Question

For a \(\mathrm{d^4}\) metal ion in an octahedral field, the correct electronic configuration is :

• A. \(t_{2g}^{4}e_{g}^{0}\) when \(\Delta_0 \lt P\)
• B. \(t_{2g}^{3}e_{g}^{1}\) when \(\Delta_0 \gt P\)
• C. \(e_{g}^{2}t_{2g}^{2}\) when \(\Delta_0 \lt P\)
• D. \(t_{2g}^{3}e_{g}^{1}\) when \(\Delta_0 \lt P\)

Hint:

For \(\Delta_0 \lt P\) (weak field) electrons occupy \(e_g\) orbitals before pairing; for \(\Delta_0 \gt P\) they pair in \(t_{2g}\).

Answer 1

The Correct Option is D

For d⁴ in an octahedral field:
If \(\Delta_0 \gt P\) (strong field) electrons pair up: \(t_{2g}^{4}e_g^{0}\) (low spin).
If \(\Delta_0 \lt P\) (weak field) the fourth electron goes to the higher \(e_g\) level: \(t_{2g}^{3}e_g^{1}\) (high spin).
Option (1) pairs the low-spin configuration with the wrong condition, option (2) pairs the high-spin configuration with the wrong condition; option (3) has \(e_g^2t_{2g}^2\), which is not the ground configuration; option (4) \(t_{2g}^{3}e_g^{1}\) with \(\Delta_0 \lt P\) is correct.
Answer: (4) \(t_{2g}^{3}e_{g}^{1}\) when \(\Delta_0 \lt P\)