QJEEM21Ch1a3f: JEE Main 2021 Chemistry - Class 12 JEEM21Ch1a3f Solution: Arrange the following metal complex/compounds in the increasing …

(A) ${}_{58}Ce \to [Xe]4{f^2}5{d^0}6{s^2}$

In complex $C{e^{ + 4}} \to [Xe]4{f^0}5{d^0}6{s^0}$

There is no unpaired electron, so, $\mu$ _m = 0 BM

(B) ${}_{64}Gd \to [Xe]4{f^7}5{d^1}6{s^2}$

In complex, ${}_{64}G{d^{2 + }} \to [Xe]4{f^7}5{d^1}$

There are 8 unpaired electrons.

$\therefore$ ${\mu _m} = \sqrt {8(8 + 2)} = \sqrt {80} = 8.94BM$

(C) ${}_{63}Eu \to [Xe]4{f^9}5{d^0}6{s^0}$

In complex ${}_{63}E{u^{ + 3}} \to [Xe]4{f^6}5{d^0}6{s^0}$

contain six unpaired electrons.

So, ${\mu _m} = \sqrt {6(6 + 2)} = \sqrt {48} BM = 6.93BM$

Hence, order of spin only magnetic moment

B > C > A .