Question

magnetic moment of Fe3+

Answer 1

U=√(n(n+2))
=√(5(5+2))
=√35 BM or 5.9 BM

n= Number of unpaired electrons in outer most orbital.

BM stands for Bohr Magneton, which is the SI unit for magnetic moment.

Answer 2

Answer: 5.92 BM.

Explanation:

The electronic configuration of $Fe$ with atomic number of 26 is:  

$1s^22s^22p^63s^23p^63d^64s^2$

The electronic configuration of $Fe^{3+}$ formed by loss of three electrons is:  

$1s^22s^22p^63s^23p^63d^5$ containing 5 unpaired electrons.

Formula used for magnetic moment :

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

where,

$\mu$ = magnetic moment

n = number of unpaired electrons  = 5

Now put all the given values in the above formula, we get the unpaired electrons.

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

$\mu=\sqrt{5(5+2)}=5.92BM$

Therefore, the magnetic moment of $Fe^{3+}$ is 5.92 BM.