Question

On the basis of the following observations made with aqueous solutions, assign secondary valences to metals in the following compounds:

Formula	Moles of $AgCl$ precipitated per mole of the compounds with excess $AgN{O}_3$
(i) $PdC{l}_2.4N{H}_3$	$2$
(ii) $NiC{l}_2.6H_{2}O$	$2$
(iii) $PtC{l}_4.2HCl$	$0$
(iv) $CoC{l}_3.4N{H}_3$	$1$
(v) $PtC{l}_2.2N{H}_3$	$0$

Answer 1

Moles of $AgCl$ precipitated per mole of the compound is equal to the primary valences of the metal.

So,

(i) $PdC{l}_2.4N{H}_3$ with $2$ moles of $AgCl$ precipitated means secondary valence of $Pd$ is $4\Rightarrow \left[Pd\right(N{H}_3{)}_4]C{l}_2$

(ii) $NiC{l}_2.6H_{2}O$ with $2$ moles of $AgCl$ precipitated means secondary valence of $Ni$ is $6\Rightarrow \left[Ni\right(H_{2}O{)}_6]C{l}_2$

(iii) $PtC{l}_4.2HCl$ with $0$ moles of $AgCl$ precipitated means secondary valence of $Pt$ is $6\Rightarrow \left[PtC{l}_4\right(HCl{)}_2]$

(iv) $CoC{l}_3.4N{H}_3$ precipitates $1$ mole of $AgCl$. So, secondary valence of $Co$ is $6\Rightarrow \left[Co\right(N{H}_3{)}_{4}C{l}_2]Cl$

(v) $PtC{l}_2.2N{H}_3$ precipitates $0$ mole of $AgCl$. So, secondary valence of $Pt$ is $4\Rightarrow \left[PtC{l}_2\right(N{H}_3{)}_2]$