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

The Correct Option is B

\(Ni\) 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 \(sp^3d^2\)