Q. Least paramagnetic property is shown by

A) Fe
B) Mn
C) Ni
D) Cu
Explanation

(d)
The element having unpaired electron is paramagnetic. More the number of unpaired electrons, more will be paramagnetic character.
Mn (25)=$1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{2},3d^{5}$
$\therefore\$5 unpaired electrons
Fe (26)= $1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{2},3d^{6}$
$\therefore\$4 unpaired electrons
Ni (28)= $1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{2},3d^{8}$
$\therefore\$2 unpaired electrons
Cu (29)= $1s^{2},2s^{2},2p^{6},3s^{2},3p^{6},4s^{2},3d^{10}$
$\therefore\$1 unpaired electrons
$\therefore\$Mn has maximum and Cu has least paramagnetic property.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.