Question

Select the correct statement.

• A. \([Ni(CN)_4]^{2-}\) is diamagnetic whereas \([Ni(CO)_4]\) is paramagnetic.
• B. \([Ni(CN)_4]^{2-}\) and \([Ni(CO)_4]\) both are diamagnetic.
• C. \([Ni(CN)_4]^{2-}\) is \(sp^3\) hybridised and square planar whereas \([Ni(CO)_4]\) is \(dsp^2\) hybridised and tetrahedral.
• D. \([Ni(CN)_4]^{2-}\) is paramagnetic and \(dsp^2\) hybridised whereas \([Ni(CO)_4]\) is diamagnetic and \(sp^3\) hybridised.

Hint:

CO and CN$^-$ are strong field ligands $\longrightarrow$ pairing $\longrightarrow$ diamagnetic complexes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Magnetic properties and geometry are determined by the oxidation state of the central metal, its electronic configuration, and the strength of the ligands (Crystal Field Theory).
Step 2: Detailed Explanation:
1. \([Ni(CN)_{4]^{2-}\):} Ni is in +2 state (\(3d^{8}\)). \(CN^{-}\) is a strong field ligand. It causes pairing of the 8 d-electrons into 4 pairs, leaving one d-orbital vacant. Hybridization is \(dsp^{2}\). Since all electrons are paired, it is diamagnetic and square planar.
2. \([Ni(CO)_{4]\):} Ni is in 0 state (\(3d^{8} 4s^{2}\)). \(CO\) is a strong field ligand. It causes the 4s electrons to shift into the 3d subshell, resulting in a \(3d^{10}\) configuration. Hybridization is \(sp^{3}\). Since all electrons are paired, it is diamagnetic and tetrahedral.

Comparing with the options, statement B correctly identifies that both are diamagnetic.
Step 3: Final Answer:
Both complexes are diamagnetic.