# GSEB Solutions Class 11 Chemistry Chapter 9 Hydrogen

Gujarat Board GSEB Textbook Solutions Class 11 Chemistry Chapter 9 Hydrogen Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Chemistry Chapter 9 Hydrogen

### GSEB Class 11 Chemistry Hydrogen Text Book Questions and Answers

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Hydrogen is the first element in the periodic table, It has the electronic configuration ls1. It is similar to alkali metal (ns1) of group I. It shows resemblance with alkali metals of group I of the periodic table. So it can be placed above the alkali metals in group I of the periodic table.

On the other hand the electronic configuration of hydrogen shows that it is short of one electron to the nearest noble gas configuration (He) having the electronic configuration ls2. Like halogens it forms covalent bonds (H2, Cl2, Br2: etc.) as well as ionic bonds (example Na+ H). It forms H+ ion by giving one electron and hydride ion (H ) by gaining one electron.

On the basis of its electronic configuration (ls2) hydrogen is placed with other ns1 elements namely alkali metals in the group I as well as in group 17 of the periodic table. Thus the position of hydrogen in the periodic table is anomalous. Hydrogen with so many unique characteristics is, therefore best placed separately in the periodic table of elements.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Hydrogen has three isotopes:

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
The H-H bond dissociation enthalpy is the highest for a single bond between two atoms of an element. It is due to this reason that dissociation of dihydrogen into its atoms is only – 0.081 % around 2000 K.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’ be increased ?
The process of producing ‘syn-gas’ from coal is called ‘coal gasification. The mixture of CO and H2 is called SYNGAS.

The production of dihydrogen can be increased by reacting CO of syngas mixtures with steam in the presence of iron chromate as catalyst.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Electrolysis of water containing an electrolyte (15-25% of acid or alkali) is the best method available to manufacture dihydrogen where electricity is cheap. The presence of an electrolyte will make the solution conducting. Platinum electrodes are used in the process.

Question 6.
Complete the following reactions:

Question 7.
Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Owing to its high bond dissociation enthalpy of H-H bond (435.88 kJ mol-1) dihydrogen is quite stable. It is the highest for a single bond between two atoms of any element. It is because of this factor that the dissociation of dihydrogen into its atoms is only 0.081 % around 2000 K which increases to 95.5% at 5000 K. Also it is relatively inert at room temperature due to the high H-H bond enthalpy.

Question 8.
What do you understand by
(i) electron-deficient
(ii) electron-precise
(iii) electron-rich compounds of hydrogen?
Provide justification with suitable examples.
Compounds of hydrogen with other elements are called hydrides. They are:

• Electron-deficient
• Electron-precise and
• Electron rich hydrides.

(i) An electron – deficient hydride has too few electrons for writing its conventional Lewis structure. Diborane (B2H6) is an example. In fact all elements of group 13 will form electron-deficient compounds. They act as Lewis acids, i.e., electron acceptors.

(ii) Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. Methane (CH4), ethane (C2H6) are; examples of electron-precise compounds. In fact, all elements of group 14 form such compounds which are tetrahedral in geometry.

(iii) Electron-rich hydrides have excess electron(s) which are present as lone pairs. Elements of group 15 – 17 form such compound. (NH3) has one lone pair, H2O has two and HF has three lone pairs of electrons). They will behave as Lewis bases, i.e., electron-donors.

Question 9.
What characteristics do you expect from an electron- deficient hydride with respect to its structure and chemical reactions?
An electron deficient hydride has an atom of the element which has not completed its octet. For example B in BF3 & in B2H6.

It has a tendency to accept a pair of electrons from a molecule having one or more lone pair of electrons. Any substance which accepts a pair of electrons, according to Lewis concept, is an acid. Therefore such compounds of group 13 with hydrogen like BCl3, BF3, AlCl3 etc are Lewis Acid.s

Example:

A coordinate bond is formed between the donor and acceptor.

Question 10.
Do you expect the carbon hydrides of the type (CnH2n+2) to act as ‘Lewis’ acid or base ? Justify your answer.
Carbon hydrides of the type CnH2n+2 are CH4, C2H66 etc. in which number of electrons present are just sufficient to write down their conventional Lewis structures.

C has neither extra electrons nor less electrons. Such compounds of the formula CnH2n+2 are called ELECTRON-PRECISE compounds. They will act neither as Lewis acids nor as Lewis bases.

Question 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.
Non-stoichiometric hydrides or interstitial hydrides are of the type LaH2.87, YbH2.55, TiH1.5-1.8, ZrH1.3-1.75, VH0.50 etc.

They are deficient in hydrogen. These are formed by many d-block and f-block elements. These hydrides are not formed by alkali metals. Saline hydrides are stoichiometric, d-block or f-block elements have partially filled d or f subshells due to which hydrogen atoms occupy interstices in the metal lattices.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
The property of adsorption of hydrogen on transition metals in metallic hydrides is widely used in catalytical reduction/ hydrogenation reactions for the preparation of large number of compounds. Some of the metals like Pd, Pt can accommodate a very large volume of hydrogen and therefore can be used as its storage media. This property has high potential for hydrogen storage and as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Atomic hydrogen and oxy-hydrogen torches find use for cutting and welding purposes. Atomic hydrogen

atoms (produced by dissociation of dihydrogen with the help of an electric arc) are allowed to recombine on the surface to be cut/ welded to generate temperature of 4000 K.

Question 14.
Among NH3, H2O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Out of NH3, H2O and HF, HF will be having the highest magnitude of hydrogen bonding because out of N, O, F, fluorine has the maximum value of electronegativity.

Question 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.
Saline hydrides like NaH react violently with water to liberate H2 gas
NaH(s) + H2O(l) → NaOH(aq) + H2(g); ∆H = – Q
Because of the exothermic nature of the reaction, the evolved dihydrogen gas (H2) catches fire. The fire so produced cannot be extinguished by CO2 because it gets reduced by the hot metal hydride.

Question 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 energy.
(iv) NaH, MgH2 and H2O in order of increasing reducing property.
(i) BeH2 < CaH2 < TiH2
(ii) LiH < NaH < CsH
(iii) F-F
(iv) H2O < MgH2 < NaH.

Question 17.
Compare the structures of H2O and H2O2.
Structure of H2O molecule : Water in the gaseous form is a bent molecule (v-shaped) with H-O-H bond angle of 104.5° and O-H bond length of 95.7 pm. It is polar in nature.

Structure of H2O2 : It has a non-planar structure. The molecular dimensions in the gas phase and solid phase are shown in Fig. given below. The two oxygen atoms are joined by a single electron pair bond.

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Water acts both as an acid as well as a base and so it is called amphoteric in character. For example, it can act as an acid with NH3 and a base towards H2S. In general, water can act as a base towards acids stronger than itself and as an acid towards a base stronger than it. Thus, in terms of its amphoteric nature, auto-protolysis of water may be represented as follows. Two molecules of water react with each other through proton transfer. One molecule of H20 is converted to H3O+ while the other is converted to OH.

Question 19.
Consider the reaction of water with F and suggest, in terms of oxidation and reduction which species are oxidised/ reduced.

In these reactions water acts as a reducing agent and hence itself gets oxidised to either O2 or O3
F2(g), on the other hand, is reduced to F.

Question 20.
Complete the following chemical reactions.
(i) PbS(s) + H2O2(aq) →
(ii) Mn O4 (aq) + H2O2(aq) →
(iii) CaO(s) + H2O(g) →
(iv) AlCl3(g) +H2O (l) →
(v) Ca3N2(s) + H2O (l) →
Classify the above into (a) hydrolysis (b) redox and (c) hydration reactions.
(i)
It is a redox reaction in which PbS is oxidised to PbSO4 and H2O2 is reduced to H2O.

(ii) 2MnO4 + 3H2O → 2MnO2 + 2H2O + 2OH + 3O2
2MnO4 is reduced to MnO2 by H2O2 in alkaline medium. It is a redox reaction.
2MnO4 + 6H+ + 5H2O2 → 2MN2+ + 8H2O + 5O2
MnO4 is reduced to MN2+ by H2O2 in acidic medium.

(iii) CaO(s) + H2O(l) → Ca(OH)2(aq)
It is a hydrolysis reaction
CaO is hydrolysed to Ca(OH)2 by water.

(iv) AlCl3 (s) + H2O(l) → [Al(OH2)6]3+ (aq) + 3Cl(aq)
It is a hydration reaction
AlCl3 is hydrated to [Al(OH2)6]3+

(v) Ca3N2 (s) + 6H2O(l) → 3Ca(OH)2(aq) + \begin{aligned}&2 \mathrm{NH}_{3}(\mathrm{~g})\\ &\text { ammonia } \end{aligned}
It is a hydrolysis reaction
Ca3N2 is hydrolysed to Ca(OH)2 and NH3
Thus (i) and (ii) are redox reactions, (iii) & (v) are hydrolysis reactions and (iv) is a hydration reaction.

Question 21.
Describe the structure of the common form of ice.
At atmospheric pressure, ice crystallises in the normal hexagonal form. In it each oxygen atom is tetrahedrally surrounded by four oxygen atoms, i.e., there is a hydrogen atom between each pair of oxygen atoms. This gives ice an open cage like structure. Each oxygen is surrounded by four hydrogen atoms, two by strong covalent bonds (shown by solid lines) and two by weak hydrogen bonds shown by dotted lines. There exists a number of vacant spaces in the crystal lattice and hence the density of ice is lower than that of liquid water.

Question 22.
What causes the temporary and permanent hardness of water?
Rain water is almost pure (may contain some dissolved gases from the atmosphere). Being a good solvent when it flows on the surface of the earth, it dissolves many salts. Presence of calcium and magnesium salts in the form of hydrogen carbonate, chloride and sulphate in water makes it hard.

Presence of Ca(HCO3)2 and Mg(HCO3)2 is responsible for temporary hardness of water. Presence of calcium and magnesium chlorides and sulphates [CaCl2, MgCl2, CaSO4, MgSO4] in water is responsible for permanent hardness of water.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Nowadays hard water is softened by using synthetic cation exchangers. This method is more efficient than zeolite process. Cation exchangers contain large organic molecule witH-SO3H group. Ion exchange resin (RSO3H) is changed to RNa by treating it with NaCl. RNa reacts with metal ion of calcium/magnesium present in hard water to make the water soft. Here R is resin anion.
2RNa(s) + M2+(aq) → R2M(s) + 2Na+(aq)
The resin can be regenerated by adding aqueous NaCl solution. Pure de-mineralised (de-ionized) water free from all soluble mineral salts is obtained by passing water successively through a cation exchange (in the H+ form) and an anion-exchange (in the OH form)

In this exchange, H+ exchanges for Na+, Ca2+, Mg2+ and other cations present in water. This process makes the water acidic. In the next exchange :
RNH2(S) + H2O(l) ⇌ RN H+3.OH(aq)
RN H+3 . OH(s) + X (aq) ⇌ RNH+3 . X(s) + OH (aq)
OH exchanges for anions Cl, HCO3, SO2-4 etc. present in water. OH ions, thus, liberated neutralize the H+ ions set free in the first exchange.
H+(aq) + OH(aq) → H2O(l)
The exhausted cation and anion exchange resin beds are regenerated by treatment with dilute acid and NaOH solution respectively.

Question 24.
Write chemical reactions to show amphoteric nature of water.
A substance is termed AMPHOTERIC if it can show both, the acid and basic behaviour. Water is an amphoteric substance.
It has the ability to act as an acid as well as a base with NH3 (which is a base) it acts as an acid.
NH3(aq) + H2O(l) ⇌ NH+4 (aq) + OH(aq)
It acts as a base with H2S which is a Bronsted acid.
H2S(aq) + H2O(l) ⇌ H3O+ (aq) + HS (aq)

Question 25.
Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Hydrogen peroxide behaves as an oxidizing agent as well as a reducing agent in both acidic and alkaline medium. However H2O2 is a powerful oxidizing agent, but a weak reducing agent:

(a) Oxidizing character : H2O2 acts as an oxidizing agent both in acidic as well as in alkaline medium.

(i) In acidic medium : It oxidises FeSO4 to Fe2(SO4)3
2FeSO4 + H2SO4 + H2O2 → Fe2(SO4)3 + 2H2O
It liberates iodine from potassium iodide
2KI + H2SO4 + H2O2 → K2SO4 + I2 + 2H2O
PbS + 4H2O2 → PbSO4 + 4H2O

(ii) In alkaline medium : It oxidises sulphites to sulphates and nitrites to nitrates
Na2SO3 + H2O2 → Na2SO4 + H2O
KNO2 + H2O2 → KNO3 + H2O
It oxidies benzene to phenol in alkaline medium

(b) Reducing Character : In the presence of strong oxidizing agents, H2O2 behaves as a reducing agent both in acidic and basic medium,
(i) In acidic medium
It reduces acidified KMnO4to Mn2+
2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 8H2O + 5O2
or 2MnO4 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2.
It reduces acidified K2Cr2O7 solution (orange colour) to chromium salts (green colour)

(ii) Reducing action of H2O2 in alkaline medium : It oxidises KMnO4 to MnO2 in alkaline medium.
2KMnO4(aq) + 3H2O2(aq) → 2MnO2 + 2KOH + 3O2 + 2H2O
It reduces ferric salts (Fe3+) to ferrous salts (Fe2+)
2Fe3+(aq) + H2O2(aq) + 2OH (aq) → 2Fe2+ + O2(g) + 2H2O(l)
It reduces alkaline ferricyanide to ferrocyanide
2K3[Fe(CN)6] + 2KOH → 2K4Fe(CN)6 + 2H2O(l) + O2 + H2O2
or 2[Fe(CN)6]3- + 2OH (aq) + H2O2 → 2[Fe(CN)6]4- + 2H2O(l) + O2
It reduces metallic oxides to metal in alkaline medium
Ag2O + H2O2 → 2Ag(s) + H2O + O2.

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Water which does not contain cations and anions is called ‘demineralised’ water. It is soft water. Dimineralised water is obtained the same way as soft water and is obtained from hard water. Demineralised or deionised water is obtained by passing hard water first through a cation exchange resin (RCOOH or RSO3H) which removes Ca2+ and Mg2+ ions from hard water by exchanging them with H+ ions and then passing through an anion exchange resin (RN+ H3OH) which removes Cl and SO2-4 ions present in hard water by exchanging them with OH ions.

Question 27.
Is demineralised or distilled water useful for drinking purposes ? If not, how can it be made useful?
No, demineralised or distilled water is not suitable for drinking purposes. It can be made useful by treating it chemically.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Photosynthesis in plants from CO2 and H2O in the presence of sunlight and green colouring material called chlorophyll to form carbohydrates and supply of dioxygen is an important usefulness of water.

Similarly glucose metabolism in our body to give out CO2 and H2O vapours with the release of energy is an important application of water

The above reaction is exothermic. Energy released is utilized by human body to do work.

Question 29.
What properties of water make it useful as a solvent ? What types of compounds can it (i) dissolve and (ii) hydrolyse?
Water because of its high dielectric constant (78.39) has the ability to dissolve most of the inorganic (ionic) compounds) and is, therefore, regarded as a universal solvent. Whereas the solubility of ionic compounds takes place due to ion-dipole interactions (i.e. solvation of ions), the solubility of covalent compounds such as alcohols, amines, urea, glucose, sugar etc. takes place due to tendency of these molecules to form hydrogen bonds with water.

(i) It can dissolve both ionic compounds as well as covalent compounds which can form hydrogen bonds with water.
Ionic compounds whose lattice energy is lower than hydration energy get dis, olved in water.

(ii) Water can hydrolyse many oxides (metallic and non-metallic) hydrides, carbiues, nitrides, phosphides and many other salts e.g.

• CaO(s) + H2O(l) → Ca(OH)2
• SO2(g) + H2O(l) → H2SO3(aq)
• CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g)
• SiCl4(l) + 4H2O(l) → SiO2.2H2O(s) + 4HCl(aq)
• Al4C3 + 12H2O(l) → 4Al(OH)3 + 3CH4
• Ca3P2(s) + 6H2O → 3Ca(OH)2 + 2PH3

Question 30.
Knowing the properties of H2O and D2O do you think that D2O can be used for drinking purposes?
Heavy water (D2O) inspite of having the usual properties of ordinary water (H2O) is TOXIC and injurious to human beings, plants, and animals since it slows down the rates of reactions occurring in them. Thus heavy water does not support life so well as does ordinary water and is not fit for drinking.

Question 31.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
The interaction between cations and anions of the salt with OH ions and H+ ions furnished by water to yield the original acid and base is called hydrolysis.

Water can easily hydrolyse oxides, halides of non-metals. It can also hydrolyse carbides, nitrides, phosphides of same metals.

In hydration process/reactions, water has strong ability to form compounds with some metal salts known as hydrates.
(i) Water molecules are coordinated to metal ion in a complex
[Cr(H2O)6]3+.3Cl

(ii) Water occupying interstitial sites in the crystal lattice
BaCl2.2H2O

(iii) Water molecules may be hydrogen bonded to certain oxygen containing anions.
[CU(H2O)4]2+ SO2-4 . H2O in CuSO4.5H2O.

Question 32.
How can saline hydrides remove traces of water from organic compounds?
H ion present in saline hydrides M+H is a strong Bronsted base. It reacts with traces of water present in organic compounds to liberate H2 gas
H2O + H → H2(g) + OH

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15,19,23, and 44 with dihydrogen? Compare their behaviour towards water.
The element with atomic numbers 15, belongs topp block [P] and forms hydride PH3. It is a molecular or covalent hydride. This hydride consists up of discrete covalent molecules which are held together by weak van der Waals forces of attraction and hence is called covalent a molecular hydride. The element with atomic number 19 belongs to s-block [K]. It forms a saline hydride K+H.
It reacts with H2O to liberate H2 gas.
KH(s) + H2O(l) → KOH (aq) + H2(g)
Because of the exothermic nature of reaction H2(g) liberated catches fire.

Element with atomic no. 23 belongs to transition or d-block elements (V) and forms metallic or interstitial hydride. It is a non- stoichiometric hydride VH16. The ratio of H atoms to metal atoms in such compounds is not fixed, but varies with temperature and pressure. Due to interstitial hydride formation these metals absorb large volume of hydrogen on their surface.

Element with atomic No. 44 belongs to group 8. It will not form a hydride. Elements of groups 7-9 do not form any hydride and referred to as hydride gap. Hydride of V is insoluble in water.

Question 34.
Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with (i) normal water (ii) acidified water and (iii) alkaline water ? Write equations wherever necessary.
KCl being the salt of a strong acid and a strong base does not give different products except furnishing K+(aq) and Cl (aq) in the solution. Al(III) chloride will undergo hydrolysis and will yield different products.

Question 35.
How does H2O2 behave as a bleaching agent?
The bleaching action of H2O2 is due to the nascent oxygen which it liberates on decomposition.
H2O2 → H2O + O
The nascent oxygen combines with colouring matter which, in turn, gets oxidised.
Colouring matter + O → colourless matter. It is used to bleach delicate materials like ivory, feather, silk, wool etc.

Question 36.
What do you understand by the terms
(i) hydrogen economy
(ii) hydrogenation
(iii) ‘syngas’
(iv) water-gas shift reaction
(v) fuel cell ?
(i) Hydrogen economy : Hydrogen has promising potential for use as a non-polluting fuel of the near future.
2H2(g) + O2 → 2H2O
When burnt in oxygen it does not produce any pollutants which come out while we burn coal, petrol, diesel, kerosene. The potential use of H9 as a fuel in homes, motor vehicles, running industries and generating electricity in future is called hydrogen economy.

(ii) HYDROGENATION means addition of H2 in presence of a catalyst to compounds containing multiple bonds to form saturated compounds.

For example :

(iii) SYNGAS : The mixture of CO and H2 is called syngas. Now a days syngas is produced from sewage, saw dust, scrap wood, newspapers etc. The process of producing syngas from coal is called coal gasification.

(iv) Water-gas shift reaction : The production of hydrogen by reacting CO (carbon monoxide) of syngas mixtures with steam in the presence of iron chromate as catalyst is called water-gas shift reaction.

CO2 is removed by scrubbing with sodium arsenite solution.

(v) Fuel cell: It is a device which converts the energy produced during the combustion of fuels like H2 directly into electrical energy. One such fuel cell is hydrogen-oxygen fuel cell.
The following reaction takes place in a fuel cell

The thermodynamic efficiency of fuel cells is more than 95 %.