Which of the following transformation least energy is required?

Options

A. `F_((g))^(-)to F_((g))+e^(-)`
B. `P_((g))^(-) to P_((g))+e^(-)`
C. `S_((g))^(-) to S_((g))+e^(-)`
D. `Cl_((g))^(-) to Cl_((g))+e^(-)`

Explanation

To determine which transformation requires the least energy, we need to analyze the electron affinity of the elements involved in the transformations. Here's a step-by-step breakdown of the solution:

### Step 1: Understand Electron Affinity
Electron affinity refers to the energy change that occurs when an electron is added to a neutral atom in the gas phase. A higher electron affinity means that the atom releases more energy when gaining an electron, which implies that less energy is required to remove that electron in the reverse process.

### Step 2: Identify the Elements and Their Transformations
The transformations mentioned involve the addition of electrons to the following elements:
- Fluorine (F) to F⁻
- Chlorine (Cl) to Cl⁻
- Sulfur (S) to S²⁻
- Phosphorus (P) to P³⁻

### Step 3: Arrange the Elements According to Electron Affinity
Based on periodic trends:
- As we move across a period from left to right, electron affinity generally increases due to increased nuclear charge.
- As we move down a group, electron affinity generally decreases due to increased atomic size and shielding effect.

### Step 4: Compare the Electron Affinities
1. **Chlorine (Cl)** has the highest electron affinity.
2. **Fluorine (F)** has a lower electron affinity than chlorine due to its small size causing electron-electron repulsion.
3. **Sulfur (S)** has a lower electron affinity than both chlorine and fluorine.
4. **Phosphorus (P)** has the lowest electron affinity among these elements.

### Step 5: Determine the Least Energy Required
Since the least energy is required for the transformation with the lowest electron affinity, we conclude that:
- The transformation involving **Phosphorus (P)** requires the least energy.

### Conclusion
Thus, the answer to the question is that the transformation involving **Phosphorus (P)** requires the least energy, making option number second the correct choice.

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