Question

The paramagnetic lanthanoid ion among the following is:

• A. $C{e}^{4+}$
• B. $Y{b}^{2+}$
• C. $L{u}^{3+}$
• D. $S{m}^{2+}$

Answer 1

The correct option is D $S{m}^{2+}$

$Yb⟶$ atomic number=$70$

Electronic configuration=$Xe4f^{14}6s^2$

$Y{b}^{2+}=Xe4f^{14}6s^0$ (fully filled orbitals).

$Ce⟶$ atomic number=$58$

Electronic configuration=$Xe4f^{1}5d^{1}6s^2$

$C{e}^{4+}=Xe$ (full filled orbitals).

$Lu⟶$ atomic number=$71$

Electronic configuration=$Xe4f^{14}5d^{1}6s^2$

$L{u}^{3+}=Xe4f^{14}$.

$Sm⟶$ atomic number=$62$

Electronic configuration=$Xe4f^{6}6s^2$

$S{m}^{2+}=Xe4f^{6}6s^0$.

(Refer to Image)

$S{m}^{2+}$ having $6$ unpaired $e^{-}s$ in $4f$ subshell is paramagnetic.