The orange solid on heating gives a colourless gas and a greensolid which can be reduced to metal by aluminium powder. The orange and the green solids are respectively

Options

A. `(NH_(4))_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
B. `Na_(2)Cr_(2)O_(7) and Cr_(2)O_(3)`
C. `K_(2)Cr_(2)O_(7) and CrO_(3)`
D. `(NH_(4))_(2) Cr_(2)O_(4) and CrO_(3)`

Explanation

To solve the problem, we need to identify the orange solid and the green solid based on the given information. Let's break down the steps:

### Step 1: Identify the Orange Solid
The problem states that the orange solid on heating produces a colorless gas and a green solid. A well-known orange solid that fits this description is ammonium dichromate, which has the chemical formula \( NH_4H_2Cr_2O_7 \).

### Step 2: Heating the Orange Solid
When ammonium dichromate is heated, it undergoes a decomposition reaction. The reaction can be represented as follows:

\[ NH_4H_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + H_2O \]

In this reaction, the orange solid decomposes to form chromium(III) oxide (\( Cr_2O_3 \)), which is a green solid, along with nitrogen gas (\( N_2 \)) and water vapor (\( H_2O \)). The colorless gas produced is nitrogen.

### Step 3: Identify the Green Solid
The green solid formed from the heating of ammonium dichromate is chromium(III) oxide, \( Cr_2O_3 \). This green solid can be further reduced to chromium metal using aluminum powder in a thermite reaction.

### Step 4: Conclusion
From the analysis, we conclude that:
- The orange solid is ammonium dichromate, \( NH_4H_2Cr_2O_7 \).
- The green solid is chromium(III) oxide, \( Cr_2O_3 \).

### Final Answer
The orange solid is \( NH_4H_2Cr_2O_7 \) and the green solid is \( Cr_2O_3 \).

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