sblock Elements 2 Question 9
9. The amphoteric hydroxide is
(2019 Main, 11 Jan I)
(a) $\mathrm{Be}(\mathrm{OH})_2$
(b) $\mathrm{Ca}(\mathrm{OH})_2$
(c) $\mathrm{Sr}(\mathrm{OH})_2$
(d) $\mathrm{Mg}(\mathrm{OH})_2$
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Answer:
Correct Answer: 9. (a)
Solution: For group- 2 metal hydroxides, basicity increases down the group, as:
$$ \mathrm{Be}(\mathrm{OH})_2<\mathrm{Mg}(\mathrm{OH})_2<\mathrm{Ca}(\mathrm{OH})_2<\mathrm{Sr}(\mathrm{OH})_2<\mathrm{Ba}(\mathrm{OH})_2 $$
This is because as the size of metal atom increases, $M-\mathrm{OH}$ bond length increases or $M-\mathrm{OH}$ bond become weaker thus readily breaks to release $\mathrm{OH}^{-}$ions which are responsible for the basicity of these solutions.
But $\mathrm{Be}(\mathrm{OH})_2$ shows amphoteric (basic as well as acidic) character as it reacts with acid and alkali both which is shown in the following reactions. $\mathrm{Be}(\mathrm{OH})_2$ as a base :
$$ \mathrm{Be}(\mathrm{OH})_2+2 \mathrm{HCl} \longrightarrow \mathrm{BeCl}_2+2 \mathrm{H}_2 \mathrm{O} $$
$\mathrm{Be}(\mathrm{OH})_2$ as an acid :
$$ \mathrm{Be}(\mathrm{OH})_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2\left[\mathrm{Be}(\mathrm{OH})_4\right] $$
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