Hydrogen
UNIT 9
HYDROGEN
“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
| Property | Hydrogen | Deuterium | Tritium |
|---|---|---|---|
| Relative abundance (%) | 99.985 | 0.0156 | $10^{-15}$ |
| Relative atomic mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right.$ ) | 1.008 | 2.014 | 3.016 |
| Melting point / K | 13.96 | 18.73 | 20.62 |
| Boiling point/ K | 20.39 | 23.67 | 25.0 |
| Density / gL | 0.09 | 0.18 | 0.27 |
| Enthalpy of fusion $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | 0.117 | 0.197 | - |
| Enthalpy of vaporization $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | 0.904 | 1.226 | - |
| Enthalpy of bond dissociation $/ \mathrm{kJ} \mathrm{mol}^{-1}$ at $298.2 \mathrm{~K}$ | 435.88 | 443.35 | - |
| Internuclear distance $/ \mathrm{pm}^{-1}$ | 74.14 | 74.14 | - |
| Ionization enthalpy $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | 1312 | - | - |
| Electron gain enthalpy $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | -73 | - | - |
| Covalent radius $/ \mathrm{pm}$ | 37 | - | |
| Ionic radius $\left(\mathrm{H}^{-}\right) / \mathrm{pm}$ | 208 |
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
| Source | % of Total |
|---|---|
| Oceans | 97.33 |
| Saline lakes and inland seas | 0.008 |
| Polar ice and glaciers | 2.04 |
| Ground water | 0.61 |
| Lakes | 0.009 |
| Soil moisture | 0.005 |
| Atmospheric water vapour | 0.001 |
| Rivers | 0.0001 |
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 $\mathrm{D_2} \mathrm{O}$
| Property | $\mathrm{H_2} \mathrm{O}$ | $\mathrm{D_2} \mathrm{O}$ |
|---|---|---|
| Molecular mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$ | 18.0151 | 20.0276 |
| Melting point/K | 273.0 | 276.8 |
| Boiling point/K | 373.0 | 374.4 |
| Enthalpy of formation $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | -285.9 | -294.6 |
| Enthalpy of vaporisation $(373 \mathrm{~K}) / \mathrm{kJ} \mathrm{mol}^{-1}$ | 40.66 | 41.61 |
| Enthalpy of fusion $/ \mathrm{kJ} \mathrm{mol}^{-1}$ | 6.01 | - |
| Temp of max. density/K | 276.98 | 284.2 |
| Density $(298 \mathrm{~K}) / \mathrm{g} \mathrm{cm}^{-3}$ | 1.0000 | 1.1059 |
| Viscosity/centipoise | 0.8903 | 1.107 |
| Dielectric constant/C $\mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}$ | 78.39 | 78.06 |
| Electrical conductivity $\left(293 \mathrm{~K} / \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right)$ | $5.710^{-8}$ | - |
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
| Melting point $/ \mathrm{K}$ | 272.4 | Density (liquid at $298 \mathrm{~K}) / \mathrm{g} \mathrm{cm}^{-3}$ | 1.44 |
|---|---|---|---|
| Boiling point(exrapolated) $/ \mathrm{K}$ | 423 | Viscosity $(290 \mathrm{~K}) /$ centipoise | 1.25 |
| Vapour pressure $(298 \mathrm{~K}) / \mathrm{mmHg}$ | 1.9 | Dielectric constant $(298 \mathrm{~K}) / \mathrm{C}^{2} / \mathrm{N} \mathrm{m}^{2}$ | 70.7 |
| Density (solid at $268.5 \mathrm{~K}) / \mathrm{g} \mathrm{cm}^{-3}$ | 1.64 | Electrical conductivity $(298 \mathrm{~K}) / \Omega^{-1} \mathrm{~cm}^{-1}$ | $5.110^{-8}$ |
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
| Energy released on combustion in kJ state |
Dihydrogen (in gaseous state) |
Dihydrogen (in liquid) |
LPG | $\mathbf{C H_\mathbf{4}}$ gas | Octane (in liquid state) |
|---|---|---|---|---|---|
| per mole | 286 | 285 | 2220 | 880 | 5511 |
| per gram | 143 | 142 | 50 | 53 | 47 |
| per litre | 12 | 9968 | 25590 | 35 | 34005 |
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.
EXERCISES
9.1 Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer Hydrogen is the first element of the periodic table. Its electronic configuration is $[1 s^{1}]$. Due to the presence of only one electron in its $1 s$ shell, hydrogen exhibits a dual behaviour, i.e., it resembles both alkali metals and halogens. Resemblance with alkali metals: $H: 1 s^{1}$ $Li:[He] 2 s^{1}$ Hence, it can lose one electron to form a unipositive ion. Resemblance with halogens: $H: 1 s^{1}$ $F: 1 s^{2} 2 s^{2} 2 p^{5}$ $Cl: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$ Hence, hydrogen can gain one electron to form a uninegative ion. Though hydrogen shows some similarity with both alkali metals and halogens, it differs from them on some grounds. Unlike alkali metals, hydrogen does not possess metallic characteristics. On the other hand, it possesses a high ionization enthalpy. Also, it is less reactive than halogens. Owing to these reasons, hydrogen cannot be placed with alkali metals (group I) or with halogens (group VII). In addition, it was also established that $H^{+}$ions cannot exist freely as they are extremely small. $H^{+}$ions are always associated with other atoms or molecules. Hence, hydrogen is best placed separately in the periodic table.Show Answer
Answer Hydrogen has three isotopes. They are: The mass ratio of protium, deuterium and tritium is 1:2:3.Show Answer
Answer The ionization enthalpy of hydrogen atom is very high $(1312 kJ mol^{-1})$. Hence, it is very hard to remove its only electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic $(H_2)$ molecule.Show Answer
Answer Dihydrogenis produced by coal gasification method as: $
C _{(s)}+H_2 O _{(g)} \xrightarrow{1270 K} CO _{(g)}+H _{2(g)}
$ (coal) The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst. $
CO _{(g)}+H_2 O _{(g)} \xrightarrow[\text{ Catalyst }]{673 K} CO _{2(g)}+H _{2(g)}
$ This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.Show Answer
Answer Dihydrogen is prepared by the electrolysis of acidified or alkaline water using platinum electrodes. Generally, 15 â€" $20 %$ of an acid $(H_2 SO_4)$ or a base $(NaOH)$ is used. Reduction of water occurs at the cathode as: $
2 H_2 O+2 e^{-} \longrightarrow 2 H_2+2 OH^{-}
$ $
2 OH^{-} \longrightarrow H_2 O+\frac{1}{2} O_2+2 e^{-}
$ $\therefore$ Net reaction can be represented as: $
H_2 O _{(l)} \longrightarrow H _{2(g)}+\frac{1}{2} O _{2(g)}
$ Electrical conductivity of pure water is very low owing to the absence of ions in it. Therefore, electrolysis of pure water also takes place at a low rate. If an electrolyte such as an acid or a base is added to the process, the rate of electrolysis increases. The addition of the electrolyte makes the ions available in the process for the conduction of electricity and for electrolysis to take place.Show Answer
(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) $\mathrm{Zn}(\mathrm{s})+\mathrm{NaOH}(\mathrm{aq}) \xrightarrow{\text { heat }}$
Answer (i) $H _{2(g)}+M_m O _{\text{o(s) }} \xrightarrow{\Delta} mM _{(s)}+H_2 O _{(f)}$ (ii) $\quad CO _{(g)}+2 H _{2(g)} \xrightarrow[\text{ catalyst }]{\Delta} CH_3 OH _{(f)}$ $$
\begin{equation*}
C_3 H _{8(g)}+3 H_2 O _{(g)} \xrightarrow[\text{ catalyst }]{\Delta} 3 CO _{(g)}+7 H _{2(g)} \tag{iii}
\end{equation*}
$$ (iv) $Zn _{(s)}+2 NaOH _{(o q)} \xrightarrow{\text{ beat }} Na_2 ZnO _{2(\rho q)}+H _{2(g)}$ Sodium zincateShow Answer
Answer The ionization enthalpy of $H-H$ bond is very high $(1312 kJ mol^{-1})$. This indicates that hydrogen has a low tendency to form $H^{+}$ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules $(H_2)$, hydrides with elements, and a large number of covalent bonds. Since ionization enthalpy is very high, hydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.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: 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.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. $$
\begin{aligned}
& B_2 H_6+2 NMe \longrightarrow 2 BH_3 \cdot NMe_3 \\
& B_2 H_6+2 CO \longrightarrow 2 BH_3 \cdot CO
\end{aligned}
$$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.Show Answer
Answer 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. For example: $LaH _{2.87}, YbH _{2.55}$, $TiH _{1.5-1.8}$ etc. Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions 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. Alkali metals will not form non-stoichiometric hydrides.Show Answer
Answer Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that 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.Show Answer
Answer 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.Show Answer
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)Show Answer
Answer 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: $
MH _{(s)}+H_2 O _{(a q)} \longrightarrow MOH _{(a q)}+H _{2(g)}
$ The reaction is violent and produces fire. $CO_2$ 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, thereby dousing the fire. $CO_2$ can be used in the present case as well. It is heavier than dihydrogen and will be effective in isolating the burning surface from dihydrogen and dioxygen.Show Answer
(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.
Answer (i) The electrical conductance of a molecule depends upon its ionic or covalent nature. Ionic compounds conduct, whereas covalent compounds do not. $BeH_2$ is a covalent hydride. Hence, it does not conduct. $CaH_2$ is an ionic hydride, which conducts electricity in the molten state. Titanium hydride, $TiH_2$ is metallic in nature and conducts electricity at room temperature. Hence, the increasing order of electrical conductance is as follows: $BeH_2<CaH_2<TiH_2$ (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. Electronegativity decreases down the group from Lithium to Caesium. Hence, the ionic character of their hydrides will increase (as shown below). $LiH<NaH<CsH$ (iii) Bond dissociation energy depends upon the bond strength of a molecule, which in turn depends upon the attractive and repulsive forces present in a molecule. The bond pair in D-D bond is more strongly attracted by the nucleus than the bond pair in $H-H$ bond. This is because of the higher nuclear mass of $D_2$. The stronger the attraction, the greater will be the bond strength and the higher is the bond dissociation enthalpy. Hence, the bond dissociation enthalpy of D-D is higher than $H-H$. However, bond dissociation enthalpy is the minimum in the case of F-F. The bond pair experiences strong repulsion from the lone pairs present on each F-centre. 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, $MgH_2$ and $H_2 O$ are covalent hydrides. $H_2 O$ is less reducing than $MgH_2$ since the bond dissociation energy of $H_2 O$ is higher than $MgH_2$. Hence, the increasing order of the reducing property is $H_2 O<MgH_2<NaH$.Show Answer
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.Show Answer
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: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 ( $\hat{a} €^{\prime \prime} 1$ ) oxidation state. A decrease in oxidation state indicates the reduction of fluorine. Water is oxidized from (â€" 2 ) to zero oxidation state. An increase in oxidation state indicates oxidation of water.Show Answer
(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.
Answer (i) $
PbS _{(s)}+H_2 O _{2(a q)} \to PbSO _{4(s)}+H_2 O _{(l)}
$ $H_2 O_2$ is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction. (ii) $2 MnO _{4(a q)}^{-}+5 H_2 O _{2(g)} \to 6 H _{(a q)}^{+} \to 2 Mn^{2+} _{(a q)}+8 H_2 O _{(i)}+5 O _{2(g)}$ $H_2 O _{2(a q)}$ is acting as a reducing agent in the acidic medium, thereby oxidizing $MnO _{4(a q)}^{-}$. Hence, the given reaction is a redox reaction. (iii) $
CaO _{(s)}+H_2 O _{(g)} \to Ca(OH) _{2(a q)}
$ The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis. (iv) $2 AlCl _{3(g)}+3 H_2 O _{(l)} \to Al_2 O _{3(s)}+6 HCl _{(a g)}$ The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of $AlCl_3$. $$
\begin{equation*}
Ca_3 N _{2(s)}+6 H_2 O _{(l)} \to 3 Ca(OH) _{2(a q)}+2 NH _{3(g)} \tag{v}
\end{equation*}
$$ The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of $Ca_3 N_2$.Show Answer
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.Show Answer
Answer Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates $(MHCO_3.$, where $.M=Mg, Ca)$ in water. Permanent hardness of water is because of the presence of soluble salts of calcium and magnesium in the form of chlorides in water.Show Answer
Answer The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g.,
ions respectively. Synthetic resins are of two types: Cation exchange resins are large organic molecules that contain the â $€$ " $SO_3 H$ group. The resin is firstly changed to $RNa$ (from $RSO_3 H$ ) by treating it with $NaCl$. This resin then exchanges $Na^{+}$ions with $Ca^{2+}$ and $Mg^{2+}$ ions, thereby making the water soft. $
2 RNa+M _{(a q)}^{2+} \longrightarrow R_2 M _{(s)}+2 Na _{(a q)}^{+}
$ There are cation exchange resins in $H^{+}$form. The resins exchange $H^{+}$ions for $Na^{+}, Ca^{2+}$, and $Mg^{2+}$ ions. $
2 RH+M _{(a q)}^{2+} leftarrows MR _{2(s)}+2 H _{(a q)}^{+}
$ Anion exchange resins exchange $OH^{1 e^{-}}$ions for anions like $Cl^{1 e^{\epsilon}}, HCO_3^{-}$, and $SO_4^{2 e^{2 \epsilon}}$ present in water. $$
\begin{aligned}
RNH_{2(s)}+H_2 O_{(l)} \leftarrow & RNH_3^{+} \cdot OH_{(s)}^{-} \\
& \downarrow + X_{(a q)}^{-} \\
& RNH_3^{+} \cdot X_{(s)}^{-}+OH _{(a q)}^{-}
\end{aligned}
$$ 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 $OH^{\hat{a t}}$ ions neutralize the $H^{+}$ions and deionize the water obtained.Show Answer
Answer The amphoteric nature of water can be described on the basis of the following reactions: 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. 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. In the reaction, two water molecules react as:Show Answer
Answer Hydrogen peroxide, $H_2 O_2$ acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. Reactions involving oxidizing actions are: Reactions involving reduction actions are:Show Answer
Answer Demineralised water is 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 $OH^{\hat{a} \epsilon^{\prime \prime}}$ form) resin. During the cation exchange process, $H^{+}$exchanges for $Na^{+}, Mg^{2+}, Ca^{2+}$, and other cations present in water. $$
\begin{equation*}
2 RH _{(s)}+M _{(a q)}^{2+} \leftarrow MR _{2(s)}+2 H _{(a q)}^{+} \tag{1}
\end{equation*}
$$ In the anion exchange process, $OH^{\text{ae }}$ exchanges for anions such as $CO_3^{2-}, SO_4^{2-}, Cl^{-}, HCO_3^{-}$etc. present in water. $$
\begin{align*}
& RNH _{2(s)}+H_2 O _{(l)} leftarrows RNH_3^{+} \cdot OH _{(s)}^{-} \\
& RNH_3^{+} OH _{(s)}^{-}+X _{(a q)}^{-} leftarrows RNH_3^{+} \cdot X _{(s)}^{-}+OH _{(a q)}^{-} \tag{2}
\end{align*}
$$ $OH^{at^{-}}$ions liberated in reaction (2) neutralize $H^{+}$ions liberated in reaction (1), thereby forming water. $
H _{(a q)}^{+}+OH _{(a q)}^{-} \longrightarrow H_2 O _{(l)}
$Show Answer
Answer Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants, and animals for survival. Demineralised water is free of all soluble minerals. Hence, it is not fit for drinking. It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.Show Answer
Answer Wateris essential for all forms of life. It constitutes around $65 %$ of the human body and $95 %$ of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment, and dielectric constant. The high heat of vapourization and heat of capacity of water helps in moderating the climate and body temperature of all living beings. It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.Show Answer
Answer A high value of dielectric constants $(78.39 C^{2} / Nm^{2})$ and dipole moment make water a universal solvent. Water is able to dissolve most ionic and covalent compounds. Ionic compounds dissolve in water because of the iondipole interaction, whereas covalent compounds form hydrogen bonding and dissolve in water. Water can hydrolyze metallic and non-metallic oxides, hydrides, carbides, phosphides, nitrides and various other salts. During hydrolysis, $H^{+}$and $OH^{\hat{a} \epsilon^{\prime \prime}}$ ions of water interact with the reacting molecule. Some reactions are: $$
\begin{aligned}
& CaO+H_2 O \longrightarrow Ca(OH)_2 \\
& NaH+H_2 O \longrightarrow NaOH+H_2 \\
& CaC_2+H_2 O \longrightarrow C_2 H_2+Ca(OH)_2
\end{aligned}
$$Show Answer
Answer Heavy water $(D_2 O)$ acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of $D_2 O$, 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.Show Answer
Answer Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions $(H^{+}.$and $OH^{\hat{\epsilon} \epsilon^{-}}$ions) of water molecule react with a compound to form products. For example: $NaH+H_2 O \longrightarrow NaOH+H_2$ Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example: $
CuSO_4+5 H_2 O \longrightarrow CuSO_4 \cdot 5 H_2 O
$Show Answer
Answer Saline hydrides areionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as: $
AH _{(s)}+H_2 O _{(f)} \longrightarrow AOH _{(a q)}+H _{2(g)}
$ (where, $A=Na, Ca, \ldots .$. ) When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide. The dry organic solvent distills over.Show Answer
Answer The elements of atomic numbers 15, 19, 23, and 44 are nitrogen, potassium, vanadium, and ruthenium respectively. 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. Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature. 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. 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 $(V, Ru) H<NH_3<KH$.Show Answer
Answer Potassium chloride $(KCl)$ is the salt of a strong acid $(HCl)$ and strong base $(KOH)$. Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows: $
KCl _{(s)} \xrightarrow{\text{ water }} K _{(a q)}^{+}+Cl _{(a q)}^{-}
$ In acidified and alkaline water, the ions do not react and remain as such. Aluminium (III) chloride is the salt of a strong acid $(HCl)$ and weak base $[Al(OH)_3]$. Hence, it undergoes hydrolysis in normal water. $
AlCl _{3(s)}+3 H_2 O _{(r)} \xrightarrow[\text{ Water }]{\text{ Normal }} Al(OH) _{3(s)}+3 H _{(a q)}^{+}+3 Cl _{(a q)}^{-}
$ In acidified water, $H^{+}$ions react with $Al(OH)_3$ forming water and giving $Al^{3+}$ ions. Hence, in acidified water, $AlCl_3$ will exist $$
\begin{aligned}
& \text{ as } Al _{(a q)}^{3+} \text{ and } Cl _{(a q)}^{-} \text{ions. } \\
& AlCl _{3(s)} \xrightarrow[\text{ Water }]{\text{ Acidified }} Al _{(a q)}^{3+}+3 Cl _{(a q)}^{-}
\end{aligned}
$$ In alkaline water, the following reaction takes place: $
Al(OH) _{3(s)}+\underbrace{OH _{(a q)}^{-}} _{\text{from alkaline water }} \longrightarrow[Al(OH)_4] _{(a q)}^{-}+2 H_2 O _{(l)}
$Show Answer
Answer $H_2 O_2$ 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.Show Answer
(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction (v) fuel-cell?
Show Answer
Answer
(i) Hydrogen economy
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 ecoâe"friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.
(ii) Hydrogenation
Hydrogenation is the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati, ghee etc.
(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.
For example :
(iv) Water shift reaction
It is a reaction of carbon monoxide of syngas mixture with steam in the presence of a catalyst as:
This reaction is used to increase the yield of dihydrogen obtained from the coal gasification reaction as:
(v) Fuel cells
Fuel cells are devices for producing electricity from fuel in the presence of an electrolyte. Dihydrogen can be used as a fuel in these cells. 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.
