The pair(s) of diamagnetic ions is(are)

The correct options are:
1. ${\mathrm{La}}^{\mathrm{3+}},{\mathrm{Ce}}^{\mathrm{4+}}$
2. ${\mathrm{Yb}}^{\mathrm{2+}},{\mathrm{Lu}}^{\mathrm{3+}}$

To determine which pair(s) of ions are diamagnetic, we need to examine their electronic configurations. An ion is diamagnetic if it has no unpaired electrons (i.e., all of its electronic shells or subshells are either completely filled or completely empty). If there are any unpaired electrons, the ion is paramagnetic.

Let us write down the electronic configurations of the lanthanide elements involved, followed by their corresponding ionic states:

1. Lanthanum (La, Z = 57):
The ground-state electronic configuration of neutral La is:
$\left[\mathrm{Xe}\right]5d^{1}6s^2$
For the ${\mathrm{La}}^{\mathrm{3+}}$ ion, three electrons are removed (two from the 6s orbital and one from the 5d orbital):
${\mathrm{La}}^{\mathrm{3+}}=\left[\mathrm{Xe}\right]4f^0$
Since the $4f$ subshell is completely empty ($f^0$), there are 0 unpaired electrons. Therefore, ${\mathrm{La}}^{\mathrm{3+}}$ is diamagnetic.

2. Cerium (Ce, Z = 58):
The ground-state electronic configuration of neutral Ce is:
$\left[\mathrm{Xe}\right]4f^{1}5d^{1}6s^2$
For the ${\mathrm{Ce}}^{\mathrm{4+}}$ ion, four electrons are removed (two from 6s, one from 5d, and one from 4f):
${\mathrm{Ce}}^{\mathrm{4+}}=\left[\mathrm{Xe}\right]4f^0$
Since the $4f$ subshell is completely empty ($f^0$), there are 0 unpaired electrons. Therefore, ${\mathrm{Ce}}^{\mathrm{4+}}$ is diamagnetic.

Hence, the pair ${\mathrm{La}}^{\mathrm{3+}},{\mathrm{Ce}}^{\mathrm{4+}}$ is diamagnetic.

3. Ytterbium (Yb, Z = 70):
The ground-state electronic configuration of neutral Yb is:
$\left[\mathrm{Xe}\right]4f^{14}6s^2$
For the ${\mathrm{Yb}}^{\mathrm{2+}}$ ion, two electrons are removed from the 6s orbital:
${\mathrm{Yb}}^{\mathrm{2+}}=\left[\mathrm{Xe}\right]4f^{14}$
Since the $4f$ subshell is completely filled with 14 electrons ($f^{14}$), all electrons are paired, leaving 0 unpaired electrons. Therefore, ${\mathrm{Yb}}^{\mathrm{2+}}$ is diamagnetic.

4. Lutetium (Lu, Z = 71):
The ground-state electronic configuration of neutral Lu is:
$\left[\mathrm{Xe}\right]4f^{14}5d^{1}6s^2$
For the ${\mathrm{Lu}}^{\mathrm{3+}}$ ion, three electrons are removed (two from 6s and one from 5d):
${\mathrm{Lu}}^{\mathrm{3+}}=\left[\mathrm{Xe}\right]4f^{14}$
Since the $4f$ subshell is completely filled ($f^{14}$), there are 0 unpaired electrons. Therefore, ${\mathrm{Lu}}^{\mathrm{3+}}$ is diamagnetic.

Hence, the pair ${\mathrm{Yb}}^{\mathrm{2+}},{\mathrm{Lu}}^{\mathrm{3+}}$ is also diamagnetic.