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 $AgNO _3$
i	$PdCl _2 \cdot 4 NH _3$	2
ii	$NiCl _2 \cdot 6 H _2 O$	2
iii	$PtCl _4 \cdot 2 HCl$	0
iv	$CoCl _3 \cdot 4\,NH _3$	1
v	$PtCl _4 \cdot 2NH _3$	0

Answer 1

Moles of $AgCl$ precipitated per mole of the compound are equal to the primary valences of the metal.
So,
$\left(\right.i\left.\right)\left(PdCl\right)_{2}.4\left(NH\right)_{3}$ ​ with two moles of $AgCl$ precipitated means secondary valence of $Pd$ is $4\Rightarrow \left[\right.Pd\left(\right. \left(NH\right)_{3} \left.\right)_{4}\left]\right.\left(Cl\right)_{2}$ ​
$\left(\right.ii\left.\right)\left(NiCl\right)_{2}.6H_{2}O$ with two moles of $AgCl$ precipitated means secondary valence of $Ni$ is $6\Rightarrow \left[\right.Ni\left(\right. H_{2} O \left.\right)_{6}​\left]\right.\left(Cl\right)_{2}$ ​
$\left(\right.iii\left.\right)\left(PtCl\right)_{4}​.2HCl$ with zero moles of $AgCl$ precipitated means secondary valence of $Pt$ is $6\Rightarrow H_{2}\left[\right.PtCl_{6}\left]\right.$
$\left(\right.iv\left.\right)\left(CoCl\right)_{3}​.4\left(NH\right)_{3}$ ​ precipitates one mole of $AgCl$ . So, secondary valence of $Co$ is $6\Rightarrow \left[\right.Co\left(\right. \left(NH\right)_{3} \left.\right)_{4}\left(Cl\right)_{2}\left]\right.Cl$
$\left(\right.v\left.\right)\left(PtCl\right)_{2}​.2\left(NH\right)_{3}$ precipitates zero mole of $AgCl$ . So, secondary valence of $Pt$ is $4\Rightarrow \left[\right.\left(PtCl\right)_{2}\left(\right. \left(NH\right)_{3} \left.\right)_{2}\left]\right.$