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Na3[V(CN)6] d configuration
Accepted Answer
The d-configuration of vanadium in Na3[V(CN)6] is d².
Here's why:
1. Vanadium's Electronic Configuration: Vanadium (V) has an atomic number of 23, so its ground state electronic configuration is [Ar] 3d³ 4s².
2. Oxidation State: In Na3[V(CN)6], vanadium is in the +3 oxidation state. This means it loses three electrons.
3. CN- Ligand: The cyanide ion (CN-) is a strong field ligand, meaning it causes a large splitting of the d orbitals.
4. d-Configuration: Removing three electrons from the neutral vanadium atom, and considering the strong field nature of CN-, the d-configuration of V³⁺ in this complex is d² (two electrons in the lower energy d orbitals).
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