Chemistry - f-block Question with Solution | TestHub

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

In complex ${\mathrm{Ce}}^{4+}\to \left[\mathrm{Xe}\right]4{\mathrm{f}}^{0}5{ \mathrm{d}}^{0}6{ \mathrm{s}}^0$

there is no unpaired electron so ${\mu }_{\mathrm{m}}=0$

(b) ${{}_{64}\mathrm{Gd}}^{3+}\to \left[\mathrm{Xe}\right]4{\mathrm{f}}^{7}5{ \mathrm{d}}^{0}6{ \mathrm{s}}^0$

contain seven unpaired electrons so, ${\mu }_{\mathrm{m}}=\sqrt{7\left(7+2\right)}=\sqrt{63} \mathrm{B}.\mathrm{M}.$

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

Hence, order of spin only magnetic movement 

$\mathrm{b}>\mathrm{c}>\mathrm{a}$