{2(g)}+2H_{2}O_{(l)}\rightarrow 4H^{+}_{(aq)}+4F^{-}_{(aq)}+O_{2(g)}$

Water is oxidised from (–2) to a zero oxidation state. An increase in oxidation state indicates oxidation of water.

Fluorine is reduced from zero to (–1) oxidation state. A decrease in oxidation state indicates the reduction of fluorine.

Q 9.20

Complete the following chemical reactions.

Ans:

(i) $PbS_{(g)}+H_{2}O_{2(aq)}\rightarrow$

(ii) $MnO^{-} _{4aq}+H_{2}O_{2(aq)}\rightarrow$

(iii) $CaO_{(g)}+H_{2}O_{(g)}\rightarrow$

(iv) $AlCl_{3(g)}+H_{2}O_{(l)}\rightarrow$

(v) $Ca_{3}N_{2(g)}+H_{2}O_{(l)}\rightarrow$

Classify the above into (a) Hydrolysis, (b) Redox and (c) Hydration reactions.

Ans:

(i)

$PbS_{(g)}+4H_{2}O_{2(aq)}\rightarrow$ $PbSO _{4(s)}+4H_{2}O_{(l)}$

H2O2 acts as an oxidising agent in the reaction. Hence, it is a redox reaction

(ii)

$2MnO^{-}_{4(aq)}+5H_{2}O_{2(aq)}+6H^{+}_{(aq )}\rightarrow 2Mn^{2+}_{(aq)}+8H_{2}O_{(l)}+5O_{2(g)}$ $H_{2}O_{2(aq)}$ acts as a reducing agent in the acidic medium, thereby oxidising $MnO^{-}_{4(aq)}$

Hence, the given reaction is a redox reaction.

(iii)

$CaO_{(g)}+H_{2}O_{(g)}\rightarrow Ca(OH)_{2(aq)}$

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Hence, the given reaction is hydrolysis.

(iv)

$2AlCl_{3(g)}+3H_{2}O_{(l)}\rightarrow Al_{2}O_{3(s)}+6HCl_{(aq)}$

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Thus, the given reaction represents the hydrolysis of AlCl3.

(v)

$Ca_{3}N_{2(s)}+6H_{2}O_{(l)}\rightarrow 3Ca\left ( OH \right )_{2(aq)}+2NH_{3(g)}$

The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Thus, the given reaction represents the hydrolysis of Ca3N2.

Q 9.21

Describe the structure of the common form of ice.

Ans:

Generally, ice is the crystalline form of water. It is visible in a hexagonal form if it is crystallised at atmospheric pressure. When the temperature is very low, it condenses to cubic form.

3 –D structure of ice:

It has hydrogen bonding and a highly ordered structure. Each of the oxygen atoms is surrounded tetrahedrally by 4 other oxygen atoms at a distance of 276 pm. The structure of ice also contains wide holes that can hold molecules of particular sizes.

Q 9.22

What causes the temporary and permanent hardness of water?

Ans:

Due to the presence of soluble salts of magnesium and calcium in the form of chlorides in the water, hardness remains permanent in water.

Due to the presence of soluble salts of calcium and magnesium in the form of hydrogen carbonates in the water, hardness remains temporary in water.

Q 9.23

Discuss the principle and method of softening of hard water with synthetic ion exchange resins.

Ans:

The process of treating the permanent hardness of water using synthetic resins is generally based on the exchange of anions and cations present in water by OH  and H+ ions, respectively.

The two types of synthetic resins are as follows:

1.  Anion exchange resins
2. Cation exchange resins

Cation exchange resins are large organic molecules that contain the –SO3H group. Firstly, the resin gets changed into RNa by treating it with NaCl. This resin exchanges Na+ ions with Ca2+ ions and Mg2+ ions, thereby making the water soft.

$2RNa+M^{2+}_{(aq)}\rightarrow R_{2}M_{(s)}+2Na^{+}_{(aq)}$

There are cation exchange resins in H+ form. The resins exchange H+ ions for Na+, Ca2+, and Mg2+ ions.

$2RH+M^{2+}_{(aq)}\rightleftharpoons MR_{2(s)}+2H^{+}_{(aq)}$

Anion exchange resins exchange OH ions for anions like Cl, $HCO^{-}_{3}$  and $SO_{4}^{2-}$ present in water.

$RNH_{2(s)}+H_{2}O_{(l)}\rightleftharpoons RNH^{+}_{3}.OH^{-}_{(s)}$ $RNH^{+}_{3}.X^{-}_{(s)}+OH^{-}_{aq)}$

During the whole process, first, the water passes through the cation exchange process. The water which is obtained after this process is free from mineral cations and naturally acidic. This acidic water is then passed through the anion exchange process where OH ions neutralise the H+ ions and de-ionise the water obtained.

Q 9.24

Write chemical reactions to show the amphoteric nature of water.

Ans:

The amphoteric nature of water can be described on the basis of the following reactions:

1) Reaction with H2S

The reaction takes place as follows:

$H_{2}O_{(l)}+H_{2}S_{(aq)}\rightleftharpoons H_{3}O^{+}_{(aq)}+HS^{-}_{(aq)}$

In the forward reaction, $H_{2}O_{(l)}$ accepts a proton from $H_{2}S_{(aq)}$. Therefore, it acts as a bronsted base.

2) Reaction with NH3

The reaction takes place as follows:

$H_{2}O_{(l)}+NH_{3(aq)}\rightleftharpoons OH^{-}_{(aq)}+NH^{+}_{4(aq)}$

In the forward reaction, $H_{2}O_{(l)}$ denotes its proton to $NH_{3(aq)}$. Therefore, it acts as bronsted acid.

3) Self-ionisation of water

2 water molecules react in this reaction as,

$H_{2}O_{(l)}+H_{2}O_{(l)}\rightleftharpoons H_{3}O^{+}_{(aq)}+OH^{-}_{(aq)}$

Q 9.25

Write chemical reactions to justify that hydrogen peroxide can function as an oxidising as well as a reducing agent.

Ans:

Hydrogen peroxide acts as an oxidising agent as well as a reducing agent in both alkaline medium and acidic mediums.

The reaction which is involved in oxidising actions are given below:

(i)  $Mn^{2+}+H_{2}O_{2}\rightarrow Mn^{4+}+2OH^{-}$

(ii) $2Fe^{2+}+H_{2}O_{2}\rightarrow 2Fe^{3+}+2OH^{-}$

(iii) $2Fe^{2+}+2H^{+}+H_{2}O_{2}\rightarrow 2Fe^{3+}+2H_{2}O$

(iv)  $PbS+4H_{2}O_{2}\rightarrow PbSO_{4}+4H_{2}O$

The reaction which is involved in reduction actions are given below:

(i) $I_{2}+H_{2}O_{2}+2OH^{-}\rightarrow 2I^{-}+2H_{2}O+O_{2}$

(ii) $2MnO^{-}_{4}+3H_{2}O_{2}\rightarrow 2MnO_{2}+3O_{2}+2H_{2}O+2OH^{-}$

(iii)  $2MnO^{-}_{4}+6H^{+}+5H_{2}O_{2}\rightarrow 2Mn^{2+}+8H_{2}O+5O_{2}$

(iv) $HOCl+H_{2}O_{2}\rightarrow H_{3}O^{+}+Cl^{-}+O_{2}$

Q 9.26

What is meant by ‘demineralised’ water, and how can it be obtained?

Ans:

This water is free from all the soluble mineral salts, and it doesn’t contain any cation or anion. It is obtained successively by passing the water through anion exchange and cation exchange resin.

During the cation exchange process, H+ exchanges for

→ Ca2+

→ Na+

→ Mg2+

and other cations present in the water.

Q 9.27

Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?

Ans:

Water is essential for our life. It consists of many dissolved nutrients that are required for us and also for plants and animals.  Demineralised water is free from all soluble minerals, so it cannot be used for drinking purposes.

After adding desired minerals in specific amounts that are required for growth, this water can be made useful.

Q 9.28

Describe the usefulness of water in the biosphere and biological systems.

Ans:

Water is very necessary for all forms of life, which constitute 65% of the human body and 95% of plants. It plays a vital role in the biosphere due to its

→ Thermal conductivity

→ Dipole moment

→ Specific heat

→ Dielectric constant and

→ Surface tension.

For moderating the body temperature of all living beings and the atmospheric climate,

(i)  The heat of capacity and

(ii) The heat of vapourisation helps a lot.

It acts as a carrier of different nutrients which are required by animals and plants for various metabolic reactions.

Q 9.29

What properties of water make it useful as a solvent? What types of compound can it (i) dissolve and (ii) hydrolyse?

Ans:

A high value of dipole moment and dielectric constants (78.39 C2 /Nm2) makes water a universal solvent. Water is able to dissolve most covalent and ionic compounds. Owing to the ion-dipole interaction, ionic compounds dissolve in water, whereas covalent compounds form hydrogen bonding and dissolve in water. Water can hydrolyse

→ metallic and non-metallic oxides

→ nitrides

→  phosphides

→ carbides

→ hydrides

and various other salts. During hydrolysis, H+ and OH  ions of water interact with the reacting molecule.

Certain reactions are given below:

$CaC_{2}+H_{2}O\rightarrow C_{2}H_{2}+Ca(OH)_{2}$ $CaO+H_{2}O\rightarrow Ca(OH)_{2}$ $NaH+H_{2}O\rightarrow NaOH+H_{2}$

Q 9.30

Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes?

Ans:

D2O is known as heavy water, which acts as a moderator (slows down the rate of reaction). Due to this property, it cannot be used for drinking purposes because it slows down

(i) catabolic reaction and

(ii) anabolic reaction

that takes place in the body, which leads to casualty.

Q 9.31

What is the difference between the terms ‘hydrolysis’ and ‘hydration’?

Ans:

Hydration:

The addition of 1 or more molecules to a molecule or ion, which results in the formation of hydrated compounds, is known as hydration.

For example, $CuSO_{4}+5H_{2}O\rightarrow CuSO_{4}.5H_{2}O$

Hydrolysis:

The chemical reaction in which hydroxide ions and hydrogen of water molecules react with a compound to form products is called hydrolysis.

For example, $NaH+H_{2}O\rightarrow NaOH+H_{2}$ .

Q 9.32

How can saline hydrides remove traces of water from organic compounds?

Ans:

Naturally, saline hydrides are ionic. Saline hydrides react with water which results in the formation of metal hydroxide along with hydrogen gas liberation. It is represented as,

$AH_{(s)}+H_{2}O_{(l)}\rightarrow AOH_{(aq)}+H_{2(g)}$

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide. The dry organic solvent distils over.

Q 9.33

What do you expect the nature of hydrides is if formed by elements of atomic numbers 1519, 23 and 44 with dihydrogen? Compare their behaviour with water.

Ans:

The elements of atomic number 15 are phosphorus, 19 is potassium, 23 is vanadium and 44 is ruthenium.

Hydride of Phosphorus

Hydride of phosphorus (PH3) is covalent in nature. Due to the presence of excess electrons as a lone pair on Phosphorus, it is electron rich.

Hydride of potassium

Due to the high electropositive nature of potassium, the dihydrogen forms ionic hydrides along with potassium. Naturally, it is non–volatile and crystalline.

Vanadium belongs to the d-block in the periodic table. The metals of d-block form non-stoichiometric or metallic hydrides. Hydrides of vanadium are naturally metallic and have a deficiency of hydrogen.

Hydrides of Ruthenium

Ruthenium belongs to the d-block in the periodic table.

Ruthenium does not form hydrides on account of its low affinity for hydrogen in its normal oxidation states.

The behaviour of hydrides towards the water

Potassium hydride is an ionic compound that reacts violently with water to produce H2 gas.

$KH_{(s)}+H_{2}O_{(aq)}\rightarrow KOH_{(aq)}+H_{2(g)}$

Q 9.34

Do you expect different products in solution when aluminium (III) chloride and potassium chloride are treated separately with (i) alkaline water, (ii) acidified water, and (iii) normal water? Write equations wherever necessary.

Ans:

Potassium chloride (KCl) is the salt of a strong acid (HCl) and a strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows:

$KCl_{(s)}\rightarrow K^{+}_{(aq)}+Cl^{-}_{(aq)}$

In acidified and alkaline water, the ions do not react and remain as such. Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)3]. Hence, it undergoes hydrolysis in normal water.

$AlCl_{3(s)}+3H_{2}O_{(l)}\rightarrow Al(OH)_{3(s)}+3H^{+}_{(aq)}+3Cl^{-}_{aq}$

In acidified water, H+ ions react with Al(OH)3, forming water and giving Al3+ ions. Hence, in acidified water, AlCl3 will exist as $Al^{3+}_{(aq)}$ and $Cl^{-}_{(aq)}$

In alkaline water, the following reaction takes place:

Al(OH)3 + OH → [Al(OH)4]

Q 9.35

How does H2 O2 behave as a bleaching agent?

Ans:

Hydrogen peroxide acts as a strong oxidising agent both in basic and acidic media. When added to a cloth, it breaks the chemical bonds of the chromophores (colour-producing agents). Hence, the visible light is not absorbed and the cloth gets whitened.

Q 9.36

What do you understand by the terms:

(i) Hydrogen economy, (ii) Hydrogenation, (iii) ‘syngas’, (iv) Water-gas shift reaction and (v) Fuel cell?

Ans:

(i) Hydrogen economy

Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. The hydrogen economy is a technique of using dihydrogen in an efficient way. It involves the transportation and storage of dihydrogen in the form of liquid or gas. It is about the transmission of this energy in the form of dihydrogen.

(ii) Hydrogenation

The process of adding dihydrogen to another reactant is known as hydrogenation. It is used to reduce a compound in the presence of a suitable catalyst.

For example, the hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as ghee and vanaspati.

(iii) ‘syngas’

Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced by the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

$CnH_{2n+2}+nH_{2}o\rightarrow nCO+(3n+1)H_{2}$

E.g.

$CH_{4(g)}+H_{2}O_{(g)}\rightarrow CO_{(g)}+3H_{2(g)}$

(iv) Water-gas shift reaction

It is a reaction of carbon monoxide of syngas mixture with steam in the presence of a catalyst as:

$CO_{(g)}+H_{2}O_{(g)}\rightarrow CO_{2(g)}+H_{2(g)}$

This reaction is used to increase the yield of dihydrogen obtained from the coal gasification reaction as:

$C_{(s)}+H_{2}O_{(g)}\rightarrow CO_{(g)}+H_{2(g)}$

(v) Fuel cell

Fuel cells are devices for producing electricity from fuel in the presence of an electrolyte. Dihydrogen can be used as fuel in these cells. It is preferred over other fuels because it is eco-friendly and releases greater energy per unit mass of fuel as compared to gasoline and other fuels.

 Also Access NCERT Exemplar for Class 11 Chemistry Chapter 9 CBSE Notes for Class 11 Chemistry Chapter 9

### Some of the important topics covered in NCERT Solutions for Class 11 Chemistry Chapter 9 are as follows:

• Dihydrogen (H2)
• Preparation and Properties of Dihydrogen
• Hydrides
• Water Ex
• Hydrogen Peroxide (H2O2)
• Heavy Water (D2O)
• Dihydrogen as a Fuel

The NCERT Solutions at BYJU’S are designed to offer several benefits to students. Firstly, students get the facility of learning according to the latest CBSE Class 11 Chemistry Syllabus right from their comfort zone. By downloading the solutions in PDF, students can learn, practise, and revise different Chemistry topics from their homes or from anywhere. The solutions are easily accessible, and students can view the solutions on BYJU’S website, or they can access BYJU’s – The Learning App.

Another benefit is that students get a personalised learning experience, where they can learn about certain difficult topics at a slower pace without taking any stress or without worrying about deadlines. BYJU’S also provides the best subject experts to guide students to learn the concepts and topics in a simple and interesting manner.

Further, if students have any doubts while going through the NCERT Class 11 Chemistry Solutions, they can always approach BYJU’S responsive support team to clear their doubts. Besides, students can bring in all their queries regarding Chemistry, as well as other subjects, including Physics, Maths and Biology.

## Frequently Asked Questions on NCERT Solutions for Class 11 Chemistry Chapter 9

### What are the physical properties of hydrogen according to NCERT Solutions for Class 11 Chemistry Chapter 9?

The physical properties of hydrogen are  listed below:
Colourless, odourless, neutral gas
Less soluble in water
Highly inflammable
Burns with a blue flame
Very low boiling points

### Is it necessary to learn all the questions provided in NCERT Solutions for Class 11 Chemistry Chapter 9?

Yes, to attempt all types of questions in the annual exams, it is necessary to attempt all the questions provided in NCERT Solutions for Class 11 Chemistry Chapter 9. These solutions are provided in student-friendly language to help you in solving difficult problems in an easier way. These solutions are formulated by expert faculty at BYJU’S, having good knowledge of the subject.

### Are NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen important from an exam perspective?

Yes, all the NCERT Solutions for Class 11 Chemistry are important from the board exam perspective. Chapter 9 of NCERT Solutions for Class 11 Chemistry explains hydrogen and its properties of hydrogen. Students can view the solutions online as well as download them on BYJU’S website. These solutions are available in PDF for free access.
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