An aqueous solution of a salt X turns blood red on treatment with $$SCN^-$$ and blue on treatment with $$K_4[Fe(CN)_6]$$, X also gives a positive chromyl chloride test. The salt X is:

First we analyse the observations one by one and match them with the behaviour of the salts given in the options.

Observation 1: The aqueous solution of the unknown salt X turns blood red when treated with $$SCN^-$$ ions.

We recall the characteristic test for the ferric ion $$Fe^{3+}$$. The reaction formula is first stated:

$$Fe^{3+} + 3\,SCN^- \rightarrow [Fe(SCN)_3]$$

The complex $$[Fe(SCN)_3]$$ possesses an intense blood-red colour. Therefore, the appearance of a blood-red colour proves that $$Fe^{3+}$$ ions are present in the solution. Hence X must contain iron in the +3 oxidation state.

Observation 2: The same solution turns blue when $$K_4[Fe(CN)_6]$$, that is, potassium ferrocyanide, is added.

The formula for the formation of Turnbull’s (or Prussian) blue is stated next:

$$3\,Fe^{3+} + 2\,[Fe(CN)_6]^{4-} \rightarrow Fe_3[Fe(CN)_6]_2 \;(\text{blue precipitate})$$

This blue precipitate once again confirms that the cation present is $$Fe^{3+}$$, because only ferric ions give this intense blue colour with the ferrocyanide ion. So both observations consistently point to the presence of $$Fe^{3+}$$.

Observation 3: The salt X gives a positive chromyl chloride test.

We first state the essence of the chromyl chloride test:

If a solid chloride containing the anion $$Cl^-$$ is heated with solid potassium dichromate $$K_2Cr_2O_7$$ and concentrated sulphuric acid, deep-red vapours of chromyl chloride $$CrO_2Cl_2$$ are produced. These vapours, when passed into alkaline medium and then acidified with acetic acid, yield a yellow precipitate of $$PbCrO_4$$ with lead acetate, confirming $$CrO_4^{2-}$$ and therefore $$CrO_2Cl_2$$. The overall sign is the appearance of red fumes.

Only chlorides (and only those chlorides in which the cation does not form volatile chlorides under the same conditions) yield this positive test. Hence the anion in X must be $$Cl^-$$.

Combining all the deductions so far:

→ The cation is $$Fe^{3+}$$ (proved twice).
→ The anion is $$Cl^-$$ (proved by chromyl chloride test).

Therefore the salt must be $$FeCl_3$$.

Now we match with the options given:

A. $$FeCl_3$$ has both $$Fe^{3+}$$ and $$Cl^-$$   ✓
B. $$Fe(NO_3)_3$$ lacks $$Cl^-$$, so it cannot give the chromyl chloride test.
C. $$CuCl_2$$ contains $$Cl^-$$ but its cation is $$Cu^{2+}$$, which neither forms the blood-red thiocyanate complex nor the blue ferrocyanide precipitate characteristic of $$Fe^{3+}$$.
D. $$Cu(NO_3)_2$$ lacks both $$Fe^{3+}$$ and $$Cl^-$$.

Only option A satisfies all three observations.

Hence, the correct answer is Option A.