Question

Find the value of $\frac{X}{Y}$ for the complex $\left[CoB{r}_2\right(CN\left)\right(N{O}_2\left)\right(H_{2}O\left)\right(N{H}_3\left)\right]$
where:
X = Number of compounds when both $B{r}^{-}$ are at cis position.
Y = Number of compounds when both $B{r}^{-}$ are at trans position.

Answer 1

$\left[CO\right(Br{)}_2\left(CN\right)\left(N{O}_2\right)\left(H_{2}O\right)\left(N{H}_3\right)]$
$N{O}_2^{-}\text{and}C{N}^{-}\text{both are ambidentate ligands.}$
$B{r}^{-}\text{at trans position (optically inactive).}$
$B{r}^{-}\text{at cis position (optical active complex)}$
The number of geometrical isomers keeping $B{r}^{-}$ at cis position is 6 and all of them are optically active.
So there are twelve cis isomers.
$\therefore X=12Y=3\text{(optical inactive complex)}$

$\frac{X}{Y}=\frac{12}{3}=\frac{4}{1}$
$\Rightarrow \frac{X}{Y}=4$