Chemistry - f-block Question with Solution | TestHub

(a) ${}_{58}\mathrm{Ce}\to [\mathrm{Xe}]4{\mathrm{f}}^{2}5{\mathrm{d}}^{0}6{\mathrm{s}}^2$ In complex ${\mathrm{Ce}}^{4+}\to [\mathrm{Xe}]4{\mathrm{f}}^{0}5{\mathrm{d}}^{0}6{\mathrm{s}}^0$ there is no unpaired electron so ${\mu }_{\mathrm{m}}=0$ (b) ${}_{64}{\mathrm{Gd}}^{3+}\to [\mathrm{Xe}]4{\mathrm{f}}^{7}5{\mathrm{d}}^{0}6{\mathrm{s}}^0$ contain seven unpaired electrons so, ${\mu }_{\mathrm{m}}=\sqrt{7(7+2)}=\sqrt{63}\text{ B.M. }$ (c) ${}_{63}{\mathrm{Eu}}^{3+}\to \left[{}_{54}\mathrm{Xe}\right]4{\mathrm{f}}^{6}5{\mathrm{d}}^{0}6{\mathrm{s}}^0$ contain six unpaired electron

so, ${\mu }_{\mathrm{m}}=\sqrt{6(6+2)}=\sqrt{48}$ B.M.

Hence, order of spin only magnetic movement $b>c>a$