Question

Match the complexes in Column-I with their hybridisation in Column-II.

Column-I     Column-II
A. \([ \mathrm{Ni(CO)_4} ]\)      1. \(dsp^2\)
B. \([ \mathrm{Ni(CN)_4} ]^{2-}\)      2. \(sp^3\)
C. \([ \mathrm{Ni(NH_3)_6} ]^{2+}\)      3. \(d^2sp^3\)
D. \([ \mathrm{Fe(CN)_6} ]^{3-}\)      4. \(sp^3d^2\)

• A. A-1, B-2, C-3, D-4
• B. A-2, B-1, C-4, D-3
• C. A-1, B-3, C-2, D-4
• D. A-2, B-3, C-1, D-4

Hint:

Strong field ligands → inner orbital (dsp$^2$, d$^2$sp$^3$); weak field → outer orbital (sp$^3$d$^2$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Hybridization in coordination compounds is determined by the coordination number and the nature of the ligands (Strong field vs Weak field), which influences electron pairing in the $d$-orbitals.
Step 3: Detailed Explanation:
1. \([Ni(CO)_{4]\)}: Ni is in 0 oxidation state ($3d^{8}4s^{2}$). $CO$ is a very strong ligand and causes pairing of $4s$ electrons into $3d$, resulting in $3d^{10}$. The empty $4s$ and $4p$ orbitals hybridize. Hybridization: \(sp^{3}\) (A-2).
2. \([Ni(CN)_{4]^{2-}\)}: Ni is in $+2$ state ($3d^{8}$). $CN^{-}$ is a strong field ligand and causes pairing of the two unpaired $d$ electrons. This leaves one $3d$ orbital vacant. Hybridization: \(dsp^{2}\) (B-1).
3. \([Ni(NH_{3})_{6]^{2+}\)}: Ni is in $+2$ state ($3d^{8}$). Coordination number is 6. With 8 electrons, it cannot form an inner orbital complex ($d^{2}sp^{3}$) as only two $d$-orbitals are needed. It uses outer $d$-orbitals. Hybridization: \(sp^{3}d^{2}\) (C-4).
4. \([Fe(CN)_{6]^{3-}\)}: Fe is in $+3$ state ($3d^{5}$). $CN^{-}$ is a strong field ligand and causes pairing, leaving two $3d$ orbitals vacant for hybridization. Hybridization: \(d^{2}sp^{3}\) (D-3).
Step 4: Final Answer:
The correct match is A-2, B-1, C-4, D-3.