Question

The following observations are made on the basis of aqueous solutions. How many complexes are formed with secondary valence on the metal ion as $6$ ?

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

Answer 1

Answer: 3

Moles of $AgCl$ precipitated per mole of the compound are equal to the primary valences of the metal.
So,
$(i){\left(PdCl\right)}_2.4{\left(NH\right)}_3$ ​ with two moles of $AgCl$ precipitated means secondary valence of $Pd$ is $4\Rightarrow [Pd({\left(NH\right)}_3{)}_4]{\left(Cl\right)}_2$ ​
$(ii){\left(NiCl\right)}_2.6H_{2}O$ with two moles of $AgCl$ precipitated means secondary valence of $Ni$ is $6\Rightarrow [Ni(H_{2}O{)}_6]{\left(Cl\right)}_2$ ​
$(iii){\left(PtCl\right)}_4.2HCl$ with zero moles of $AgCl$ precipitated means secondary valence of $Pt$ is $6\Rightarrow H_2[PtC{l}_6]$
$(iv){\left(CoCl\right)}_3.4{\left(NH\right)}_3$ ​ precipitates one mole of $AgCl$ . So, secondary valence of $Co$ is $6\Rightarrow [Co({\left(NH\right)}_3{)}_4{\left(Cl\right)}_2]Cl$
$(v){\left(PtCl\right)}_2.2{\left(NH\right)}_3$ precipitates zero mole of $AgCl$ . So, secondary valence of $Pt$ is $4\Rightarrow [{\left(PtCl\right)}_2({\left(NH\right)}_3{)}_2]$