Consider the following complex compounds
(i) $[C u (N H_{3})_{4}] [C u (N O_{2})_{4}]$ (ii) $[C r (p y)_{6}] [C r (S C N)_{6}]$ (III) $[C o (N H_{3})_{5} (N O_{2})] [P t (S C N)_{4}]$
If $x_{1}, x_{2}, x_{3}$ are more possible coordination isomers of given complex compounds respectively then calculate value of $x_{1}^{2} + x_{2}^{2} + x_{3}^{2}$ .

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Solution

$[M_{1}^{2 +} A_{4}]^{2 +} [M_{1}^{2 +} B_{4}]^{2 -} x_{z} = 1$
$[M_{2}^{3 +} A_{6}]^{3 +} [M_{2}^{3 +} B_{6}]^{3 -} x_{2} = 2$
$[M_{2}^{3 +} C (A)_{s}]^{2 +} [M_{1} B_{4}]^{2 -} x_{3} = 6$
So, $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 1^{2} + 2^{2} + 6^{2} = 41$ .

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