The spin-only magnetic moment, calculated via \( \mu = \sqrt{n(n+2)} \, \text{BM} \), quantifies the magnetic moment based on \( n \), the count of unpaired electrons.
Electronic Configurations and Resulting Magnetic Moments:
- \( d^5 \) (high spin): \( n = 1 \), yielding \( \mu = \sqrt{3} \approx 1.73 \, \text{BM} \).
- \( d^3 \) (low or high spin): \( n = 3 \), resulting in \( \mu = \sqrt{15} \approx 3.87 \, \text{BM} \).
- \( d^4 \) (low spin): \( n = 4 \), yielding \( \mu = \sqrt{24} \approx 4.89 \, \text{BM} \).
- \( d^4 \) (high spin): \( n = 2 \), resulting in \( \mu = \sqrt{8} \approx 2.82 \, \text{BM} \).
Therefore, the \( d^4 \) electron configuration under strong ligand field conditions yields a magnetic moment of \( \mu = 2.82 \, \text{BM} \).
Final Answer: \[ \boxed{d^4 \, (\text{in strong ligand fields})} \]