In K,Rb and Cs, the decreasing order of reducing power in gaseous state is:-

Options

A. `K gt Cs gt Rb`
B. `Cs gt Rb gt K`
C. `K gt Cs gt Rb`
D. `Rb gt Cs gt K`

Answer

To determine the decreasing order of reducing power in the gaseous state for K (potassium), Rb (rubidium), and Cs (cesium), we need to analyze the properties of these alkali metals.

### Step-by-Step Solution:

1. **Understanding Reducing Power**:
- Reducing power refers to the ability of an element to donate electrons. The easier it is for an element to lose an electron, the higher its reducing power.

2. **Ionization Energy Trend**:
- In the alkali metals (Group 1), ionization energy decreases as we move down the group from K to Rb to Cs. This means that it requires less energy to remove an electron from cesium than from potassium.

3. **Comparing Ionization Energies**:
- Potassium (K) has the highest ionization energy among the three.
- Rubidium (Rb) has a lower ionization energy than potassium but higher than cesium.
- Cesium (Cs) has the lowest ionization energy, meaning it can lose its outermost electron most easily.

4. **Conclusion on Reducing Power**:
- Since reducing power is directly related to the ability to lose electrons, we conclude that:
- Cs > Rb > K
- Therefore, the decreasing order of reducing power in the gaseous state is:
- **Cs > Rb > K**

5. **Selecting the Correct Option**:
- Based on our analysis, the correct answer is option B: **Cs > Rb > K**.

### Final Answer:
The decreasing order of reducing power in the gaseous state is: **Cs > Rb > K**.

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