Question

Match List-I with List-II

| List-I
(Complex)| | List-II
(Hybridization)
---|---|---|---
A.| Ni(CO)4| I.| sp3
B.| [Ni(CN)4]2-| II.| sp3d2
C.| [Co(CN)6]3-| III.| d2sp3
D.| [CoF6]3-| IV.| dsp2

Choose the correct answer from the options given below:

• A. A-IV, B-I, C-III, D-II
• B. A-I. B-IV, C-III, D-II
• C. A-I. B-IV, C-II, D-III
• D. A-IV, B-I, C-II. D-III

Answer 1

Answer: (B)

A-I. B-IV, C-III, D-II

Answer 2

$Ni$ is in zero oxidation state in $Ni(CO)_4$ so the electronic configuration of $Ni$ is $3d^84s^2$. As $CO$ is a strong ligand, it pushes all the electrons in the 3d orbital, therefore the hybridisation of $Ni(CO)_4$ is $sp^3$ and it has tetrahedral geometry. It is diamagnetic due to the absence of unpaired electrons.

In $[Ni(CN)_4]^{2-}$, there is $Ni^{2+}$ ion for which the electronic configuration in the valence shell is $3d^84s^0$.

In presence of strong field $CN^-$ ions, all the electrons are paired up.

The empty, 3d, 3s and two 4p Orbitals undergo $dsp^2$ hybridization to make bonds with $CN^-$ ligands in square planar geometry.

The atomic number of $Co$ is 27 and its valence shell electronic configuration is $3d^74s^2$.

$Co$ is in +3 oxidation state in the complex $[CoF_6]^{3-}$.

$[CoF_6]^{3-}$ is $sp^3d^2$ hybridized and it is octahedral in shape.

Another $Co^{3+}$ complex, $[Co(CN)_6]^{3-}$, is diamagnetic and has no unpaired electrons. The hybrid orbitals used to form this complex are $d^2sp^3$.

$NiCO_4$ Hybridisation $sp^3$

$NiCN_4^{2-}$ Hybridisation $dsp^2$

$CoCN_6^{3-}$ Hybridisation $d^2sp^3$

$CoF_6^{3-}$ Hybridisation $sp^3d^2$