Question

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

• A. 5.92
• B. 5.26
• C. Zero
• D. 5.0

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 :
$M{n}^{2+}:\left[Ar\right]3d^5$


In aqueous solution it exists as ${\left[Mn{\left(H_{2}O\right)}_6\right]}^{2+}$ . Since $H_{2}O$ is a weak field ligand, it does not cause pairing of unpaired electrons .

So , its spin only magnetic moment is


$\mu =\sqrt{n(n+2)}=\sqrt{5\times 7}=5.92B.M$