Number of molecules having bond order 2 from the following molecules is _____ : $$C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$$

We need to find the number of molecules with bond order 2 from: $$C_2, O_2, Be_2, Li_2, Ne_2, N_2, He_2$$.

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 bonding electrons and $$N_a$$ is the number of antibonding electrons.

Since $$C_2$$ (12 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\pi_{2p})^4$$, the bond order is $$(8-4)/2 = 2$$. ✓

Now $$O_2$$ (16 electrons) has configuration $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^2$$, so the bond order is $$(10-6)/2 = 2$$. ✓

For $$Be_2$$ (8 electrons), the bond order is $$(4-4)/2 = 0$$. ✗

For $$Li_2$$ (6 electrons), the bond order is $$(4-2)/2 = 1$$. ✗

For $$Ne_2$$ (20 electrons), the bond order is $$(10-10)/2 = 0$$. ✗

For $$N_2$$ (14 electrons), the bond order is $$(10-4)/2 = 3$$. ✗

For $$He_2$$ (4 electrons), the bond order is $$(2-2)/2 = 0$$. ✗

Therefore the molecules with bond order 2 are $$C_2$$ and $$O_2$$.

This gives the answer $$\boxed{2}$$.