A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted the following observations:
(i) A and D both form blue-violet colour with ninhydrin.
(ii) Lassaigne extract of C gives positive $$AgNO_3$$ test and negative $$Fe_4[Fe(CN)_6]_3$$ test.
(iii) Lassaigne extract of B and D gives positive sodium nitroprusside test.
Based on these observations which option is correct?

We have four unknown samples A, B, C and D which are known to be taken from the list Aspartame, Saccharin, Sucralose and Alitame. The task is to match each unknown with the correct sweetener by using the classical qualitative tests mentioned in the statement.

First we look at the ninhydrin observation. The ninhydrin reagent gives a characteristic blue-violet (Ruhemann’s purple) colour whenever a compound possesses a free $$\;-NH_2\;$$ or $$\;-NH-\;$$ group that is α to a carbonyl, exactly as found in amino acids and peptides. Therefore

$$\text{blue-violet colour with ninhydrin}\;\Longrightarrow\;\text{compound contains an amino (peptide) group}.$$

The data say that both A and D give this colour. In the family of artificial sweeteners provided:

Aspartame $$\bigl($$L-Aspartyl-L-phenylalanine-methyl-ester$$\bigr)$$ is a dipeptide; it definitely contains a free $$\;-NH_2\;$$ group.
Alitame $$\bigl($$L-Aspartyl-D-alanine amide of a thietanyl group$$\bigr)$$ is also a peptide derivative and possesses the same group.
Saccharin is a sulfonimide and has no free amino group.
Sucralose is a chlorinated sugar without any nitrogen at all.

Hence

$$A,D \;\text{(ninhydrin +)} \;\;\Rightarrow\;\; A,D \in \{\text{Aspartame, Alitame}\}.$$

Now we pass to sample C. Its Lassaigne filtrate shows

(i) Positive silver nitrate test: $$Ag^+ + X^- \longrightarrow AgX\downarrow\quad(X = Cl^-,Br^-,I^-).$$

(ii) Negative Prussian blue test with $$Fe_4[Fe(CN)_6]_3,$$ which is the classical confirmation for nitrogen.

So C contains a halogen but does not contain nitrogen. Surveying the list:

Sucralose ≡ C$$_{12}$$H$$_{19}$$Cl$$_3$$O$$_8$$ (three chlorine atoms, no nitrogen) ✔️
Aspartame, Saccharin and Alitame all contain nitrogen (and no halogen).

Therefore

$$C = \text{Sucralose}.$$

Finally, the Lassaigne extract of B and D gives a positive sodium nitroprusside test. The underlying reaction is

$$Na_2[Fe(CN)_5NO] + S^{2-} \;\longrightarrow\; [Fe(CN)_5NOS]^{4-}\;(\text{violet}).$$

Hence sulfur is present in both B and D.

Looking again at the molecular compositions:

Saccharin - contains a sulfonyl (S=O)$$_2$$ group ✔️
Alitame - contains a thietane (four-membered ring with S) ✔️
Aspartame - no sulfur ✖️
Sucralose - no sulfur ✖️

We already fixed $$D$$ as either Aspartame or Alitame, but sulfur positivity immediately removes Aspartame from $$D$$. Therefore

$$D = \text{Alitame}.$$

Because $$D$$ is Alitame, the other ninhydrin-positive sample $$A$$ must be the remaining peptide sweetener, namely Aspartame:

$$A = \text{Aspartame}.$$

We have also determined $$C = \text{Sucralose}$$. The only sweetener still unassigned is Saccharin, so that must be $$B$$; moreover $$B$$ was indeed sulfur-positive, perfectly consistent:

$$B = \text{Saccharin}.$$

The complete matching is therefore

$$ \begin{aligned} A &\;:\; \text{Aspartame},\\ B &\;:\; \text{Saccharin},\\ C &\;:\; \text{Sucralose},\\ D &\;:\; \text{Alitame}. \end{aligned} $$

This ordering is exactly the one listed in Option A.

Hence, the correct answer is Option A.