Question:
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25 , in it's aqueous solution?
Correct Option: 1
Solution:
Electronic configuration of divalent metal ion having atomic number 25 is
Total number of unpaired electrons $=5$
$\mu($ Magnetic moment $)=\sqrt{\mathrm{n}(\mathrm{n}+2)} \mathrm{BM}$
where $\mathrm{n}=$ number of unpaired $\mathrm{e}^{-}$
$\therefore \mu=\sqrt{5(5+2)}=\sqrt{35} \mathrm{BM}=5.92 \mathrm{BM}$