Question

Magnetic moment of $A^{3+}$ ion is $5.48\times {10}^{-23}J/T$ (and $9.27\times {10}^{-24}J/T=1BM$). If $A$ belongs to $4$th period, then element $A$ is:

• A. $Fe$
• B. $Ni$
• C. $Co$
• D. $M$

Answer 1

The correct option is A $Fe$

Magnetic moment $=5.48\times {10}^{-23}J/T$
$\frac{5.48\times {10}^{-23}}{9.27\times {10}^{-24}}=5.92BM$
Also as we know,
magnetic moment $=\sqrt{n(n+2)}$
$\therefore \sqrt{n(n+2)}=5.92=\sqrt{35}$
$\therefore n=5$,i.e., no.of unpaired electron in $A^{3+}=5$
$\therefore$ Element is ${}_{26}\text{Fe}$ and
${}_{26}{\text{Fe}}^{3+}$ is $:1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5$.