In the reaction sequence
`C_(6)H_(5)CHO overset(NH_(2)OH//H^(o+))underset(Delta)rarr(X) overset(P_(2)O_(5)//Delta)rarr (Y) overset(H_(2)O//H^(o+))rarr(Z)`
(X), (Y) and (Z) respectively be

Options

A. `C_(6)H_(5)CH=N-OH, C_(6)H_(5)CN, C_(6)H_(5)COOH`
B. `C_(6)H_(5)CH=NOH, C_(6)H_(5)CONH_(2),C_(6)H_(5)CONH_(2)`
C. `C_(6)H_(5)-CH=NOH, C_(6)H_(5)COOH, C_(6)H_(5)CONH_(2)`
D. `C_(6)H_(5)-CH=NOH, C_(6)H_(5)COOH, C_(6)H_(5)CN`

Answer

To solve the reaction sequence given in the question, we will analyze each step carefully.

### Step-by-Step Solution:

1. **Identify the starting compound**:
The starting compound is C₆H₅CHO, which is benzaldehyde.

2. **First Reaction with NH₂OH**:
- Benzaldehyde reacts with hydroxylamine (NH₂OH) under heating conditions.
- This reaction forms an oxime by the elimination of water (H₂O).
- The structure of the oxime formed will be C₆H₅C=NOH.
- Therefore, the product (X) is:
\[ X = C_6H_5C=NOH \]

3. **Second Reaction with P₂O₅**:
- The oxime (X) is then treated with phosphorus pentoxide (P₂O₅) and heated.
- P₂O₅ is a dehydrating agent, which removes water from the oxime, converting it into a nitrile.
- The product formed will be C₆H₅C≡N (benzonitrile).
- Therefore, the product (Y) is:
\[ Y = C_6H_5C≡N \]

4. **Third Reaction with H₂O**:
- The nitrile (Y) undergoes hydrolysis in the presence of water (H₂O) and an acid (H⁺).
- This hydrolysis converts the nitrile into a carboxylic acid.
- The product formed will be C₆H₅COOH (benzoic acid).
- Therefore, the product (Z) is:
\[ Z = C_6H_5COOH \]

### Final Products:
- (X) = C₆H₅C=NOH (Benzaldehyde oxime)
- (Y) = C₆H₅C≡N (Benzonitrile)
- (Z) = C₆H₅COOH (Benzoic acid)

### Summary of the Products:
- **X** = C₆H₅C=NOH
- **Y** = C₆H₅C≡N
- **Z** = C₆H₅COOH