“Hydrogen, the most abundant element in the universe and the third most abundant on the surface of the globe, is being visualised as the major future source of energy.”

Hydrogen has the simplest atomic structure among all the elements around us in Nature. In atomic form it consists of only one proton and one electron. However, in elemental form it exists as a diatomic $\left(\mathrm{H_2}\right)$ molecule and is called dihydrogen. It forms more compounds than any other element. Do you know that the global concern related to energy can be overcome to a great extent by the use of hydrogen as a source of energy? In fact, hydrogen is of great industrial importance as you will learn in this unit.

9.1 POSITION OF HYDROGEN IN THE PERIODIC TABLE

Hydrogen is the first element in the periodic table. However, its placement in the periodic table has been a subject of discussion in the past. As you know by now that the elements in the periodic table are arranged according to their electronic configurations.

Hydrogen has electronic configuration $1 s^{1}$. On one hand, its electronic configuration is similar to the outer electronic configuration ( $n s^{1}$ ) of alkali metals , which belong to the first group of the periodic table. On the other hand, like halogens (with $n s^{2} n p^{5}$ configuration belonging to the seventeenth group of the periodic table), it is short by one electron to the corresponding noble gas configuration, helium $\left(1 s^{2}\right)$. Hydrogen, therefore, has resemblance to alkali metals, which lose one electron to form unipositive ions, as well as with halogens, which gain one electron to form uninegative ion. Like alkali metals, hydrogen forms oxides, halides and sulphides. However, unlike alkali metals, it has a very high ionization enthalpy and does not possess metallic characteristics under normal conditions. In fact, in terms of ionization enthalpy, hydrogen resembles more with halogens, $\Delta_{i} H$ of $\mathrm{Li}$ is $520 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~F}$ is $1680 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and that of $\mathrm{H}$ is $1312 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Like halogens, it forms a diatomic molecule, combines with elements to form hydrides and a large number of covalent compounds. However, in terms of reactivity, it is very low as compared to halogens.

Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals and halogens, it differs from them as well. Now the pertinent question arises as where should it be placed in the periodic table? Loss of the electron from hydrogen atom results in nucleus $\left(\mathrm{H}^{+}\right)$of $\sim 1.510^{-3} \mathrm{pm}$ size. This is extremely small as compared to normal atomic and ionic sizes of 50 to $200 \mathrm{pm}$. As a consequence, $\mathrm{H}^{+}$does not exist freely and is always associated with other atoms or molecules. Thus, it is unique in behaviour and is, therefore, best placed separately in the periodic table (Unit 3).

9.2 DIHYDROGEN, $\mathrm{H_2}$

9.2.1 Occurrence

Dihydrogen is the most abundant element in the universe ($70 \%$ of the total mass of the universe) and is the principal element in the solar atmosphere. The giant planets Jupiter and Saturn consist mostly of hydrogen. However, due to its light nature, it is much less abundant ($0.15 \%$ by mass) in the earth’s atmosphere. Of course, in the combined form it constitutes $15.4 \%$ of the earth’s crust and the oceans. In the combined form besides in water, it occurs in plant and animal tissues, carbohydrates, proteins, hydrides including hydrocarbons and many other compounds.

9.2.2 Isotopes of Hydrogen

Hydrogen has three isotopes: protium, ${ _1}^{1} \mathrm{H}$, deuterium,,${ _1}^{2} \mathrm{H}$ or D and tritium, ${ _1}^{3} \mathrm{H}$ or T. Can you guess how these isotopes differ from each other? These isotopes differ from one another in respect of the presence of neutrons. Ordinary hydrogen, protium, has no neutrons, deuterium (also known as heavy hydrogen) has one and tritium has two neutrons in the nucleus. In the year 1934, an American scientist, Harold C. Urey, got Nobel Prize for separating hydrogen isotope of mass number 2 by physical methods.

The predominant form is protium. Terrestrial hydrogen contains $0.0156 \%$ of deuterium mostly in the form of HD. The tritium concentration is about one atom per $10^{18}$ atoms of protium. Of these isotopes, only tritium is radioactive and emits low energy $\beta^{-}$particles ( $t, 12.33$ years).

Table 9.1 Atomic and Physical Properties of Hydrogen

Since the isotopes have the same electronic configuration, they have almost the same chemical properties. The only difference is in their rates of reactions, mainly due to their different enthalpy of bond dissociation (Table 9.1). However, in physical properties these isotopes differ considerably due to their large mass differences.

9.3 PREPARATION OF DIHYDROGEN, $\mathrm{H_2}$

There are a number of methods for preparing dihydrogen from metals and metal hydrides.

9.3.1 Laboratory Preparation of Dihydrogen

(i) It is usually prepared by the reaction of granulated zinc with dilute hydrochloric acid.

$\mathrm{Zn}+2 \mathrm{H}^{+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{H_2}$

(ii) It can also be prepared by the reaction of zinc with aqueous alkali.

$$ \begin{aligned} & \mathrm{Zn}+2 \mathrm{NaOH} \rightarrow \underset{\text { Sodium zincate }}{\mathrm{Na_2} \mathrm{ZnO_2}} +\mathrm{H_2} \\ \end{aligned} $$

9.3.2 Commercial Production of Dihydrogen below:

The commonly used processes are outlined below:

(i) Electrolysis of acidified water using platinum electrodes gives hydrogen.

$$ 2 \mathrm{H_2} \mathrm{O}(1) \xrightarrow[\text { Traces of acid } / \text { base }]{\text { Electrolyis }} 2 \mathrm{H_2}(\mathrm{~g})+\mathrm{O_2}(\mathrm{~g}) $$

(ii) High purity (>99.95 %) dihydrogen is obtained by electrolysing warm aqueous barium hydroxide solution between nickel electrodes.

(iii) It is obtained as a byproduct in the manufacture of sodium hydroxide and chlorine by the electrolysis of brine solution. During electrolysis, the reactions that take place are:

at anode: $2 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl_2}(\mathrm{~g})+2 \mathrm{e}^{-}$

at cathode: $2 \mathrm{H_2} \mathrm{O}$ (l) $+2 \mathrm{e}^{-} \rightarrow \mathrm{H_2}(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq})$

The overall reaction is

$$ \begin{gathered} 2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \\ \downarrow \\ \mathrm{Cl_2}(\mathrm{~g})+\mathrm{H_2}(\mathrm{~g})+2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) \end{gathered} $$

(iv) Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yields hydrogen.

$\mathrm{C_\mathrm{n}} \mathrm{H_2 \mathrm{n} 2} \quad \mathrm{nH_2} \mathrm{O} \quad \underset{\mathrm{Ni}}{1270 \mathrm{~K}} \quad \mathrm{nCO} \quad\left(\begin{array}{lll}2 \mathrm{n} & 1\end{array}\right) \mathrm{H_2}$

e.g.,

$\mathrm{CH_4}(\mathrm{~g})+\mathrm{H_2} \mathrm{O}(\mathrm{g}) \xrightarrow[N i]{1270 \mathrm{~K}} \mathrm{CO}(\mathrm{g})+3 \mathrm{H_2}(\mathrm{~g})$

The mixture of $\mathrm{CO}$ and $\mathrm{H_2}$ is called water gas. As this mixture of $\mathrm{CO}$ and $\mathrm{H_2}$ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or ‘syngas’. Nowadays ‘syngas’ is produced from sewage, saw-dust, scrap wood, newspapers etc. The process of producing ‘syngas’ from coal is called ‘coal gasification’.

$\mathrm{C}(\mathrm{s})+\mathrm{H_2} \mathrm{O}(\mathrm{g}) \xrightarrow{1270 \mathrm{~K}} \mathrm{CO}(\mathrm{g})+\mathrm{H_2}(\mathrm{~g})$

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

$\mathrm{CO}(\mathrm{g})+\mathrm{H_2} \mathrm{O}(\mathrm{g}) \xrightarrow[\text { catalyst }]{673 \mathrm{~K}} \mathrm{CO_2}(\mathrm{~g})+\mathrm{H_2}(\mathrm{~g})$

This is called water-gas shift reaction. Carbon dioxide is removed by scrubbing with sodium arsenite solution.

Presently $\sim 77 \%$ of the industrial dihydrogen is produced from petro-chemicals, $18 \%$ from coal, $4 \%$ from electrolysis of aqueous solutions and $1 \%$ from other sources.

9.4 PROPERTIES OF DIHYDROGEN

9.4.1 Physical Properties

Dihydrogen is a colourless, odourless, tasteless, combustible gas. It is lighter than air and insoluble in water. Its other physical properties alongwith those of deuterium are given in Table 9.1.

9.4.2 Chemical Properties

The chemical behaviour of dihydrogen (and for that matter any molecule) is determined, to a large extent, by bond dissociation enthalpy. The $\mathrm{H}-\mathrm{H}$ bond dissociation enthalpy is the highest for a single bond between two atoms of any element. What inferences would you draw from this fact? It is because of this factor that the dissociation of dihydrogen into its atoms is only $\sim 0.081 \%$ around $2000 \mathrm{~K}$ which increases to $95.5 \%$ at $5000 \mathrm{~K}$. Also, it is relatively inert at room temperature due to the high $\mathrm{H}-\mathrm{H}$ bond enthalpy. Thus, the atomic hydrogen is produced at a high temperature in an electric arc or under ultraviolet radiations. Since its orbital is incomplete with $1 s^{1}$ electronic configuration, it does combine with almost all the elements. It accomplishes reactions by (i) loss of the only electron to give $\mathrm{H}^{+}$, (ii) gain of an electron to form $\mathrm{H}^{-}$, and (iii) sharing electrons to form a single covalent bond.

The chemistry of dihydrogen can be illustrated by the following reactions:

Reaction with halogens: It reacts with halogens, $\mathrm{X_2}$ to give hydrogen halides, $\mathrm{HX}$, $\mathrm{H_2}(\mathrm{~g})+\mathrm{X_2}(\mathrm{~g}) \rightarrow 2 \mathrm{HX}(\mathrm{g}) \quad(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I})$

While the reaction with fluorine occurs even in the dark, with iodine it requires a catalyst.

Reaction with dioxygen: It reacts with dioxygen to form water. The reaction is highly exothermic.

$2 \mathrm{H_2}(\mathrm{~g})+\mathrm{O_2}(\mathrm{~g}) \xrightarrow{\text { catalyst or heating }} 2 \mathrm{H_2} \mathrm{O}(\mathrm{l})$;

$$ \Delta H^{\ominus}=-285.9 \mathrm{~kJ} \mathrm{~mol}^{-1} $$

Reaction with dinitrogen: With dinitrogen it forms ammonia.

$$ \begin{aligned} & & 3 \mathrm{H_2}(\mathrm{~g})+\mathrm{N_2}(\mathrm{~g}) \xrightarrow{\text { 673K, 200atm }} 2 \mathrm{NH_3}(\mathrm{~g}) ; \\ & & \Delta H^{\ominus}=-92.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$

This is the method for the manufacture of ammonia by the Haber process.

Reactions with metals: With many metals it combines at a high temperature to yield the corresponding hydrides (section 9.5)

$\mathrm{H_2}(\mathrm{~g})+2 \mathrm{M}(\mathrm{g}) \rightarrow 2 \mathrm{MH}(\mathrm{s})$

where $\mathrm{M}$ is an alkali metal

Reactions with metal ions and metal oxides: It reduces some metal ions in aqueous solution and oxides of metals (less active than iron) into corresponding metals.

$$ \begin{aligned} & \mathrm{H_2}(\mathrm{~g})+\mathrm{Pd}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Pd}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \\ & \mathrm{yH_2}(\mathrm{~g})+\mathrm{M_\mathrm{x}} \mathrm{O_\mathrm{y}}(\mathrm{s}) \rightarrow \mathrm{xM}(\mathrm{s})+\mathrm{yH_2} \mathrm{O}(\mathrm{l}) \end{aligned} $$

Reactions with organic compounds: It reacts with many organic compounds in the presence of catalysts to give useful hydrogenated products of commercial importance. For example : (i) Hydrogenation of vegetable oils using nickel as catalyst gives edible fats (margarine and vanaspati ghee)

(ii) Hydroformylation of olefins yields aldehydes which further undergo reduction to give alcohols.

$$ \begin{aligned} & \mathrm{H_2}+\mathrm{CO}+\mathrm{RCH}=\mathrm{CH_2} \rightarrow \mathrm{RCH_2} \mathrm{CH_2} \mathrm{CHO} \\ & \mathrm{H_2}+\mathrm{RCH_2} \mathrm{CH_2} \mathrm{CHO} \rightarrow \mathrm{RCH_2} \mathrm{CH_2} \mathrm{CH_2} \mathrm{OH} \end{aligned} $$

Problem 9.1

Comment on the reactions of dihydrogen with (i) chlorine, (ii) sodium, and (iii) copper(II) oxide

Solution

(i) Dihydrogen reduces chlorine into chloride $\left(\mathrm{Cl}^{-}\right)$ion and itself gets oxidised to $\mathrm{H}^{+}$ion by chlorine to form hydrogen chloride. An electron pair is shared between $\mathrm{H}$ and $\mathrm{Cl}$ leading to the formation of a covalent molecule.

(ii) Dihydrogen is reduced by sodium to form $\mathrm{NaH}$. An electron is transferred from $\mathrm{Na}$ to $\mathrm{H}$ leading to the formation of an ionic compound, $\mathrm{Na}^{+} \mathrm{H}^{-}$.

(iii) Dihydrogen reduces copper(II) oxide to copper in zero oxidation state and itself gets oxidised to $\mathrm{H_2} \mathrm{O}$, which is a covalent molecule.

9.4.3 Uses of Dihydrogen

• The largest single use of dihydrogen is in the synthesis of ammonia which is used in the manufacture of nitric acid and nitrogenous fertilizers.
• Dihydrogen is used in the manufacture of vanaspati fat by the hydrogenation of polyunsaturated vegetable oils like soyabean, cotton seeds etc.
• It is used in the manufacture of bulk organic chemicals, particularly methanol.

$$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H_2}(\mathrm{~g}) \xrightarrow[\text { catalyst }]{\text { cobalt }} \mathrm{CH_3} \mathrm{OH}(\mathrm{l}) $$

• It is widely used for the manufacture of metal hydrides (Section 9.5)
• It is used for the preparation of hydrogen chloride, a highly useful chemical.
• In metallurgical processes, it is used to reduce heavy metal oxides to metals.
• 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 welded to generate the temperature of $4000 \mathrm{~K}$.
• It is used as a rocket fuel in space research.
• Dihydrogen is used in fuel cells for generating electrical energy. It has many advantages over the conventional fossil fuels and electric power. It does not produce any pollution and releases greater energy per unit mass of fuel in comparison to gasoline and other fuels.

9.5 HYDRIDES

Dihydrogen, under certain reaction conditions, combines with almost all elements, except noble gases, to form binary compounds, called hydrides. If ’ $\mathrm{E}$ ’ is the symbol of an element then hydride can be expressed as $\mathrm{EH_\mathrm{x}}$ (e.g., $\mathrm{MgH_2}$ ) or $\mathrm{E_\mathrm{m}} \mathrm{H_\mathrm{n}}$ (e.g., $\mathrm{B_2} \mathrm{H_6}$ ).

The hydrides are classified into three categories :

(i) Ionic or saline or saltlike hydrides

(ii) Covalent or molecular hydrides

(iii) Metallic or non-stoichiometric hydrides

9.5.1 Ionic or Saline Hydrides

These are stoichiometric compounds of dihydrogen formed with most of the s-block elements which are highly electropositive in character. However, significant covalent character is found in the lighter metal hydrides such as $\mathrm{LiH}, \mathrm{BeH_2}$ and $\mathrm{MgH_2}$. In fact $\mathrm{BeH_2}$ and $\mathrm{MgH_2}$ are polymeric in structure. The ionic hydrides are crystalline, non-volatile and nonconducting in solid state. However, their melts conduct electricity and on electrolysis liberate dihydrogen gas at anode, which confirms the existence of $\mathrm{H}^{-}$ion.

$2 \mathrm{H}^{-}($melt $) \xrightarrow{\text { anode }} \mathrm{H_2}(\mathrm{~g})+2 \mathrm{e}^{-}$

Saline hydrides react violently with water producing dihydrogen gas.

$\mathrm{NaH}(\mathrm{s})+\mathrm{H_2} \mathrm{O}(\mathrm{aq}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\mathrm{H_2}(\mathrm{~g})$

Lithium hydride is rather unreactive at moderate temperatures with $\mathrm{O_2}$ or $\mathrm{Cl_2}$. It is, therefore, used in the synthesis of other useful hydrides, e.g.,

$$ \begin{aligned} & 8 \mathrm{LiH}+\mathrm{Al_2} \mathrm{Cl_6} \rightarrow 2 \mathrm{LiAlH_4}+6 \mathrm{LiCl} \\ & 2 \mathrm{LiH}+\mathrm{B_2} \mathrm{H_6} \rightarrow 2 \mathrm{LiBH_4} \end{aligned} $$

9.5.2 Covalent or Molecular Hydride

Dihydrogen forms molecular compounds with most of the $p$-block elements. Most familiar examples are $\mathrm{CH_4}, \mathrm{NH_3}, \mathrm{H_2} \mathrm{O}$ and $\mathrm{HF}$. For convenience hydrogen compounds of nonmetals have also been considered as hydrides. Being covalent, they are volatile compounds.

Molecular hydrides are further classified according to the relative numbers of electrons and bonds in their Lewis structure into :

(i) electron-deficient, (ii) electron-precise, and (iii) electron-rich hydrides.

An electron-deficient hydride, as the name suggests, has too few electrons for writing its conventional Lewis structure. Diborane $\left(\mathrm{B_2} \mathrm{H_6}\right)$ is an example. In fact all elements of group 13 will form electron-deficient compounds. What do you expect from their behaviour? They act as Lewis acids i.e., electron acceptors.

Electron-precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 form such compounds (e.g., $\mathrm{CH_4}$ ) which are tetrahedral in geometry.

Electron-rich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 form such compounds. $\left(\mathrm{NH_3}\right.$ has 1- lone pair, $\mathrm{H_2} \mathrm{O}-2$ and $\mathrm{HF}-3$ lone pairs). What do you expect from the behaviour of such compounds? They will behave as Lewis bases i.e., electron donors. The presence of lone pairs on highly electronegative atoms like $\mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ in hydrides results in hydrogen bond formation between the molecules. This leads to the association of molecules.

Problem 9.2

Would you expect the hydrides of $\mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ to have lower boiling points than the hydrides of their subsequent group members? Give reasons.

Solution

On the basis of molecular masses of $\mathrm{NH_3}$, $\mathrm{H_2} \mathrm{O}$ and $\mathrm{HF}$, their boiling points are expected to be lower than those of the subsequent group member hydrides. However, due to higher electronegativity of $\mathrm{N}, \mathrm{O}$ and $\mathrm{F}$, the magnitude of hydrogen bonding in their hydrides will be quite appreciable. Hence, the boiling points $\mathrm{NH_3}, \mathrm{H_2} \mathrm{O}$ and $\mathrm{HF}$ will be higher than the hydrides of their subsequent group members.

9.5.3 Metallic or Non-stoichiometric (or Interstitial) Hydrides

These are formed by many $d$-block and $f$-block elements. However, the metals of group 7, 8 and 9 do not form hydride. Even from group 6 , only chromium forms $\mathrm{CrH}$. These hydrides conduct heat and electricity though not as efficiently as their parent metals do. Unlike saline hydrides, they are almost always nonstoichiometric, being deficient in hydrogen. For example, $\mathrm{LaH_2.87}, \mathrm{YbH_2.55}, \mathrm{TiH_1.5-1.8}, \mathrm{ZrH_1.3-1.75}, \mathrm{VH_0.56}, \mathrm{NiH_0.6-0.7}, \mathrm{PdH_0.6-0.8}$ etc. In such hydrides, the law of constant composition does not hold good.

Earlier it was thought that in these hydrides, hydrogen occupies interstices in the metal lattice producing distortion without any change in its type. Consequently, they were termed as interstitial hydrides. However, recent studies have shown that except for hydrides of Ni, Pd, Ce and Ac, other hydrides of this class have lattice different from that of the parent metal. The property of absorption of hydrogen on transition metals is widely used in catalytic reduction / hydrogenation reactions for the preparation of large number of compounds. Some of the metals (e.g., 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.

Problem 9.3

Can phosphorus with outer electronic configuration $3 s^{2} 3 p^{3}$ form $\mathrm{PH_5}$ ?

Solution

Although phosphorus exhibits +3 and +5 oxidation states, it cannot form $\mathrm{PH_5}$. Besides some other considerations, high $\Delta_{\mathrm{a}} H$ value of dihydrogen and $\Delta_{e q} H$ value of hydrogen do not favour to exhibit the highest oxidation state of $\mathrm{P}$, and consequently the formation of $\mathrm{PH_5}$.

9.6 WATER

A major part of all living organisms is made up of water. Human body has about $65 \%$ and some plants have as much as $95 \%$ water. It is a crucial compound for the survival of all life forms. It is a solvent of great importance. The distribution of water over the earth’s surface is not uniform. The estimated world water supply is given in Table 9.2

Table 9.2 Estimated World Water Supply

9.6.1 Physical Properties of Water

It is a colourless and tasteless liquid. Its physical properties are given in Table 9.3 along with the physical properties of heavy water.

The unusual properties of water in the condensed phase (liquid and solid states) are due to the presence of extensive hydrogen bonding between water molecules. This leads to high freezing point, high boiling point, high heat of vaporisation and high heat of fusion in comparison to $\mathrm{H_2} \mathrm{~S}$ and $\mathrm{H_2} \mathrm{Se}$. In comparison to other liquids, water has a higher specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant, etc. These properties allow water to play a key role in the biosphere.

Table 9.3 Physical Properties of $\mathrm{H_2} \mathrm{O}$ and $\mathrm{D_2} \mathrm{O}$

The high heat of vaporisation and heat capacity are responsible for moderation of the climate and body temperature of living beings. It is an excellent solvent for transportation of ions and molecules required for plant and animal metabolism. Due to hydrogen bonding with polar molecules, even covalent compounds like alcohol and carbohydrates dissolve in water.

9.6.2 Structure of Water

In the gas phase water is a bent molecule with a bond angle of $104.5^{\circ}$, and $\mathrm{O}-\mathrm{H}$ bond length of 95.7 pm as shown in Fig 9.1(a). It is a highly

polar molecule, (Fig 9.1(b)). Its orbital overlap picture is shown in Fig. 9.1(c). In the liquid phase water molecules are associated together by hydrogen bonds.

The crystalline form of water is ice. At atmospheric pressure ice crystallises in the hexagonal form, but at very low temperatures it condenses to cubic form. Density of ice is less than that of water. Therefore, an ice cube floats on water. In winter season ice formed on the surface of a lake provides thermal insulation which ensures the survival of the aquatic life. This fact is of great ecological significance.

9.6.3 Structure of Ice

Ice has a highly ordered three dimensional hydrogen bonded structure as shown in Fig. 9.2. Examination of ice crystals with

X-rays shows that each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of $276 \mathrm{pm}$.

Hydrogen bonding gives ice a rather open type structure with wide holes. These holes can hold some other molecules of appropriate size interstitially.

9.6.4 Chemical Properties of Water

Water reacts with a large number of substances. Some of the important reactions are given below.

(1) Amphoteric Nature: It has the ability to act as an acid as well as a base i.e., it behaves as an amphoteric substance. In the Brönsted sense it acts as an acid with $\mathrm{NH_3}$ and a base with $\mathrm{H_2} \mathrm{~S}$.

$$ \begin{array}{lll} \mathrm{H_2} \mathrm{O}(\mathrm{l})+\mathrm{NH_3}(\mathrm{aq}) \rightarrow& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{NH_4}^{+}(\mathrm{aq}) \\ \mathrm{H_2} \mathrm{O}(\mathrm{l})+\mathrm{H_2} \mathrm{~S}(\mathrm{aq}) \rightarrow & \mathrm{H_3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HS}^{-}(\mathrm{aq}) \end{array} $$

The auto-protolysis (self-ionization) of water takes place as follows :

$$ \begin{aligned} & \underset{ \substack {\text{acid -1} \\ \text{(acid)} }}{\mathrm{H_2} \mathrm{O}(\mathrm{l})} + \underset{\substack {\text{base -1} \\ \text{(base)} }}{\mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow} \quad \underset{\substack{\text{acid-2} \\ \text{(conjugate acid)}}}{\mathrm{H_3} \mathrm{O}^{+}(\mathrm{aq})}+ \underset{\substack{\text{base -1 }\\ \text{(conjugate base)}}}{\mathrm{OH}^{-}(\mathrm{aq})} \\ \end{aligned} $$

(2) Redox Reactions Involving Water: Water can be easily reduced to dihydrogen by highly electropositive metals.

$$ 2 \mathrm{H_2} \mathrm{O}(\mathrm{l})+2 \mathrm{Na}(\mathrm{s}) \rightarrow 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H_2}(\mathrm{~g}) $$

Thus, it is a great source of dihydrogen.

Water is oxidised to $\mathrm{O_2}$ during photosynthesis.

$6 \mathrm{CO_2}(\mathrm{~g})+12 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C_6} \mathrm{H_12} \mathrm{O_6}(\mathrm{aq})+6 \mathrm{H_2} \mathrm{O}(\mathrm{l}) +6 \mathrm{O_2}(\mathrm{~g})$

With fluorine also it is oxidised to $\mathrm{O_2}$.

$2 \mathrm{~F_2}(\mathrm{~g})+2 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow 4 \mathrm{H}^{+}$(aq) $+4 \mathrm{~F}^{-}(\mathrm{aq})+\mathrm{O_2}(\mathrm{~g})$

(3) Hydrolysis Reaction: Due to high dielectric constant, it has a very strong hydrating tendency. It dissolves many ionic compounds. However, certain covalent and some ionic compounds are hydrolysed in water.

$\mathrm{P_4} \mathrm{O_10}(\mathrm{~s})+6 \mathrm{H_2} \mathrm{O}(1) \rightarrow 4 \mathrm{H_3} \mathrm{PO_4}(\mathrm{aq})$

$\mathrm{SiCl_4}(\mathrm{l})+2 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{SiO_2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq})$

$$ \mathrm{N}^{3-}(\mathrm{s})+3 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NH_3}(\mathrm{~g})+3 \mathrm{OH}^{-}(\mathrm{aq}) $$

(4) Hydrates Formation: From aqueous solutions many salts can be crystallised as hydrated salts. Such an association of water is of different types viz.,

(i) coordinated water e.g.,

$\left[\mathrm{Cr}\left(\mathrm{H_2} \mathrm{O}\right)_{6}\right]^{3+} 3 \mathrm{Cl}^{-}$

(ii) interstitial water e.g., $\mathrm{BaCl_2} \cdot 2 \mathrm{H_2} \mathrm{O}$

(iii) hydrogen-bonded water e.g.,

$$ \left[\mathrm{Cu}\left(\mathrm{H_2} \mathrm{O}\right)_{4}\right]^{2+} \mathrm{SO_4}^{2-} \cdot \mathrm{H_2} \mathrm{O} \text { in } \mathrm{CuSO_4} \cdot 5 \mathrm{H_2} \mathrm{O} $$

Problem 9.4

How many hydrogen-bonded water molecule(s) are associated in $\mathrm{CuSO_4} \cdot 5 \mathrm{H_2} \mathrm{O}$ ?

Solution

Only one water molecule, which is outside the brackets (coordination sphere), is hydrogen-bonded. The other four molecules of water are coordinated.

9.6.5 Hard and Soft 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 hydrogencarbonate, chloride and sulphate in water makes water ‘hard’. Hard water does not give lather with soap. Water free from soluble salts of calcium and magnesium is called Soft water. It gives lather with soap easily.

Hard water forms scum/precipitate with soap. Soap containing sodium stearate $\left(\mathrm{C_{17}} \mathrm{H_{35}} \mathrm{COONa}\right)$ reacts with hard water to precipitate out $\mathrm{Ca} / \mathrm{Mg}$ stearate.

$2 \mathrm{C_{17}} \mathrm{H_{35}} \mathrm{COONa}(\mathrm{aq})+\mathrm{M}^{2+}(\mathrm{aq}) \rightarrow$

$\left(\mathrm{C_{17}} \mathrm{H_{35}} \mathrm{COO}\right)_{2} \mathrm{M} \downarrow+2 \mathrm{Na}^{+}(\mathrm{aq}) ; \mathrm{M}$ is $\mathrm{Ca} / \mathrm{Mg}$

It is, therefore, unsuitable for laundry. It is harmful for boilers as well, because of deposition of salts in the form of scale. This reduces the efficiency of the boiler. The hardness of water is of two types: (i) temporary hardness, and (ii) permanent hardness.

9.6.6 Temporary Hardness

Temporary hardness is due to the presence of magnesium and calcium hydrogencarbonates. It can be removed by :

(i) Boiling: During boiling, the soluble $\mathrm{Mg} (\mathrm{HCO_3} )_2$ is converted into insoluble $\mathrm{Mg}(\mathrm{OH})_2$ and $\mathrm{Ca} \left(\mathrm{HCO_3} \right)_2$ is changed to insoluble $\mathrm{CaCO_3}$. It is because of high solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ as compared to that of $\mathrm{MgCO_3}$, that $\mathrm{Mg}(\mathrm{OH})_2$ is precipitated. These precipitates can be removed by filtration. Filtrate thus obtained will be soft water.

$$ \begin{aligned} & \mathrm{Mg}\left(\mathrm{HCO_3}\right)_2 \xrightarrow{\text { Heating }} \mathrm{Mg}(\mathrm{OH})_2 \downarrow+2 \mathrm{CO_2} \uparrow \\ & \mathrm{Ca}\left(\mathrm{HCO_3}\right)_2 \xrightarrow{\text { Heating }} \mathrm{CaCO_3} \downarrow+\mathrm{H_2} \mathrm{O}+\mathrm{CO_2} \uparrow \end{aligned} $$

(ii) Clark’s method: In this method calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off.

$$ \begin{aligned} \mathrm{Ca} \left(\mathrm{HCO_3} \right)_2+\mathrm{Ca}(\mathrm{OH})_2 & \rightarrow 2 \mathrm{CaCO_3} \downarrow+2 \mathrm{H_2} \mathrm{O} \end{aligned} $$

$$ \begin{aligned} \mathrm{Mg} \left(\mathrm{HCO_3} \right)_2+2 \mathrm{Ca}(\mathrm{OH})_2 & \rightarrow 2 \mathrm{CaCO_3} \downarrow + \mathrm{Mg}(\mathrm{OH})_2 \downarrow+2 \mathrm{H_2} \mathrm{O}\\ \end{aligned} $$

9.6.7 Permanent Hardness

It 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. It can be removed by the following methods:

(i) Treatment with washing soda (sodium carbonate): Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.

$$ \begin{aligned} \mathrm{MCl_2}+\mathrm{Na_2} \mathrm{CO_3} \rightarrow \mathrm{MCO_3} \downarrow+ & 2 \mathrm{NaCl} \\ & (\mathrm{M}=\mathrm{Mg}, \mathrm{Ca}) \\ \mathrm{MSO_4}+\mathrm{Na_2} \mathrm{CO_3} \rightarrow \mathrm{MCO_3} \downarrow+ & \mathrm{Na_2} \mathrm{SO_4} \end{aligned} $$

(ii) Calgon’s method: Sodium hexametaphosphate $\left(\mathrm{Na_6} \mathrm{P_6} \mathrm{O_18}\right)$, commercially called ‘calgon’, when added to hard water, the following reactions take place. $\mathrm{Na_6} \mathrm{P_6} \mathrm{O_18} \rightarrow 2 \mathrm{Na}^{+}+\mathrm{Na_4} \mathrm{P_6} \mathrm{O_18}^{2-}$

$$ (\mathrm{M}=\mathrm{Mg}, \mathrm{Ca}) $$

$\mathrm{M}^{2+}+\mathrm{Na_4} \mathrm{P_6} \mathrm{O_18}^{2-} \rightarrow\left[\mathrm{Na_2} \mathrm{MP_6} \mathrm{O_18}\right]^{2-}+2 \mathrm{Na}^{+}$

The complex anion keeps the $\mathrm{Mg}^{2+}$ and $\mathrm{Ca}^{2+}$ ions in solution.

(iii) Ion-exchange method: This method is also called zeolite/permutit process. Hydrated sodium aluminium silicate is zeolite/permutit. For the sake of simplicity, sodium aluminium silicate $\left(\mathrm{NaAlSiO_4}\right)$ can be written as $\mathrm{NaZ}$. When this is added in hard water, exchange reactions take place.

$$ \begin{array}{r} 2 \mathrm{NaZ}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \rightarrow \mathrm{MZ_2}(\mathrm{~s})+2 \mathrm{Na}^{+}(\mathrm{aq}) \\ (\mathrm{M}=\mathrm{Mg}, \mathrm{Ca}) \end{array} $$

Permutit/zeolite is said to be exhausted when all the sodium in it is used up. It is regenerated for further use by treating with an aqueous sodium chloride solution.

$\mathrm{MZ_2}(\mathrm{~s})+2 \mathrm{NaCl}(\mathrm{aq}) \rightarrow 2 \mathrm{NaZ}(\mathrm{s})+\mathrm{MCl_2}(\mathrm{aq})$

(iv) Synthetic resins method: Nowadays hard water is softened by using synthetic cation exchangers. This method is more efficient than zeolite process. Cation exchange resins contain large organic molecule with - $\mathrm{SO_3} \mathrm{H}$ group and are water insoluble. Ion exchange resin $\left(\mathrm{RSO_3} \mathrm{H}\right)$ is changed to $\mathrm{RNa}$ by treating it with $\mathrm{NaCl}$. The resin exchanges $\mathrm{Na}^{+}$ions with $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ ions present in hard water to make the water soft. Here $\mathrm{R}$ is resin anion.

$2 \mathrm{RNa}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \rightarrow \mathrm{R_2} \mathrm{M}(\mathrm{s})+2 \mathrm{Na}^{+}(\mathrm{aq})$

The resin can be regenerated by adding aqueous $\mathrm{NaCl}$ 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 $\mathrm{H}^{+}$form) and an anionexchange (in the $\mathrm{OH}^{-}$form) resins:

$$ 2 \mathrm{RH}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \rightarrow \quad \mathrm{MR_2}(\mathrm{~s})+2 \mathrm{H}^{+}(\mathrm{aq}) $$

In this cation exchange process, $\mathrm{H}^{+}$exchanges for $\mathrm{Na}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$ and other cations present in water. This process results in proton release and thus makes the water acidic. In the anion exchange process:

$$ \begin{array}{rr} \mathrm{RNH_2}(\mathrm{~s})+\mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow \quad \mathrm{RNH_3}^{+} \cdot \mathrm{OH}^{-}(\mathrm{s}) \\ \mathrm{RNH_3}^{+} \cdot \mathrm{OH}^{-}(\mathrm{s})+\mathrm{X}^{-}(\mathrm{aq}) \rightarrow \quad \mathrm{RNH_3}^{+} \cdot \mathrm{X}^{-}(\mathrm{s}) +\mathrm{OH}^{-}(\mathrm{aq}) \\ \end{array} $$

$\mathrm{OH}^{-}$exchanges for anions like $\mathrm{Cl}^{-}, \mathrm{HCO_3}^{-}, \mathrm{SO_4}^{2-}$ etc. present in water. $\mathrm{OH}^{-}$ions, thus, liberated neutralise the $\mathrm{H}^{+}$ions set free in the cation exchange.

$$ \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H_2} \mathrm{O}(\mathrm{l}) $$

The exhausted cation and anion exchange resin beds are regenerated by treatment with dilute acid and alkali solutions respectively.

9.7 HYDROGEN PEROXIDE $\left(\mathrm{H_2} \mathrm{O_2}\right)$

Hydrogen peroxide is an important chemical used in pollution control treatment of domestic and industrial effluents.

9.7.1 Preparation

It can be prepared by the following methods.

(i) Acidifying barium peroxide and removing excess water by evaporation under reduced pressure gives hydrogen peroxide.

$$ \begin{array}{r} \mathrm{BaO_2} \cdot 8 \mathrm{H_2} \mathrm{O}(\mathrm{s})+\mathrm{H_2} \mathrm{SO_4}(\mathrm{aq}) \rightarrow \mathrm{BaSO_4}(\mathrm{~s})+ \mathrm{H_2} \mathrm{O_2}(\mathrm{aq})+8 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \end{array} $$

(ii) Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide.

$$ \begin{aligned} & 2 \mathrm{HSO_4}^{-}(\mathrm{aq}) \xrightarrow{\text { Electrolysis }} \mathrm{HO_3} \mathrm{SOOSO_3} \mathrm{H}(\mathrm{aq}) \\ & \xrightarrow{\text { Hydrolysis }} 2 \mathrm{HSO_4}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{H_2} \mathrm{O_2}(\mathrm{aq}) \end{aligned} $$

This method is now used for the laboratory preparation of $\mathrm{D_2} \mathrm{O_2}$.

$$ \mathrm{K_2} \mathrm{~S_2} \mathrm{O_8}(\mathrm{~s})+2 \mathrm{D_2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{KDSO_4}(\mathrm{aq})+\mathrm{D_2} \mathrm{O_2}(\mathrm{l}) $$

(iii) Industrially it is prepared by the autooxidation of 2-alklylanthraquinols.

$ \begin{aligned} \text{ 2- ethylanthraquinol }\underset{\mathrm{H_2} / \mathrm{Pd}}{\stackrel{\mathrm{O_2} \text { (air) }}{\rightleftarrows}} \mathrm{H_2} \mathrm{O_2}+ \\ & \text{(oxidised product)} \end{aligned} $

In this case $1 \% \mathrm{H_2} \mathrm{O_2}$ is formed. It is extracted with water and concentrated to $\sim 30 \%$ (by mass) by distillation under reduced pressure. It can be further concentrated to $\sim 85 \%$ by careful distillation under low pressure. The remaining water can be frozen out to obtain pure $\mathrm{H_2} \mathrm{O_2}$.

9.7.2 Physical Properties

In the pure state $\mathrm{H_2} \mathrm{O_2}$ is an almost colourless (very pale blue) liquid. Its important physical properties are given in Table 9.4.

$\mathrm{H_2} \mathrm{O_2}$ is miscible with water in all proportions and forms a hydrate $\mathrm{H_2} \mathrm{O_2} \cdot \mathrm{H_2} \mathrm{O}$ (mp 221K). A $30 \%$ solution of $\mathrm{H_2} \mathrm{O_2}$ is marketed as ‘100 volume’ hydrogen peroxide. It means that one millilitre of $30 \% \mathrm{H_2} \mathrm{O_2}$ solution will give $100 \mathrm{~mL}$ of oxygen at STP. Commercially marketed sample is $10 \mathrm{~V}$, which means that the sample contains $3 \% \mathrm{H_2} \mathrm{O_2}$.

Problem 9.5

Calculate the strength of 10 volume solution of hydrogen peroxide.

Solution

10 volume solution of $\mathrm{H_2} \mathrm{O_2}$ means that $1 \mathrm{~L}$ of this $\mathrm{H_2} \mathrm{O_2}$ solution will give $10 \mathrm{~L}$ of oxygen at STP

$2 \mathrm{H_2} \mathrm{O_2}(\mathrm{l}) \rightarrow \mathrm{O_2}(\mathrm{~g})+\mathrm{H_2} \mathrm{O}(\mathrm{l})$

$234 \mathrm{~g} \quad 22.7 \mathrm{~L}$ at STP

$68 \mathrm{~g}$

On the basis of above equation $22.7 \mathrm{~L}$ of $\mathrm{O_2}$ is produced from $68 \mathrm{~g} \mathrm{H_2} \mathrm{O_2}$ at STP

$10 \mathrm{~L}$ of $\mathrm{O_2}$ at STP is produced from

$$ \frac{6810}{22.7} \mathrm{~g}=29.9 \mathrm{~g} \quad 30 \mathrm{~g} \mathrm{H_2} \mathrm{O_2} $$

Therefore, strength of $\mathrm{H_2} \mathrm{O_2}$ in 10 volume $\mathrm{H_2} \mathrm{O_2}$ solution $=30 \mathrm{~g} / \mathrm{L}=3 \% \mathrm{H_2} \mathrm{O_2}$ solution

Table 9.4 Physical Properties of Hydrogen Peroxide

9.7.3 Structure

Hydrogen peroxide has a non-planar structure. The molecular dimensions in the gas phase and solid phase are shown in Fig 9.3

9.7.4 Chemical Properties

It acts as an oxidising as well as reducing agent in both acidic and alkaline media. Simple reactions are described below.

(i) Oxidising action in acidic medium

$$ \begin{aligned} & 2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+ \mathrm{H_2} \mathrm{O_2}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{H_2} \mathrm{O}(1) \\ & \mathrm{PbS}(\mathrm{s})+4 \mathrm{H_2} \mathrm{O_2}(\mathrm{aq}) \rightarrow \mathrm{PbSO_4}(\mathrm{~s})+4 \mathrm{H_2} \mathrm{O}(\mathrm{l}) \end{aligned} $$

(ii) Reducing action in acidic medium

$$ \begin{aligned} & 2 \mathrm{MnO_4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H_2} \mathrm{O_2} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H_2} \mathrm{O}+5 \mathrm{O_2} \\ & \mathrm{HOCl}+\mathrm{H_2} \mathrm{O_2} \rightarrow \mathrm{H_3} \mathrm{O}^{+}+\mathrm{Cl}^{-}+\mathrm{O_2} \end{aligned} $$

(iii) Oxidising action in basic medium

$$ \begin{aligned} & 2 \mathrm{Fe}^{2+}+\mathrm{H_2} \mathrm{O_2} \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{OH}^{-} \\ & \mathrm{Mn}^{2+}+\mathrm{H_2} \mathrm{O_2} \rightarrow \mathrm{Mn}^{4+}+2 \mathrm{OH}^{-} \end{aligned} $$

(iv) Reducing action in basic medium

$$ \begin{aligned} & \mathrm{I_2}+\mathrm{H_2} \mathrm{O_2}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{I}^{-}+2 \mathrm{H_2} \mathrm{O}+\mathrm{O_2} \\ & 2 \mathrm{MnO_4}^{-}+3 \mathrm{H_2} \mathrm{O_2} \rightarrow 2 \mathrm{MnO_2}+3 \mathrm{O_2}+ 2 \mathrm{H_2} \mathrm{O}+2 \mathrm{OH}^{-} \end{aligned} $$

9.7.5 Storage

$\mathrm{H_2} \mathrm{O_2}$ decomposes slowly on exposure to light.

$$ 2 \mathrm{H_2} \mathrm{O_2}(\mathrm{l}) \rightarrow 2 \mathrm{H_2} \mathrm{O}(\mathrm{l})+\mathrm{O_2}(\mathrm{~g}) $$

In the presence of metal surfaces or traces of alkali (present in glass containers), the above reaction is catalysed. It is, therefore, stored in wax-lined glass or plastic vessels in dark. Urea can be added as a stabiliser. It is kept away from dust because dust can induce explosive decomposition of the compound.

9.7.6 Uses

Its wide scale use has led to tremendous increase in the industrial production of $\mathrm{H_2} \mathrm{O_2}$. Some of the uses are listed below:

(i) In daily life it is used as a hair bleach and as a mild disinfectant. As an antiseptic it is sold in the market as perhydrol.

(ii) It is used to manufacture chemicals like sodium perborate and per-carbonate, which are used in high quality detergents.

(iii) It is used in the synthesis of hydroquinone, tartaric acid and certain food products and pharmaceuticals (cephalosporin) etc.

(iv) It is employed in the industries as a bleaching agent for textiles, paper pulp, leather, oils, fats, etc.

(v) Nowadays it is also used in Environmental (Green) Chemistry. For example, in pollution control treatment of domestic and industrial effluents, oxidation of cyanides, restoration of aerobic conditions to sewage wastes, etc.

9.8 HEAVY WATER, $\mathrm{D_2} \mathrm{O}$

It is extensively used as a moderator in nuclear reactors and in exchange reactions for the study of reaction mechanisms. It can be prepared by exhaustive electrolysis of water or as a by-product in some fertilizer industries. Its physical properties are given in Table 9.3. It is used for the preparation of other deuterium compounds, for example:

$$ \begin{aligned} & \mathrm{CaC_2}+2 \mathrm{D_2} \mathrm{O} \rightarrow \mathrm{C_2} \mathrm{D_2}+\mathrm{Ca}(\mathrm{OD})_2 \\ & \mathrm{SO_3}+\mathrm{D_2} \mathrm{O} \rightarrow \mathrm{D_2} \mathrm{SO_4} \\ & \mathrm{Al_4} \mathrm{C_3}+12 \mathrm{D_2} \mathrm{O} \rightarrow 3 \mathrm{CD_4}+4 \mathrm{Al}(\mathrm{OD})_3 \end{aligned} $$

9.9 DIHYDROGEN AS A FUEL

Dihydrogen releases large quantities of heat on combustion. The data on energy released by combustion of fuels like dihydrogen, methane, LPG etc. are compared in terms of the same amounts in mole, mass and volume, are shown in Table 9.5.

From this table it is clear that on a mass for mass basis dihydrogen can release more energy than petrol (about three times). Moreover, pollutants in combustion of dihydrogen will be less than petrol. The only pollutants will be the oxides of dinitrogen (due to the presence of dinitrogen as impurity with dihydrogen). This, of course, can be minimised by injecting a small amount of water into the cylinder to lower the temperature so that the reaction between dinitrogen and dioxygen may not take place. However, the mass of the containers in which dihydrogen will be kept must be taken into consideration. A cylinder of compressed dihydrogen weighs about 30 times as much as a tank of petrol containing the same amount of energy. Also, dihydrogen gas is converted into liquid state by cooling to $20 \mathrm{~K}$. This would require expensive insulated tanks. Tanks of metal alloy like $\mathrm{NaNi_5}, \mathrm{Ti}-\mathrm{TiH_2}$, $\mathrm{Mg}-\mathrm{MgH_2}$ etc. are in use for storage of dihydrogen in small quantities. These limitations have prompted researchers to search for alternative techniques to use dihydrogen in an efficient way.

In this view Hydrogen Economy is an alternative. The basic principle of hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. Advantage of hydrogen economy is that energy is transmitted in the form of dihydrogen and not as electric power. It is for the first time in the history of India that a pilot project using dihydrogen as fuel was launched in October 2005 for running automobiles. Initially $5 \%$ dihydrogen has been mixed in CNG for use in four-wheeler vehicles. The percentage of dihydrogen would be gradually increased to reach the optimum level.

Nowadays, it is also used in fuel cells for generation of electric power. It is expected that economically viable and safe sources of dihydrogen will be identified in the years to come, for its usage as a common source of energy.

Table 9.5 The Energy Released by Combustion of Various Fuels in Moles, Mass and Volume

Summary

Hydrogen is the lightest atom with only one electron. Loss of this electron results in an elementary particle, the proton. Thus, it is unique in character. It has three isotopes, namely : protium $\left( _1^1 \mathrm{H}\right)$, deuterium (D or $\left. _1^2 \mathrm{H}\right)$ and tritium $\left(\mathrm{T}\right.$ or $\left. _1^3 \mathrm{H}\right)$. Amongst these three, only tritium is radioactive. Inspite of its resemblance both with alkali metals and halogens, it occupies a separate position in the periodic table because of its unique properties.

Hydrogen is the most abundant element in the universe. In the free state it is almost not found in the earth’s atmosphere. However, in the combined state, it is the third most abundant element on the earth’s surface.

Dihydrogen on the industrial scale is prepared by the water-gas shift reaction from petrochemicals. It is obtained as a byproduct by the electrolysis of brine.

The $\mathrm{H}-\mathrm{H}$ bond dissociation enthalpy of dihydrogen $\left(435.88 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ is the highest for a single bond between two atoms of any elements. This property is made use of in the atomic hydrogen torch which generates a temperature of $\sim 4000 \mathrm{~K}$ and is ideal for welding of high melting metals.

Though dihydrogen is rather inactive at room temperature because of very high negative dissociation enthalpy, it combines with almost all the elements under appropriate conditions to form hydrides. All the type of hydrides can be classified into three categories: ionic or saline hydrides, covalent or molecular hydrides and metallic or non-stoichiometric hydrides. Alkali metal hydrides are good reagents for preparing other hydride compounds. Molecular hydrides (e.g., $\mathrm{B_2} \mathrm{H_6}, \mathrm{CH_4}, \mathrm{NH_3}, \mathrm{H_2} \mathrm{O}$ ) are of great importance in day-to-day life. Metallic hydrides are useful for ultrapurification of dihydrogen and as dihydrogen storage media.

Among the other chemical reactions of dihydrogen, reducing reactions leading to the formation hydrogen halides, water, ammonia, methanol, vanaspati ghee, etc. are of great importance. In metallurgical process, it is used to reduce metal oxides. In space programmes, it is used as a rocket fuel. In fact, it has promising potential for use as a non-polluting fuel of the near future (Hydrogen Economy).

Water is the most common and abundantly available substance. It is of a great chemical and biological significance. The ease with which water is transformed from liquid to solid and to gaseous state allows it to play a vital role in the biosphere. The water molecule is highly polar in nature due to its bent structure. This property leads to hydrogen bonding which is the maximum in ice and least in water vapour. The polar nature of water makes it: (a) a very good solvent for ionic and partially ionic compounds; (b) to act as an amphoteric (acid as well as base) substance; and (c) to form hydrates of different types. Its property to dissolve many salts, particularly in large quantity, makes it hard and hazardous for industrial use. Both temporary and permanent hardness can be removed by the use of zeolites, and synthetic ion-exchangers.

Heavy water, $\mathrm{D_2} \mathrm{O}$ is another important compound which is manufactured by the electrolytic enrichment of normal water. It is essentially used as a moderator in nuclear reactors.

Hydrogen peroxide, $\mathrm{H_2} \mathrm{O_2}$ has an interesting non-polar structure and is widely used as an industrial bleach and in pharmaceutical and pollution control treatment of industrial and domestic effluents.

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

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

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

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

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

9.6 Complete the following reactions:

(i) $4\mathrm{H_2}(\mathrm{~g})+\mathrm{M_\mathrm{m}} \mathrm{O_\mathrm{o}}(\mathrm{s}) \xrightarrow{\Delta}$

(ii) $\mathrm{CO}(\mathrm{g})+\mathrm{H_2}(\mathrm{~g}) \xrightarrow[\text { catalyst }]{\Delta}$

(ii) $\mathrm{C_3} \mathrm{H_8}(\mathrm{~g})+3 \mathrm{H_2} \mathrm{O}(\mathrm{g}) \xrightarrow[\text { catalyst }]{\Delta}$

(iv) $\mathrm{Zn}(\mathrm{s})+\mathrm{NaOH}(\mathrm{aq}) \xrightarrow{\text { heat }}$

9.7 Discuss the consequences of high enthalpy of $\mathrm{H}-\mathrm{H}$ bond in terms of chemical reactivity of dihydrogen.

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.

Show Answer

Answer

Molecular hydrides are classified on the basis of the presence of the total number of electrons and bonds in their Lewis structures as:

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

An electron-deficient hydride has very few electrons, less than that required for representing its conventional Lewis structure e.g. diborane $(B_2 H_6)$. In $B_2 H_6$, 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.

An electron-precise hydride has a sufficient number of electrons to be represented by its conventional Lewis structure e.g. $CH_4$. The Lewis structure can be written as:

Four regular bonds are formed where two electrons are shared by two atoms.

An electron-rich hydride contains excess electrons as lone pairs e.g. $NH_3$

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?

Show Answer

Answer

An electron-deficient hydride does not have sufficient electrons to form a regular bond in which two electrons are shared by two atoms e.g., $B_2 H_6, Al_2 H_6$ etc.

These hydrides cannot be represented by conventional Lewis structures. $B_2 H_6$, for example, 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.

$$ \begin{aligned} & B_2 H_6+2 NMe \longrightarrow 2 BH_3 \cdot NMe_3 \\ & B_2 H_6+2 CO \longrightarrow 2 BH_3 \cdot CO \end{aligned} $$

9.10 Do you expect the carbon hydrides of the type $\left(\mathrm{C_\mathrm{n}} \mathrm{H_2 \mathrm{n}+2}\right)$ to act as ‘Lewis’ acid or base? Justify your answer.

Show Answer

Answer

For carbon hydrides of type $C_n H _{2 n+2}$, the following hydrides are possible for

$$ \begin{gathered} n=1 \Rightarrow CH_4 \\ n=2 \Rightarrow C_2 H_6 \\ n=3 \Rightarrow C_3 H_8 \\ \vdots \\ \quad \vdots \end{gathered} $$

For a hydride to act as a Lewis acid i.e., electron accepting, it should be electron-deficient. Also, for it to act as a Lewis base i.e., electron donating, it should be electron-rich.

Taking $C_2 H_6$ as an example, the total number of electrons are 14 and the total covalent bonds are seven. Hence, the bonds are regular $2 e^{at}-2$ centered bonds.

Hence, hydride $C_2 H_6$ has sufficient electrons to 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.

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

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

9.14 Among $\mathrm{NH_3}, \mathrm{H_2} \mathrm{O}$ and $\mathrm{HF}$, which would you expect to have highest magnitude of hydrogen bonding and why?

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Answer

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 $HF>H_2 O>NH_3$.

But, the actual order is $H_2 O>HF>NH_3$.

Although fluorine is more electronegative than oxygen, the extent of hydrogen bonding is higher in water. There is a shortage of hydrogens in HF, whereas there are exactly the right numbers of hydrogens in water. As a result, only straight chain bonding takes place. On the other hand, oxygen forms a huge ring-like structure through its high ability of hydrogen bonding.

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

(i)

(ii) (iii)

9.15 Saline hydrides are known to react with water violently producing fire. $\mathrm{Can} \mathrm{CO_2}$, a well known fire extinguisher, be used in this case? Explain.

9.16 Arrange the following

(i) $\mathrm{CaH_2}, \mathrm{BeH_2}$ and $\mathrm{TiH_2}$ in order of increasing electrical conductance.

(ii) $\mathrm{LiH}, \mathrm{NaH}$ and $\mathrm{CsH}$ in order of increasing ionic character.

(iii) $\mathrm{H}-\mathrm{H}, \mathrm{D}-\mathrm{D}$ and $\mathrm{F}-\mathrm{F}$ in order of increasing bond dissociation enthalpy.

(iv) $\mathrm{NaH}, \mathrm{MgH_2}$ and $\mathrm{H_2} \mathrm{O}$ in order of increasing reducing property.

9.17 Compare the structures of $\mathrm{H_2} \mathrm{O}$ and $\mathrm{H_2} \mathrm{O_2}$.

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Answer

In gaseous phase, water molecule has a bent form with a bond angle of $104.5^{\circ}$. The O-H bond length is $95.7 pm$. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is $111.5^{\circ}$ and $90.2^{\circ}$ respectively.

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

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Answer

Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion $(OH^{\hat{a} \epsilon^{*}})$ and a hydronium ion $(H_3 O^{+})$.

The reaction involved can be represented as: $H_2 O _{(i)}+$ $H_2 O _{(l)}$ $H_3 O^{+} _{(aq)}$ $+OH _{(a q)}^{-}$

Hydronium ion Hydroxide ion

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:

9.19 Consider the reaction of water with $\mathrm{F_2}$ and suggest, in terms of oxidation and reduction, which species are oxidised/reduced.

Show Answer

Answer

The reaction between fluorine and water can be represented as:

$ 2 F _{2(g)}+2 H_2 O _{(l)} \longrightarrow 4 H _{(a q)}^{+}+4 F _{(a q)}^{-}+O _{2(g)} $

This is an example of a redox reaction as water is getting oxidized to oxygen, while fluorine is being reduced to fluoride ion.

The oxidation numbers of various species can be represented as:

Oxidation

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

Water is oxidized from (-2 ) to zero oxidation state. An increase in oxidation state indicates oxidation of water.

9.20 Complete the following chemical reactions.

(i) $\mathrm{PbS}(\mathrm{s})+\mathrm{H_2} \mathrm{O_2}(\mathrm{aq}) \rightarrow$

(ii) $\mathrm{MnO_4}^{-}(\mathrm{aq})+\mathrm{H_2} \mathrm{O_2}(\mathrm{aq}) \rightarrow$

(iii) $\mathrm{CaO}(\mathrm{s})+\mathrm{H_2} \mathrm{O}(\mathrm{g}) \rightarrow$

(v) $\mathrm{AlCl_3}(\mathrm{~g})+\mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow$

(vi) $\mathrm{Ca_3} \mathrm{~N_2}(\mathrm{~s})+\mathrm{H_2} \mathrm{O}(\mathrm{l}) \rightarrow$

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

9.21 Describe the structure of the common form of ice.

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Answer

Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.

The three-dimensional structure of ice is represented as:

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of $276 pm$. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

9.22 What causes the temporary and permanent hardness of water ?

9.23 Discuss the principle and method of softening of hard water by synthetic ionexchange resins.

9.24 Write chemical reactions to show the amphoteric nature of water.

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Answer

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

1. Reaction with $H_2 S$

The reaction takes place as:

In the forward reaction, $H_2 O _{(l)}$ accepts a proton from $H_2 S _{(a q)}$. Hence, it acts as a Lewis base.

2. Reaction with $NH_3$

The reaction takes place as:

In the forward reaction, $H_2 O _{(j)}$ denotes its proton to $NH _{3(a q)}$. Hence, it acts as a Lewis acid.

3. Self-ionization of water

In the reaction, two water molecules react as:

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

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

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

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

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

9.30 Knowing the properties of $\mathrm{H_2} \mathrm{O}$ and $\mathrm{D_2} \mathrm{O}$, do you think that $\mathrm{D_2} \mathrm{O}$ can be used for drinking purposes?

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

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

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.

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Answer

The elements of atomic numbers 15, 19, 23, and 44 are nitrogen, potassium, vanadium, and ruthenium respectively.

1.) Hydride of nitrogen

Hydride of nitrogen $(NH_3)$ is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on nitrogen.

2.) Hydride of potassium

Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature.

3.) Hydrides of Vanadium and Ruthenium

Both vanadium and ruthenium belong to the d-block of the periodic table. The metals of d-block form metallic or non-stoichiometric hydrides. Hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen.

4.) Behaviour of hydrides towards water

Potassium hydride reacts violently with water as:

$ KH _{(s)}+H_2 O _{(a q)} \longrightarrow KOH _{(\alpha q)}+H _{2(g)} $

Ammonia $(NH_3)$ behaves as a Lewis base and reacts with water as:

$ H_2 O _{(i)}+NH _{3(a q)} \longleftrightarrow OH _{(a q)}^{-}+NH _{4(a q)}^{+} $

Hydrides of vanadium and Ruthenium do not react with water. Hence, the increasing order of reactivity of the hydrides is $(V, Ru) H<NH_3<KH$.

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.

9.35 How does $\mathrm{H_2} \mathrm{O_2}$ behave as a bleaching agent?

9.36 What do you understand by the terms:

(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction (v) fuel-cell?