$$\dot{C}l + CH_4 \rightarrow A + B$$. $$A$$ and $$B$$ in the above atmospheric reaction step are

We need to identify the products A and B in the atmospheric reaction: $$\dot{C}l + CH_4 \rightarrow A + B$$

The reaction involves a chlorine free radical ($$\dot{C}l$$) reacting with methane ($$CH_4$$). This is the propagation step of the free radical substitution (halogenation) of methane.

In the propagation step of free radical halogenation, the chlorine radical abstracts a hydrogen atom from methane:

The chlorine radical ($$\dot{C}l$$) has an unpaired electron and is highly reactive. It attacks the C-H bond in methane, removing a hydrogen atom to form HCl (a stable molecule), while leaving behind a methyl radical ($$\dot{C}H_3$$).

From the reaction, $$A = \dot{C}H_3$$ (methyl radical) and $$B = HCl$$ (hydrogen chloride).

Option A: $$C_2H_6$$ and $$Cl_2$$ — Incorrect. This would require recombination, not a propagation step.

Option B: $$\dot{C}HCl_2$$ and $$H_2$$ — Incorrect. The chlorine radical abstracts H, it does not insert into the C-H bond.

Option C: $$\dot{C}H_3$$ and $$HCl$$ — Correct. This matches the propagation step.

Option D: $$C_2H_6$$ and $$HCl$$ — Incorrect. $$C_2H_6$$ cannot form from a single methane molecule in one step.

Hence, the correct answer is Option C: $$\dot{C}H_3$$ and $$HCl$$.