Question

Which of the following lanthanoid ions is diamagnetic? (Atoms, $Ce=58,Sm=62,Eu=63,Yb=70$)

• A. ${Yb}^{2+}$
• B. ${Ce}^{2+}$
• C. ${Sm}^{2+}$
• D. ${Eu}^{2+}$

Answer 1

Answer: (A)

${Yb}^{2+}$

Lanthanide ion with no unpaired electron is diamagnetic in nature.

$C{e}_{58}$ =$4f^{2}5d^{0}6s^2$

$C{e}^{2+}=\left[xe\right]4f^2$

$S{m}_{62}=4f^{6}5d^{0}6s^2$

$s{m}^{+}=\left[xe\right]4f^6$

$E{u}_{63}=\left[xe\right]4f^{7}5d^{0}6s^2$

$E{u}^{+}=\left[xe\right]4f^7\left(sevenunpairedelectron\right)$

$Y{b}_{70}=\left[xe\right]4f^{14}5d^{0}6s^2$

$Y{b}^{+}=\left[xe\right]4f^{14}\left(Nounpairedelectron\right)$

Hence $Y{b}^{+}$ is diamagnetic, due to the absence of the unpaired electron.