A stronger reducing agent is

Options

A. `H_(2)O`
B. `H_(2)S`
C. `H_(2)Se`
D. `H_(2)Te`

Answer

To determine which compound is the stronger reducing agent among the given options (H2O, H2S, H2Se, and H2Te), we can analyze the reducing properties of each compound based on the elements involved and their oxidation states.

### Step-by-Step Solution:

1. **Understanding Reducing Agents**:
- A reducing agent is a substance that donates electrons to another substance, reducing the oxidation state of that substance while itself being oxidized. The stronger the reducing agent, the more readily it donates electrons.

2. **Analyzing the Given Compounds**:
- The compounds given are H2O (water), H2S (hydrogen sulfide), H2Se (hydrogen selenide), and H2Te (hydrogen telluride). All these compounds contain hydrogen and elements from the oxygen family (Group 16).

3. **Oxidation States**:
- In H2O, oxygen has an oxidation state of -2.
- In H2S, sulfur has an oxidation state of -2.
- In H2Se, selenium has an oxidation state of -2.
- In H2Te, tellurium has an oxidation state of -2.
- All these compounds have hydrogen in the +1 oxidation state.

4. **Comparing the Elements**:
- The ability of these compounds to act as reducing agents increases as we move down the group in the periodic table. This is because the elements become larger and the bond strength with hydrogen decreases, making it easier for them to lose electrons.
- Therefore, the reducing strength increases in the order: H2O < H2S < H2Se < H2Te.

5. **Conclusion**:
- Among the given options, H2S (hydrogen sulfide) is the strongest reducing agent. This is because sulfur can donate electrons more readily than oxygen, and it is in a lower oxidation state compared to the other elements in the options.

### Final Answer:
The stronger reducing agent is **H2S (hydrogen sulfide)**.

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