The decreasing order of the first ionization energy (in kJ `mol^(-1)`) of He, Mg and Na is `HegtMggtNa`. The increasing order of the `2^(nd)` ionization energy (in kJ `mol^(-1)`) of these elements will be

Options

A. `NaltMgltHe`
B. `MgltNaltHe`
C. `MgltHeltNa`
D. `NaltHeltMg`

Explanation

To determine the increasing order of the second ionization energy for Helium (He), Magnesium (Mg), and Sodium (Na), we can follow these steps:

### Step 1: Understand the Concept of Ionization Energy
Ionization energy is the energy required to remove an electron from an atom. The first ionization energy refers to the energy needed to remove the first electron, while the second ionization energy refers to the energy needed to remove a second electron after the first has already been removed.

**Hint:** Remember that the first ionization energy is generally lower than the second because the atom becomes more positively charged after losing an electron, making it harder to remove another one.

### Step 2: Write the Electron Configurations
- **Helium (He)**: Atomic number 2, electron configuration: 1s²
- **Magnesium (Mg)**: Atomic number 12, electron configuration: 1s² 2s² 2p⁶ 3s²
- **Sodium (Na)**: Atomic number 11, electron configuration: 1s² 2s² 2p⁶ 3s¹

**Hint:** Knowing the electron configurations helps in understanding how many electrons are present in the outer shell and how stable the remaining configuration will be after ionization.

### Step 3: Analyze the First Ionization Energy
From the video transcript, we know the decreasing order of first ionization energy is:
- He > Mg > Na

This is because:
- Helium has a complete outer shell and a small atomic size, making it very difficult to remove an electron.
- Magnesium has two electrons in its outermost shell (3s²), which are easier to remove than Helium.
- Sodium has one electron in its outermost shell (3s¹), which is the easiest to remove.

**Hint:** The more stable the electron configuration after the removal of an electron, the higher the ionization energy.

### Step 4: Determine the Second Ionization Energy
- **For Helium (He)**: After removing one electron, it becomes He⁺ (1s¹), which is still stable. Removing the second electron will require a lot of energy.
- **For Magnesium (Mg)**: After removing one electron (Mg → Mg⁺), it becomes 1s² 2s² 2p⁶ 3s¹. The second ionization will remove the remaining 3s electron, which is still relatively easier compared to Helium.
- **For Sodium (Na)**: After removing one electron (Na → Na⁺), it becomes 1s² 2s² 2p⁶, which has a complete octet. The second ionization energy will involve removing an electron from a stable noble gas configuration, which requires a lot of energy.

**Hint:** The more stable the resulting ion after the first ionization, the lower the second ionization energy will be.

### Step 5: Establish the Increasing Order of Second Ionization Energy
Based on the above analysis:
- **He** has the highest second ionization energy because removing the second electron from a stable configuration requires a lot of energy.
- **Na** has the next highest second ionization energy because it has a stable octet after the first ionization.
- **Mg** has the lowest second ionization energy because its remaining electron is easier to remove compared to the others.

Thus, the increasing order of the second ionization energy is:
**Na < Mg < He**

**Final Answer:** The increasing order of the second ionization energy is Na < Mg < He.