The energy needed for `Li_((g))to Li_((g))^(+3)+3e^(-)` is `1.96xx10^(4)"KJ mole"^(-1)`. If the first ionisation energy of Li is 520 KJ `"mole"^(-1)` calculate second ionisation energy for Li. Given `IE_(1)` for `H = 2.18xx10^(-18)"J atom"^(-1)`

Options

A. `5270 "KJ mole"^(-1)`
B. `3210"KJ mole"^(-1)`
C. `7270 "KJ mole"^(-1)`
D. `9290"KJ mole"^(-1)`

Explanation

To solve the problem, we need to find the second ionization energy (IE2) of lithium (Li) using the given data. The total energy required to convert lithium (Li) to lithium ion with a +3 charge (Li³⁺) is provided, along with the first ionization energy (IE1) of lithium.

### Step-by-Step Solution:

1. **Understand Ionization Energies**:
- The first ionization energy (IE1) is the energy required to remove the first electron from a neutral atom.
- The second ionization energy (IE2) is the energy required to remove the second electron from a singly charged ion (Li⁺).
- The third ionization energy (IE3) is the energy required to remove the third electron from a doubly charged ion (Li²⁺).

2. **Write the Ionization Energy Equations**:
- The total energy required to convert Li to Li³⁺ can be expressed as:
\[ \text{Total Energy} = \text{IE1} + \text{IE2} + \text{IE3} \]
- Given:
- Total Energy = \(1.96 \times 10^4 \, \text{kJ/mol}\)
- IE1 = \(520 \, \text{kJ/mol}\)

3. **Calculate IE3**:
- For the third ionization energy (IE3), since Li²⁺ behaves like a hydrogen-like atom, we can calculate it using the formula for hydrogen-like atoms:
\[ \text{IE3} = 13.6 \times Z^2 \times \frac{1}{n^2} \, \text{eV} \]
where \(Z\) is the atomic number and \(n\) is the principal quantum number. For Li²⁺, \(Z = 3\) and \(n = 1\):
\[ \text{IE3} = 13.6 \times 3^2 \times \frac{1}{1^2} = 13.6 \times 9 = 122.4 \, \text{eV} \]
- Convert eV to kJ/mol:
\[ \text{IE3} = 122.4 \, \text{eV} \times 96.49 \, \text{kJ/mol/eV} = 11810.376 \, \text{kJ/mol} \]

4. **Substitute Values into the Total Energy Equation**:
- Now, we can substitute the known values into the total energy equation:
\[ 1.96 \times 10^4 = 520 + \text{IE2} + 11810.376 \]

5. **Solve for IE2**:
- Rearranging the equation to solve for IE2:
\[ \text{IE2} = 1.96 \times 10^4 - 520 - 11810.376 \]
- Calculate:
\[ \text{IE2} = 19600 - 520 - 11810.376 = 7270.624 \, \text{kJ/mol} \]
- Rounding off gives:
\[ \text{IE2} \approx 7270 \, \text{kJ/mol} \]

### Final Answer:
The second ionization energy (IE2) for lithium is approximately **7270 kJ/mol**.