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Question

Match list-I (Complexes) with list - II (Hybridization of Central Atom) and select the correct answer using the codes given below the lists :

List - I List - II
A.[Ni(CO)4] 1. sp3
B. [Ni(CN)4]2 2. dsp2
C. [Fe(CN)6]4 3. sp3d2
D. [MnF6]4 4. d2sp3
5. sp2d

A
A-1, B-3, C-2, D-4
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B
A-5, B-2, C-4, D-3
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C
A-5, B-3, C-2, D-4
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D
A-1, B-2, C-4, D-3
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Solution

The correct option is D A-1, B-2, C-4, D-3
Ni atom in [Ni(CO)4] is sp3 hybridized. This results in a tetrahedral geometry.

Ni atom in [Ni(CN)4]2 is dsp2 hybridized. This results in square planar geometry. Here it is showing dsp2 hybridization because here CN is strong field ligand hence it forms an inner complex.

Fe atom in Fe(CN)6]4 is d2sp3 hybridized. This results in an octahedral geometry. Here it is showing d2sp3 hybridization because here again cyanide group is present which is strong field ligand which forms inner complexes.

Mn atom in [MnF6]4 is sp3d2 hybridized. This results in octahedral geometry.

Therefore, the correct option is D.

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