1. Write the electronic configuration of neutral Nd

Neodymium, $\mathrm{Nd}$, has atomic number $60$.

Its ground-state electronic configuration is:

$$\mathrm{Nd} = [Xe] \, 4f^4 6s^2$$

2. Forming $\mathrm{Nd^{2+}}$

To form $\mathrm{Nd^{2+}}$, two electrons are removed from the neutral atom.

In lanthanides, the outermost electrons are removed first, i.e. the $6s$ electrons are lost before the $4f$ electrons.

So,

$$\mathrm{Nd} = [Xe] \, 4f^4 6s^2$$

Removing two $6s$ electrons gives:

$$\mathrm{Nd^{2+}} = [Xe] \, 4f^4$$

3. Match with the options

• A: $\mathrm{4f^4 6s^2}$ → neutral Nd, not $\mathrm{Nd^{2+}}$
• B: $\mathrm{4f^3}$ → incorrect
• C: $\mathrm{4f^4}$ → correct
• D: $\mathrm{4f^2 6s^2}$ → incorrect

4. Final Answer

Therefore,

$$\boxed{\mathrm{Nd^{2+}} = 4f^4}$$

So the correct option is C.