Question

$\left[M\right(N{H}_3{)}_2\left(py{)}_2\right(N{O}_2{)}_2\left]\underset{-N{O}_2}{\overset{+N{H}_3}{\to }}\right[M\left(N{H}_3{)}_3\right(py{)}_2\left(N{O}_2\right)]$. Difference between possible geometrical isomers of reactant and product is:

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Reactant $\left[M\right(N{H}_3{)}_2\left(py{)}_2\right(N{O}_2{)}_2]$ is of type $\left[M{a}_2{b}_2{c}_2\right]$ and product $\left[M\right(N{H}_3{)}_3\left(py{)}_2\right(N{O}_2\left)\right]$ is of type $\left[M{a}_3{b}_{2}c\right]$.
We know
$\left[M{a}_2{b}_2{c}_2\right]$ type of complexes show $5$ geometrical isomers.



And complexes of type $\left[M{a}_3{b}_{2}c\right]$ show $3$ geometrical isomers.


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