`MSO_4 overset(NH_4OH)to darr underset("White")X underset("excess")overset(NH_4OH)to Y overset(H_2S) to darr Z`
Here M and Z are

Options

A. `Fe, FeS`
B. `Cu, ZnS`
C. `Al, Al_2S_3`
D. `Zn,ZnS`

Explanation

To solve the question, we need to analyze the reactions step by step.

### Step 1: Identify M in the reaction MSO₄ + NH₄OH → X
- The compound MSO₄ indicates a metal sulfate, where M is a metal ion. Common metal ions that form sulfate salts include Cu²⁺, Zn²⁺, and others.
- When MSO₄ reacts with NH₄OH (ammonium hydroxide), it typically forms a precipitate of the corresponding metal hydroxide, which is often white in color.

### Step 2: Determine the identity of X
- The white precipitate formed (X) is likely to be a metal hydroxide. For example, if M is Zn, then the reaction would yield Zn(OH)₂, which is a white precipitate.

### Step 3: Reaction of X with excess NH₄OH to form Y
- The excess NH₄OH can dissolve the precipitate of X, forming a soluble complex. For instance, Zn(OH)₂ can react with excess NH₄OH to form a complex ion such as [Zn(NH₃)₄]²⁺.

### Step 4: Reaction of Y with H₂S to form Z
- When the complex Y reacts with H₂S (hydrogen sulfide), it typically leads to the formation of a metal sulfide, which is often a colored precipitate. For example, if Y is [Zn(NH₃)₄]²⁺, the reaction with H₂S would yield ZnS, which is a white precipitate.

### Conclusion
- From the analysis, we can conclude that:
- M is likely Zn (if we assume MSO₄ is ZnSO₄).
- Z is ZnS, which is the product formed after the reaction with H₂S.

### Final Answer
- M = Zn
- Z = ZnS