The correct statement with respect to dinitrogen is:

First, let us recall the electronic configuration of a nitrogen atom. A single N atom has the ground-state configuration $$1s^2\,2s^2\,2p^3$$.

When two nitrogen atoms combine to give the dinitrogen molecule $$N_2$$, we construct its molecular orbital (MO) diagram. For homonuclear diatomic molecules up to atomic number 14, the order of MOs is

$$\sigma(1s),\; \sigma^*(1s),\; \sigma(2s),\; \sigma^*(2s),\; \pi(2p_x)=\pi(2p_y),\; \sigma(2p_z),\; \pi^*(2p_x)=\pi^*(2p_y),\; \sigma^*(2p_z).$$

Each nitrogen contributes 7 electrons, so $$N_2$$ has $$2 \times 7 = 14$$ electrons in total. We now fill the MOs in the order above:

$$\sigma(1s)^2,\; \sigma^*(1s)^2,\; \sigma(2s)^2,\; \sigma^*(2s)^2,\; \pi(2p_x)^2,\; \pi(2p_y)^2,\; \sigma(2p_z)^2.$$

There are no electrons left for any antibonding $$\pi^*$$ or $$\sigma^*$$ orbitals. Hence all electrons are paired.

According to MO theory, a molecule with all paired electrons is diamagnetic, not paramagnetic. Therefore the statement “$$N_2$$ is paramagnetic in nature” is wrong, making Option A incorrect.

Next, we look at the reactivity toward dioxygen $$O_2$$. At ordinary temperature ($$25^\circ\text{C}$$) the reaction

$$N_2 + O_2 \rightarrow 2\,NO$$

does not occur to any measurable extent because the triple bond in $$N_2$$ (bond order 3) is very strong (bond dissociation energy ≈ $$941\ \text{kJ mol}^{-1}$$). Notice that appreciable formation of $$NO$$ requires an electric spark or very high temperature (≈ $$2000-3000^\circ\text{C}$$). Hence the claim that $$N_2$$ “can combine with dioxygen at $$25^\circ\text{C}$$” is false. So Option B is wrong.

Liquid dinitrogen (commonly called liquid nitrogen) boils at $$77\ \text{K}$$. Owing to this very low temperature it is widely used for cryosurgery, cryopreservation, and instant tissue freezing. The given sentence “liquid dinitrogen is not used in cryosurgery” contradicts actual medical practice, so Option C is also incorrect.

Finally, consider whether $$N_2$$ can be used as an inert diluent. Because $$N_2$$ is chemically inert under normal conditions, it is frequently employed to blanket or dilute reactive gases and volatile chemicals, preventing unwanted oxidation or combustion. Common laboratory and industrial procedures (e.g., handling of pyrophoric organometallics) often replace air with a stream of dry nitrogen precisely for this reason. Therefore the statement “it can be used as an inert diluent for reactive chemicals” is correct.

We have examined every option: only Option D withstands scrutiny.

Hence, the correct answer is Option 4.