The decreasing order of the second ionization potentials of `K, Ca` and `Ba` is

Options

A. `K gt Ca gt Ba`
B. `Ca gt Ba gt K `
C. `Ba gt K gt Ca`
D. `K gt Ba gt Ca `

Answer

To determine the decreasing order of the second ionization potentials of potassium (K), calcium (Ca), and barium (Ba), we can follow these steps:

### Step 1: Understand Ionization Potential
Ionization potential (or ionization energy) is the energy required to remove an electron from an atom or ion. The second ionization potential refers to the energy required to remove a second electron after the first one has already been removed.

### Step 2: Write the Electronic Configurations
- **Potassium (K)**: Atomic number = 19
- Electronic configuration: \( [Ar] 4s^1 \)

- **Calcium (Ca)**: Atomic number = 20
- Electronic configuration: \( [Ar] 4s^2 \)

- **Barium (Ba)**: Atomic number = 56
- Electronic configuration: \( [Xe] 6s^2 \)

### Step 3: Determine the Configuration After First Ionization
- For **K**: After removing one electron, K becomes \( K^+ \) with configuration \( [Ar] \).
- For **Ca**: After removing one electron, Ca becomes \( Ca^+ \) with configuration \( [Ar] 4s^1 \).
- For **Ba**: After removing one electron, Ba becomes \( Ba^+ \) with configuration \( [Xe] 6s^1 \).

### Step 4: Analyze the Stability of the Resulting Ions
- The second ionization potential will be higher for ions with a more stable electron configuration after the first ionization.
- \( K^+ \) has a noble gas configuration \( [Ar] \) which is very stable.
- \( Ca^+ \) has \( [Ar] 4s^1 \) and is less stable than \( K^+ \).
- \( Ba^+ \) has \( [Xe] 6s^1 \) and is also less stable than \( K^+ \).

### Step 5: Compare Sizes and Nuclear Charge
- **Potassium (K)** is in the 4th period, **Calcium (Ca)** is also in the 4th period, and **Barium (Ba)** is in the 6th period.
- As we move down the group from K to Ba, the atomic size increases, which means the outer electrons are further from the nucleus and experience less nuclear charge.
- Therefore, \( Ca^+ \) will have a higher second ionization potential than \( Ba^+ \) because it is smaller and has a higher effective nuclear charge.

### Step 6: Conclusion
Based on the stability of the resulting ions and their sizes, we can conclude the decreasing order of the second ionization potentials:
1. \( K^+ \) (highest)
2. \( Ca^+ \)
3. \( Ba^+ \) (lowest)

Thus, the decreasing order of the second ionization potentials is:
**K > Ca > Ba**