Metal `X` is found in earth's crust. This metal forms a reddish brown substance when exposed to moist air. When a blue coloured solution `Y` is stored in a container made of `X`, the solution turns green and a reddish brown metal `Z` gets deposited on the container. Here `X, Y` and `Z` respectively are

Options

A. `Cu, CuSO_4, Fe`
B. `Zn, ZnSO_4, Fe`
C. `Fe, CuSO_4, Cu`
D. `Cu, FeSO_4, Fe`

Explanation

To solve the question, we need to identify the metals and solutions based on the clues provided.

### Step 1: Identify Metal X
The question states that metal X is found in the Earth's crust and forms a reddish-brown substance when exposed to moist air. This description matches **copper (Cu)**, which forms copper(II) oxide (CuO) or copper(II) hydroxide (Cu(OH)₂) that appears reddish-brown.

### Step 2: Identify Solution Y
The blue-colored solution Y is likely to be **copper(II) sulfate (CuSO₄)**, which is known to have a blue color in its aqueous form.

### Step 3: Understand the Reaction
When the blue solution (copper(II) sulfate) is stored in a container made of metal X (copper), a reaction occurs. The copper in the container displaces the copper ions in the solution, leading to the formation of a green solution. This green color indicates the formation of copper(I) sulfate (Cu₂SO₄) or the presence of copper(I) ions.

### Step 4: Identify Metal Z
The reddish-brown metal Z that gets deposited on the container is also **copper (Cu)**. This is because the copper ions from the solution are reduced and deposit as solid copper on the surface of the container.

### Conclusion
Thus, we can summarize:
- Metal X is **Copper (Cu)**
- Solution Y is **Copper(II) sulfate (CuSO₄)**
- Metal Z is **Copper (Cu)**

### Final Answer:
- X = Copper (Cu)
- Y = Copper(II) sulfate (CuSO₄)
- Z = Copper (Cu)