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 compound with excess $AgNO_3$
$(i) PdCl_2 \cdot 4NH_3$	$2$
$(ii) NiCl_2 \cdot 6H_2O$	$2$
$(iii) PtCl_4 \cdot 2HCl$	$0$
$(iv) CoCl_3 \cdot 4NH_3$	$1$
$(v) PtCl_2 \cdot 2NH_3$	$0$

(A) The secondary valence corresponds to the coordination number of the central metal atom.
$(i) PdCl_2 \cdot 4NH_3$ gives $2$ moles of $AgCl$,meaning $2$ $Cl^-$ ions are outside the coordination sphere. The formula is $[Pd(NH_3)_4]Cl_2$. Secondary valence = $4$.
$(ii) NiCl_2 \cdot 6H_2O$ gives $2$ moles of $AgCl$,meaning $2$ $Cl^-$ ions are outside. The formula is $[Ni(H_2O)_6]Cl_2$. Secondary valence = $6$.
$(iii) PtCl_4 \cdot 2HCl$ gives $0$ moles of $AgCl$,meaning all $Cl^-$ ions are inside. The formula is $H_2[PtCl_6]$. Secondary valence = $6$.
$(iv) CoCl_3 \cdot 4NH_3$ gives $1$ mole of $AgCl$,meaning $1$ $Cl^-$ ion is outside. The formula is $[Co(NH_3)_4Cl_2]Cl$. Secondary valence = $6$.
$(v) PtCl_2 \cdot 2NH_3$ gives $0$ moles of $AgCl$,meaning all $Cl^-$ ions are inside. The formula is $[Pt(NH_3)_2Cl_2]$. Secondary valence = $4$.