Question

Use hund's rule to derive the electronic configuration of $C{e}^{+3}$ ion, and calculate its magnetic moment on the basis of 'spin only' formula. ( For Ce, Z = 58)

Answer 1

Electronic configuration of $Ce:$

$Ce:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{1}5d^{1}6s^2$

$\therefore$ Electronic configuration of $C{e}^{+3}$ using hund's rule is,

$C{e}^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^1$

or, $C{e}^{3+}{:}_{54}\left[Xe\right]4f^1$

No. of unpaired electrons in $C{e}^{+3}=1=n$

$\therefore$ Magnetic moment, $\mu =\sqrt{n(n+2)}$ $B.M$

$⟹\mu =\sqrt{1(1+2)}$ $B.M$

$⟹\mu =\sqrt{3}$ $B.M$

$⟹\mu =1.73$ $B.M$