Question

The calculated spin magnetic moment of Fe(II) is

• A. 4.90 B.M.
• B. 6.90 B.M.
• C. 5.91 B.M.

Answer 1

The correct option is A 4.90 B.M.

Fe(II) has 4 unpaired electrons.
$\sqrt{n(n+2)}=\sqrt{4(4+2)}$
$=\sqrt{24}$ = 4.90 B. M.