Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature? (Given Z for Nd = 60, Yb = 70, La = 57, Ce = 58)

We begin by recalling that the magnetic character of an ion depends on the number of unpaired electrons present. The spin-only magnetic moment of any ion is given by the formula

$$\mu=\sqrt{n(n+2)}\;\text{BM}$$

where $$n$$ represents the number of unpaired electrons. If $$n=0$$, then $$\mu=0$$ and the species is diamagnetic. Hence, to obtain a diamagnetic lanthanide in the $$+2$$ oxidation state we must look for a completely filled 4f subshell (i.e. $$4f^{14}$$) or a completely empty 4f subshell (i.e. $$4f^{0}$$). A fully filled subshell is especially stable and gives $$n=0$$, leading to diamagnetism.

Next, let us write the ground-state electronic configurations of the given lanthanide atoms (using the xenon core $$[\,\text{Xe}\,]$$ for convenience):

$$\text{La (Z = 57)} : [\,\text{Xe}\,]\,5d^{1}6s^{2}$$

$$\text{Ce (Z = 58)} : [\,\text{Xe}\,]\,4f^{1}5d^{1}6s^{2}$$

$$\text{Nd (Z = 60)} : [\,\text{Xe}\,]\,4f^{4}6s^{2}$$

$$\text{Yb (Z = 70)} : [\,\text{Xe}\,]\,4f^{14}6s^{2}$$

When these atoms form the $$+2$$ oxidation state, the two $$6s$$ electrons are removed first (because they are the outermost and highest in energy). We therefore obtain:

$$\text{La}^{2+} : [\,\text{Xe}\,]\,5d^{1}$$

$$\text{Ce}^{2+} : [\,\text{Xe}\,]\,4f^{1}5d^{1}$$

$$\text{Nd}^{2+} : [\,\text{Xe}\,]\,4f^{4}$$

$$\text{Yb}^{2+} : [\,\text{Xe}\,]\,4f^{14}$$

Now we count the unpaired electrons $$n$$ in each case:

• $$\text{La}^{2+}$$ has the configuration $$5d^{1}$$. The single electron in a d-orbital is unpaired, so $$n = 1$$, giving $$\mu = \sqrt{1(1+2)} = \sqrt{3} \ne 0$$. Hence La2+ is paramagnetic.

• $$\text{Ce}^{2+}$$ has one electron in 4f and one in 5d, i.e. $$4f^{1}5d^{1}$$. Thus $$n = 2$$, so $$\mu = \sqrt{2(2+2)} = \sqrt{8} \ne 0$$. Ce2+ is paramagnetic.

• $$\text{Nd}^{2+}$$ possesses the configuration $$4f^{4}$$. Four unpaired electrons give $$n = 4$$, and $$\mu = \sqrt{4(4+2)} = \sqrt{24} \ne 0$$, making Nd2+ paramagnetic.

• $$\text{Yb}^{2+}$$, however, has $$4f^{14}$$, a completely filled 4f subshell. All electrons are paired, so $$n = 0$$ and consequently $$\mu = 0$$. Therefore Yb2+ is diamagnetic.

Comparing all four species, only ytterbium in the $$+2$$ oxidation state satisfies both required conditions: existence of the $$+2$$ oxidation state and diamagnetic behaviour.

Hence, the correct answer is Option B.