Magnetic moment from given CFSE value Spin magnetic moment of a octahedral complex having CFSE = $– 0.8∆$ and surrounded by weak field ligands can be:
A. $\sqrt{15} \space \mu_B$
B. $\sqrt{8} \space \mu_B$
C. (A) &amp; (B) both
D. None of the above

The CFSE value given in the above question corresponds to only $d^2$ configuration for weak field ligands. Hence, according to me, the value of spin magnetic moment should be only $\sqrt{8} \space \mu_B$.
For the answer to be $\sqrt{15} \space \mu_B$, the configuration would have to be $d^7$ requiring the CFSE value given in the question to have an additional 2P(twice of pairing energy) but that is not the case.
Hence answer should be option (B) but the source of the question says that the answer is option (C).
I am unable to understand the consideration of both $d^2$ and $d^7$ configurations.
Can anyone please explain?
Even for weak ligands, the electrons will eventually have to pair up, once the number becomes more than $5$. So, $\mathrm{d^7}$ is indeed possible with a weak field ligand. Consider $\ce{[CoCl6]^{4-}}$, for example.