The pair of lanthanoid ions which are diamagnetic is

The correct answer is Ce⁴⁺ and Yb²⁺.

To determine which pair of lanthanoid ions is diamagnetic, we need to analyze their electronic configurations. An ion is diamagnetic if it contains no unpaired electrons (i.e., all of its electronic shells are either completely filled or completely empty).

Let us write down the electronic configurations of the ions in the correct option:

1. Cerium ion (Ce⁴⁺):
The atomic number of Cerium (Ce) is 58.
Its ground state electronic configuration is:

$\text{Ce}=\left[\text{Xe}\right]4{\text{f}}^{1}5{\text{d}}^{1}6{\text{s}}^2$

To form the tetravalent Ce⁴⁺ ion, it loses four electrons (two from the 6s orbital, one from the 5d orbital, and one from the 4f orbital):

${\text{Ce}}^{4+}=\left[\text{Xe}\right]4{\text{f}}^0$

Since the 4f orbital is completely empty (f⁰ configuration), there are no unpaired electrons present. Therefore, Ce⁴⁺ is diamagnetic.

2. Ytterbium ion (Yb²⁺):
The atomic number of Ytterbium (Yb) is 70.
Its ground state electronic configuration is:

$\text{Yb}=\left[\text{Xe}\right]4{\text{f}}^{14}6{\text{s}}^2$

To form the divalent Yb²⁺ ion, it loses two electrons from the outer 6s orbital:

${\text{Yb}}^{2+}=\left[\text{Xe}\right]4{\text{f}}^{14}$

Since the 4f subshell is completely filled (f¹⁴ configuration), all 14 electrons are fully paired in their orbitals. Therefore, Yb²⁺ is also diamagnetic.

Because both Ce⁴⁺ and Yb²⁺ have closed-shell electronic configurations with zero unpaired electrons, the pair is diamagnetic.