Q. Number of electrons in 3$d$-orbital of $V^{2 +},\ {Cr}^{2 +},\ {Mn}^{2 +}$, and ${Fe}^{2 +}$ are 3, 4, 5 and 6 respectively. Which of the following ions will have largest value of magnetic moment $(\mu)$?
A) $V^{2 +}$
B) ${Cr}^{2 +}$
C) ${Mn}^{2 +}$
D) ${Fe}^{2 +}$
Explanation
(c)
Magnetic moment $(\mu) = \sqrt{n(n + 2)}\$BM$\$where,$\$'$n$' is the number of unpaired electrons.
_{23}$V^{2 +}$=[Ar]$3d^{3}$ ($n$=3)
_{24}${Cr}^{2 +}$=[Ar]$3d^{4}$ ($n$=4)
_{25}${Mn}^{2 +}$=[Ar]$3d^{5}$ ($n$=5)
_{26}${Fe}^{2 +}$=[Ar]$3d^{6}$ ($n$=4)
Hence magnetic moment will be maximum for ${Mn}^{2 +}$ (equal to 5.92 BM).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.