$CN^-$ is a strong field ligand. This is due to the fact that :-
A
It is a pseudohalide
B
It can accept electrons from metal species
C
It forms high spin complexes with metal species
D
It carries negative charge
The answer is B.
Solution
A strong field ligand produces a large crystal field splitting ($\Delta_o$). $CN^-$ donates a lone pair from carbon to the metal (a $\sigma$ bond). In addition, it has empty $\pi^*$ antibonding orbitals that can accept electron density back from the filled metal $t_{2g}$ d orbitals (back bonding, $\pi$-acceptor behaviour). This metal-to-ligand $\pi$ back donation lowers the energy of the $t_{2g}$ orbitals and increases $\Delta_o$. That is why $CN^-$ (like CO) is at the strong end of the spectrochemical series. Being a pseudohalide or carrying a negative charge does not make a ligand strong field ($Cl^-$ and $F^-$ are negative but weak field). Strong field ligands form low spin complexes, not high spin ones. So $CN^-$ is a strong field ligand because it can accept electrons from the metal species ($\pi$-acceptor).
ConceptPi-acceptor ligands are strong field
Coordination Compounds in past papers
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