Question

The formula of carbonate of a metal M is M₂CO₃.
(a) What will be the formula of its iodide ?
(b) What will be the formula of its nitride ?
(c) What will be the formula of its phosphate ?

Answer 1

Let us first work out the valency of the metal M (we know that carbonate ion has a valency of −2):
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{CO}}_3^{2-}\\ \mathrm{Valency}& 1& 2\end{array}$
It is clear that M has a valency of 1.

(a) Formula of iodide of M:
$\begin{array}{ccc}\mathrm{Element}& \mathrm{M}& \mathrm{I}\\ \mathrm{Valency}& 1& 1\end{array}$
Crossing the valencies, formula of the iodide of M is MI.

(b) Formula of nitride of M:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{N}}^{3-}\\ \mathrm{Valency}& 1& -3\end{array}$
Crossing the valencies, formula of the nitride of M is M₃N.

(c) Formula of phosphate of M:
$\begin{array}{ccc}\mathrm{Element}/\mathrm{Ion}& \mathrm{M}& {\mathrm{PO}}_4^{3-}\\ \mathrm{Valency}& 1& -3\end{array}$
Crossing the valencies, formula of the phosphate of M is M₃PO₄.