(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

$$B > C > A$$.