NCERT Solutions for Class 11 Chemistry Chapter 9: Hydrogen
9.1. Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Ans. In the modern periodic table, Hydrogen is the first element but its position is not clear. Due to the presence of only one electron in its 1s shell [1s1], it show similarity with both alkali metals (first group elements) and Halogens (Group 17 elements).
(a) Resemblance with alkali metals: Like alkali metals, hydrogen contains one valence electron in its valency shell i.e., its electronic configuration is similar to alkali metals. For example
H : 1s1
Similarly, Li : [He] 2s1 and Na: [Ne]3s1
Like all alkali metals, it combines with electronegative elements to form oxides, halides, and sulphides. Also, it can lose one electron to form a unipositive ion. However, unlike alkali metals, it has a very high ionisation enthalpys and does not possess metallic characteristics under normal conditions.
(b) Resemblance with halogens: As hydrogen has one electronic valence shell, it requires one electron to complete their octets like halogens.
H : 1s1
Similarly, F : 1s2 2s2 2p5
Like halogens, hydrogen can gain one electron to form a uni-negative ion. Also, it forms a diatomic molecule and several covalent compounds. Like halogens, it forms
a diatomic molecule, reacts with elements to form hydrides and a large number of covalent compounds. However, it is less reactive than halogens.
Thus, on the basis of above facts, hydrogen cannot be placed with alkali metals (group I) or with halogens (group VII). Hence, hydrogen is best placed separately in the periodic table.
9.2. Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Ans. Hydrogen has three isotopes. They are:
$$\text{(i) Protium,} _1^1 \text{H}\\ \text{(ii) Deuterium,} _1^2 \text {H or D, and}\\ \text{(iii) Tritium or T}, _1^3 \text {H}$$
The mass ratio of protium, deuterium and tritium is 1 : 2 : 3.
9.3. Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Ans. Hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions due to its high ionisation enthalpy (1312 kJ mol–1). Hence, it is very difficult to remove its valence electron. Therefore, it cannot exist in monoatomic form. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic (H2) molecule.
9.4. How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Ans. The process of producing a mixture of carbon monoxide and hydrogen from coal is called coal gasification. This mixture is also called synthesis gas or syn gas.
$$\text{C(s)+H}_2\text{O(g)}\xrightarrow{1270k} \text{CO(g)+H}_2(g)$$
The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) of syn gas mixture with steam in the presence of iron chromate as a catalyst at 673 K.
$$\text{CO(g)+H}_2\text{O(g)}\xrightarrow[\text{Catalyst}]{\text{FeCrO}_4,673 k}\text{CO}_2\text{(g)}+H_2(g)$$
This reaction is called the water-gas shift reaction. This is because with the removal of CO2 the reaction shifts in the forward direction and thus the production of dihydrogen will be increased. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.
9.5. Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Ans. Dihydrogen is prepared by the electrolysis of acidified or alkaline water using platinum electrodes. Generally, 15 – 20% of an acid (H2SO4) or a base (NaOH) is used.
At Cathode: At cathode, reduction of water occurs and hydrogen gas is released.
$$2\text{H}_2\text{O} + 2\text{e}^-\to \text{H}_2 + 2\text{OH}^–$$
At Anode: At the anode, oxidation of OH- ions takes place
$$2\text{OH}^-\rightarrow \text{H}_2\text{O} +\frac{1}{2}\text{O}_2+2\text{e}^-$$
The net reaction can be represented as:
$$2\text{OH}^-\rightarrow \text{H}_2O +\frac{1}{2}\text{O}_2+2\text{e}^-$$
9.6. Complete the following reactions:
$$(i)\text{H}_2\text{(g)}+\text {M}_\text{m}\text{O}_\text{O(s)}\xrightarrow{\Delta}\\\text{(ii) CO(g)} +\text{H}_2\text{(g)} \xrightarrow[\text{catalyst}] {\Delta}\\\text{(iii) C}_3 \text{H}_8\text{(g)} + 3\text{H}_3\text{O(g)} \xrightarrow[\text{catalyst}]{\Delta}\\\text{(iv)Zn(s)} + \text{NaOH(aq)} \xrightarrow{\text{heat}}\\ \textbf{Ans.} \text{(i) H}_2\text{(g)}+ \text{M}_\text{m} \text{O}_\text{O(s)}\xrightarrow{\Delta} \text{mM(s)}+ \text{H}_2\text{O}(1)\\\text{(ii) CO(g)} + 2\text{H}_2\text{(g)} \xrightarrow[\text{catalyst}] {\Delta} \text{CH}_3\text{OH}(1)\\\text{(iii) C}_3 \text{H}_8\text{(g)} + 3\text{H}_2\text{O(g)} \xrightarrow[\text{catalyst}]{\Delta}3\text{CO(g)}+7\text{H}_2\text{(g)}\\\text{(iv)Zn(s)} + \text{NaOH(aq)} \xrightarrow{\text{heat}} \text{Na}_2\text{ZnO}_2\text{(aq) sodium}\space \text{zincate} +\text{H}_2\text{(g)}$$
9.7. Discuss the consequences of high enthalpy of H–H bond in terms of chemical reactivity of dihydrogen.
Ans. The ionisation enthalpy of H–H bond is very high (1312 kJ mol–1). This indicates that:
(i) Hydrogen has a low tendency to form H+ ions.
(ii) Its ionisation enthalpy value is comparable to that of halogens. Thus, it forms diatomic molecules (H2), hydrides with elements, and a large number of covalent bonds.
(iii) Also, due to high ionisation enthalpy, hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals under normal conditions.
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.
Ans. Dihydrogen forms molecular hydrides with most of the p-block elements.
1. Electron-deficient hydrides: Electron-deficient compounds or hydrides are formed by elements of group 13. Electron deficient hydride as the name suggests, has very few electrons, less than that required for writing its conventional Lewis structure. They act as Lewis acids i.e., electron acceptors. In diborane (B2H6) there are six bonds in all, out of which only four bonds are regular two centered - two electron bonds. The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.
2. Electron-precise hydrides: All elements of group 14 form electron precise hydrides. An electron precise hydride has the required number of electrons to write their conventional Lewis structures. For example; CH4 has tetrahedral geometry.The Lewis structure can be written as:
Four regular bonds are formed where two electrons are shared by two atoms.
3. Electron-rich hydrides: Elements of group 15-17 form such compounds. Electron-rich hydrides have excess electrons which are present as lone pairs. For example (NH3 has 1- lone pair, H2O has 2 lone pairs and HF has 3 lone pair). They behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like N, O and F in hydrides results in hydrogen bond formation between the molecules. This results in the association of molecules.
Four regular bonds are formed where two electrons are shared by two atoms.
3. Electron-rich hydrides: Elements of group 15-17 form such compounds. Electron-rich hydrides have excess electrons which are present as lone pairs. For example (NH3 has 1- lone pair, H2O has 2 lone pairs and HF has 3 lone pair). They behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like N, O and F in hydrides results in hydrogen bond formation between the molecules. This results in the association of molecules.
There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.
9.9. What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Ans. Elements of group 13 will form electron-deficient compounds or hydrides. Electron deficient hydride as the name suggests, has very few electrons for writing its conventional Lewis structure.
For example, B2H6, contains four regular bonds and two three centered-two electron bond. Its structure can be represented as:
Since these hydrides are electron-deficient, they have a tendency to accept electrons. Hence, they act as Lewis acids, i.e., electron acceptors.
$$\text{B}_2\text{H}_6+2\text{NMe}\rightarrow 2\text{BH}_3\text{NMe}\\\text{B}_2\text{H}_6+2\text{CO}\rightarrow2\text{BH}_3\text{CO}$$
9.10. Do you expect the carbon hydrides of the type (CnH2n+2) to act as ‘Lewis’ acid or base? Justify your answer.
Ans. For carbon hydrides of type CnH2n+2, the following hydrides are possible for :
n = 1 ⇒ CH4
n = 2 ⇒ C2H6
n = 3 ⇒ C3H8
Taking the example of C2H6,
Total number of valence
electrons = 14
Total covalent bonds = 7.
Hence, the bonds are regular 2e–-2C centered bonds.
Thus, hydride C2H6 has sufficient electrons and can be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets. Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.
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.
Ans. Non-Stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition. These hydrides conduct heat and electricity though not as parent metal do. For example: LaH2.87, YbH2.55, TiH1.5 – 1.8 etc.
Alkali metals form stoichiometric hydrides. The hydrides are ionic in nature and have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed. Thus alkali metals will not form non-stoichiometric hydrides
9.12. How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Ans. Metallic hydrides are hydrogen deficient i.e., they do not hold the law of constant composition. In the hydrides of Ni, Pd, Ce, and Ac, hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.
9.13. How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Ans. Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol–1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.
9.14. Among NH3, H2O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Ans. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine and oxygen, the increasing order of their electronegativities are N < O < F
Hence, the expected order of the extent of
hydrogen bonding is HF > H2O > NH3.
But, the actual order is H2.O > HF > NH3.
This can be explained as follows
Although fluorine is more electronegative than oxygen, the extent of hydrogen bonding is higher in water. There is a shortage of hydrogens in HF, as a result, only straight chain bonding takes place.
In case of water, there are exactly the right numbers of hydrogens. Therefore, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.
In case of ammonia, the extent of hydrogen bonding is limited because nitrogen has only one lone pair. Therefore, it cannot satisfy all hydrogens.
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., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:
$$\text{MH(s)} + \text{H}_2\text{O(aq)} \rightarrow \text{MOH(aq)} + \text{H}_2\text{(g)}$$
The reaction is violent and produces fire. CO2 is heavier than dioxygen. It is used as a fire extinguisher because it covers the fire as a blanket and inhibits the supply of dioxygen. Also in this case, CO2 can be used because it is heavier than dihydrogen and will be effective in isolating the burning surface from dihydrogen and dioxygen.
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 property.
Ans. (i) The electrical conductance of a molecule, depends upon its ionic or covalent nature. Ionic compounds conduct electricity and
covalent compounds do not. Among the given hydrides,
BeH2 = A covalent hydride.
CaH2 = An ionic hydride, and conducts electricity in the molten state.
Titanium hydride, TiH2 - metallic in nature and conducts electricity at room temperature.
Hence, the increasing order of electrical conductance is as follows:
BeH2 < CaH2 < TiH2
(ii) The ionic character of a bond is dependent on the electronegativities of the atoms involved. The higher the difference between the electronegativities of atoms, the smaller is the ionic character. Thus, electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase as:
LiH < NaH < CsH
(iii) Bond dissociation energy depends upon the bond strength of a molecule.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.
9.17. Compare the structures of H2O and H2O2.
Ans. A molecule of water consists of two hydrogen atoms joined to an oxygen atom by covalent bonds. The oxygen atom has six electrons (1s22s22p4) in its outermost shell. The ’s’ and ‘p’ orbital of the valence shell are sp3 hybridised to form four sp3 hybrid orbital oriented tetrahedrally around the oxygen atom. Oxygen is bonded to the two hydrogen atoms by two O-H covalent bonds, and there are two lone-pairs of electrons on the oxygen atom. Due to the presence of two lone-pairs of electrons on the O atom, the H-O-H bond angle is 104.5°, which is slightly less than the tetrahedral angle of 109°28’. Therefore, the structure of water molecule is an angular or bent structure. The O–H bond length is 95.7 pm. The structure can be represented as :
Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.
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 two water molecules react to produce a hydroxide ion (OH–) and a hydronium ion (H3O+). The reaction involved can be represented as:
$$\text{H}_2\text{O}(1)+\text{H}_2\text{O}(1) \rightleftharpoons\underset{\text{Hydronium}\space \text{ion}} {\text{H}_3\text{O}^+\text{(aq)}}+\underset{\text{hydroxide}\space \text{ion}}{\text{OH}^-\text{(aq)}}$$
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:
$$\underset{\text{(acid)}}{\text{H}_2\text{O}(1)}+\underset{\text{(base)}}{\text{H}_2\text{O}(1)}\rightleftharpoons \underset{(\text{conjugate}\space \text{acid})}{\text{H}_3\text{O}^+\text{(aq)}}+\underset{\text{conjugate}\space \text{base}}{\text{OH}^-\text{(aq)}}$$
9.19. Consider the reaction of water with F2 andsuggest, in terms of oxidation and reduction, which species are oxidized/reduced.
Ans. The reaction between fluorine and water can be represented as:
$$2\text{F}_2\text{(g)}+2\text{H}_2\text{O}(1)\rightarrow 4\text{H}^+\text{(aq)}4\text{F}^-\text{(aq)}+\text{O}_2\text{(g)}$$
This is an example of a redox reaction as water is getting oxidised to oxygen, while fluorine is being reduced to fluoride ion.
The oxidation numbers of various species can be represented as:
Fluorine is reduced from zero to (– 1) oxidation state. A decrease in oxidation state indicates the reduction of fluorine.Water is oxidised from (– 2) to zero oxidation state. An increase in oxidation state indicates oxidation of water.
9.20. Complete the following chemical reactions.
(i) PbS(s+)H_2O_2(aq)
(ii) MnO_4\space^-(aq)+ H_2O_2(aq)
(iii) CaO(s)+H_2O(g)
(iv) AICI_3(g)+H_2O(1)
(v) Ca_3N_2(s)+6H_2O(1)
Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.
Ans. (i) PbS(s) + 4H2O2(aq) → PbSO4(s) + 4H2O(l) H2O2 is acting as an oxidising agent in the reaction. Hence, it is a redox reaction.
$$\text{(ii)} 2\text{MnO}_4\space^-\text{(aq)}+5\text{H}_2\text{O}_2\text{(aq)}+6\text{H}^+\text{(aq)}\rightarrow2\text{Mn}^{2+}\text{(aq)}+8\text{H}_2\text{O}(1)+5\text{O}_2\text{(g)}$$
H2O2 is acting as a reducing agent in the acidic medium, thereby oxidising MnO2– . Hence, the given reaction is a redox reaction.
$$\text{(iii) CaO(s)}+\text{H}_2\text{O(g)}\rightarrow \text{Ca}.\text{(OH)}_2\text{(aq)}$$
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Thus, the given reaction is an example of hydrolysis reaction.
$$\text{(iv) AICI}_3\text{(g)}+3\text{H}_2\text{O}(1)\rightarrow[\text{AI(OH)}_3]^{3+}\text{(aq)}+3\text{HCI}^-\text{(aq)}$$
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Thus, the given reaction represents hydrolysis of AlCl3.
$$\text{(v) Ca}_3\text{N}_2\text{(s)}+6\text{H}_2\text{O}(1)\rightarrow3\text{Ca(OH)}_2\text{(aq)}+2\text{NH}_3\text{(g)}$$
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. Thus, the given reaction represents hydrolysis of Ca3N2.
9.21. Describe the structure of the common form of ice.
Ans. Ice is the crystalline form of water. It takes a hexagonal form if crystallised at atmospheric pressure, but condenses to cubic form if the temperature is very low. The three-dimensional structure of ice is represented as:
Ice is crystalline form of water. Ice has a highly ordered three dimensional hydrogen bonded structure. Examination of ice crystals with X-rays shows that each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. One hydrogen atom lies in between each pair of oxygen atoms. Thus, each and every hydrogen atom is covalently bonded to an oxygen atom and linked to another oxygen atom by a hydrogen bond. This arrangement leads to a packing with large open spaces and results in a lower density of ice than that of liquid water and ice floats on it. The structure also contains wide holes that can hold molecules of appropriate size interstitially.
9.22. What causes the temporary and permanent hardness of water?
Ans. The temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO3, where M = Mg, Ca) in water. Permanent hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of chlorides and sulphates in water. Permanent hardness is not removed by boiling.
9.23. Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Ans. The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na+, Ca2+, Mg2+ etc.) and anions (e.g., Cl– , SO42–, HCO3–etc.) present in water by H+ and OH–ions respectively.
Synthetic resins are of two types:
(1) Cation exchange resins : Cation exchange resins are large organic molecules that contain the –SO3H group. The resin is firstly changed to RNa (from RSO3H) by treating it with NaCl. This resin then exchanges Na+ ions with Ca2+ and Mg2+ ions, thereby making the water soft.
$$2\text{RH}+\text{M}^2\text{(aq)} \rightleftharpoons \text{MR}_2\text{(s)}+2\text{H}^+\text{(aq)}$$
(2) Anion exchange resins:Anion exchange resins exchange OH– ions for anions like Cl–, HCO3– , and SO42– present in water.
$$\text{RNH}_2\text{(s)}+ \text{H}_2\text{O}(1)\rightleftharpoons \underset{\darr+\text{X}^-\text{(aq)RNH}_3\space ^+.\text{X}^-\text{(s)}+\text{OH}^-\text{(aq)}}{\text{RNH}_3\space^+.\text{OH}^-\text{(s)}}$$
During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature. This acidic water is then passed through the anion exchange process where OH– ions neutralise the H+ ions and de-ionise the water obtained.
9.24. Write chemical reactions to show the amphoteric nature of water.
Ans. It has the ability to act as an acid as well as a base i.e., it behaves as an amphoteric substance. In the Bronsted sense it acts as an acid with NH3 and a base with H2S.
$$\text{H}_2\text{O}(1)+\text{NH}_3\text{(aq)} \rightarrow \text{NH}_4\space ^+\text{(aq)}+\text{OH}^-\text{(aq)}\\\text{H}_2\text{O}(1)+\text{H}_2\text{S(aq)} \rightarrow \text{H}_3\text{O}^+ +\text{HS}^-$$
The auto-protolysis (self-ionization) of water takes place as follows
$$\text{H}_2\text{O}((1)+\text{H}_2\text{O}(1) \rightarrow \text{H}_3\text{O}^+ +\text{OH}^-\text{(aq)}$$
9.25. Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Ans. Hydrogen peroxide, H2O2 acts as an oxidising as well as a reducing agent in both acidic and alkaline media
Reactions involving oxidising actions are:
$$(1)2\text{Fe}^{2+}+2\text{H}^++\text{H}_2\text{O}_2\rightarrow 2\text{Fe}^{3+}+2\text{H}_2\text{O}\\(2) \text{MN}^{2+}+\text{H}_2\text{O}_2 \rightarrow \text{Mn}^{4+}+2\text{OH}^-\\(3)\text {PbS}+4\text{H}_2\text{O}_2\rightarrow \text{PbSO}_4+4\text{H}_2\text{O}\\(4) 2\text{F}^{2+}+\text{H}_2\text{O}_2\rightarrow2\text{Fe}^{3+}+2\text{OH}^-$$
Reactions involving reduction actions are :
$$(1) 2\text{MnO}_4\space^-+6\text{H}^-+5\text{H}_2\text{O}_2\rightarrow2\text{Mn}^{2+}+8\text{H}_2\text{O}+5\text{O}_2\\(2) \text{I}_2+\text{H}_2\text{O}_2+2\text{OH}^-\rightarrow 2\text{I}^-+2\text{H}_2\text{O}+\text{O}_2\\(3) \text{HOCI}+\text{H}_2\text{O}_2 \rightarrow \text{H}_3\text{O}^++\text{CI}^-+\text{O}_2\\(4) 2\text{MnO}_4\space^-+3\text{H}_2\text{O}_2 \rightarrow 2\text{MnO}_2+3\text{O}_2+2\text{H}_2\text{O}+2\text{OH}^-$$
9.26. What is meant by ‘demineralised’ water and how can it be obtained?
Ans. ‘‘Demineralised water’’ means water free from all soluble mineral salts. It does not contain any anions or cations. Demineralised water is obtained by passing water successively through a cation exchange (in the H+ form) and an anion exchange (in the OH– form) resin. 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) resins. In this cation exchange process, H+ exchanges for Na+, Ca2+, Mg2+and other cations present in water. This process results in proton release and thus makes the water acidic. During the cation exchange process, H+ exchanges for Na+, Mg2+, Ca2+, and other cations present in water.
$$2\text{RH(s)}+\text{M}^2+\text{(aq)}\rightleftharpoons \text{MR}_2\text{(s)}+2\text{H}^+\text{(aq)}...(1)$$
In the anion exchange process, OH– exchanges for anions like Cl–, HCO–, SO4–2 etc. present in water.
$$\text{RNH}_2\text{(s)}+\text{H}_2\text{O(I)} \rightleftharpoons \text{RNH}_3^+.\text{OH}^-\text{(s)}\\\text{RNH}_2\text{(s)}+\text{H}_2\text{O(I)} \rightleftharpoons \text{RNH}_3\space ^+.\text{X}^-\text{(s)}+\text{OH}^-\text{(aq)}...\text{(2)}$$
OH– ions, thus, liberated neutralise the H+ ions set free in the cation exchange.
$$\text{H}^+\text{(aq)}+\text{OH}^-\text{(aq)}\rightarrow \text{H}_2\text{O(I)}$$
9.27. Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Ans. Demineralised water cannot be used for drinking purposes as it is free of all soluble minerals and nutrients that are required by human beings, plants, and animals for survival. It can be made useful by addition of desired minerals in specific amounts, which are important for the growth, of plant and survival of human beings and animals.
9.28. Describe the usefulness of water in biosphere and biological systems.
Ans. The importance of water is:
(i) Water is essential for life. It constitutes around 65% of the human body and 95% of plants.
(ii) It is a universal solvent and plays an important role in biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant.
(iii)It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.
(iv) The high heat of vaporisation and heat of capacity of water helps in moderating the climate and body temperature of all living beings. .
(v) Various biochemical reactions takes place in the presence of water.
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 dielectric constants (78.39 C2/Nm2) and dipole moment make water a universal solvent.
Water is able to dissolve both ionic and covalent compounds. Ionic compounds dissolve in water due to the presence of ions. Covalent compounds form hydrogen bonding and dissolve in water. For example, compounds such as sugar, glucose and ethyl alcohol form hydrogen bond with water and dissolve in it. Water can hydrolyse metallic and non-metallic oxides, hydrides, carbides, phosphides, nitrides and various other salts. During hydrolysis, H+ and OH– ions of water interact with the reacting molecule.
For example :
$$\text{P}_4\text{O}_{10}+6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 (\text{hydrolysis}\space \text{reaction})\\\text{CaH}_2+2\text{H}_2\text{O}\rightarrow \text{Ca(OH)}_2+2\text{H}_2(\text{hydrolysis}\space \text{reaction})$$
9.30. Knowing the properties of H2O and D2O, do you think that D2O can be used for drinking purposes?
Ans. Heavy water (D2O) acts as a moderator, i.e., it slows the rate of a reaction. It cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions taking place in the body and lead to a casualty.
9.31. What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Ans. Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+ and OH– ions) of water molecule react with a compound to form products. It is the process by which chemical compounds are broken apart by the addition of water. It comes from the Greek words for “water” and “separation.”
$$\text{H}_2\text{O} \xrightleftharpoons{\text{hydrolysis}} \text{H}^++\text{OH}^-$$
For example:
$$\text{NaCl}+\text{H}_2\text{O} \rightarrow \text{NaOH}+ \text{HCl}$$
Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. These water molecules are known as water of hydration or crystallisation. For example: Glauber’s salt (sodium sulphate decahydrate, Na2SO4.10H2O); Here, the salt contains 10 molecules of water as water of crystallisation.
$$\text{Na}_2\text{SO}_4.10\text{H}_2\text{O(s)} \rightarrow \text{Na}_2\text{SO}_4.10\text{H}_2\text{O}(l)$$
9.32. How can saline hydrides remove traces of water from organic compounds?
Ans. Saline hydrides are ionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction can be represented as:
$$\text{AH(s)}+ \text{H}_2\text{O(l)} \rightarrow \text{AOH(aq)} + \text{H}_2\text{(g)} (\text{where}, \text{A}=\text{Na, Ca}, ....... )$$
When an organic solvent is added to it, it react with water present in it. As a result of chemical reaction, hydrogen escapes into the atmosphere. The metallic hydroxide is left behind. The dry organic solvent distills over.
9.33. 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.
Ans. The elements with atomic numbers 15, 19, 23, and 44 are Phosphorous, Potassium, Vanadium, and Ruthenium respectively.
(1) Hydride of Phosphorous : Hydride of phosphorous (PH3) is a covalent molecule. It is an electron-rich hydride as it has excess electrons as a lone pair of electrons on phosphorous. It is slightly soluble in water. Its structure can be represented as :
(2) Hydride of potassium: As potassium is an electropositive element, it forms an ionic hydride with hydrogen. It is crystalline and non-volatile in nature. Potassium hydride reacts violently with water giving Potassium hydroxide. The reaction can be represented as:
$$\text{KH(s)}+\text{H}_2\text{O(aq)}\rightarrow \text{KOH(aq)}+ \text{H}_2\text{(g)}$$
(3) Hydrides of Vanadium and Ruthenium: As Vanadium and Ruthenium belong to the d-block of the periodic table. These metals of d-block form metallic or non-stoichiometric hydrides. Therefore, hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen. Hydrides of vanadium and Ruthenium do not react with water.
Behaviour of hydrides towards water :
Potassium hydride reacts violently with water as:
$$\text{KH(s)}+\text{H}_2\text{O(aq)} \rightarrow \text{KOH(aq)}+ \text{H}_2\text{(g)}$$
Phosporus (PH3) is a covalent hydride andslightly soluble in water.
Hydrides of vanadium and Ruthenium do not react with water. Hence, the increasing
order of reactivity of the hydrides is (V, Ru) H < PH3 < KH.
9.34. 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.
Ans. Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). It is neutral in nature. It does not undergo hydrolysis in normal water. It dissociates into its constituent ions as follows:
$$\text{KCl(s)}\xrightarrow{\text{water}} \text{K}^+\text{(aq)}+ \text{Cl}^-\text{(aq)}$$
In acidified and alkaline water also, 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]. Thus, in normal water, it hydrolyses as:
$$\text{AlCl}_3\text{(s)}+3\text{H}_2\text{O(l)}\xrightarrow[\text{water}]{\text{normal}}\text{Al(OH)}_3\text{(s)}$$
In acidified water, H+ ions react with Al(OH)3 and forms water. Also, it gives Al3+ ions. Hence, in acidified water, AlCl3 will exist as Al3+ and Cl– ions.
$$\text{AlCl}_3\text{(s)}\xrightarrow[\text{water}]{\text{acidified}}\text{Al}^{3+}\text{(aq)}+3\text{Cl}^-\text{(aq)}$$
In alkaline water, the following reaction takes place :
$$\underset{\text{From}\space \text{alkaline}\space \text{water}}{\text{Al(OH)}_3\text{(s)}+\text{OH}^-\text{(aq)}}\rightarrow \underset {\text{(aq)}}{[\text{Al(OH)}_4]^-}$$
9.35. How does H2O2 behave as a bleaching agent?
Ans. H2O2 or hydrogen peroxide acts as a strong oxidizing agent both in acidic and basic 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.
9.36. What do you understand by the terms: (i) Hydrogen economy (ii) Hydrogenation (iii) ‘Syngas’ (iv) Water-gas shift reaction (v) Fuel-cell ?
Ans. (i) Hydrogen economy : Hydrogen economy is about the transmission of energy in the form of dihydrogen. Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas. Dihydrogen releases more energy than petrol and is more eco–friendly. Hence, it can be used in fuel cells to generate electric power.
(ii) Hydrogenation: It is a chemical reaction between molecular hydrogen (H2) and another compound or element, usually in the presence of a catalyst such as nickel, palladium or platinum. The process is commonly employed to reduce or saturate organic compounds.
For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati, ghee etc. A simple hydrogenation reaction is:
$$\text{H}_2\text{C}=\text{CH}_2+\text{H}_2\rightarrow \text{CH}_3\text{CH}_3$$
(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 on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.
$$C_nH_{2n+2}+nH_2O\xrightarrow[Ni]{120.0K}nCO+(2n+1)H_2$$
For example,
$$\text{CH}_4\text{(g)}+\text{H}_2\text{O(g)}\xrightarrow[\text{Ni}]{1270\text{k}}\underset{\text{Syngas}}{\underbrace{\text{CO(g)}+3\text{H}_2\text{(g)}}}$$
The process of producing syngas from coal is called ‘coal gasification’.
$$\text{C(s)}+\text{H}_2\text{O(g)}\xrightarrow[\text{Ni}]{1270\text{k}}\text{CO(g)}+\text{H}_2\text{(g)}$$
(iv)Water shift reaction: It is a reaction of carbon monoxide of syngas mixture with steam in the presence of iron chromate as catalyst. The reaction can be represented as :
$$\text{C(s)}+ \text{H}_2\text{O(g)}\xrightarrow[\text{FeCrO}_4]{673\text{k}}\text{CO(g)}+\text{H}_2\text{(g)}$$
The reaction is used to increase the yield of dihydrogen obtained from the coal gasification reaction.
$$\text{C(s)}+\text{H}_2\text{O(g)}\rightarrow \text{CO(g)}+\text{H}_2\text{(g)}$$
(v) Fuel cells: A fuel cell is a device which is used to produce electricity from fuel in the presence of an electrolyte. It converts hydrogen and oxygen into water, and in the process also creates electricity. It is an electro-chemical energy conversion device that produces electricity, water, and heat. Dihydrogen can be used as a fuel in these cells.. Every fuel cell has two electrodes, one positive, called the anode, and one negative, called the cathode. These are separated by an electrolyte barrier. Fuel goes to the anode side, while oxygen (or just air) goes to the cathode side. When both of these chemicals hit the electrolyte barrier, they react, split off their electrons, and create an electric current. A chemical catalyst speeds up the reactions here. 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.
NCERT Solutions for Class 11 Chemistry Chapter 9 Free PDF Download
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