1. Test being used

   The sodium carbonate extract is first acidified with dilute $\mathrm{HNO_3}$ and then treated with $\mathrm{AgNO_3}$. This is the standard test for halide ions.

   The relevant precipitates are:

   $\mathrm{AgCl} \;\text{white}$ $\mathrm{AgBr} \;\text{pale\ yellow / cream}$ $\mathrm{AgI} \;\text{yellow}$

2. Solubility in $\mathrm{NH_4OH}$

   Their behavior with ammonium hydroxide is:

   • $\mathrm{AgCl}$: soluble readily in dilute $\mathrm{NH_4OH}$
   • $\mathrm{AgBr}$: soluble with difficulty in concentrated $\mathrm{NH_4OH}$
   • $\mathrm{AgI}$: insoluble in $\mathrm{NH_4OH}$

3. Match with observation

   The observation given is:

   • pale yellow precipitate
   • soluble with difficulty in $\mathrm{NH_4OH}$

   This matches silver bromide, $\mathrm{AgBr}$.

   Therefore the acidic radical present is:

   $\mathrm{Br^-}$

4. Check options

   • A: $\mathrm{I^-}$ $\rightarrow$ gives yellow $\mathrm{AgI}$, insoluble in $\mathrm{NH_4OH}$ ❌
   • B: $\mathrm{CO_3^{2-}}$ $\rightarrow$ removed on acidification with $\mathrm{HNO_3}$ ❌
   • C: $\mathrm{Cl^-}$ $\rightarrow$ gives white $\mathrm{AgCl}$, readily soluble in $\mathrm{NH_4OH}$ ❌
   • D: $\mathrm{Br^-}$ $\rightarrow$ gives pale yellow $\mathrm{AgBr}$, soluble with difficulty in $\mathrm{NH_4OH}$ ✅

Hence, the correct answer is D.