The spin-only magnetic moment value of an octahedral complex among $C o C l_{3} .4 N H_{3}, N i C l_{2} .6 H_{2} O$ and $P t C l_{4} .2 H C l$ , which upon reaction with excess of $A g N O_{3}$ gives 2 moles of $A g C l$ is B.M. (Nearest Integer)

3.0

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3.00

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Solution

$C o C l_{3} .4 N H_{3} A g N O_{3} - --- \to excess [C o (N H_{3})_{4} . C l_{2}] + A g C l$

$N i C l_{2} .6 H_{2} O A g N O_{3} - --- \to excess [N i (H_{2} O)_{6}]^{2 +} + 2 A g C l$

$P t C l_{4} .2 H C l \to [P t C l_{6}]^{4 -} + No AgCl ppt$

$[N i^{2 +} (H_{2} O_{6})] = d^{8} = t_{2 g}^{6} e_{g}^{2} = 2 unpaired electrons$

$Magnetic moment = \sqrt{2 (2 + 2)}$
$= \sqrt{8} \approx 3$

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