`FeCr_(2)O_(4)+Na_(2)CO_(3)+O_(2)overset("Fusion")to[X]underset(H_(2)O)overset(H^(+))[Y]underset(H_(2)O_(2))overset(H^(+))to[Z]`
Whichf of the following statement is true for the compounds [X], [Y] and [Z]?

Options

A. In all three compounds , the chromium is in +6 oxidation state.
B. [Z] is a deep blue - violet coloured compound which decomposes rapidly in aqueous solution into Cr3+ and dioxygen
C. Saturated solution of [Y] gives bright organge compound, chromic anhydride, with concentrated H20SO4.
D. All of the above.

Answer

To solve the given question, we need to analyze the reactions step by step and determine the oxidation states and properties of the compounds formed.

### Step 1: Identify the Reactants and Products
The reactants are:
- \( \text{FeCr}_2\text{O}_4 \) (chromite)
- \( \text{Na}_2\text{CO}_3 \) (sodium carbonate)
- \( \text{O}_2 \) (oxygen)

The products formed after fusion will be denoted as [X], [Y], and [Z].

### Step 2: Reaction to Form [X]
When \( \text{FeCr}_2\text{O}_4 \) reacts with \( \text{Na}_2\text{CO}_3 \) and \( \text{O}_2 \), the product [X] is formed:
- The reaction can be represented as:
\[ \text{FeCr}_2\text{O}_4 + \text{Na}_2\text{CO}_3 + \text{O}_2 \rightarrow \text{Na}_2\text{CrO}_4 + \text{FeO} + \text{CO}_2 \]
- Here, [X] is \( \text{Na}_2\text{CrO}_4 \).

### Step 3: Determine the Oxidation State of Chromium in [X]
In \( \text{Na}_2\text{CrO}_4 \):
- The oxidation state of oxygen is -2 (4 oxygens give -8).
- The total charge from sodium is +2 (2 sodiums give +2).
- Therefore, for chromium:
\[ \text{Charge on Cr} = +6 \quad (\text{since } +6 - 8 + 2 = 0) \]
- Thus, chromium in [X] is in the +6 oxidation state.

### Step 4: Reaction to Form [Y]
Next, [X] reacts in an acidic medium to form [Y]:
- The reaction can be represented as:
\[ \text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Cr}_2\text{O}_7^{2-} + 2\text{Na}^+ + \text{H}_2\text{O} \]
- Here, [Y] is \( \text{Cr}_2\text{O}_7^{2-} \) (dichromate ion).

### Step 5: Determine the Oxidation State of Chromium in [Y]
In \( \text{Cr}_2\text{O}_7^{2-} \):
- The total charge from oxygen is -14 (7 oxygens give -14).
- The total charge is -2 (for the dichromate ion).
- Therefore, for chromium:
\[ 2\text{Cr} - 14 = -2 \implies 2\text{Cr} = 12 \implies \text{Cr} = +6 \]
- Thus, chromium in [Y] is also in the +6 oxidation state.

### Step 6: Reaction to Form [Z]
Finally, [Y] reacts in the presence of \( \text{H}_2\text{O}_2 \) to form [Z]:
- The reaction can be represented as:
\[ \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow \text{CrO}_5 + 2\text{H}_2\text{O} \]
- Here, [Z] is \( \text{CrO}_5 \) (chromium pentoxide).

### Step 7: Determine the Oxidation State of Chromium in [Z]
In \( \text{CrO}_5 \):
- The structure can be viewed as \( \text{Cr(O}_2\text{O}_2)_2 \) (peroxide).
- The total charge from the 5 oxygens is -10 (4 from peroxides and 1 from normal).
- Therefore, for chromium:
\[ \text{Charge on Cr} = +6 \quad (\text{as it cannot exceed +6}) \]
- Thus, chromium in [Z] is also in the +6 oxidation state.

### Conclusion
Now we can evaluate the statements:
1. In all three compounds, chromium is in the +6 oxidation state. **True**
2. [Z] is a deep blue violet colored compound which decomposes rapidly in aqueous solution into \( \text{Cr}^{3+} \) and \( \text{O}_2 \). **True**
3. Saturated solution of [Y] gives bright orange compound chromic anhydride with concentrated \( \text{H}_2\text{SO}_4 \). **True**

Since all statements are true, the correct answer is:
**Option 4: All of the above are correct.**