The stability of peroxide and superoxide of alkali metals increase as we go down to group. Explain giving reason.
The stability of peroxide or superoxide increases as the size of metal ion increases i.e.,
$$\mathrm{KO}_2<\mathrm{RbO}_2<\mathrm{CsO}_2$$
The reactivity of alkali metals toward oxygen to form different oxides is due to strong positive field around each alkali metal cation. $\mathrm{Li}^{+}$is the smallest, it does not allow $\mathrm{O}^{2-}$ ion to react with $\mathrm{O}_2$ further. $\mathrm{Na}^{+}$is larger than Li , its positive field is weaker than $\mathrm{Li}^{+}$. It cannot prevent the conversion of $\mathrm{O}^{2-}$ into $\mathrm{O}_2^{2-}$.
The largest $\mathrm{K}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$ions permit $\mathrm{O}_2^{2-}$ ion to react with $\mathrm{O}_2$ forming superoxide ion $\mathrm{O}_2^{-}$.
$\underset{\text { Oxide }}{\mathrm{O}_2^{2-}} \xrightarrow{\frac{1}{2} \mathrm{O}_2} \underset{\text { Peroxide }}{\mathrm{O}^{2-}} \xrightarrow{\mathrm{O}_2} \underset{\text { Superoxide }}{2 \mathrm{O}_2^{-}}$
Futhermore, increased stability of the peroxide or superoxide with increase in the size of metal ion is due to the stabilisation of large anions by larger cations through lattice energy effect.
When water is added to compound $(A)$ of calcium, solution of compound $(B)$ is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds $A, B, C$ and $D$. Explain why the milkiness disappears in the last step.
Appearance of milkiness on passing $\mathrm{CO}_2$ in the solution of compound $B$ indicates that compound $B$ is lime water and compound $C$ is $\mathrm{CaCO}_3$. Since, compound $B$ is obtained by adding $\mathrm{H}_2 \mathrm{O}$ to compound $A$, therefore, compound $A$ is quicklime, CaO .
The reactions are as follows
(i) $\underset{\substack{\text { Calcium } \\ \text { oxide } \\ \text { (A) }}}{\mathrm{CaO}}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { Lime water } \\(B)}}{\mathrm{Ca}(\mathrm{OH})_2}$
(ii) $\underset{(B)}{\mathrm{Ca}(\mathrm{OH})_2}+\mathrm{CO}_2 \longrightarrow \underset{\substack{\text { Calcium carbonate } \\ \text { (Milkiness) }}}{\mathrm{CaCO}_3}+\underset{(\mathrm{C})}{\mathrm{H}_2 \mathrm{O}}$
(iii) When excess of $\mathrm{CO}_2$ is passed, milkiness disappears due to the formation of soluble calcium bicarbonate $(D)$.
$\underset{\substack{\text { Milkiness } \\(\mathrm{C})}}{\mathrm{CaCO}_3}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\left.\text { Calcium bicarbonate } \\ \text { (Soluble in } \mathrm{H}_2 \mathrm{O}\right) \\(D)}}{\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2}$
Lithium hydride can be used to prepare other useful hydrides. Beryllium hydride is one of them. Suggest a route for the preparation of beryllium hydride starting from lithium hydride. Write chemical equations involved in the process.
$\mathrm{BeH}_2$ can be prepared from the corresponding halides by the reduction with complex alkali metal hydrides such as lithium aluminium hydride $\mathrm{LiAlH}_4$.
$$\begin{aligned} & 8 \mathrm{LiH}+\mathrm{Al}_2 \mathrm{Cl}_6 \longrightarrow 2 \mathrm{LiAlH}_4+6 \mathrm{LiCl} \\ & 2 \mathrm{BeCl}_2+\mathrm{LiAlH}_4 \longrightarrow 2 \mathrm{BeH}_2+\mathrm{LiCl}+\mathrm{AlCl}_3 \end{aligned}$$
An element of group 2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid.
The alkaline earth metals burn in oxygen to form monoxide $M \mathrm{O}$. BeO is essentially covalent in nature, other being ionic in nature.
BeO is amphoteric while other oxides are basic in nature and react with water to form sparingly soluble hydroxides.
BeO dissolves both in acid and alkalis to give salt and is amphoteric.
$$\mathrm{BeO}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\substack{\text { Bendllium } \\ \text { hydroxide }}}{\mathrm{Be}(\mathrm{OH})_2}$$
$\mathrm{Be}(\mathrm{OH})_2$ is an amphoteric hydroxide, dissolving in both acids and alkalies. With alkalies it dissolves to form the tetrahydroxidoberyllate $\left(Z^{-}\right)$ anion with sodium hydroxide solution.
$$ 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{Be}(\mathrm{OH})_2(\mathrm{~s}) \longrightarrow \underset{\text { Sodium tetra hydroxidoberyllate }}{\mathrm{Na}_2 \mathrm{Be}(\mathrm{OH})_4(\mathrm{aq})} $$
With acids, it forms beryllium salts.
$\mathrm{Be}(\mathrm{OH})_2+\underset{\substack{\text { Sulphuric } \\ \text { acid }}}{\mathrm{H}_2 \mathrm{SO}_4} \longrightarrow \underset{\substack{\text { Beryllium } \\ \text { sulphate }}}{\mathrm{BeSO}_4}+2 \mathrm{H}_2 \mathrm{O}$
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and aminoacids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Yellow colour flame in flame test indicates that the alkali metal must be sodium. It reacts with $\mathrm{O}_2$ to form a mixture of sodium peroxide, $\mathrm{Na}_2 \mathrm{O}_2$ and sodium oxide $\mathrm{Na}_2 \mathrm{O}$.
$\begin{gathered}4 \mathrm{Na}+\mathrm{O}_2 \xrightarrow{\Delta} 2 \mathrm{Na}_2 \mathrm{O} \text { (Minor) } \\ 2 \mathrm{Na}_2 \mathrm{O}+\mathrm{O}_2 \xrightarrow{\Delta} 2 \mathrm{Na}_2 \mathrm{O}_2 \text { (Major) } \\ 2 \mathrm{Na}+\mathrm{O}_2 \xrightarrow{\Delta} \mathrm{Na}_2 \mathrm{O}_2\end{gathered}$
Ionisation enthalpy of sodium is low. When sodium metal or its salt is heated in Bunsen flame, the flame energy causes an excitation of the outermost electron which on reverting back to its initial position gives out the absorbed energy as visible light and complementary colour of absorbed colour from the light radiation is seen. That's why sodium imparts yellow colour to the flame.