Question

Which of the following compound (s) show(s) optical isomerism?
$\text{en- Ethylenediamine}\text{bn- Butylenediamine}\text{gly- Gylcinate}$

• A. $\left[Pt\right(bn{)}_2{]}^{2+}$
• B. $\left[CrC{l}_2\right(en{)}_2{]}^{+}$
• C. $\left[Co\right(en{)}_3\left]\right[Co{F}_6]$
• D. $\left[Zn\right(gly{)}_2]$

Answer 1

Answer: (B), (C), (D)

$\left[Zn\right(gly{)}_2]$

(a). $\left[Pt\right(bn{)}_2{]}^{2+}$ has square planar complex because of its $ds{p}^2$ hybridisation and hence, molecular plane is the plane of symmetry, it does not show optical isomerism.

(b). Cis form of $\left[CrC{l}_2\right(en{)}_2{]}^{+}$ is optically active, because the $en$ ligands are at adjacent position and are at different plane. Lack of plane of symmetry result in optical isomerism.
trans - form of this compound is optically inactive due to plane of symmetry.


(c). The complex ion $\left[Co\right(en{)}_3{]}^{3+}$ is optically active due to the absence of plane of symmetry.


(d). $\left[Zn\right(gly{)}_2]$ is a outer orbital complex due to its fully filled d-orbital so the hybridisation of the complex is $s{p}^3$ and it form tetrahedral complex.
Tetrahedrals have no plane of symmetry so it show optical isomerism. P