Assertion (A) NaCl reacts with concentrated $\mathrm{H}_2 \mathrm{SO}_4$ to give colourless fumes with pungent smell. But on adding $\mathrm{MnO}_2$ the fumes become greenish yellow.
Reason $(\mathrm{R}) \mathrm{MnO}_2$ oxidises HCl to chlorine gas which is greenish yellow.
Assertion (A) $\mathrm{SF}_6$ cannot be hydrolysed but $\mathrm{SF}_4$ can be.
Reason (R) Six F-atoms in $\mathrm{SF}_6$ prevent the attack of $\mathrm{H}_2 \mathrm{O}$ on sulphur atom of $\mathrm{SF}_6$.
An amorphous solid " $A$ " burns in air to form a gas " $B$ " which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $\mathrm{KMnO}_4$ solution and reduces $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$. Identify the solid " A " and the gas " B " and write the reactions involved.
Since, the by-product of roasting of sulphide ore is $\mathrm{SO}_2$, so A is $\mathrm{S}_8$ ' $A$ ' $=\mathrm{S}_8{ }^{\prime} \cdot B$ ' $=\mathrm{SO}_2$
Reactions
(i) $\mathrm{S}_8+8 \mathrm{O}_2 \xrightarrow{\Delta} 8 \mathrm{SO}_2$
(ii) $\mathrm{Ca}(\mathrm{OH})_2+\mathrm{SO}_2 \longrightarrow \mathrm{CaSO}_3+\mathrm{H}_2 \mathrm{O}$
(iii) $\underset{\text { (violet) }}{2 \mathrm{MnO}_4^{-}}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}+\underset{\text { (Colourless) }}{2 \mathrm{Mn}^{2+}}$
(iv) $2 \mathrm{Fe}^{3+}+\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Fe}^{2-}+\mathrm{SO}_4^{2-}+4 \mathrm{H}^{+}$
On heating lead (II) nitrate gives a brown gas " $A$ ". The gas " $A$ " on cooling changes to colourless solid " $B$ ". Solid " $B$ " on heating with NO changes to a blue solid ' $C$ '. Identify ' $A$ ', ' $B$ ' and ' $C$ ' and also write reactions involved and draw the structures of ' $B$ ' and ' $C$ '.
$\mathrm{Pb}\left(\mathrm{NO}_3\right)_2$ on heating produces a brown coloured gas which may be $\mathrm{NO}_2$. Since, on reaction with $\mathrm{N}_2 \mathrm{O}_4$ and on heating it produces $\mathrm{N}_2 \mathrm{O}_3$ and $\mathrm{N}_2 \mathrm{O}_4$ respectively.
Structures
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen $\left(\mathrm{H}_2\right)$ in the presence of a catalyst gives another gas (C) which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.
The main constituents of air are nitrogen (78%) and oxygen (21%). Only $\mathrm{N}_2$ reacts with three moles of $\mathrm{H}_2$ in the presence of a catalyst to give $\mathrm{NH}_3$ (ammonia) which is a gas having basic nature. On oxidation, $\mathrm{NH}_3$ gives $\mathrm{NO}_2$ which is a part of acid rain. So, the compounds $A$ to $D$ are as
$$A=\mathrm{NH}_4 \mathrm{NO}_2 ; B=\mathrm{N}_2 ; C=\mathrm{NH}_3 ; D=\mathrm{HNO}_3$$
Reactions involved can be given, as
(i) $$\mathop {N{H_4}N{O_2}}\limits_{[A]} \buildrel \Delta \over \longrightarrow \mathop {{N_2}}\limits_{[B]} + 2{H_2}O$$
(ii) $$\mathop {{N_2}}\limits_{[B]} + 3{H_2}$$ $\rightleftharpoons$ $$\mathop {2N{H_3}}\limits_{[C]} $$
(iii) $$4N{H_3} + 5{O_2}\buildrel {Oxidation} \over \longrightarrow 4NO + 6{H_2}O$$
(iv) $\mathrm{2NO+O_2 \rightarrow 2NO_2}$
(v) $$3N{O_2} + {H_2}O \to \mathop {2HN{O_3}}\limits_{[D]} + NO$$