On the basis of the following observations made with aqueous solutions, how many of them have 6 secondary valencies?

Formula	Moles of AgCl precipitated per mole of the compounds with excess ${\mathrm{AgNO}}_3$•
$\text{ (a) }{\mathrm{PdCl}}_2\cdot 4{\mathrm{NH}}_3$	2
$\text{ (b) }{\mathrm{NiCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$	2
$\text{ (c) }{\mathrm{PtCl}}_4-2\mathrm{HCl}$	0
$\text{ (d) }{\mathrm{CoCl}}_3-4{\mathrm{NH}}_3$	1
$\text{ (e) }{\mathrm{PtCl}}_2\cdot 2{\mathrm{NH}}_3$	0

 

The complex which has the 6 secondary valencies are,
$\begin{array}{l}\left[\mathrm{Pd}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2\left[{\mathrm{Ni}}_2{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2\\ {\mathrm{H}}_2\left[{\mathrm{PtCl}}_6\right],\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}\\ \left[{\mathrm{Pt}}^2{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right]\end{array}$
The primary valences of the metal are represented by the number of moles of AgCl precipitated per mole of the compound.
So, ${\mathrm{PdCl}}_2\cdot 4{\mathrm{NH}}_3$ precipitated with 2 moles of AgCl indicates that the secondary valence of Pd is 4 in
$\left[\mathrm{Pd}\right({\mathrm{NH}}_3{)}_4]{\mathrm{Cl}}_2$.
(ii) ${\mathrm{NiCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$ precipitated with 2 moles of AgCl indicates that the secondary valence of Ni is 6 in 
$\left[\mathrm{Ni}\right({\mathrm{H}}_2\mathrm{O}{)}_6]{\mathrm{Cl}}_2$.
(iii) ${\mathrm{PtCl}}_{4}2\mathrm{HCl}$ precipitated with 0 moles of AgCl indicates that the secondary valence of Pt is 6 in 
$\left[{\mathrm{PtCl}}_4\right(\mathrm{HCl}{)}_2]$.
(iv) 1 mole of AgCl is precipitated by  ${\mathrm{CoCl}}_3\cdot 4{\mathrm{NH}}_3$. As a result, the secondary valence of Co is 6. 0 mole of AgCl is precipitated by $\left[\mathrm{Co}\right({\mathrm{NH}}_3{)}_4{\mathrm{Cl}}_2]\mathrm{Cl}.$
(v) ${\mathrm{PtCl}}_2.2{\mathrm{NH}}_3$ as a result, the secondary valence of Pt is 4 in $\left[{\mathrm{PtCl}}_2\right({\mathrm{NH}}_3{)}_2].$