Which of the following electronic configurations would exhibit the lowest ionisation energy?

Options

A. `1s^2`
B. `1s^2 , 2s^2 2p^2`
C. `1s^2 , 2s^2 2p^6`
D. `1s^2 , 2s^2 2p^6,3s^1`

Answer

To determine which electronic configuration exhibits the lowest ionization energy, we need to analyze the factors that influence ionization energy:

### Step 1: Understand Ionization Energy
Ionization energy is the minimum amount of energy required to remove an electron from an atom in its gaseous state. Factors that affect ionization energy include:
- The size of the atom: Larger atoms generally have lower ionization energies because the outer electrons are farther from the nucleus and experience less electrostatic attraction.
- Electron configuration stability: Atoms with stable electron configurations (like noble gases) require more energy to remove an electron.

### Step 2: Analyze the Given Electronic Configurations
Assuming we have the following electronic configurations to analyze:
1. 1s²
2. 2s² 2p⁶
3. 3s¹
4. 3p⁶

We will evaluate each configuration based on the factors mentioned above.

### Step 3: Evaluate Each Configuration
1. **1s²**: This configuration is fully filled and stable. It will have a high ionization energy because removing an electron would disrupt this stable configuration.

2. **2s² 2p⁶**: This is also a fully filled configuration (like a noble gas). It will also have a high ionization energy for the same reasons as 1s².

3. **3s¹**: This configuration has one electron in the outermost shell (3rd shell). The size of the atom is larger than those in the first two configurations, and since it has only one electron in the outer shell, it is relatively easy to remove. Thus, it will have a low ionization energy.

4. **3p⁶**: Similar to the previous configurations, this is a fully filled configuration and will have a high ionization energy.

### Step 4: Conclusion
Among the given configurations, **3s¹** will exhibit the lowest ionization energy due to its larger atomic size and the presence of only one electron in the outermost shell, making it easier to remove.

### Final Answer
The electronic configuration that exhibits the lowest ionization energy is **3s¹**.

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