Which of the following complexes is/are correctly matched with their geometries?

[C o C l_{4}]^{2 -} - t e t r a h e d r a l

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[C o (P M e_{3})_{4}] - t e t r a h e d r a l

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[C u (C N)_{4}]^{3 -} - t e t r a h e d r a l

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[F e (C O)_{4}]^{2 -} - s q u a r e

planar

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Solution

The correct options are
A $[C o C l_{4}]^{2 -} - t e t r a h e d r a l$
B $[C o (P M e_{3})_{4}] - t e t r a h e d r a l$
C $[C u (C N)_{4}]^{3 -} - t e t r a h e d r a l$
Square planar geometry is due to $d s p^{2}$ hybridization and tetrahedral geometry is due to $s p^{3}$ hybridization. The complexes in A,B and C undergo $s p^{3}$ hybridization. Hence they have tetrahedral geometry.
Thus the geometries of complexes in A,B and C are correctly matched.
For the complex in D, due to presence of strong field CO ligand, electron paring results in tetrahedral geometry.

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Similar questions

N i^{2 +}, P t^{2 +}

Z n^{2 +},

The geometries of the ammonia complexes of $N i^{2 +}, P t^{2 +} a n d Z n^{2 +},$ respcetively, are

[N i (C O)_{4}]

[P t (N H_{3})_{4}]^{2 +} [C o (S C N)_{4}]^{2 -} [C u (e n)_{2}]^{2 +} [H g I_{4}]^{2 -}

[N i (C O)_{4}]

[N i C l_{4}]^{2 -}

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