Question

What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?

• A. $5.92$
• B. $5.0$
• C. $5.26$
• D. $0$

Answer 1

The correct option is A $5.92$

The element having atomic number 25 is manganese. The electronic configuration of $M{n}^{2+}$ is $M{n}^{2+}:3d^5$

In aqueous solution it exists as $\left[Mn\right(H_{2}O{)}_6{]}^{2+}$. Since $H_{2}O$ is a weak field ligand, it does not cause pairing of unpaired electrons. So, its spin only magnetic moment is
$\mu =\sqrt{n(n+2)}B.M$
n = no. of unpaired electrons.
$\mu =\sqrt{5\times 7}=5.92BM$