Match List-I with List-II List-I Complex List-II HYbridization (A) [C…

The correct answer is option 4. (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

List-I Complex	List-II HYbridization
(A) \([CoF_6]^{3-}\)	(IV) \(sp^3d^2\)
(B) \([Co(NH_3)_6]^{3+}\)	(III) \(d^2sp^3\)
(C) \([NiCl_4]^{2-}\)	(I) \(sp^3\)
(D) \([Ni(CN)_4]^{2-}\)	(II) \(dsp^2\)

(A) \([CoF_6]^{3-}\):

The atomic number of Co is 27 and its electronic configuration is \(_{18}Ar]4s^2 3d^7\)

Here, \(Co\) is in \(+3\) oxidation state, so the electronic configuration will be

since \(F^-\) is a weak field ligand it will form an outer orbital complex so, the hybridization will be

Hence the hybridization is \(sp^3d^2\) hybridization.

(B) \([Co(NH_3)_6]^{3+}\)

The atomic number of Co is 27 and its electronic configuration is \(_{18}Ar]4s^2 3d^7\)

Here, \(Co\) is in \(+3\) oxidation state, so the electronic configuration will be

Ammonia is placed in the middle of the spectrochemical series. It is a weak ligand but in presence of octahedral complexes containing a central metal atom belonging to 3d series has a coordination number of 3, ammonia acts as a strong ligand. For example, in the complex \([Co(NH_3)6]^{3+}\), Cobalt belongs to 3d series and has the oxidation number 3, in this case ammonia acts as a strong ligand and leads to pairing.

Hence the hybridization is \(d^2sp^3\) hybridization.

(C) \([NiCl_4]^{2-}\)

The atomic number of Ni is 28 and its electronic configuration is \(_{18}Ar]4s^2 3d^8\)

Here, \(Ni\) is in \(+2\) oxidation state, so the electronic configuration will be

As chlorine is a weak field ligand, it is not able to pair the electrons in Ni and thus d orbitals do not participate in hybridization. So, only 4s and 4p orbitals participate.

So, the hybridization is \(sp^3\)

 (D) \([Ni(CN)_4]^{2-}\)

The atomic number of Ni is 28 and its electronic configuration is \(_{18}Ar]4s^2 3d^8\)

Here, \(Ni\) is in \(+2\) oxidation state, so the electronic configuration will be

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.

So, the hybridization is \(dsp^2\).

Hence the correct answer is (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II).