Match List-I with List-II.

List-I (Molecule)	List-II (Bond order)
(a) $$Ne_2$$	(i) 1
(b) $$N_2$$	(ii) 2
(c) $$F_2$$	(iii) 0
(d) $$O_2$$	(iv) 3

Choose the correct answer from the options given below:

We need to determine the bond order of each molecule using molecular orbital theory. The bond order is calculated as $$\text{Bond Order} = \frac{N_b - N_a}{2}$$, where $$N_b$$ is the number of electrons in bonding molecular orbitals and $$N_a$$ is the number of electrons in antibonding molecular orbitals.

(a) $$Ne_2$$: Each neon atom has 10 electrons (electronic configuration: $$1s^2 \, 2s^2 \, 2p^6$$), so $$Ne_2$$ has 20 electrons total. The molecular orbital configuration is $$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^2 (\sigma^*_{2p_z})^2$$$. This gives $$N_b = 10$$ and $$N_a = 10$$, so the bond order is $$\frac{10 - 10}{2} = 0$$. This matches (iii).

(b) $$N_2$$: Each nitrogen atom has 7 electrons, so $$N_2$$ has 14 electrons. The molecular orbital configuration is $$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2$$$. Here $$N_b = 10$$ and $$N_a = 4$$, giving bond order $$= \frac{10 - 4}{2} = 3$$. This matches (iv).

(c) $$F_2$$: Each fluorine atom has 9 electrons, so $$F_2$$ has 18 electrons. The molecular orbital configuration is $$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^2$$$. Here $$N_b = 10$$ and $$N_a = 8$$, giving bond order $$= \frac{10 - 8}{2} = 1$$. This matches (i).

(d) $$O_2$$: Each oxygen atom has 8 electrons, so $$O_2$$ has 16 electrons. The molecular orbital configuration is $$$(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^1 (\pi^*_{2p_y})^1$$$. Here $$N_b = 10$$ and $$N_a = 6$$, giving bond order $$= \frac{10 - 6}{2} = 2$$. This matches (ii).

The correct matching is (a)$$\to$$(iii), (b)$$\to$$(iv), (c)$$\to$$(i), (d)$$\to$$(ii), which corresponds to Option (2).