0
$\begingroup$

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) & (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?

$\endgroup$
  • $\begingroup$ The total spin for $d^2$ must be S=1 or 0 which using $M=\sqrt{4S(S+1)}$ gives $\sqrt{8}$ or zero, and for $d^7$ the spin is $S=3/2$ or $1/2$ for low spin so $\sqrt{15} $ or $ \sqrt{3}$, so if the answer is (3) you must have an octahedral complex can be either S=1 ( two unpaired electrons) or S=3/2 (three unpaired) with little difference in energy between them. $\endgroup$ – porphyrin Apr 24 at 14:31
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ yes I understand that. But for d7, the value should be CFSE = –0.8Δ + 2P. In the question as the 2P is not given, I feel the configuration can only be restricted to d2. $\endgroup$ – suhridi sen Apr 25 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.