Given `DeltaH` for the process `Li(g)rarrLi^(+3)(g) + 3e^(-)` is `19800kJ//"mole" & IE_(1)` for `Li` is `520` then `IE_(2) & IE_(1)` of `Li^(+)` are respectively (approx value) :-

Options

A. `11775,7505`
B. `19280,520`
C. `11775,19280`
D. data insufficient

Explanation

To solve the problem, we need to determine the second ionization energy (IE2) of lithium ion (Li⁺) and the first ionization energy (IE1) of lithium ion (Li⁺). We are given the following information:

1. The enthalpy change (ΔH) for the process:
\[ \text{Li(g)} \rightarrow \text{Li}^{3+}(g) + 3e^- \]
is \(19800 \, \text{kJ/mole}\).

2. The first ionization energy (IE1) for lithium (Li) is \(520 \, \text{kJ/mole}\).

### Step-by-Step Solution:

**Step 1: Understand the Ionization Process**
- The ionization process involves removing electrons from an atom or ion. For lithium, the first ionization energy (IE1) is the energy required to remove the first electron to form Li⁺:
\[ \text{Li(g)} \rightarrow \text{Li}^+(g) + e^- \]
- The second ionization energy (IE2) is the energy required to remove a second electron from Li⁺ to form Li²⁺:
\[ \text{Li}^+(g) \rightarrow \text{Li}^{2+}(g) + e^- \]

**Step 2: Relate the Ionization Energies to the Total Enthalpy Change**
- The total enthalpy change for the complete ionization of lithium to Li³⁺ can be expressed as:
\[ \Delta H = \text{IE1} + \text{IE2} + \text{IE3} \]
- We know that:
\[ \Delta H = 19800 \, \text{kJ/mole} \]
- Given that:
\[ \text{IE1} = 520 \, \text{kJ/mole} \]

**Step 3: Calculate the Remaining Energy**
- We can express the total energy for the second and third ionizations as:
\[ \text{IE2} + \text{IE3} = \Delta H - \text{IE1} \]
- Substituting the known values:
\[ \text{IE2} + \text{IE3} = 19800 - 520 = 19280 \, \text{kJ/mole} \]

**Step 4: Determine the Values of IE2 and IE3**
- We assume that the second ionization energy (IE2) is greater than the first ionization energy (IE1) of lithium. Thus, we can make an assumption about the values of IE2 and IE3.
- A reasonable assumption is that IE2 is significantly larger than IE1, and we can try some values to find a suitable pair that adds up to 19280.

**Step 5: Check Possible Values**
- Let's try \(IE2 = 11775 \, \text{kJ/mole}\) and \(IE3 = 7505 \, \text{kJ/mole}\):
\[ IE2 + IE3 = 11775 + 7505 = 19280 \, \text{kJ/mole} \]
- This combination satisfies the equation.

### Final Answer:
- The ionization enthalpy (IE1) of Li⁺ is \(11775 \, \text{kJ/mole}\) and the ionization enthalpy (IE2) of Li⁺ is \(7505 \, \text{kJ/mole}\).