Chapter 09 Hydrogen
Multiple Choice Questions (MCQs)
1. Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(a) Its tendency to lose an electron to form a cation
(b) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration
(c) Its low negative electron enthalpy value
(d) Its small size
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.Show Answer
(a) Ionisation enthalpy of hydrogen resembles that of alkali metals
(b) Its reactivity is similar to halogens
(c) It resembles both alkali metals and halogens
(d) 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 (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.Show Answer
(a) $\mathrm{LiH}>\mathrm{NaH}>\mathrm{CsH}>\mathrm{KH}>\mathrm{RbH}$
(b) $\mathrm{LiH}<\mathrm{NaH}<\mathrm{KH}<\mathrm{RbH}<\mathrm{CsH}$
(c) $\mathrm{RbH}>\mathrm{CsH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{LiH}$
(d) $\mathrm{NaH}>\mathrm{CsH}>\mathrm{RbH}>\mathrm{LiH}>\mathrm{KH}$
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 $$
\mathrm{LiH}<\mathrm{NaH}<\mathrm{KH}<\mathrm{RbH}<\mathrm{CsH}
$$Show Answer
(a) $B_{2} H_{6}$
(b) $\mathrm{NH}_{3}$
(c) $\mathrm{H}_{2} \mathrm{O}$
(d) $\mathrm{CH}_{4}$
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.Show Answer
(a) protium
(b) deuterium
(c) tritium
(d) hydronium
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.Show Answer
(i) $H_{2} O_{2}+2 HI \longrightarrow I_{2}+2 H_{2} O$
(ii) $HOCl+H_{2} O_{2} \longrightarrow H_{3} O^{+}+Cl^{-}+O_{2}$
Which of the following statements is correct about $H_{2} O_{2}$ with reference to these reactions? Hydrogen peroxide is
(a) an oxidising agent in both (i) and (ii)
(b) an oxidising agent in (i) and reducing agent in (ii)
(c) a reducing agent in (i) and oxidising agent in (ii)
(d) a reducing agent in both (i) and (ii)
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.Show Answer
(a) $PbO_{2}$
(b) $BaO_{2} \cdot 8 H_{2} O+O_{2}$
(c) $MnO_{2}$
(d) $TiO_{2}$
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 $H_{2} SO_{4}$. $$ \underset{\text{peroxide}}{\underset{Hydrated barium}{BAO_2 . 8H_2O(s)}} + H_2SO_4 (aq) \longrightarrow BASO_4 (s + )\underset{\text{peroxide}}{\underset{Hydrated barium}{ H_2O_2(aq)}} + 8H_2O (l) $$Show Answer
(a) $2 MnO_{4}^{-}+6 H^{+}+5 H_{2} O_{2} \longrightarrow 2 Mn^{2+}+8 H_{2} O+5 O_{2}$
(b) $2 Fe^{3+}+2 H^{+}+H_{2} O_{2} \longrightarrow 2 Fe^{2+}+2 H_{2} O+O_{2}$
(c) $2 I^{-}+2 H^{+}+H_{2} O_{2} \longrightarrow I_{2}+2 H_{2} O$
(d) $KIO_{4}+H_{2} O_{2} \longrightarrow KIO_{3}+H_{2} O+O_{2}$
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.,Show Answer
(a) $2[Fe(CN)_6]^{4-} +2 H^+ + H_2 O_2 \longrightarrow 2[Fe(CN)_6]^{3-} +2 H_2 O$
(b) $I_{2}+H_{2} O_{2}+2 OH^{-} \longrightarrow 2 I^{-}+2 H_{2} O+O_{2}$
(c) $Mn^{2+}+H_{2} O_{2} \longrightarrow Mn^{4+}+2 OH^{-}$
(d) $PbS+4 H_{2} O_{2} \longrightarrow PbSO_{4}+4 H_{2} O$
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 $$
\begin{aligned}
I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\
2 MnO_{4}^{-}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-}
\end{aligned}
$$Show Answer
(a) an oxidising agent
(b) a reducing agent
(c) both an oxidising and a reducing agent
(d) neither oxidising nor reducing agent
Answer (c) Hydrogen peroxide acts as an oxidising as well as reducing agent in both acidic and alkaline media.Show Answer
(a) $CH_4(g)+H_{2} O(g) \xrightarrow[\text { Ni }]{1270k} CO(g)+3H_{2}(~g)$
(b) $C(s)+H_{2} O(g) \xrightarrow{1270 ~K} CO(g)+H_{2}(~g)$
(c) $CO(g)+H_{2} O(g) \xrightarrow[\text { Catalyst }]{673 ~K} CO_{2}(~g)+H_{2}(~g)$
(c) $C_2H_6+2H_{2} O \xrightarrow[\text { Ni }]{1270 k} 2CO+5H_{2}$
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.Show Answer
(a) sodium sulphate and water
(b) sodium sulphate and oxygen
(c) sodium sulphate, hydrogen and oxygen
(d) sodium sulphate and hydrogen peroxide
Answer (d) When sodium peroxide is treated with dilute sulphuric acid, we get sodium sulphate and hydrogen peroxide $$
Na_{2} O_{2}+\text { dil. } H_{2} SO_{4} \longrightarrow Na_{2} SO_{4}+H_{2} O_{2}
$$Show Answer
(a) water
(b) sulphuric acid
(c) hydrochloric acid
(d) fused sodium peroxide
Answer (b) Peroxodisulphate, obtained by electrolytic oxidation of acidified sulphate solutions at high current density, on hydrolysis yields hydrogen peroxide. $$
\begin{aligned}
2 HSO_{4}^{-}(aq) \xrightarrow{\text { Electrolysis }} HO_{3} SOOSO_{3} H(aq) \xrightarrow{\text { Hydrolysis }} 2 HSO_{4}^{-}(aq) & \\
& +2 H^{+}(aq)+H_{2} O_{2} \text { (aq) }
\end{aligned}
$$Show Answer
(a) $CH_{4}(~g)+H_{2} O(g) \xrightarrow{1270 ~K} CO(g)+H_{2}(~g)$
(b) $CO(g)+H_{2} O(g) \xrightarrow[\text { Catalyst }]{673 ~K} CO_{2}(~g)+H_{2}(~g)$
(c) $C_{n} H_{2 n+2}+n H_{2} O(g) \xrightarrow{1270 ~K} n CO+(2 n+1) H_{2}$
(d) $CO(g)+2 H_{2}(g) \xrightarrow[\text { Catalvst }]{\text { Cobalt }} CH_{3} OH(l)$
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.Show Answer
(a) $\mathrm{Ca}^{2+}$
(b) $\mathrm{Na}^{+}$
(c) $\mathrm{Cl}^{-}$
(d) $\mathrm{K}^{+}$
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.Show Answer
(a) $Ca_{3}\left(PO_{4}\right)_{2}$
(b) $Na_{3} PO_{4}$
(c) $Na_{6} P_{6} O_{18}$
(d) $Na_{2} HPO_{4}$
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 $Show Answer
(a) Groups 7, 8, 9
(b) Group 13
(c) Groups 15, 16, 17
(d) Group 14
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.Show Answer
(a) group 6
(b) group 7
(c) group 8
(d) group 9
Show Answer
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.
Multiple Choice Questions (More Than One Options)
19. Which of the following statements are not true for hydrogen?
(a) It exists as diatomic molecule
(b) It has one electron in the outermost shell
(c) It can lose an electron to form a cation which can freely exist
(d) It forms a large number of ionic compounds by losing an electron
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.Show Answer
(a) water gas
(b) syn gas
(c) producer gas
(d) industrial gas
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’.Show Answer
(a) Heavy water is used as a moderator in nuclear reactor
(b) Heavy water is more effective as solvent than ordinary water
(c) Heavy water is more associated than ordinary water
(d) Heavy water has lower boiling point than ordinary water
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.Show Answer
(a) Hydrogen has three isotopes of which protium is the most common
(b) Hydrogen never acts as cation in ionic salts
(c) Hydrogen ion, $\mathrm{H}^{+}$, exists freely in solution
(d) Dihydrogen does not act as a reducing agent
Answer $(a, b)$ Among the three isotopes of hydrogen, protonium is the most common. In ionic salts, hydrogen exists as hydride $\left(\mathrm{H}^{-}\right)$.Show Answer
(a) Water is known to be a universal solvent
(b) Hydrogen bonding is present to a large extent in liquid water
(c) There is no hydrogen bonding in the frozen state of water
(d) Frozen water is heavier than liquid water
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.Show Answer
(a) chlorides of $\mathrm{Ca}$ and $\mathrm{Mg}$ in water
(b) sulphates of $\mathrm{Ca}$ and $\mathrm{Mg}$ in water
(c) hydrogen carbonates of $\mathrm{Ca}$ and $\mathrm{Mg}$ in water
(d) carbonates of alkali metals in 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.Show Answer
(a) Elements of group 15 form electron deficient hydrides
(b) All elements of group 14 form electron precise hydrides
(c) Electron precise hydrides have tetrahedral geometries.
(d) Electron rich hydrides can act as Lewis acids.
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.Show Answer
(a) Hydrides of group 13 act as Lewis acids
(b) Hydrides of group 14 are electron deficient hydrides
(c) Hydrides of group 14 act as Lewis acids
(d) Hydrides of group 15 act as Lewis bases
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.Show Answer
(a) Metallic hydrides are deficient of hydrogen
(b) Metallic hydrides conduct heat and electricity
(c) Ionic hydrides do not conduct electricity in solid state
(d) lonic hydrides are very good conductors of electricity in solid state
Show Answer
Answer
$(a, b, c)$
The ionic hydrides are crystalline, non-volatile and non-conducting in solid state. However, their molten state conduct electricity.
Short Answer Type Questions
28. How can production of hydrogen from water gas be increased by using water gas shift reaction?
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.Show Answer
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 hydridesShow Answer
Molecular hydrides
Metallic hydrides
These are mainly formed by p-block
elements and some s-block elementsThese are formed by group 3, 4, $5(\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Y}, \mathrm{Zr}$,
(Be and $\mathrm{Mg}$ ).
$\mathrm{Nb}$, La, $\mathrm{Hf}, \mathrm{Ta}$, $\mathrm{Ac}$ etc., $10,11,12$ (Pd, Cu, Zn etc.)
Those are usually volatile compounds
having low melting and boiling point.These are hard, have a metallic lustre.
It conduct electricity.
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.Show Answer
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 iceShow Answer
(i) $PbS(s)+H_{2} O_{2}(a q) \longrightarrow$
(ii) $\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \xrightarrow[\text { catalyst }]{\text { Cobalt }}$
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. $$
CO(g)+2 H_{2}(g) \xrightarrow[\text { catalyst }]{\text { Cobalt }} CH_{3} OH(l)
$$Show Answer
(i) Lakes freeze from top towards bottom.
(ii) Ice floats on water.
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.Show Answer
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., $$ \underset{Acid_1}{H_2O(l)} + \underset{Base_2}{NH_3(aq)} \longrightarrow \underset{Acid_2}{NH_4^+(aq)} + \underset{Base_1}{OH^-(aq)} $$ $$ \underset{Base_1}{H_2O(l)} + \underset{Acid_2}{H_2S(aq)} \longrightarrow \underset{Acid_1}{H_3O^+(aq)} + \underset{Base_2}{HS^-(aq)} $$Show Answer
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. $$
R \stackrel{+}{N} H_{3} \cdot OH^{-}(s)+X^{-}(aq) \rightleftharpoons R \stackrel{+}{N} H_{3} \cdot X^{-}(s)+OH^{-}(aq)
$$Show Answer
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.Show Answer
Answer Heavy water is prepared by prolonged electrolysis of water. Comparison of physical properties of heavy water with those of ordinary water is as followsShow Answer
Property
$\mathbf{H}_{\mathbf{2}} \mathbf{0}$
$\mathbf{D}_{\mathbf{2}} \mathbf{0}$
Molecular mass $\left(\mathrm{g} \mathrm{mol}^{-1}\right)$
18.0151
20.0276
Melting point $(\mathrm{K})$
273.0
276.8
Boiling point $(\mathrm{K})$
373.0
374.4
Enthalpy of formation $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$
-285.9
-294.6
Enthalpy of vaporisation $-373 \mathrm{~K}\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
40.66
41.61
Enthalpy of fusion $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$
6.01
-
Temperature of max. density $(\mathrm{K})$
276.98
284.2
Density at 298 $\mathrm{K}\left(\mathrm{g} \mathrm{cm}{ }^{-3}\right)$
1.0000
1.1059
Viscosity $($ centipoise $)$
0.8903
1.107
Dielectric constant $\left(\mathrm{C}^{2} / \mathrm{Nm}^{2}\right)$
78.39
78.06
Electrical conductivity at $298 \mathrm{~K}\left(\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\right)$
$5.7 \times 10^{-8}$
-
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 $BaO_{2}$. $$
BaO_{2}+D_{2} SO_{4} \longrightarrow BaSO_{4}+D_{2} O_{2}
$$Show Answer
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.Show Answer
(ii) $H_{2} O_{2}$ is a better oxidising agent than water. Explain.
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. $$
PdS+4 H_{2} O_{2} \rightarrow PbSO_{4}+4 H_{2} O
$$Show Answer
$\mathbf{H}_{\mathbf{2}} \mathbf{0}$ | $\mathbf{D}_{\mathbf{2}} \mathbf{0}$ | |
---|---|---|
Melting point/K | 373.0 | 374.4 |
Enthalpy of vaporisation at $(373 \mathrm{~K}) / \mathrm{kJ} \mathrm{mol}^{-1}$ | 40.66 | 41.61 |
Viscosity/centipoise | 0.8903 | 1.107 |
On the basis of this data explain in which of these liquids intermolecular forces are stronger?
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 $H_{2} O$.Show Answer
$\mathbf{H}_{\mathbf{2}} \mathbf{0}$
$\mathbf{D}_{\mathbf{2}} \mathbf{0}$
Melting point/K
373.0
374.4
Enthalpy of vaporisation at $(373 \mathrm{~K}) / \mathrm{kJ} \mathrm{mol}^{-1}$
40.66
41.61
Viscosity/centipose
0.8903
1.107
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 $D_{2}$.Show Answer
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.Show Answer
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.Show Answer
(i) Write the name of this compound.
(ii) Explain why such precautions are taken for storing this chemical.
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. $$
2 H_{2} O_{2}(l) \stackrel{h v}{\longrightarrow} 2 H_{2} O(l)+O_{2}(g)
$$Show Answer
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.Show Answer
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.Show Answer
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 $\left(496 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$.Show Answer
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.Show Answer
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.Show Answer
Answer The Lewis structure of hydrogen peroxide isShow Answer
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. $$
2 KMnO_{4}+3 H_{2} O_{2} \longrightarrow 2 MnO_{2}+2 KOH+3 O_{2}+2 H_{2} O
$$Show Answer
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 matterShow Answer
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.Show Answer
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? Answer Dilute solutions of $H_{2} O_{2}$ cannot be concentrated by heating because it decomposes much below its boiling point. $$
2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2}
$$ 1 % $H_{2} O_{2}$ 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 $H_{2} O_{2}$. 57. Why is hydrogen peroxide stored in wax lined bottles? Answer Hydrogen peroxide is decomposed by rough surfaces of glass, alkali oxides present in it and light to form $H_{2} O$ and $O_{2}$. $$
2 H_{2} O_{2} \longrightarrow 2 H_{2} O+O_{2}
$$ To prevent this decomposition, hydrogen peroxide is usually stored in paraffin wax coated plastic or teflon bottles. 58. Why does hard water not form lather with soap? Answer Hard water contains salts of calcium and magnesium ions. Hard water does not give lather with soap and forms scum/precipitate with soap. Soap containing sodium stearate $\left(C_{17} H_{35} COONa\right)$ reacts with hard water to precipitate out as $Ca / Mg$ stearate. $$
2 C_{17} H_{35} COONa(aq)+M^{2+}(aq) \longrightarrow\left(C_{17} H_{35} COO\right)_{2} M \downarrow+2 Na^{+}(aq)
$$ (where, $M$ is $Ca / Mg$ ) It is therefore, unsuitable for laundry. 59. Phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why? Answer $H_{2} SO_{4}$ acts as a catalyst for decomposition of $H_{2} O_{2}$. Therefore, some weaker acids such as $H_{3} PO_{4}, H_{2} CO_{3}$ is preferred over $H_{2} SO_{4}$ for preparing $H_{2} O_{2}$ from peroxides. $3 \mathrm{BaO}_2+2 \mathrm{H}_3 \mathrm{PO}_4 \longrightarrow \underset{\substack{\text { (Insoluble) }}}{\mathrm{Ba}_3\left(\mathrm{PO}_4\right)_2}+3 \mathrm{H}_2 \mathrm{O}_2$ 60. How will you account for $104.5^{\circ}$ bond angle in water? 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. Answer Fluorine is a strong oxidising agent, it oxidises $H_{2} O$ to $O_{2}$ or $O_{3}$. The reactions are as follows $$
\begin{aligned}
2 ~F_{2}(g)+2 H_{2} O(l) \longrightarrow O_{2}(g)+4 H^{+}(aq)+4 ~F^{-}(aq) \\
3 ~F_{2}(g)+3 H_{2} O(l) \longrightarrow O_{3}(g)+6 H^{+}(aq)+6 ~F^{-}(aq)
\end{aligned}
$$ 62. Write two reactions to explain amphoteric nature of water. Answer Water has the ability to act as an acid as well as base, i.e., it behaves as an amphoteric substance. From the Bronsted Lowry theory, it acts as an acid with $NH_{3}$ and a base with $H_{2} ~S$. $$
\begin{gathered}
H_{2} O(l)+NH_{3}(aq) \longrightarrow OH^{-}(aq)+NH_{4}^{+}(aq) \\
H_{2} O(l)+H_{2} ~S(aq) \longrightarrow H_{3} O^{+}(aq)+HS^{-}(aq)
\end{gathered}
$$ The auto - protolysis (self-ionisation) of water takes place. The reaction are as follows $ \underset{Acid}{H_2O(l)}+ \underset{Base}{H_2O(l)} \longrightarrow \underset{\text{Conjugate acid}}{H_3O^+}(aq) + \underset{\text{Conjugate base}}{OH^-(aq)}$ 63. Correlate the items listed in Column I with those listed in Column II. Find out as many correlations as you can. Answer A. $\rightarrow(9,10)$ B. $\rightarrow(4,5,7,8,9)$ C. $\rightarrow(6)$ D. $\rightarrow(1,3)$ E. $\rightarrow(2,4)$ F. $\rightarrow(5)$ A. Synthesis gas - Synthesis of methanol -Mixture of $\mathrm{Co}$ and $\mathrm{H}_{2}$ B. Dihydrogen — Reducing agent -Stoichiometric compounds of s-block elements $-Zn+NaOH$ $-Zn+$ dil. $H_{2} SO_{4}$ -Synthesis of methanol C. Heavy water - Prolonged electrolysis of water D. Calgon $-Na_{2}\left[Na_{4}\left(PO_{3}\right)_{6}\right]$ -Softening of water E. Hydrogen peroxide - Oxidising agent — Reducing agent F. Salt like hydrides - Stoichiometric compounds of s-block elements. 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.Show Answer
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Matching The Columns
Column I
Column II
A.
Synthesis gas
1.
$Na_{2}\left[Na_{4}\left(PO_{3}\right)_{6}\right]$
B.
Dihydrogen
2.
Oxidising agent
C.
Heavy water
3.
Softening of water
D.
Calgon
4.
Reducing agent
E.
Hydrogen peroxide
5.
Stoichiometric compounds of s-block elements
F.
Salt like hydrides
6.
Prolonged electrolysis of water
7.
$\mathrm{Zn}+\mathrm{NaOH}$
8.
Zn + dil. $H_{2} SO_{4}$
9.
Synthesis of methanol
10.
Mixture of $\mathrm{CO}$ and $\mathrm{H}_{2}$
Show Answer
Column I
Column II
A.
$\mathrm{H}$
1.
used in the name of perhydrol.
B.
$\mathrm{H}_{2}$
2.
can be reduced to dihydrogen by $\mathrm{NaH}$.
C.
$\mathrm{H}_{2} \mathrm{O}$
3.
can be used in hydroformylation of olefin.
D.
$H_{2} O_{2}$
4.
can be used in cutting and welding.
Column I | Column II | ||
---|---|---|---|
A. | Electrolysis of water produces | 1. | atomic reactor |
B. | Lithium aluminium hydride is used as |
2. | polar molecule |
C. | Hydrogen chloride is a | 3. | recombines on metal surface to generate high temperature |
D. | Heavy water is used in | 4. | reducing agent |
E. | Atomic hydrogen | 5. | hydrogen and oxygen |
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.Show Answer
Column I | Column II | ||
---|---|---|---|
A. | Hydrogen peroxide is used as a | 1. | zeolite |
B. | Used in Calgon method | 2. | perhydrol |
C. | Permanent hardness of hard | 3. | sodium hexametaphosphate water is removed by |
4. | propellant |
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. 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.Show Answer
Assertion and Reason
Reason (R) Washing soda reacts with soluble magnesium and calcium sulphate to form insoluble carbonates.
(a) Statements $\mathrm{A}$ and $\mathrm{R}$ both are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) $\mathrm{A}$ is correct but $\mathrm{R}$ is not correct
(c) $\mathrm{A}$ and $\mathrm{R}$ both are correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
(d) $A$ and $R$ both are false
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} $Show Answer
Reason (R) Platinum and palladium can absorb large volumes of hydrogen.
(a) Statements $\mathrm{A}$ and $\mathrm{R}$ both are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
(b) $\mathrm{A}$ is correct but $\mathrm{R}$ is not correct
(c) $\mathrm{A}$ and $\mathrm{R}$ both are correct but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
(d) $A$ and $R$ both are false
Show Answer
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.
Long Answer Type Questions
69. Atomic hydrogen combines with almost all elements but molecular hydrogen does not. Explain.
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.Show Answer
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 $$
\begin{aligned}
NaOH+D_{2} O & \longrightarrow NaOD+HOD \\
HCl+D_{2} O & \longrightarrow DCl+HOD \\
NH_{4} Cl+D_{2} O & \longrightarrow NH_{3} DCl+HOD
\end{aligned}
$$Show Answer
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.Show Answer
(ii) Illustrate oxidising, reducing and acidic properties of hydrogen peroxide with equations.
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 $$
\begin{aligned}
I_{2}+H_{2} O_{2}+2 OH^{-} & \longrightarrow 2 I^{-}+2 H_{2} O+O_{2} \\
2 MnO_{4}+3 H_{2} O_{2} & \longrightarrow 2 MnO_{2}+3 O_{2}+2 H_{2} O+2 OH^{-}
\end{aligned}
$$Show Answer
(ii) Calculate the mass of oxygen which will be liberated by the decomposition of $200 \mathrm{~mL}$ of this solution.
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 $\frac{32 \times 34}{68}=16 ~g O_{2}$Show Answer
(i) Suggest possible structure of $A$.
(ii) Write chemical equations for its decomposition reaction in light.
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) $2 H_{2} O_{2} \xrightarrow[\text { Sunlight }]{\text { hv }} 2 H_{2} O+O_{2}$Show Answer
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. $$
8 LiH+Al_{2} Cl_{6} \longrightarrow 2 LiAlH_{4}+6 LiCl
$$Show Answer
Show Answer
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.
$$ \mathrm{Na}^{+} \mathrm{H}^{-}(l) \xrightarrow{\text { Electrolysis }} \underset{\text { At cathode }}{2 \mathrm{Na}(l)}+\underset{\text { At anode }}{\mathrm{H}_{2}(g)} $$