Question

$A$ and $B$ are two ores of copper.
$A$ undergoes calcination to form a solid, $C{O}_{2\left(g\right)}$ and $O_{2\left(g\right)}$
$B$ undergoes roasting to form a solid and a gas $C$ which turns acidified $K_{2}C{r}_2{O}_7$ to green solution.
$B$ also reacts with dil $HCl$ to form a solid and gas $D$ which turns lead acetate solution black. $D$ also reacts with $C$ to form colloidal sulphur in the presence of moisture.

The nature of the gas ($D$) and hybridization of the central atom is respectively:

• A. Weakly acidic, $s{p}^3$
• B. Strongly acidic, $s{p}^3$
• C. Weakly acidic, $s{p}^2$
• D. Strongly acidic, $s{p}^2$

Answer 1

The correct option is A Weakly acidic, $s{p}^3$

A = $Cu\left(C{O}_3\right)$ and B = $CuS$

Reactions involved are:

Calcination of A:
$\underset{\left(A\right)}{Cu\left(C{O}_3\right)}\to \underset{\left(solid\right)}{CuO}+C{O}_2+O_2$

Roasting of B:
$\underset{\left(B\right)}{CuS}\to \underset{\left(solid\right)}{CuO}+\underset{\left(C\right)}{S{O}_2}$

$\underset{\left(C\right)}{S{O}_2}+K_{2}C{r}_2{O}_7\to K_2(SO{)}_4+\underset{\left(green\right)}{C{r}^{3+}}$

$\underset{\left(B\right)}{CuS}+dil.HCl\to \underset{\left(solid\right)}{CuC{l}_2}+\underset{\left(D\right)}{H_{2}S}$

'$D$' is $H_{2}S$.

$H_{2}S$ is weakly acidic and according to VSEPR theory is tetrahedral in geometry and hence hybridisation is $s{p}^3$.