Answer

(b) Hydrogen resembles halogens in many respects for which several factors are responsible. The most important is hydrogen like halogens accept an electron readily to achieve nearest inert gas configuration.

Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals and halogens.

Answer

(d) $\mathrm{H}^{+}$ion always get associated with other atoms or molecules. The reason is that loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to small size it cannot exist free.

Answer

(b) Metal hydrides are ionic, covalent or molecular in nature. lonic character increases as the size of the atom increases or the electronegativity of the atom decreases. The correct order of increasing ionic character is

$$ \mathrm{LiH}<\mathrm{NaH}<\mathrm{KH}<\mathrm{RbH}<\mathrm{CsH} $$

Answer

(d) Electron-precise hydrides contain exact number of electrons to form normal covalent bonds. e.g., $-\mathrm{CH}_{4}$ which has tetrahedral in geometry.

Thinking Process

To solve this problem, the point kept in mind that nucleids with $n / p$ (neutron/proton) ratio $>1.5$ are usually radioactive.

Answer

(c) The radioactive isotope of hydrogen is tritium. For tritium $(n=3, p=1)$, therefore $n / p$ ratio is 3.

Thinking Process

(i) Reducing agents are those substance (atoms, ions or molecules) which can readily lose electrons to other substance.

(ii) Oxidising agents are those substance (atoms, ions or molecules) which can readily accept electrons from other substance.

Answer

(b)

(i)

Thus, here $H_{2} O_{2}$ oxidises $HI$ into $I_{2}$ hence, it behaves as oxidising agent.

Answer

(b) Oxides such as $BaO_{2}, Na_{2} O_{2}$ etc; which contain peroxide linkage (i.e., $-O-O$ or $O_{2}^{2-}$ ) on treatment with dilute $H_{2} SO_{4}$ give $H_{2} O_{2}$ but dioxides $(O=M=O$, where $M$ is the metal atom) such as $PbO_{2}, MnO_{2}, TiO_{2}$ do not give $H_{2} O_{2}$ on treatment with dilute $H_{2} SO_{4}$.

$$ \underset{\text{peroxide}}{\underset{Hydrated barium}{BAO_2 . 8H_2O(s)}} + H_2SO_4 (aq) \longrightarrow BASO_4 (s + )\underset{\text{peroxide}}{\underset{Hydrated barium}{ H_2O_2(aq)}} + 8H_2O (l) $$

Answer

(c) The reaction in which $H_{2} O_{2}$ is reduced i.e., oxidation state of oxygen decreases from -1 to -2 depicts the oxidising nature of $H_{2} O_{2}$. e.g.,

Answer

(b) $H_{2} O_{2}$ acts as an oxidising as well as reducing agent in alkaline media. The given below reaction show the reducing action in basic medium

$$ \begin{aligned} I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\ 2 MnO_{4}^{-}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-} \end{aligned} $$

Answer

(c) Hydrogen peroxide acts as an oxidising as well as reducing agent in both acidic and alkaline media.

Answer

(c) The process of producing syn gas or synthesis gas from coal is called ‘coal gasification’.

$\underset{\text { Coal }}{\mathrm{C}(\mathrm{s})}+\underset{\text { Steam }}{\mathrm{H} _2 \mathrm{O}(g)} \xrightarrow[\mathrm{Ni}]{1270 \mathrm{~K}} \underbrace{\mathrm{CO}(g)+\mathrm{H} _2(g)} _{\text {Syn gas }}$

The production of hydrogen can be increased by reacting carbon monoxide of the syn gas with steam in the presence of iron chromate as a catalyst at $673 \mathrm{~K}$.

$\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g) \xrightarrow{\mathrm{FeCrO}_4, 673 \mathrm{~K}} \mathrm{CO}_2(g)+\mathrm{H}_2(g)$

$\mathrm{CO}_{2}$ is removed by scrubbing with a solution of sodium arsenite.

Answer

(d) When sodium peroxide is treated with dilute sulphuric acid, we get sodium sulphate and hydrogen peroxide

$$ Na_{2} O_{2}+\text { dil. } H_{2} SO_{4} \longrightarrow Na_{2} SO_{4}+H_{2} O_{2} $$

Answer

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

$$ \begin{aligned} 2 HSO_{4}^{-}(aq) \xrightarrow{\text { Electrolysis }} HO_{3} SOOSO_{3} H(aq) \xrightarrow{\text { Hydrolysis }} 2 HSO_{4}^{-}(aq) & \\ & +2 H^{+}(aq)+H_{2} O_{2} \text { (aq) } \end{aligned} $$

Answer

(d) The water gas is the combination of carbon monoxide and hydrogen.

$$ CO(g)+2 H_{2}(g) \xrightarrow[\text { Catalyst }]{\text { Cobalt }} CH_{3} OH(l) $$

It is an example of water gas used in the synthesis of methanol.

Answer

(a) Bicarbonates, chlorides and sulphates of $\mathrm{Ca}$ and $\mathrm{Mg}$ are responsible for the hardness of water.

Note Hard water forms scum/precipitate with soap. Soap containing sodium stearate $\left(C_{17} H_{35} COONa\right)$ reacts with hard water to precipitate out $Ca/ Mg$ stearate. $2 C_{17} H_{35} COONa(aq)+M^{2+}(aq) \longrightarrow\left(C_{17} H_{35} COO\right)_{2} M \downarrow$

$$ +2 \mathrm{Na}^{+}(\mathrm{aq})(\text { where, } \mathrm{M}=\mathrm{Ca} / \mathrm{Mg}) $$

It is unsuitable for laundry and boilers.

Answer

(c) For water softening, sodium hexametaphosphate is used. The chemical formula is

$ \underset{\text{(From hard water)}}{2CaCl_2} + \underset{\text{Sodium hexametaphosphate}}{Na_2[Na_4(PO_3)_6]} \longrightarrow \underset{\text{Complex salt (soluble)}}{Na_2[Ca_2(PO_3)_6]} + 4NaCl $

Answer

(a) Dihydrogen forms molecular compounds with most of the p-block elements. Most familiar examples are $CH_{4}, NH_{3}, H_{2} O$ and $HF$. For convenience hydrogen compounds of non-metals have also been considered as hydrides.

Answer

(a) Only one element of group 6, i.e., Cr forms hydride.

Note Metallic (or interstitial) hydrides 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.

Answer

$(c, d)$

$\mathrm{H}^{+}$does not exist freely and is always associated with other atoms or molecules.

Like alkali metals, hydrogen forms oxides, halides and sulphides. However, unlike alkali metals, it has a very high ionisation enthalpy and does not possess metallic characteristics under normal conditions.

Answer

$(a, b)$

Dihydrogen can be prepared on commercial scale by different methods. Reaction of steam on hydrocarbons or coke at high temperatures in the presence of catalyst yield hydrogen.

$$ C_nH_{2n+2} + nH_2O +2 \xrightarrow[Ni]{1270 K} nCO_2 + (2n+1)H_2$$

e.g. $$ CH_2(g) + H_2O(g) \xrightarrow[Ni]{1270 k} CO(g) + 3H_2(g) $$

The mixture of $CO$ and $H_{2}$ is called water gas. As this mixture of $CO$ and $H_{2}$ is used for the synthesis of methanol and a number of hydrocarbons, it is also called synthesis gas or ‘syn gas’.

Answer

$(a, c)$

Heavy water is used as moderator in nuclear reactor. Boiling point of heavy water is higher than ordinary water and it is not as effective in the form of solvent as water due to its low dielectric constant.

Answer

$(a, b)$

Among the three isotopes of hydrogen, protonium is the most common. In ionic salts, hydrogen exists as hydride $\left(\mathrm{H}^{-}\right)$.

Answer

(c, $d)$

There is $\mathrm{H}$-bonding even in frozen state of water, i.e., ice is lighter than liquid water.

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.

Answer

$(a, b)$

Permanent hardness 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.

Answer

$(b, c)$

Electron precise compounds have the required number of electrons to write their conventional Lewis structures. All elements of group 14 forms electron-precise compounds (e.g., $\mathrm{CH}_{4}$ ) which are tetrahedral in geometry.

Answer

$(a, d)$

All elements of group 13 will form electrondeficient compounds which acts as Lewis acids.

All elements of group 14 will form electronprecise compounds.

Electronrich hydrides have excess electrons which are present as lone pairs. Elements of group 15-17 forms such compounds. $NH_{3}$ has 1-lone pair, $H_{2} O-2$ and $HF-3$ lone pairs act as Lewis bases.

Answer

$(a, b, c)$

The ionic hydrides are crystalline, non-volatile and non-conducting in solid state. However, their molten state conduct electricity.

Answer

Water gas is produced when superheated steam is passed over red hot coke or coal at $1270 \mathrm{~K}$ in presence of nickel as catalyst.

$ \underset{\text{Coke}}{C(s)} + \underset{\text{Steam}}{H_2O(g)} +121.3 KJ \xrightarrow[Nickel]{1270 k} \underbrace{CO(g)+ H_2(g)}_{\text{Water gas}}$

It is inconvinient to obtain pure $H_{2}$ from water gas as $CO$ is difficult to remove. Hence, to increase the production of $H_{2}$ from water gas, $CO$ is oxidised to $CO_{2}$ by mixing it with more steam and passing the mixture over $FeCrO_{4}$ catalyst at $673 ~K$.

$ \underbrace{CO(g)+ H_2(g)}_{\text{Water gas}} + \underset{\text{Steam}}{H_2O(g)} \xrightarrow[FeCrO_4]{673 K} CO_2(g) + 2H_2(g) $

This is called water-gas shiff reaction. Carbon dioxide is removed by scrubbing with mixture of sodium arsenite solution or by passing the mixture through water under $30 ~atm$ pressure when $CO_{2}$ dissolves leaving behind $H_{2}$ which is collected.

Answer

Metallic/interstitial hydrides are formed by many $d$-block and $f$-block elements. These hydrides conduct heat and electricity.

Unlike saline hydride, they are almost always non-stoichiometric, being deficient in hydrogen. e.g., $LaH_{2.87}, YbH_{2.55}, TiH_{1.5-1.8}, ZrH_{1.3-1.75}, VH_{0.56}, NiH_{0.6-0.7}, PdH_{0.6-0.8}$ etc. In such hydrides, the law of constant composition does not hold good.

Comparision between molecular and metallic hydrides

Answer

$H_{2} O$ - Covalent or molecular hydride (electron rich hydride).

$B_{2} H_{6}$ - Covalent or molecular hydride (electron deficient hydride).

$NaH$ - Ionic or saline hydride.

Note Molecular hydrides are further classified according to the relative number of electrons and bonds in their Lewis structures.

(i) Electron deficient hydride has too few electrons for writing its conventional Lewis structure.

(ii) Electron precise compounds have the required number of electrons to write their conventional Lewis structures.

(iii) Electron rich hydrides have excess electrons which are present as lone pairs.

Answer

In ice, molecules of $\mathrm{H}_{2} \mathrm{O}$ are not packed so closely as in liquid water. There exists vacant spaces in the crystal lattice. This results in larger volume and lower density (density $=$ mass/volume).

In other words, density of ice is lower than liquid water and hence ice floats on water.

Hexagonal honey comb structure of ice

Answer

(i) When $\mathrm{PbS}$ react with hydrogen peroxide, then $\mathrm{PbSO}_{4}$ and water are formed.

$$ PbS(s)+4 H_{2} O_{2}(aq) \longrightarrow PbSO_{4}+4 H_{2} O $$

(ii) When carbon mono-oxide reacts with hydrogen in the presence of cobalt catalyst, then methanol is formed.

$$ CO(g)+2 H_{2}(g) \xrightarrow[\text { catalyst }]{\text { Cobalt }} CH_{3} OH(l) $$

Answer

(i) Density of ice is less than that of liquid water. During severe winter, the temperature of lake water keeps on decreasing. Since, cold water is heavier, therefore, it moves towards bottom of the lake and warm water from the bottom moves towards surface. This process continues. The density of water is maximum at $277 \mathrm{~K}$.

Therefore, any further decrease in temperature of the surface water will decrease in density. The temperature of surface water keeps on decreasing and ultimately it freezes.

Thus, the ice layer at lower temperature floats over the water below it. Due to this, freezing of water into ice takes place continuously from top towards bottom.

(ii) Density of ice is less than that of liquid water, so it floats over water.

Answer

Auto-protolysis means self ionisation of water.

$ \underset{\text{Acid }_1}{H _2O(l)} + \underset{Base _2}{H _2O(l)} \leftrightharpoons \underset{Acids_2}{H_3O^+(aq)} + \underset{Base_1}{OH^-(aq)} $

Due to auto-protolysis, water is amphoteric in nature. It reacts with both acids and bases.

e.g., $$ \underset{Acid_1}{H_2O(l)} + \underset{Base_2}{NH_3(aq)} \longrightarrow \underset{Acid_2}{NH_4^+(aq)} + \underset{Base_1}{OH^-(aq)} $$

$$ \underset{Base_1}{H_2O(l)} + \underset{Acid_2}{H_2S(aq)} \longrightarrow \underset{Acid_1}{H_3O^+(aq)} + \underset{Base_2}{HS^-(aq)} $$

Answer

Water which is free from all soluble minerals salts is called demineralised water. Demineralised water is obtained by passing water successively through a cation exchange and an anion exchange resins.

In cation exchanger, $Ca^{2+}, Mg^{2+}, Na^{+}$and other cations present in water are removed by exchanging them with $H^{+}$ions while in anion exchanger, $Cl^{-}, HCO_{3}^{-}, SO_{4}^{2-}$, etc., present in water are removed by exchanging them with $OH^{-}$ions.

$$ \underset{\text { (Released in cation exchanger) }}{\mathrm{H}^{+}} \underset{\text { (Released in anion exchanger) }}{\mathrm{OH}^{-}} \longrightarrow \mathrm{H}_{2} \mathrm{O} $$

Synthetic ion exchange resins are of two types.

Cation exchange resins contain large organic molecule with $\mathrm{SO}_{3} \mathrm{H}$ group and are water soluble. It is changed to $R \mathrm{Na}$ by treating it with $\mathrm{NaCl}$. The resin $R \mathrm{Na}$ exchanges $\mathrm{Mg}^{2+}$ and $\mathrm{Ca}^{2+}$ ions present in hard water to make the water soft.

$$ 2 \mathrm{RNa}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{R}_{2} \mathrm{M}(\mathrm{s})+2 \mathrm{Na}^{+}(\mathrm{aq}) \quad\left(M=\mathrm{Ca}^{2+} \text { or } \mathrm{Mg}^{2+}\right) $$

The resin can be regenerated by passing $\mathrm{NaCl}$ (aqueous solution) in it.

Pure demineralised (deionised) water is obtained by passing water successively through a cation exchange and anion exchange resins. In the cation exchange process,

$$ 2 R \mathrm{H}(\mathrm{s})+\mathrm{M}^{2+}(\mathrm{aq}) \rightleftharpoons \underset{\begin{array}{l} \text { (Cation exchange } \\ \text { resin in the } \mathrm{H}^{+} \text {form) } \end{array}}{M R_{2}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})} $$

$\mathrm{H}^{+}$exchanges for $\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}$ and other cations present in water.

This process results in proton release and thus, makes the water acidic. In the anion exchange process

$$ R NH_{2}(s)+H_{2} O(l) \rightleftharpoons R \stackrel{+}{N} H_{3} \cdot OH^{-}(s) $$

$R \stackrel{+}{N} H_{3} \cdot OH^{-}$is substituted ammonium hydroxide anion exchange resin.

$$ R \stackrel{+}{N} H_{3} \cdot OH^{-}(s)+X^{-}(aq) \rightleftharpoons R \stackrel{+}{N} H_{3} \cdot X^{-}(s)+OH^{-}(aq) $$

Answer

Molecular hydrides are classified according to the relative numbers of electrons and bonds in Lewis structure as follow

(i) Electron deficient hydrides These type of hydrides contain central atom with incomplete octet. These are formed by 13 group elements, e.g., $BH_{3}$, $AlH_{3}$, etc. To complele their octet they generally exist in polymeric forms such as $B_{2} H_{6}, ~B_{4} H_{10},\left(AlH_{3}\right)_{n}$ etc. These hydrides act as Lewis acids.

(ii) Electron precise hydrides These hydrides have exact number of electrons required to form normal covalent bonds. These are formed by 14 group elements, e.g., $CH_{4}, SiH_{4}$, etc. These are tetrahedral in shape.

(iii) Electron rich hydrides These hydrides contain central atom with excess electrons, which are present as ione pairs.

These are formed by 15,16 and 17 group elements, e.g., $NH_{3}, H_{2} O, HF$, etc. These hydrides act as Lewis bases.

Answer

Heavy water is prepared by prolonged electrolysis of water. Comparison of physical properties of heavy water with those of ordinary water is as follows

Answer

The one chemical reaction for the preparation of $D_{2} O_{2}$ is by the action of $D_{2} SO_{4}$ dissolved in water over $BaO_{2}$.

$$ BaO_{2}+D_{2} SO_{4} \longrightarrow BaSO_{4}+D_{2} O_{2} $$

Answer

By definition, 5 volumes $H_{2} O_{2}$ solution means that $1 ~L$ of this $H_{2} O_{2}$ solution on decomposition produces $5 ~L of _{2}$ at STP.

$$ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} $$

$$ 2 \times 34 g \longrightarrow 22.7L \quad at \quad STP $$

If $22.7 LO_{2}$ at STP will be obtained from $H_{2} O_{2}=68 ~g$

$\therefore 5 ~L$ of $O_{2}$ at STP will be obtained from $H_{2} O_{2}=\frac{68 \times 5}{22.7} ~g=14.98=15 ~g$

$\therefore$ Strength of $H_{2} O_{2}$ in 5 volume $H_{2} O_{2}$ solution $=15 ~g ~L^{-1}$.

$\Rightarrow \quad$ Percentage strength of $H_{2} O_{2}$ solution $=\frac{15}{1000} \times 100=1.5 $%

Therefore, strength of $H_{2} O_{2}$ in 5 volume $H_{2} O_{2}$ solution $=15 ~g / L=1.5 $% $H_{2} O_{2}$ solution.

Answer

(i) $H_{2} O_{2}$ has a non-planar structure. The molecular dimensions in the gas phase and solid phase are given below

(a) gas phase

(b) solid phase

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~k $, dihedral angle is $90.2^{\circ}$.

(ii) $H_{2} O_{2}$ is better oxidising agent than water as discussed below

(a) $H_{2} O_{2}$ oxidises an acidified solution of $KI$ to give $I_{2}$ which gives blue colour with starch solution but $H_{2} O$ does not.

$$ \begin{gathered} 2 KI +H_{2} SO_{4}+H_{2} O_{2} \\ ~K_{2} SO_{4}+2 H_{2} O+I_{2} \end{gathered} $$

(b) $H_{2} O_{2}$ turns black PbS to white $PbSO_{4}$ but $H_{2} O$ does not.

$$ PdS+4 H_{2} O_{2} \rightarrow PbSO_{4}+4 H_{2} O $$

Thinking Process

The data given in the question shows that melting point, enthalpy of vaporisation and viscosity of $D_{2} O$ is more than that of $H_{2} O$. The intermolecular force is directly proportional to these three parameters.

Answer

Given that,

From this data, it is concluded that the values of melting point, enthalpy of vaporisation and viscosity depend upon the intermolecular forces of attraction.

Since, their values are higher for $D_{2} O$ as compared to those of $H_{2} O$, therefore, intermolecular forces of attraction are stronger in $D_{2} O$ than in $H_{2} O$.

Answer

The isotope of hydrogen which contains one proton and one neutron is deuterium (D). Therefore, when dideuterium reacts with dioxygen, heavy water $\left(\mathrm{D}_{2} \mathrm{O}\right)$ is produced.

$$ \underset{\text { Dideuterium }}{2 D_{2}(g)}+\underset{\text { Dioxygen }}{O_{2}(g)} \xrightarrow{\text { Heat }} \underset{\begin{array}{c} \text { Deuterium oxide } \\ \text { (Heavy water) } \end{array}}{2 D_{2} O} $$

The reactivity of $H_{2}$ and $D_{2}$ towards oxygen will be different. Since, the $D-D$ bond is stronger than $H-H$ bond, therefore, $H_{2}$ is more reactive than $D_{2}$.

Answer

$F$ is smaller and more electronegative than $Cl$, so it forms stronger $H$-bonds as compared to $Cl$. As the consequence, more energy is needed to break the $H$-bonds in $HF$ than $HCl$ and hence the boiling point of $HF$ is higher than that of $HCl$.

That’s why $\mathrm{HF}$ is liquid and $\mathrm{HCl}$ is a gas.

Answer

The first element of the periodic table is hydrogen and its molecular form is dihydrogen $\left(H_{2}\right)$. When $H_{2}$ reacts with $O_{2}$, water is formed. Water is a liquid at room temperature. When liquid water freezes, it expands to form ice.

Density of ice is lower than that of liquid water and hence ice floats over water. Water is amphoteric in nature. It acts as a base in presence of strong acids and as an acid in presence of strong bases.

$$ \underset{Base_1}{H_2O(l)} + \underset{Acid_2}{H_2S(aq)} \longrightarrow \underset{Acid_1}{H_3O^+(aq)} + \underset{Base_2}{HS^-(aq)} $$

$$ \underset{Acid_1}{H_2O(l)} + \underset{Base_2}{NH_3(aq)} \longrightarrow \underset{Acid_2}{NH_4^+(aq)} + \underset{Base_1}{OH^-(aq)} $$

Due to amphoteric character, water undergoes self ionisation as shown below

$ \underset{\text{Acid }_1}{H _2O(l)} + \underset{Base _2}{H _2O(l)} \leftrightharpoons \underset{\substack{Acids_2 \\ \text{(Conjugate} \\ \text{acid)}}}{H_3O^+(aq)} + \underset{\substack{Base_1 \\ \text{(Conjugate} \\ \text{base)}}}{OH^-(aq)}$

This self ionisation of water is called auto-protolysis or autoionisation.

Answer

(i) The name of the compound is hydrogen peroxide, $H_{2} O_{2}$. It acts as an oxidising agent as well as reducing agent in both acidic and basic medium.

(ii) $H_{2} O_{2}$ decomposes slowly on exposure to light and dust particles. In the presence of metal surfaces or traces of alkali present in glass containers, the decomposition of $H_{2} O_{2}$ is catalysed.

It is, therefore, stored in wax lined glass or plastic vessels in dark. Urea is added as a negative catalyst or stabiliser to check its decomposition.

$$ 2 H_{2} O_{2}(l) \stackrel{h v}{\longrightarrow} 2 H_{2} O(l)+O_{2}(g) $$

Answer

Hydrogen resembles alkali metals, i.e., $\mathrm{Li}, \mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}$ and $\mathrm{Fr}$ of group I of the periodic table in the following respects

(i) Like alkali metals, hydrogen also contain one electron in its outermost (valence) shell and exhibit +1 oxidation state.

(ii) Like alkali metals, hydrogen also loses its only electron to form hydrogen ion, i.e., $\mathrm{H}^{+}$ (proton).

(iii) Like alkali metals, hydrogen combines with electronegative elements (non-metals) such as oxygen, halogens and sulphur forming their oxides, halides and sulphides respectively.

(iv) Like alkali metals, hydrogen also acts as a strong reducing agent.

Answer

Hydrogen has one electron which it can either lose or gain or share to acquire noble gas, i.e., helium gas configuration.

Therefore, in principle, it can form either ionic or covalent bonds. But the ionisation enthalpy of hydrogen is very high ( $\left.1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ and its electron gain enthalpy is only slightly negative $\left(-73 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$.

From this consequence, it does not have a high tendency to form ionic bonds but rather prefers to form only covalent bonds.

Answer

The ionisation enthalpy of hydrogen higher than that of sodium. Both hydrogen and sodium have one electron in the valence shell. But the size of hydrogen is much smaller than that of sodium and hence, the ionisation enthalpy of hydrogen is much higher $\left(1312 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ than that of sodium $\left(496 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$.

Answer

Basic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen. Hydrogen is a gas at room temperature.

However, by cooling and applying high pressure, gaseous $H_{2}$ can be converted into liquid $H_{2}$ which has much smaller volume and hence can be transported easily. Thus, the basic property of hydrogen which is useful for hydrogen economy is that it can be converted into a liquid by cooling under high pressure.

Answer

Following are the importance of heavy water

(i) It is extensively used as a moderator in nuclear reactors.

(ii) It is used as a tracer compound in the study of reaction mechanism.

(iii) It is used for the preparation of other deuterium compounds such as $CD_{4}, D_{2} SO_{4}$, etc.

Answer

The Lewis structure of hydrogen peroxide is

Answer

Following are the chemical equation of $H_{2} O_{2}$ in which it behaves as an oxidising as well as reducing agent

(i) $H_{2} O_{2}$ oxidises acidified $KI$ to iodine.

$$ 2 KI+H_{2} O_{2}+H_{2} SO_{4} \longrightarrow I_{2}+K_{2} SO_{4}+2 H_{2} O $$

(ii) $H_{2} O_{2}$ reduces $KMnO_{4}$ to $MnO_{2}$ in alkaline medium.

$$ 2 KMnO_{4}+3 H_{2} O_{2} \longrightarrow 2 MnO_{2}+2 KOH+3 O_{2}+2 H_{2} O $$

Answer

The bleaching action of hydrogen peroxide is due to the nascent oxygen which it liberates on decomposition.

$$ H_{2} O_{2} \longrightarrow H_{2} O+[O] $$

The nascent oxygen combines with colouring matter, in turn, gets oxidised. Thus, the bleaching action of $H_{2} O_{2}$ is due to the oxidation of colouring matter by nascent oxygen. It is used for the bleaching of delicate materials like ivory, feathers, silk, wool etc.

Colouring matter $+[\mathrm{O}] \longrightarrow$ Colourless matter

Answer

Oxygen is more electronegative $(\mathrm{EN}=3.5)$ than hydrogen $(\mathrm{EN}=2.1)$ hence, $\mathrm{O}-\mathrm{H}$ bond is polar. In the water molecule, two polar $\mathrm{O}-\mathrm{H}$ bonds are present which are held together at an angle of $104.5^{\circ}$. Due to the resultant of these two dipoles, water molecule is polar and has an dipole moment of 1.84 Debye.

Answer

Water show high boiling point as compared to hydrogen sulphide due to high electronegativity of oxygen ( $\mathrm{EN}=3.5$ ). Water undergoes extensive $\mathrm{H}$ - bonding as a result of which water exists as associated molecule.

For breaking these hydrogen bond, a large amount of energy is needed and hence the boiling point of $\mathrm{H}_{2} \mathrm{O}$ is high.

In other words, due to lower electronegativity of $S(E N=2.5)$, hydrogen sulphide do not undergo $H$-bonding. Consequently, $H_{2} ~S$ exists as discrete molecule and hence its boiling point is much lower than that of $H_{2} O$. Thats why $H_{2} ~S$ is a gas at room temperature. obtained?

57. Why is hydrogen peroxide stored in wax lined bottles?

58. Why does hard water not form lather with soap?

59. Phosphoric acid is preferred over sulphuric acid in preparing hydrogen

peroxide from peroxides. Why?

60. How will you account for $104.5^{\circ}$ bond angle in water?

Show Answer

Answer

In water, oxygen has $s p^{3}$-hybridisation and the bond angle of $\mathrm{HOH}$ should have been $109^{\circ} 28^{\prime}$. In $\mathrm{H}_{2} \mathrm{O}$, the oxygen atom is surrounded by two shared pairs and two lone pairs of electrons. From VSEPR theory, lone pair - lone pair repulsions are stronger than bond pair-bond pair repulsions.

As a result, the bond angle of $\mathrm{HOH}$ in water slightly decreases from the regular tetrahedral angle of $109^{\circ} .28^{\prime}$ to $104.5^{\circ}$.

61. Write redox reaction between fluorine and water.

62. Write two reactions to explain amphoteric nature of water.

Matching the Columns

63. Correlate the items listed in Column I with those listed in Column II. Find out as many correlations as you can.

64. Match Column I with Column II for the given properties/applications mentioned therein.

A. $\rightarrow(4)$

B. $\rightarrow(3)$

C. $\rightarrow$ (2)

D. $\rightarrow(1)$

A. Atomic hydrogen $(\mathrm{H})$ can be used in cutting and welding.

B. Dihydrogen $\left(\mathrm{H}_{2}\right)$ can be used in hydroformylation of olefin.

C. Water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ can be reduced to dihydrogen by $\mathrm{NaH}$.

D. Hydrogen peroxide $\left(H_{2} O_{2}\right)$ used in the name of perhydrol.

Answer

A. $\rightarrow(5)$

B. $\rightarrow(4)$

C. $\rightarrow$ (2)

D. $\rightarrow(1)$

E. $\rightarrow$ (3)

A. Electrolysis of water produce hydrogen and oxygen.

B. Lithium aluminium hydride is used as reducing agent.

C. Hydrogen chloride is a polar molecule.

D. Heavy water is used in atomic reactor as moderator.

E. Atomic hydrogen recombines on metal surface to generate high temperature.

Answer

A. $\rightarrow(2,4)$

B. $\rightarrow$ (3)

C. $\rightarrow(1,3)$

A. Hydrogen peroxide is used as a perhydrol and propellant.

B. Sodium hexametaphosphate is used in Calgon method.

C. Permanent hardness of hard water is removed by zeolite and sodium hexametaphosphate.

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason ( $\mathrm{R}$ ) is given. Choose the correct option out of the options given below in each question.

Answer

(a) Statements of assertion and reason both are correct and reason is the correct explanation of assertion.

$ Na_2CO_3 + \underset{\substack{\text{or }\\ \text{ CaSO }_4 \\ \text{( from hard } \\ \text{ water )}}}{MgSO_4} \longrightarrow Na_2SO_4 +\underset{\substack{\text{or}\\ \text{CaCO}_3 \\ \text{insoluble}}}{MgCO_3} $

Answer

(a) Statements of assertion and reason both are correct and reason is the correct explanation of assertion. Since, metals like Pd and Pt adsorbs a large volume of hydrogen, hence, these are used as a storage media for it.

Answer

Atomic hydrogen is highly unstable. Since, the electronic configuration of atomic hydrogen is $1 s^{1}$, it needs one more electron to complete its configuration and gain stability. Therefore, atomic hydrogen is very reactive and combines with almost all the elements.

It, however, reacts in three different ways i.e.,

(i) by loss of its single electron to form $\mathrm{H}^{+}$,

(ii) by gain of one electron to form $\mathrm{H}^{-}$and

(iii) by sharing its electron with other atoms to form single covalent bonds. In contrast, the bond dissociation energy form $\mathrm{H}-\mathrm{H}$ bond is very high ( $435.88 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ). As a result, molecular hydrogen is almost inert at room temperature and hence reacts only with a few elements.

Answer

(i) $\mathrm{D}_{2} \mathrm{O}$ can be prepared by prolonged electrolysis of water.

(ii) Physical properties

(a) $\mathrm{D}_{2} \mathrm{O}$ is colourless, odourless, tasteless liquid. It has maximum density $-1.1073 \mathrm{~g} \mathrm{~mL}^{-1}$ at $11.6^{\circ} \mathrm{C}$ (Maximum density of water at $4^{\circ} \mathrm{C}$ ).

(b) Solubility of salts in heavy water is less than in ordinary water because it is more viscous than ordinary water.

(c) Nearly, all physical constants of $D_{2} O$ are higher than $H_{2} O$. It is due to the greater nuclear mass of deuterium atom than $H$-atom and stronger $H$-bonding in $D_{2} O$ than $H_{2} O$.

(iii) Exchange reactions of hydrogen with deuterium

$$ \begin{aligned} NaOH+D_{2} O & \longrightarrow NaOD+HOD \\ HCl+D_{2} O & \longrightarrow DCl+HOD \\ NH_{4} Cl+D_{2} O & \longrightarrow NH_{3} DCl+HOD \end{aligned} $$

Answer

(i) Industrially, $H_{2} O_{2}$ is prepared by the auto-oxidation of 2-alkylanthraquinols.

$$ \text { 2-ethylanthraquinol } \underset{H_{2} / Pd}{\stackrel{O_{2} / \text { (air) }}{\rightleftarrows}} H_{2} O_{2}+\text { Oxidised product } $$

In this case, 1 % $H_{2} O_{2}$ 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 $H_{2} O_{2}$. (ii) $H_{2} O_{2}$ has a non-planar structure.

The molecular dimensions in the gas phase and solid phase are shown below

(a)

(b)

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~K$, dihedral angle is $90.2^{\circ}$

In the gas phase, $\mathrm{H}_{2} \mathrm{O}$ is a bent molecule with a bond angle of $104.5^{\circ}$ and $\mathrm{O}-\mathrm{H}$ bond length of $95.7 \mathrm{pm}$ as shown below

(a)

(b)

(a) The bent structure of water;

(b) The water molecule as a dipole and

(c) The orbital overlop picture in water molecule.

(iii) Following are the three important uses of $H_{2} O_{2}$

(a) 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.

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

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

Answer

(i) $H_{2} O_{2}$ is industrially manufactured by the auto-oxidation of 2alkylanthraquinols

2-ethylanthraquinol $ \xleftrightharpoons[H_2/Pd]{O_2/(air)} H_2O_2 +$ Oxidised product

In this case, 1 % $ H_{2} 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 $H_{2} O_{2}$.

(ii) $H_{2} O_{2}$ acts as an oxidising as well as reducing agent in both acidic and alkaline media. Following reactions are described below

(a) Oxidising action in acidic medium

$$ \begin{array}{r} 2 Fe^{2+}(aq)+2 H^{+}(aq)+H_{2} O_{2}(aq) \longrightarrow 2 Fe^{3+}(aq)+2 H_{2} O(l) \\ PbS(s)+4 H_{2} O_{2}(aq) \longrightarrow PbSO_{4}(~s)+4 H_{2} O(l) \end{array} $$

(b) Reducing action in acidic medium

$$ \begin{aligned} 2 MnO_{4}^{-}+6 H^{+}+5 H_{2} O_{2} & \longrightarrow 2 Mn^{2+}+8 H_{2} O+5 O_{2} \\ HOCl+H_{2} O_{2} & \longrightarrow H_{3} O^{+}+Cl^{-}+O_{2} \end{aligned} $$

(c) Oxidising action in basic medium

$$ \begin{aligned} 2 Fe^{2+}+H_{2} O_{2} & \longrightarrow 2 Fe^{3+}+2 OH^{-} \\ Mn^{2+}+H_{2} O_{2} & \longrightarrow Mn^{4+}+2 OH^{-} \end{aligned} $$

(d) Reducing action in basic medium

$$ \begin{aligned} I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\ 2 MnO_{4}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-} \end{aligned} $$

Answer

(i) Molar mass of $H_{2} O_{2}=34 ~g ~mol^{-1}$

$1 ~L$ of $5 M$ solution of $H_{2} O_{2}$ will contain $34 \times 5 ~g H_{2} O_{2}$

$2 ~L$ of $5 M$ solution of $H_{2} O_{2}$ will contain $34 \times 5 \times 2=340 ~g H_{2} O_{2}$

Mass of $H_{2} O_{2}$ present in $2 ~L$ of 5 molar solution $=340 ~g$

(ii) $0.2 ~L$ (or $200 ~mL$ ) of $5 M$ solution will contain

$$ \begin{gathered} \frac{340 \times 0.2}{2}=34 ~g H_{2} O_{2} \\ 2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2} \end{gathered} $$

$68 ~g H_{2} O_{2}$ on decompósitíoñiquil give $32 ~g O_{2} \quad 2 \times 16=32 ~g$

$\therefore 34 ~g H_{2} O_{2}$ on decomposition will give $\frac{32 \times 34}{68}=16 ~g O_{2}$

Answer

Since, a colourless liquid ‘A’ contains only hydrogen and oxygen and decomposes slowly on exposure to light but is stabilised by addition of urea, therefore, liquid $A$ may be hydrogen peroxide.

(i) The structure of $H_{2} O_{2}$ is

(a) Gas phase

5&top_left_x=818)

(b) Solid phase

(a) $H_{2} O_{2}$ structure in gas phase, dihedral angle is $111.5^{\circ}$.

(b) $H_{2} O_{2}$ structure in solid phase at $110 ~k$, dihedral angle is $90.2^{\circ}$.

(ii) $2 H_{2} O_{2} \xrightarrow[\text { Sunlight }]{\text { hv }} 2 H_{2} O+O_{2}$

Answer

It is LiH because it has significant covalent character due to the smallest alkali metal, Li. $\mathrm{LiH}$ is very stable. It is almost unreactive towards oxygen and chlorine.

It reacts with $Al_{2} Cl_{6}$ to form lithium aluminium hydride.

$$ 8 LiH+Al_{2} Cl_{6} \longrightarrow 2 LiAlH_{4}+6 LiCl $$

Answer

Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid.

$$ 2 \mathrm{Na}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{Na}^{+} \mathrm{H}^{-} $$

It reacts violently with water to produce $\mathrm{H}_{2}$ gas

$$ 2 NaH+2 H_{2} O \longrightarrow 2 NaOH+2 H_{2} $$

In solid state, $\mathrm{NaH}$ does not conduct electricity. On electrolysis, in its molten state it gives $\mathrm{H}_{2}$ at anode and $\mathrm{Na}$ at cathode.

$$ \mathrm{Na}^{+} \mathrm{H}^{-}(l) \xrightarrow{\text { Electrolysis }} \underset{\text { At cathode }}{2 \mathrm{Na}(l)}+\underset{\text { At anode }}{\mathrm{H}_{2}(g)} $$