Question

Which of the following ion is more paramagnetic?

• A. $L{a}^{3+}(Z=57)$
• B. $L{u}^{3+}(Z=71)$
• C. $Y{b}^{3+}(Z=70)$
• D. $S{m}^{3+}(Z=62)$

Answer 1

The correct option is D $S{m}^{3+}(Z=62)$

Both $Y{b}^{3+}$ and $S{m}^{3+}$ contain unpaired electrons. Hence, they are paramagnetic.

On the other hand, no unpaired electron is present in $L{a}^{3+}$ and $L{u}^{3+}$. Hence, they are diamagnetic.

The electronic configurations are as follow:

$Y{b}^{3+}(X=70)\left[Xe\right]4f^{13}5d^{0}6s^0$ contains 1 unpaired electron.

$S{m}^{3+}(X=62)\left[Xe\right]4f^{5}5d^{0}6s^0$ contains 5 unpaired electrons.

$L{a}^{3+}(X=57)\left[Xe\right]4f^{0}5d^{0}6s^0$ contains 0 unpaired electrons.

$L{u}^{3+}(X=71)\left[Xe\right]4f^{14}5d^{0}6s^0$ contains 0 unpaired electrons.