Question

What is the magnetic moment for $M{n}^{2+}$ ion?

• A. 5.92 B.M.
• B. 2.80 B.M
• C. 8.95 B.M.
• D. 9.55 B.M.

Answer 1

Answer: (A)

$M{n}^{2+}:3d^5,$ no. of unpaired electrons $(n)=5$
Magnetic moment $(\mu )=\sqrt{n(n+2)}B.M.$
$=\sqrt{5(5+2)}B.M.=\sqrt{35}$
$=5.92B.M.$