The correct answer is option 'B', H2O.

Explanation:
When p-methylbenzenediazonium chloride reacts with water (H2O), it undergoes hydrolysis to form p-cresol.

Hydrolysis of p-methylbenzenediazonium chloride:
p-Methylbenzenediazonium chloride contains a diazonium group (-N2+) attached to the benzene ring. When it reacts with water, the diazonium group gets replaced by a hydroxyl group (-OH) through hydrolysis.

The hydrolysis reaction can be represented as follows:
C6H4N2Cl + H2O → C6H4OH + N2 + HCl

Formation of p-cresol:
The product of the hydrolysis reaction is p-cresol. p-Cresol is formed when the hydroxyl group (-OH) replaces the diazonium group (-N2+) at the para position (opposite to the methyl group) on the benzene ring.

Overall reaction:
C6H4N2Cl + H2O → C6H4OH + N2 + HCl

In this reaction, water acts as a nucleophile and attacks the electrophilic diazonium group, leading to the displacement of nitrogen gas (N2) and the formation of p-cresol.

Other reagents:
- Cu powder (option 'A') does not have any role in this reaction. It is not involved in the hydrolysis of p-methylbenzenediazonium chloride.
- H3PO2 (option 'C') is a reducing agent and does not have the ability to hydrolyze the diazonium group. It would not convert p-methylbenzenediazonium chloride into p-cresol.
- C6H5OH (option 'D') is phenol, which is similar to p-cresol but does not have the methyl group. It is not possible to directly obtain p-cresol from p-methylbenzenediazonium chloride using phenol.

Therefore, the correct reagent to convert p-methylbenzenediazonium chloride into p-cresol is water (H2O) through hydrolysis.