The Lassaigne's extract is boiled with dil $$HNO_3$$ before testing for halogens because,

In the Lassaigne's test for halogens, the sodium fusion extract may contain $$Na_2S$$ (if sulphur is present) and $$NaCN$$ (if nitrogen is present) along with $$NaX$$ (sodium halide).

When silver nitrate is added to test for halogens, both $$Na_2S$$ and $$NaCN$$ would also react with $$AgNO_3$$:

$$Na_2S + 2AgNO_3 \rightarrow Ag_2S \downarrow (\text{black}) + 2NaNO_3$$

$$NaCN + AgNO_3 \rightarrow AgCN \downarrow (\text{white}) + NaNO_3$$

These precipitates ($$Ag_2S$$ and $$AgCN$$) would interfere with the detection of silver halides.

To avoid this interference, the Lassaigne's extract is first boiled with dilute $$HNO_3$$. The dilute nitric acid decomposes $$Na_2S$$ and $$NaCN$$:

$$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow$$

$$NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow$$

The gases $$H_2S$$ and $$HCN$$ escape on boiling, eliminating the interference. After this treatment, only the halide ions remain in solution and can be detected cleanly with $$AgNO_3$$.

The correct answer is Option 4: $$Na_2S$$ and $$NaCN$$ are decomposed by $$HNO_3$$.