Question

Which of the following coordination compounds is not likely to show optical isomerism?

• A. \([\text{NiCl}_{2}(\text{en})_{2}]\)
• B. \([\text{Co(en)}_{3}]^{3+}\)
• C. \([\text{Cu(NH}_{3})_{4}]^{2+}\)
• D. \([PtCl_{2}(en)_{2}]^{2+}\)

Hint:

A complex is optically active only if it has no plane or centre of symmetry. Square planar complexes with all four ligands in one plane are not chiral.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Optical isomerism needs a non-superimposable mirror image, so the complex must have no plane or centre of symmetry. Octahedral complexes with bidentate ligands are often chiral.

Step 2: Octahedral complexes
\([\text{Co(en)}_{3}]^{3+}\) is chiral with three chelate rings. The cis forms of \([\text{NiCl}_{2}(\text{en})_{2}]\) and \([\text{PtCl}_{2}(\text{en})_{2}]^{2+}\) are also chiral and exist as d and l forms.

Step 3: Square planar complex
\([\text{Cu(NH}_{3})_{4}]^{2+}\) is square planar. The plane containing Cu and the four N atoms is a mirror plane, so the molecule is achiral.

Final Answer:
The square planar tetraamminecopper(II) ion is not optically active. \[ \boxed{\text{(C) }[\text{Cu(NH}_{3})_{4}]^{2+}} \]