Question

Write the electronic configuration of \(d^5\) in terms of \( t_{2g} \) and \( e_g \) in an octahedral field as: 
\( \Delta_o>P \)
 

Hint:

The relative strength of \( \Delta_o \) and \( P \) determines whether a complex is high-spin or low-spin. A larger \( \Delta_o \) favors low-spin, while a smaller \( \Delta_o \) favors high-spin.

Answer 1

- (i) \( \Delta_o>P \): When the crystal field splitting energy (\( \Delta_o \)) exceeds the pairing energy (\( P \)), electrons fill the lower \( t_{2g} \) orbitals first, resulting in a \( t_{2g}^6 e_g^0 \) configuration, which is low-spin. - (ii) \( \Delta_o<P \): When \( \Delta_o \) is less than \( P \), electrons occupy the higher \( e_g \) orbitals to reduce repulsion, leading to a \( t_{2g}^4 e_g^1 \) configuration, which is high-spin. Explanation: - The spin state (high-spin or low-spin) of a complex is determined by the relationship between crystal field splitting energy (\( \Delta_o \)) and pairing energy (\( P \)). A larger \( \Delta_o \) promotes low-spin complexes, whereas a smaller \( \Delta_o \) favors high-spin complexes.