Class 11 - Chemistry - Hydrogen

Question 9.1

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

Answer.

Hydrogen has electronic configuration 1s1.

Alkali metals have electronic configuration ns1.

Outer electronic configuration of hydrogen is similar to alkali metals.

Question 9.2

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

Answer:

Isotopes of hydrogen

  • Protium 1H1
  • Deuterium 2H1
  • Tritium 3H1

Mass ratio of hydrogen isotopes is 1:2:3.

Question 9.3

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

Answer.

Hydrogen atom has only one electron and thus to achieve stable inert gas configuration of helium,

it shares its single electron with electron of other hydrogen atom to form stable diatomic molecule.

In monoatomic form, hydrogen atom has only one electron in K-shell (1s1) while in diatomic form, K-shell is complete (1s2).

 

Question 9.4

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

Answer.

The process of producing syngas from coal and steam is called coal gasification.

C(s) + H2O(g) --> CO(g) + H2(g) (temperature is 1270 K and Ni as a catalyst)

Production of H2 (dihydrogen) can be increased by reacting CO (carbon monoxide) of syngas mixture with H2O (steam)

in the presence of Fe (iron) chromate as catalyst.

CO(s) + H2O(g) --> CO2(g) + H2(g) (temperature is 673 K and in presence of catalyst)

This is Water-gas shift reaction. Carbondioxide is removed by scrubbing it with a solution of sodium arsenite.

Question 9.5

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

Answer.

The preparation of di hydrogen is by the electrolysis of acidified or alkaline water using platinum electrodes.

Generally, 15 – 20% of an acid (H2 SO 4) or a base (NaOH) is used.

At the cathode, reduction of water occurs as:

 

2H2​O+2e→2H2​+ 2OH

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

2OH --> H2O + (1/2) O2 + 2e-

Net reaction is represented as:-

H2O --> H2(g) + (1/2) O2(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 an 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.

 

 

Question 9.6

Complete the following reactions:

  • H2(g) + MmOo (s) à
  • CO (g) + H2(g) à
  • C3H8 (g) + 3H2O à
  • Zn (s) + NaOH (aq) à

Answer.

 

 Class_11_Chemistry_Hydrogen_Equation_6

 

Question 9.7

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

Answer.

Due to high bond enthalpy of H-H bond, dihydrogen molecule (H2) is quite unreactive at

room temperature . However, at high temperature(2000 K) or in presence of catalysts, it combines 

with many metals and non-metals to form respective hydrides.

 

Dissociation of hydrogen into atoms is only 0.081% which increases to 95.5% at 5000 K.

The bond dissociation energy is very high.

H2  --> H + H; ∆H= 435.9 kJmol-1

Since, it has very high bond dissociation energy it is not very reactive.

Question 9.8

What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii) electron-rich compounds of hydrogen?

Provide justification with suitable examples.

Answer.

  • Electron-deficient hydrides – Group 13 elements have hydrides which have incomplete octet.
  • Since, they have electron deficient molecules. They act as Lewis acids.
  • Electron-precise hydrides – These compounds have required number of electrons to write their conventional Lewis structures.
  • Elements of Group 14 form such compounds which have tetrahedral structure.
  • Electron-rich compounds of hydrogen – These compounds have excess electrons which present as lone pairs.
  • Elements of group 15-17 form such compounds.

 

Question 9.9

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

Answer.

To form a regular bond, an electron-deficient hydride does not have sufficient number of electrons to form normal covalent bonds.

They generally exist in polymeric forms such as B2H6, B2H10

Due to deficinecy of electrons, hydrides act as a Lewis acids and thus,

form complex entities with Lewis bases such as:

NH3 , H- ions, etc.   

 

 

B2H6 +2..NH3 --> [BH2(NH3)2]+ (BH4)-

B2H6 +2 NaH --> 2Na+[BH4]-

                           Sodium borohydride

 

Question 9.10

Do you expect the carbon hydrides of the type (CnH2n+2) to act as ‘Lewis’ acid or base? Justify your answer.

Answer.

CnH2n+2 is not a Lewis acid or Lewis base because carbon atoms have a complete octet.

So, hydrides behave a normal covalent hydrides.

 

 

Question 9.11

What do you understand by the term ‘non-stoichiometric hydrides’?

Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.

Answer.

Hydrides of d-block include Group No. 3, 4, 5 and 6.

F-block are metallic or interstitial hydrides.

This is because hydrogen occupies some interstitial sites in the metal lattice producing distortion.

These hydrides are also called as non-stoichiometric hydrides.

Law of constant composition does not hold good for them.

For example: LaH2.87, YbH2.55

Alkali metals do not form these types of hydrides because alkali metals

denote lone pair of electrons to hydrogen which form H- ion.

H- ion is formed by complete transfer of electrons. Hence, alkali metals and H atoms are in fixed stoichiometric ratio.

 

Question 9.12

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

Answer.

In some of the transition metal hydrides, hydrogen is absorbed as H-atoms.

Due to the inclusion of H-atoms, the metal lattice expands and thus becomes less stable.

Therefore, when such metallic hydride is heated, it decomposes to release hydrogen gas and very finely divided metal.

The hydrogen evolved in this way can be used as a fuel.

Thus, transition metals or their alloys can act as sponge and can be used to store and transport hydrogen to be used as a fuel.

 

 

Question 9.13

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

Answer.

Atomic hydrogen torch - When molecular hydrogen  is passed through an electric arc struck between tungsten electrodes(3773 - 4273K),

The reaction is as follows:

Class_11_Chemistry_Hydrogen_Equation_7

The life span of atomic hydrogen is about 0.3 sec and therefore, it immediately gets converted into the molecular hydrogen

liberating large amount of energy which is used for cutting and welding purposes in the form of atomic hydrogen torch.

 

  

Question 9.14

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

Answer.

HF is expected to have highest magnitude of hydrogen bonding, as fluorine is most electronegative ans also atomic size of fluorine is small.

Therefore, HF is most polar.

  

Question 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.

Answer.

Saline hydrides [i.e.,LiH, NaH etc.] react with water violently to form the corresponding metal hydroxides with the evolution of dihydrogen gas.

The dihydrogen gas so liberated undergoes spontaneous combustion causing fire.

This is because of exothermic nature of combustion reactions.

NaH(s) +  H2O--> NaOH(aq) + H2(g)

CaH2(g) +2H2O(l) --> Ca(OH) (aq) + 2H2(g) 

The fire so produced cannot be extinguished COby because it gets deuced by the hot metal hydride to formate ions. 

NaH + CO2--> HCOONa

However, sand (becuase of its stable nature) is more effective fire extinguisher in such a case.

 

 

Question 9.16

Arrange the following

  • CaH2, BeH2 and TiH2 in order of increasing electrical conductance.
  • LiH, NaH and CsH in order of increasing ionic character.
  • H-H, D-D and F-F in order of increasing bond dissociation enthalpy.
  • NaH, MgH2 and H2O in order of increasing reducing property.

Answer.

  • BeH2 < TiH2 < CaH2 [BeH2 is covalent, CaH2 is ionic, TiH2 is metallic hydride and conducts electricity at room temperature]
  • LiH < NaH < CsH [Electronegativity decreases as Li > Na > Cs]
  • F-F < H-H < D-D
  • H2O < MgH2 < NaH(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 MgHsince the bond dissociation energy of is H2O higher than MgH2).

 

Question 9.17

Compare the structures of H2O and H2O2.

Answer.

In water, oxygen is sp3-hybridised. Two half-filled sp3 –orbitals form O-H sigma bonds while other two contains lone pairs of electrons.

H-O-H makes an angle 109.5o but its experimental value comes out to be 104.5o because lone pair lone pair repulsion

is greater than base pair base pair repulsion.

Class_11_Chemistry_Hydrogen_Structure_Of_Water

On the other hand, H2O2 is a non-planar molecule. Dihedral angle between two planes is 111.5o and angle OOH is 94.8o.

 Class_11_Chemistry_Hydrogen_Dihedral_Structure_Of_Water

 

Question 9.18

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

Answer.

Auto-protolysis means self-ionisation of water.

H2O (l) + H2O(l) -->H3O+(aq) + OH-(aq)

Acid1       Base 2       Acid 2          Base 1

The reason why water is amphoteric in nature is due to auto-protolysis. Water can act as an acid as well as base.

 

Question 9.19

Consider the reaction of water with F2 and suggest, in terms of oxidation and reduction,

which species are oxidized/reduced?

Answer.

In these reactions, water acts as a reducing agent and gets oxidised to either Oor Oon the other hand,  

F2 acts as oxidising agent it gets Fion.

 2F2(g) + 2H2O(l) --> 4H+(aq) + 4F-(aq) + O2(g)

 

Question 9.20

Complete the following chemical reactions.

  • PbS (s) + H2O2 (aq) --->
  • MnO4- (aq) + H2O2 (aq) --->
  • CaO (s) + H2O (g) --->
  • AlCl3 (g) + H2O (l) --->
  • Ca3N2 (g) + H2O(l) --->

Classify the above into (a) hydrolysis (b) redox and (c) hydration reactions.

Answer:

  • PbS (s) + H2O(aq) --->  PbSO4(s) +H2O (l)

    ​ H2O2 is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction

  • MnO4- (aq) + H2O2 (aq) ---> 2Mn (aq) 2+​ + 8H2​O(l)​ +5O2(g)​ ; H2O2  is acting as a reducing agent in the acidic medium, thereby oxidizing

    MnO4-Hence, the given reaction is a redox reaction.

  • CaO (s) + H2O (g) --->Ca(OH) 2(aq)​     

    The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions.

    The given reaction is hydrolysis.

  • AlCl3 (g) + H2O (l) --->

    Al2​O3(s) +6HCl (aq)

    ​The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions.

    The given reaction represents hydrolysis of AlCl3.

  • Ca3N2 (g) + H2O(l) --->3Ca (OH) 2(aq)​ +2NH3(g)​               

     The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions.

    The given reaction represents hydrolysis of Ca3N2.

 

 

Question 9.21

Describe the structure of the common form of ice.

Answer.

Ice is the crystalline form of water. It is visible in a hexagonal form if it is crystallised  at atmospheric pressure.

At a very low temperature, it condenses to cubic form.

3D structure of ice:-

Ice has hydrogenbonding and highly ordered structure.

Each of the oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm.

 

 

Question 9.22

What causes the temporary and permanent hardness of water?

Answer.

Temporary hardness of water is due to the presence of bicarbonates of calcium and magnesium in water.

[Ca(HCO3)2 and Mg(HCO3).

Permanent hardness of water is due to the presence of soluble chlorides and sulphates of calcium and magnesium.

[CaCl2, CaSO4, MgCl2, MgSO4]

 

Question 9.23

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

Answer.

Cation exchange resins have large organic molecule with HSO3 group which are insoluble in water.

Ion exchange resin (RSO3H) is changed to RNA on treatment with NaCl.

The resin exchange Na+ ions with Ca2+ and Mg2+ ions present in hard water and make it soft.

Resins can be regenerated by adding aqueous NaCl solution.

 2RNa(s) + M2+(aq)  R2M(s) + 2Na+(aq)

Question 9.24

Write the chemical reaction to show the amphoteric nature of water.

Answer.

 

1) Reaction with H2S

The reaction takes place as:

H2O (l) +H2S (aq) ⇌H3O+ (aq) + HS

In the forward reaction, H2​O(l)  accepts a proton from H2S (aq) . Therefore, it acts as a lewis base.

2) Reaction with NH3

The reaction takes place as:

H2​O (l)​  +NH3(aq)​⇌OH (aq)​ +NH4(aq)+

In the forward reaction, H2O(l)  denotes its proton to NH3(aq) . Therefore, it acts as lewis acid.

3) Self-ionization of water

2 water molecules reacts in this reaction as,

H2​O (l)​ +H2​O(l)  ⇌H3O(aq)+  +OH(aq)

 

Question 9.25

Write the chemical reaction to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.

Answer.

Hydrogen peroxide acts as an oxidizing agent as well as reducing agent in both acidic media and alkaline media.

As an oxidizing agent in acidic medium

2Fe2+(aq) + 2H+(aq) + H2O2(aq) ---> Fe3+(aq) + 2H2O (l)

As an oxidizing agent in basic medium

Mn2+ (aq) +H2O2​ (aq)+ 2OH-(aq)→MnO2(s)+2H2O(l)

                                                  Manganese dioxide

As a reducing agent in acidic medium

 I2(s) + H2O2(aq) + 2OH-(aq) ---> 2I-(aq) + 2H2O(l) + O2(g)

As a reducing agent in basic medium

2[Fe(CN)6]3-(g) + H2O2(aq) + 2OH-(aq) à 2[Fe(CN)6]4-(aq) + 2H2O(l) + O2(g)

 

Question 9.26

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

Answer.

Demineralised water is free from all soluble mineral salts which is obtained by passing water successively

through a cation exchange and an anion exchange in the form of OH- resins.

2RH(s) + M2+ (aq) -->MR2(s) + 2H+ (aq)

 

Question 9.27

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

Answer.

Demineralised water is not fit for drinking purposes because it does not contain even useful minerals.

In order to make it useful for drinking purposes, useful minerals in proper amounts should be added to demineralised

or distilled water.                                  

 

Question 9.28

Describe the usefulness of water in biosphere and biosphere systems.

Answer.

  • Most of the part of all living system is made of water.
  • It constitutes about 65-70% of body weights of animals and plants.
  • It is a raw material for photosynthesis in plants.
  • It is an excellent fluid for transportation of minerals and nutrients in plants.
  • It regulates temperature of living beings due to high heat of vaporization.

 

Question 9.29

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

Answer.

  • Water is highly polar in nature due to which it has high dielectric constant and high dipole moment.
  • Because of these two properties, water is a universal solvent.
  • It can dissolve polar substances and also some organic compounds due to hydrogen bonding.
  • Water can hydrolyse many oxides, halides, phosphides, nitrides due to interionic attraction between them.

 

Question 9.30

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

Answer.

No, D2O is injurious to human beings, plants and animals.

It cannot be used for drinking purpose because it slows down

(i) catabolic reaction and

(ii) anabolic reaction

That takes place in the body which leads to casualty.

 

Question 9.31

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

Answer.

Hydrolysis is a chemical reaction in which a substance reacts with water under natural, acidic or alkaline conditions.

Na2CO3 + 2H2O --> 2NaOH + H2CO3

Salt 1                        Base 2    Acid 1

Hydration is the property of a chemical compound to take up molecules of water of crystallization and get hydrated.

 CuSO4 + 5 H2O (l) ---> CuSO4.5H2O (s)

Colourless            (Blue)

 

Question 9.32

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

Answer.

Saline hydrides react with water and form the corresponding metal hydroxides with the liberation of hydrogen gas.

These hydrides can be used to remove traces of water from organic compounds.

 

NaH(s) + H2O (l) --> NaOH (aq) + H2O

CaH2(s) + 2H2O (l) -->Ca (OH)2(aq) + H2(g)

 

Question 9.33

What do you expect the nature of hydride is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen?

Compare their behavior towards water.

Answer.

Atomic number – 15 (Phosphorus)

Atomic number – 19 (Potassium)

Atomic number – 23 (Vanadium)

Atomic number – 44 (Ruthenium)

Hydride of PH3 and its nature is covalent.

Hydride of KH and it is ionic in nature.

Hydride of VH is interstitial or metallic.

Hydride of Ruthenium is interstitial or metallic.

 

 

Question 9.34

Do you expect different products in solution when aluminum (III) chloride and potassium chloride treated separately

with (i) normal water (ii) acidified water, and (iii) alkaline water? Write equations wherever necessary.

Answer.

  • Normal water

AlCl3 + 3H2O --> Al(OH)3 + 3HCl

KCl will dissolve in water and ions will get hydrated.

  • KCl will be unaffected in acidified water.

 In acidic water H+ ions react with Al(OH)3 to form Al3+ ions and water.

          acidified

AlCl3    -->        Al3+ (aq) + 3Cl- (aq)

  • In alkaline water, aqueous solution of KCl is neutral, so it is unaffected.

Al(OH)3 reacts to form soluble tetrahydroxoaluminate complex.

Class_11_Chemistry_Hydrogen_Equation_2 

Question 9:35

How does H2O2 behave as a bleaching agent?

Answer.

Bleaching action of hydrogen peroxide is due to the oxidation of coloring matter by nascent oxygen.

Bleaching action of hydrogen peroxide is permanent.

 H2O2 -->H2O +[O]

 The mascent oxygen oxidise the colouring matter to colourless products.

It is used for the bleaching of delicate materials like ivory,feather, silk, wool etc.

Question 9.36

What do you understand by the terms:

  • Hydrogen economy
  • Hydrogenation
  • ‘syngas’
  • Water-gas shift reaction
  • Fuel-cell?

Answer.

  • Hydrogen economy

The main principle behind this is the storage and transportation of energy in the form of liquid or gas in dihydrogen form.

  • Hydrogenation

 It means addition of hydrogen across double and triple bonds in the presence of catalyst to form saturated compounds.

Class_11_Chemistry_Hydrogen_Equation_1

This is called water-gas shift reaction.

  • Fuel cell

It is a commercial cell in which the chemical energy is produced during the combustion of fuel.

This energy is directly converted to electricity.

A fuel cell is also used as a source of electrical energy in the space vehicles.

 

 

Share this with your friends  

Download PDF


You can check our 5-step learning process


.