A colourless liquid ' $A$ ' contains H and 0 elements only. It decomposes slowly on exposure to light. It is stabilised by mixing urea to store in the presence of light.
(i) Suggest possible structure of $A$.
(ii) Write chemical equations for its decomposition reaction in light.
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) $\mathrm{H}_2 \mathrm{O}_2$ structure in gas phase, dihedral angle is $111.5^{\circ}$.
(b) $\mathrm{H}_2 \mathrm{O}_2$ structure in solid phase at 110 K , dihedral angle is $90.2^{\circ}$.
(ii) $2 \mathrm{H}_2 \mathrm{O}_2 \xrightarrow[\text { Sunlight }]{\text { hv }} 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$
An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $\mathrm{Al}_2 \mathrm{Cl}_6$.
It is LiH because it has significant covalent character due to the smallest alkali metal, Li. LiH is very stable. It is almost unreactive towards oxygen and chlorine.
It reacts with $\mathrm{Al}_2 \mathrm{Cl}_6$ to form lithium aluminium hydride.
$$8 \mathrm{LiH}+\mathrm{Al}_2 \mathrm{Cl}_6 \longrightarrow 2 \mathrm{LiAlH}_4+6 \mathrm{LiCl}$$
Sodium forms a crystalline ionic solid with dihydrogen. The solid is non-volatile and non-conducting in nature. It reacts violently with water to produce dihydrogen gas. Write the formula of this compound and its reaction with water. What will happen on electrolysis of the melt of this solid?
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
$$2 \mathrm{NaH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+2 \mathrm{H}_2$$
In solid state, NaH does not conduct electricity. On electrolysis, in its molten state it gives $\mathrm{H}_2$ at anode and Na at cathode.
$\mathrm{Na}^{+} \mathrm{H}^{-}(l) \xrightarrow{\text { Electrolysis }} \underset{\text { At cathode }}{2 \mathrm{Na}(l)}+\underset{\text { At anode }}{\mathrm{H}_2(g)}$