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Square planar complexes are formed by d^...

Square planar complexes are formed by `d^(8)` ions with strong field ligands The crystal field splitting `Delta_(0)` is larger for the second and theird row transition elements and for more highly charged species All the complexes having `4d^(8)` and `5d^(8)` configurations are mostly square planar including those with weak field ligands such as halide ions square planar complexes can show geometrical isomerism but they do not show optical isomerism due to the presence of plane of symmetry
Among the following complexes which has a square planar geometry?
(a) `[RhCI(CO)(PPh_(3))_(2)]`
(b) `K_(3)[Cu(CN)_(4)]`
(c ) `[Ni(CO)_(4)]`
(d) `K_(2)[Zn(CN)_(4)]` .

A

`[RhCI(CO)(PPH_(3))_(2)]`

B

`K_(3)[CU(CN)_(4)]`

C

`K_(2)[Zn(CN)_(4)]`

D

`[Ni(CO)_(4)]`

Text Solution

Verified by Experts

(A), `4d^(8)` configuration as Rh is in +1 oxidation state. `4d^(2)` because of greater CFSE favours square planar geometry.
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Square planar complexes are generally formed by d^(8) ions with strong field ligands. The crystal field splitting is larger for second and third row transition elements and for more highly charged species. All the complexes having 4d^(8)" and "5d^(8) configuration are mostly square including those with weak field ligands such halide ions. Square planar complexes with coordination number four can show diastereoisomerism. Which one of the following square planar complexes will show geometrical isomerism ?