JEE Main 2021 — D And F Block Elements Question with Solution

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

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

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

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

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

There are 8 unpaired electrons.

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

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

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

contain six unpaired electrons.

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

Hence, order of spin only magnetic moment

$$B>C>A$$.