The energy required to pull the most loosely bound electron form an atom is known as ionizatino potential it is expressed in electron volts. The value of ionization potntial depends on three factors: (i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells.
Q. Incorrect order of ionisation energy is:

Options

A. `Pb(I.E.) gt Sn(I.E.)`
B. `Na^(+)(I.E.) gt Mg^(+) (I.E.)`
C. `Li^(+)(I.E.) lt O^(+) (I.E.)`
D. `Be^(+) (I.E.) lt C^(+) (I.E.)`

Answer

To determine the incorrect order of ionization energy, we need to analyze the given elements based on their atomic structure and the factors affecting ionization energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand Ionization Energy
Ionization energy is the energy required to remove the most loosely bound electron from an atom. It generally increases with an increase in nuclear charge and decreases with an increase in atomic radius and the screening effect of inner electrons.

### Step 2: Identify the Elements
From the video transcript, we have the following elements to analyze:
- Lead (Pb)
- Tin (Sn)
- Sodium (Na⁺)
- Magnesium (Mg²⁺)
- Lithium (Li⁺)
- Oxygen (O)
- Beryllium (Be)

### Step 3: Write Electronic Configurations
1. **Lead (Pb)**: Atomic number 82
- Electronic configuration: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p²
2. **Tin (Sn)**: Atomic number 50
- Electronic configuration: [Kr] 4d¹⁰ 5s² 5p²
3. **Sodium (Na⁺)**: Atomic number 11 (after losing one electron)
- Electronic configuration: 1s² 2s² 2p⁶
4. **Magnesium (Mg²⁺)**: Atomic number 12 (after losing two electrons)
- Electronic configuration: 1s² 2s² 2p⁶
5. **Lithium (Li⁺)**: Atomic number 3 (after losing one electron)
- Electronic configuration: 1s²
6. **Oxygen (O)**: Atomic number 8
- Electronic configuration: 1s² 2s² 2p⁴
7. **Beryllium (Be)**: Atomic number 4
- Electronic configuration: 1s² 2s²

### Step 4: Analyze Ionization Energies
- **Pb vs. Sn**: Pb has a higher atomic number and more electron shells, which means its outermost electrons are farther from the nucleus and experience less attraction. Therefore, Pb should have a lower ionization energy than Sn.
- **Na⁺ vs. Mg²⁺**: Both have the same electron configuration (1s² 2s² 2p⁶), but Mg²⁺ has a higher nuclear charge, thus requiring more energy to remove an electron. Therefore, Na⁺ < Mg²⁺.
- **Li⁺ vs. O**: O has a higher nuclear charge and is closer to the nucleus, so it has a higher ionization energy than Li⁺.
- **Be vs. O**: Be has a completely filled 2s subshell, while O has a half-filled 2p subshell. The filled subshell in Be makes it more stable, thus requiring more energy to remove an electron compared to O.

### Step 5: Determine the Incorrect Order
From the analysis:
- The order Pb > Sn is incorrect (Pb has lower ionization energy).
- The order Na⁺ < Mg²⁺ is correct.
- The order Li⁺ < O is correct.
- The order Be > O is correct.

### Conclusion
The incorrect order of ionization energy is Pb > Sn.