Question

Calculate the spin only magnetic moment of fe2+ ion

Answer 1

the total unpaired electron electron in Fe +2 is 2 .... and spin multiplicity is equal to (unpaired electron + 1)......so its spin multiplicity is equal to 3 ........ you can check for unpaired electron, coz I'm not sure

Answer 2

Answer : The magnetic moment is 4.89 BM

Solution :

The electronic configuration of Fe with atomic number of 26 is, $1s^22s^22p^63s^23p^63d^64s^2$

The oxidation state of Fe is 2+.

The electronic configuration of $Fe^{2+}$ with atomic number of 24 is, $1s^22s^22p^63s^23p^63d^6$

The number of unpaired electrons in Fe = 4

Formula used for magnetic moment :

$\mu=\sqrt{n(n+2)}$

where,

$\mu$ = magnetic moment

n = number of unpaired electrons

Now put all the given values in the above formula, we get the magnetic moment.

$\mu=\sqrt{4(4+2)}=4.89 BM$

Therefore, the magnetic moment is 4.89 BM