Consider the following ionsisation reaction `:`
`IE(KJ mol^(-1))" "IE(KJ mol^(-1))`
`A_((g))rarrA_((g))^(+)+e^(ө),A_(1)" "B_((g))rarrB_((g))^(+)+e^(ө),B_(1)`
`B_((g))^(+)rarrB_((g))^(+2)+e^(ө),B_(2)" "C_((g))rarrC_((g))^(+)+e^(ө),C_(1)`
`C_((g))^(+)rarrC_((g))^(+2)+e^(ө),C_(2)" "C_((g))^(+2)rarrC_((g))^(+3)+e^(ө),C_(3)`
If monovalent positive ion of `A`, divalent positive ion of `B` and trivalent positive ion of `C` have zero electron. Then incorrect order of corresponding I.P. is `:`

Options

A. `C_(3)gtB_(2)gtA_(1)`
B. `B_(1)gtA_(1)gtC_(1)`
C. `C_(3)gtC_(2)gtB_(2)`
D. `B_(2)gtC_(3)gtA_(1)`

Explanation

To solve the problem, we need to analyze the ionization energies of the given elements A, B, and C based on the information provided.

### Step-by-Step Solution:

1. **Identify the Elements:**
- We know that A is a monovalent positive ion, which means it loses one electron. This suggests that A could be Hydrogen (H), as it has one electron.
- B is a divalent positive ion, which means it loses two electrons. This suggests that B could be Helium (He), as it can lose two electrons.
- C is a trivalent positive ion, which means it loses three electrons. This suggests that C could be Lithium (Li), as it can lose three electrons.

2. **Determine Electron Configurations:**
- For A (H): After losing one electron, A+ has the configuration of 1s^0 (zero electrons).
- For B (He): After losing two electrons, B+ has the configuration of 1s^0 (zero electrons).
- For C (Li): After losing three electrons, C+ has the configuration of 1s^0 (zero electrons).

3. **Analyze Ionization Energies:**
- The first ionization energy (IE1) for A (H) will be lower than that of B (He) because He has a full shell and is more stable.
- The order of first ionization energies can be established as:
\[ IE1(B) > IE1(A) > IE1(C) \]
- For the second ionization energy (IE2) of B and the first ionization energy (IE1) of C, we can compare:
- B loses its second electron (B+ to B2+) and C loses its first electron (C to C+).
- The order will be:
\[ IE2(B) > IE1(C) \]

4. **Third Ionization Energy:**
- For C, the third ionization energy (IE3) is the energy required to remove the third electron from Li+ to Li2+.
- The order of ionization energies will be:
\[ IE3(C) > IE2(B) > IE1(A) \]

5. **Final Comparison:**
- We can summarize the ionization energies as:
- For first ionization: \( IE1(B) > IE1(A) > IE1(C) \)
- For second ionization: \( IE2(B) > IE1(C) \)
- For third ionization: \( IE3(C) > IE2(B) > IE1(A) \)

6. **Identify Incorrect Order:**
- The question asks for the incorrect order of ionization potentials. The only incorrect order based on our analysis is:
\[ B2 > C3 > A1 \]
- This order is incorrect because we established that \( C3 > B2 > A1 \).

### Conclusion:
The incorrect order of corresponding ionization potentials is \( B2 > C3 > A1 \).