Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:
$$CH_4 + I_2 \underset{\text{Reversible}}{\overset{h\nu}{\rightleftharpoons}} CH_3 - I + HI$$

First, let us recall why the photochemical iodination of methane is reversible. After the forward photochemical step

$$CH_4 + I_2 \xrightarrow{h\nu} CH_3I + HI,$$

the hydrogen iodide that is formed is a very strong reducing agent. It immediately reacts back with the product methyl iodide and reproduces the starting materials:

$$CH_3I + HI \longrightarrow CH_4 + I_2.$$

This back-reaction is fast, so the overall process settles at an equilibrium and the arrow becomes $$\rightleftharpoons.$$ To convert the system into an irreversible one we must remove, destroy or otherwise make $$HI$$ unavailable as soon as it appears; then the reverse step cannot occur and the reaction will keep moving to the right until one of the reactants is exhausted.

Now we test each reagent given in the options to see whether it can do this job.

Option A : $$HOCl$$ (hypochlorous acid)
$$HOCl$$ can indeed oxidise iodide ions, but in doing so it itself gets reduced to $$HCl$$ and water, and it works only under very dilute, neutral media. In the strongly acidic medium in which $$HI$$ exists, $$HOCl$$ disproportionates rapidly and does not survive long enough to consume all the $$HI$$ formed continuously in the photochemical reaction. Hence the equilibrium is not driven completely to the right.

Option B : dilute $$HNO_2$$ (nitrous acid)
$$HNO_2$$ is a mild oxidising agent. The reaction $$2\,HI + HNO_2 \;\longrightarrow\; I_2 + NO + H_2O$$ is slow and incomplete; the small amount of $$HI$$ removed is quickly replenished by the back-reaction. Thus the system still remains reversible.

Option C : liquid $$NH_3$$
Being a base, ammonia neutralises hydrogen iodide : $$NH_3 + HI \;\longrightarrow\; NH_4I.$$ Although free $$HI$$ disappears, the iodide ion $$(I^-)$$ that stays in the medium keeps reducing $$CH_3I$$ by $$S_N2$$ attack to reproduce methane: $$CH_3I + I^- \;\longrightarrow\; CH_3^- + I_2 \;\longrightarrow\; CH_4 + I_2.$$ Therefore the reverse path is still available and the reaction is not rendered irreversible.

Option D : concentrated $$HIO_3$$ (iodic acid)
Concentrated iodic acid is a much stronger oxidising agent for hydrogen iodide. The well-known redox reaction is

$$5\,HI + HIO_3 \;\longrightarrow\; 3\,I_2 + 3\,H_2O.$$

Observe each consequence carefully:

• The freshly produced $$HI$$ is immediately destroyed, so its concentration in the reaction vessel remains practically zero.
• Because $$HI$$ never accumulates, the reverse step $$CH_3I + HI \rightarrow CH_4 + I_2$$ becomes impossible.
• The extra $$I_2$$ formed in the redox process simply joins the starting $$I_2$$ pool and continues to react with more $$CH_4$$ under the influence of light, producing still more $$CH_3I$$. In other words, the presence of $$HIO_3$$ continuously pulls the overall reaction to the right.

Therefore, concentrated $$HIO_3$$ removes the key species $$HI$$ as soon as it forms, blocking the backward path and converting the equilibrium into a one-way (irreversible) reaction.

Hence, the correct answer is Option D.