Question

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

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

Answer 1

The correct option is A 5.92

Answer :(a)
${}_{25}Mn-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$

$M{n}^{2+}-1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^5$

$\text{Spin-only magnetic moment}=\sqrt{n(n+2)}BM$

where n is the number of unpaired electron.
Here, n = 5
Hence,
$\text{Magnetic moment}=\sqrt{5(5+2)}=\sqrt{35}$
$\text{Magnetic moment}=5.92BM$