(i) An aqueous solution of a compound (A) is acidic towards litmus and (A) sublimes at about `300^(@)C`.
(ii) (A) on treatment with an excess of `NH_(4)SCN` gives a red coloured compound (B) and on treatment with a solution of `K_(4)[Fe(CN)_(6)]` gives a blue coloured compound (C).
(iii) (A) on heating with excess of `K_(2)Cr_(2)O_(7)` in the presence of concentrated `H_(2)SO_(4)` evolves deep red vapours of (D).
(iv) On passing the vapour of (D) into a solution of NaOH and then adding the solution of acetic acid and lead acetate, a yellow precipitate of compound (E) is obtained.
Identify (A) to (E) and give chemical equations for the reactions.

Options

A. `Ni(SCN)_(2) and Ni_(2)[Fe(CN)_(6)]`
B. `Co(SCN)_(2) and Co_(2)[Fe(CN)_(6)]`
C. `[Fe(SCN)_(6)]^(3-) and Fe_(3)[Fe(CN)_(6)]_(2)`
D. `Fe(SCN)_(3) and Fe_(4)[Fe(CN)_(6)]_(3)`

Explanation

To solve the problem, we need to identify the compounds A to E based on the given reactions and characteristics. Let's break down the steps:

### Step 1: Identify Compound A
- **Given Information**: An aqueous solution of compound (A) is acidic towards litmus and sublimes at about 300°C.
- **Analysis**: The only compound that fits these criteria is **FeCl3 (Iron(III) chloride)**. It is acidic in solution and has a sublimation point around 300°C.

**Conclusion**: A = FeCl3

### Step 2: Reaction of A with NH4SCN
- **Reaction**: When FeCl3 reacts with NH4SCN, it forms a red-colored complex.
- **Chemical Equation**:
\[ \text{FeCl}_3 + 3 \text{NH}_4\text{SCN} \rightarrow \text{Fe(SCN)}_3 + 3 \text{NH}_4\text{Cl} \]
- **Conclusion**: The red-colored compound formed is **B = Fe(SCN)3**.

### Step 3: Reaction of A with K4[Fe(CN)6]
- **Reaction**: When FeCl3 reacts with K4[Fe(CN)6], it forms a blue-colored precipitate.
- **Chemical Equation**:
\[ 3 \text{FeCl}_3 + 2 \text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{Fe}_3[\text{Fe(CN)}_6]_2 + 6 \text{KCl} \]
- **Conclusion**: The blue-colored compound formed is **C = Fe3[Fe(CN)6]2 (Prussian Blue)**.

### Step 4: Heating A with K2Cr2O7 and H2SO4
- **Reaction**: Heating FeCl3 with K2Cr2O7 in the presence of concentrated H2SO4 produces red vapors.
- **Chemical Equation**:
\[ 6 \text{FeCl}_3 + 3 \text{K}_2\text{Cr}_2\text{O}_7 + 6 \text{H}_2\text{SO}_4 \rightarrow 2 \text{CrO}_2\text{Cl}_2 + 3 \text{Fe}_2(\text{SO}_4)_3 + 6 \text{KCl} + 6 \text{H}_2\text{O} \]
- **Conclusion**: The red vapors evolved are **D = CrO2Cl2 (Chromyl chloride)**.

### Step 5: Reaction of D with NaOH and Acetic Acid
- **Reaction**: Passing the vapors of D into NaOH solution and then adding acetic acid leads to the formation of a yellow precipitate.
- **Chemical Equation**:
\[ \text{CrO}_2\text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{CrO}_4 + 2 \text{NaCl} + \text{H}_2\text{O} \]
\[ \text{Na}_2\text{CrO}_4 + 2 \text{CH}_3\text{COOH} + \text{Pb(OAc)}_2 \rightarrow \text{PbCrO}_4 + 2 \text{CH}_3\text{COONa} \]
- **Conclusion**: The yellow precipitate formed is **E = PbCrO4 (Lead chromate)**.

### Summary of Identifications
- A = FeCl3
- B = Fe(SCN)3
- C = Fe3[Fe(CN)6]2 (Prussian Blue)
- D = CrO2Cl2 (Chromyl chloride)
- E = PbCrO4 (Lead chromate)