Which of the following order is correct for the property mentioned in brackets?

Options

A. `S^(2-) gt Cl^- gt K^(+) gt Ca^(2+) ` (Ionisation energy )
B. ` C < N < F < O (`2^(nd)` Ionisation energy ) '
C. B > Al > Ga > In > Tl (elctronegativity)
D. `Na^(+) gt Li^(+) gt Mg^(2+) gt Be^(2+) gt Al^(3+)` (Ionic radius )

Explanation

To solve the question regarding the correct order of properties for the given ions and elements, we will analyze each option step by step.

### Step 1: Understanding Ionization Energy
1. **Definition**: Ionization energy is the energy required to remove an electron from an isolated gaseous atom.
2. **Isoelectronic Species**: For isoelectronic species (ions with the same number of electrons), the ionization energy is influenced by the effective nuclear charge (Z_eff). A higher Z_eff means a greater attraction between the nucleus and the electrons, leading to higher ionization energy.

### Step 2: Analyzing the First Option
- The first option states: \( S^{2-} > Cl^{-} > K^{+} > Ca^{2+} \).
- **Correct Order**: The correct order based on ionization energy is \( Ca^{2+} > K^{+} > Cl^{-} > S^{2-} \).
- **Reasoning**:
- \( Ca^{2+} \) has the highest effective nuclear charge and thus the highest ionization energy.
- \( K^{+} \) has a lower effective nuclear charge than \( Ca^{2+} \) but is still higher than \( Cl^{-} \).
- \( Cl^{-} \) has a higher electron-electron repulsion due to the additional electron compared to \( S^{2-} \), making it easier to remove an electron from \( S^{2-} \).

### Step 3: Analyzing the Second Option
- The second option states: \( C < N < F < O \) for second ionization energy.
- **Correct Order**: The order is indeed \( C < N < F < O \).
- **Reasoning**:
- The second ionization energy is the energy required to remove a second electron.
- \( O \) has a stable half-filled configuration after removing one electron, making it require more energy to remove the second electron compared to \( C \), \( N \), and \( F \).

### Step 4: Analyzing the Third Option
- The third option states: \( B < Tl < In < Ga < Al \) for electronegativity.
- **Correct Order**: The correct order of electronegativity is \( Al > Ga > In > Tl > B \).
- **Reasoning**:
- Electronegativity generally decreases down a group due to increased atomic size and shielding effect. \( B \) has the highest electronegativity in the group.

### Step 5: Analyzing the Fourth Option
- The fourth option states: \( Na^{+}, Li^{+}, Mg^{2+}, Al^{3+}, Mg^{2+}, Al^{3+}, Be^{2+} \) for ionic radius.
- **Correct Order**: The correct order of ionic radius is \( Al^{3+} < Mg^{2+} < Li^{+} < Na^{+} < Be^{2+} \).
- **Reasoning**:
- Ionic radius decreases with increasing positive charge due to increased effective nuclear charge pulling the electrons closer to the nucleus.

### Conclusion
After analyzing all options, the correct order for the properties mentioned in the brackets is:

**Correct Answer**: The second option is the correct order for the property mentioned in brackets.