Question

The complex that can show fac- and mer-isomers is

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• A. $\left[Co\right(N{H}_3{)}_3\left(N{O}_2{)}_3\right]$
• B. $\left[Pt\right(N{H}_3{)}_{2}C{l}_2]$
• C. $\left[CoC{l}_2\right(en{)}_2]$
• D. $\left[Co\right(N{H}_3{)}_{4}C{l}_2{]}^{+}$

Answer 1

The correct option is A $\left[Co\right(N{H}_3{)}_3\left(N{O}_2{)}_3\right]$

Octaherdral geometry of type $\left[M{a}_3{b}_3\right]$ where a and b are unidentate ligands, shows facial and meridional isomers.
Facial isomer have three identical ligands on one triangular face.
Meridional isomer have three identical ligands in a plane bisecting the molecule.
Thus, $\left[Co\right(N{H}_3{)}_3\left(N{O}_2{)}_3\right]$ will show fac and mer isomers