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.
Answer:
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?
Answer:
Hydrogen has three isotopes:

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer:
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 ?
Answer:
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?
Answer:
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:

Answer:

Question 7.
Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
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.
Answer:
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?
Answer:
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 (C_nH_2n+2) to act as ‘Lewis’ acid or base ? Justify your answer.
Answer:
Carbon hydrides of the type C_nH_2n+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 C_nH_2n+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.
Answer:
Non-stoichiometric hydrides or interstitial hydrides are of the type LaH_2.87, YbH_2.55, TiH_1.5-1.8, ZrH_1.3-1.75, VH_0.50 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.
Answer:
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.
Answer:
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?
Answer:
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.
Answer:
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.
Answer:
(i) BeH₂ < CaH₂ < TiH₂
(ii) LiH < NaH < CsH
(iii) F-F
(iv) H₂O < MgH₂ < NaH.

Question 17.
Compare the structures of H₂O and H₂O₂.
Answer:
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?
Answer:
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.
Answer:

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 O^–₄ (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.
Answer:
(i)
It is a redox reaction in which PbS is oxidised to PbSO₄ and H₂O₂ is reduced to H₂O.

(ii) 2MnO^–₄ + 3H₂O → 2MnO₂ + 2H₂O + 2OH^– + 3O₂
2MnO^–₄ is reduced to MnO₂ by H₂O₂ in alkaline medium. It is a redox reaction.
2MnO^–₄ + 6H⁺ + 5H₂O₂ → 2MN₂₊ + 8H₂O + 5O₂
MnO^–₄ 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.
Answer:
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?
Answer:
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.
Answer:
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^–, HCO^–₃, SO^2-₄ 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.
Answer:
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.
Answer:
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
It oxidises lead sulphide (black) to lead sulphate (white)
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 2MnO^–₄ + 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?
Answer:
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 SO^2-₄ 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?
Answer:
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.
Answer:
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?
Answer:
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?
Answer:
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’?
Answer:
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⁺ SO^2-₄ . H₂O in CuSO₄.5H₂O.

Question 32.
How can saline hydrides remove traces of water from organic compounds?
Answer:
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.
Answer:
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.
Answer:
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?
Answer:
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 ?
Answer:
(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 %.