$[M (N H_{3})_{2} (p y)_{2} (N O_{2})_{2}] + N H_{3} - --- \to - N O_{2} [M (N H_{3})_{3} (p y)_{2} (N O_{2})]$ . Difference between possible geometrical isomers of reactant and product is:

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Solution

Reactant $[M (N H_{3})_{2} (p y)_{2} (N O_{2})_{2}]$ is of type $[M a_{2} b_{2} c_{2}]$ and product $[M (N H_{3})_{3} (p y)_{2} (N O_{2})]$ is of type $[M a_{3} b_{2} c]$ .
We know
$[M a_{2} b_{2} c_{2}]$ type of complexes show $5$ geometrical isomers.

And complexes of type $[M a_{3} b_{2} c]$ show $3$ geometrical isomers.

Hence, difference between possible geometrical isomers of reactant and product is $5 - 3 = 2$ .

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[M (N H_{3})_{2} (p y)_{2} (N O_{2})_{2}] + N H_{3} - --- \to - N O_{2} [M (N H_{3})_{3} (p y)_{2} (N O_{2})]

C_{3} H_{5} C l

C_{3} H_{5} C l

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