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Q. The value of the "spin only" magnetic moment for one of the following configurations is 2.84 BM. The correct one is :

AIEEEAIEEE 2005

Solution:

(a) $d^5$ in strong field
n = unpaired electron = 1
Magnetic moment $= \sqrt{n\left(n + 2\right)}BM$
$= \sqrt{3} BM = 1.73 \,BM$
$\left(b\right) d^{3}$ in strong/weak field
$n = 3$
Magnetic moment $= \sqrt{15} = 3.87\,BM$
$\left(c\right) d^{4}$ in weak field
$n = 4$
Magnetic moment $=\sqrt{24} = 4.90\,BM$
$\left(d\right) d^{4 }$ in strong field
$n = 2$
Magnetic moment $=\sqrt{8} = 2.83 \,BM$

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