Question

Which one of the following has a square planar geometry
(At nos.: Fe = 26, Co = 27, Ni = 28, Pt = 78)

• A. $[PtC{l}_4{]}^{2-}$
• B. $[CoC{l}_4{]}^{2-}$
• C. $[FeC{l}_4{]}^{2-}$
• D. $[NiC{l}_4{]}^{2-}$

Answer 1

Answer: (A)

As $C{l}^{-}$ is weak field ligand so no pairing up.
Hence it is $s{p}^3$ hybridized giving tetrahedral geometry.
Due to $C{l}^{-}$, back pairing is not observed so it will be $s{p}^3$ hybridized giving tetrahedral geometry.
Because weak ligand, back pairing is not observed so it will be $s{p}^3$ i.e. tetrahedral geometry. All the complexes of $P{t}^{2+}$ are square planar including those with weak field ligand such as halide ions thus (4) is correct.