Question

Most complex compounds have either a coordination number of 4 or 6 and, in both cases, they show stereomerism. They may show geometrical isomerism or optical isomerism or both.

List-I contains complex compounds and List-II contains the number of stereomers they can form and the type of stereomers.
$\begin{array}{cc}\text{List-I}& \text{List-II}\\ \text{(I)}K\left[PdN{H}_{3}BrClF\right]& \text{(P)}\text{Geometrical isomerism}\\ \text{(II)}\left[Co\right(N{H}_3{)}_{3}BrClF]& \text{(Q)}\text{Optical isomerism}\\ \text{(III)}\left[Cr\right(N{H}_3{)}_{3}B{r}_{2}Cl]& \text{(R)}3\\ \text{(IV)}\left[FeB{r}_2\right(en{)}_2]N{O}_3& \text{(S)}5\\ & \text{(T)}6\\ & \text{(U)}2\end{array}$
Which of the following options has the correct combination considering compounds in List-I and number of their stereoisomers and type in List-II?

• A. (II), (P)(U) and (III), (Q)(U)
• B. (II), (P)(Q)(T) and (IV), (P)(U)
• C. (II), (P)(Q)(S) and (III), (P)(R)
• D. (II), (P)(Q)(U) and (III), (P)(U)

Answer 1

The correct option is C (II), (P)(Q)(S) and (III), (P)(R)

Here GI stands for geometrical isomers and SI stands for stereoisomers.

(I) is square planar so, it can show only GI.
Square planar complex of type $\left[Mabcd\right]$ exhibit 3 GI.

(II) has 4 GI in which one is chiral so, a total of 5 SI.

(III) has 3 GI and all are optically inactive. So, a total of 3 SI.

(IV) has two GI, in which cis is Optically active.