In the molecular orbital diagram for the molecular ion, N$$_2^+$$, the number of electrons in the $$\sigma_{2p}$$ molecular orbital is:

We begin by counting the electrons that must be accommodated in the molecular orbitals. Each nitrogen atom has atomic number 7, so its ground-state electronic configuration is $$1s^2\,2s^2\,2p^3.$$

In the neutral molecule $$\mathrm{N_2}$$ the total number of electrons is therefore $$2 \times 7 \;=\;14.$$ For the molecular ion $$\mathrm{N_2^{+}}$$ one electron is removed, leaving $$14-1 \;=\;13$$ electrons to place in the molecular orbitals.

For homonuclear diatomic molecules made of atoms with atomic number $$Z\le 7$$ (that is, for $$\mathrm{B_2},\; \mathrm{C_2},\; \mathrm{N_2}$$ and their ions), experimental evidence shows that the energy ordering of the valence molecular orbitals arising from the $$2s$$ and $$2p$$ atomic orbitals is

$$\sigma_{2s}\; <\; \sigma_{2s}^\* \; <\; \pi_{2p_x} = \pi_{2p_y}\; <\; \sigma_{2p_z}\; <\; \pi_{2p_x}^\* = \pi_{2p_y}^\*\; <\; \sigma_{2p_z}^\*.$$

We now fill these molecular orbitals with the 13 electrons, obeying the Aufbau principle (lower energy orbitals are filled first), the Pauli exclusion principle (maximum two electrons per orbital with opposite spins) and Hund’s rule (for degenerate orbitals, electrons remain unpaired as long as possible).

Step-by-step filling

1. $$\sigma_{1s}$$ receives $$2$$ electrons → total used $$2$$.
2. $$\sigma_{1s}^\*$$ receives $$2$$ electrons → total used $$4$$.
3. $$\sigma_{2s}$$ receives $$2$$ electrons → total used $$6$$.
4. $$\sigma_{2s}^\*$$ receives $$2$$ electrons → total used $$8$$.
5. The degenerate pair $$\pi_{2p_x}$$ and $$\pi_{2p_y}$$ together hold $$4$$ electrons (2 in each) → total used $$12$$.
6. We have now placed 12 electrons; one electron still remains (because 13 are required). The next orbital in the energy ladder is $$\sigma_{2p_z}$$, so this remaining electron enters that orbital.

Thus the population of the $$\sigma_{2p}$$ bonding molecular orbital (specifically $$\sigma_{2p_z}$$) in $$\mathrm{N_2^{+}}$$ is

$$1\;\text{electron}.$$

Since the options list the possible counts of electrons in $$\sigma_{2p}$$ as 0, 2, 3 and 1, we select the value 1.

Hence, the correct answer is Option D.