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Q. Magnetic moment of $Cr^{2+}$ is nearest to

AIIMSAIIMS 2012The d-and f-Block Elements

Solution:

Cr2+ = 3d4, No. of unpaired electrons (n) = 4
Magnetic moment $=\sqrt{n\left(n+2\right)}$ BM
$\quad\quad\quad\quad=\sqrt{4\left(4+2\right)}=\sqrt{24}=4.89$ BM
Fe2+ = 3d6, No. of unpaired electrons (n) = 4
Magnetic moment $=\sqrt{4\left(4+2\right)}$ BM
$\quad\quad\quad\quad\quad\quad\quad=\sqrt{24}=4.89$ BM
Mn2+ = 3d5, No. of unpaired electrons (n) = 5
Magnetic moment $=\sqrt{5\left(5+2\right)}$ BM
$\quad\quad\quad\quad\quad\quad\quad=\sqrt{35}=5.91$ BM
Co2+ = 3d7, No. of unpaired electrons (n) = 3
Magnetic moment $=\sqrt{3\left(3+2\right)}$ BM
$\quad\quad\quad\quad\quad\quad\quad=\sqrt{15}=3.87$ BM
Ni2+ = 3d8, No. of unpaired electrons (n) = 2
Magnetic moment $=\sqrt{2\left(2+2\right)}$ BM
$\quad\quad\quad\quad\quad\quad\quad=\sqrt{8}=2.82$ BM