Sodium extract is heated with concentrated $$HNO_3$$ before testing for halogens because:

In Lassaigne’s test we first fuse the organic compound with metallic sodium. During this fusion the different hetero-elements present in the compound are converted into simple, water-soluble sodium salts:

$$\text{Halogen (X)} \; \rightarrow \; NaX$$ $$\text{Sulphur (S)} \; \rightarrow \; Na_2S$$ $$\text{Nitrogen (N)} \; \rightarrow \; NaCN$$

After the fusion, the water extract that contains these ions is called the sodium extract or Lassaigne filtrate. To detect halide ions we ordinarily add aqueous silver nitrate. The analytical reaction we depend upon is

$$Ag^+ + X^- \rightarrow AgX\downarrow \quad (\text{white, pale-yellow or yellow precipitate})$$

However, if the ions $$S^{2-}$$ or $$CN^-$$ are still present, they also give precipitates with silver ions:

$$2\,Ag^+ + S^{2-} \rightarrow Ag_2S\downarrow \;(\text{black})$$ $$Ag^+ + CN^- \rightarrow AgCN\downarrow \;(\text{white})$$

The black $$Ag_2S$$ or the white $$AgCN$$ would be mistaken for the halide precipitate and thus interfere with the test. Therefore, before we add silver nitrate we deliberately destroy $$S^{2-}$$ and $$CN^-$$ by boiling the sodium extract with concentrated nitric acid.

First we acidify:

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

Then the strongly oxidising, hot $$HNO_3$$ converts sulphide further to sulphur or sulphate; a simplified oxidation step can be written as

$$3\,H_2S + 8\,HNO_3 \rightarrow 3\,S\downarrow + 8\,NO_2\uparrow + 4\,H_2O$$

In the same way, cyanide is oxidised and finally broken down into harmless gaseous products:

$$2\,NaCN + 4\,HNO_3 \rightarrow 2\,NaNO_3 + 2\,CO_2\uparrow + 2\,NO\uparrow + H_2O$$

Because $$S^{2-}$$ and $$CN^-$$ are completely destroyed, no $$Ag_2S$$ or $$AgCN$$ can form later. The halide ions, on the other hand, remain unaffected by concentrated nitric acid, so they alone precipitate with $$AgNO_3$$ and the test becomes reliable.

Thus the reason for heating the sodium extract with concentrated $$HNO_3$$ is that the possible interfering ions $$S^{2-}$$ and $$CN^-$$ are decomposed and cannot give misleading precipitates.

Hence, the correct answer is Option C.