Consider the following ionization reactions with reference to Boron: $A(g)\rightarrow A^{+}(g)+e^{-} IE=A_{1}kJ/mol$ $A^{+}(g)\rightarrow A^{2+}(g)+e^{-} IE=A_{2}kJ/mol$ $A^{2+}(g)\rightarrow A^{3+}(g)+e^{-} IE=A_{3}kJ/mol$ then correct order of IE is:

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The correct order of ionization energies (IE) for the ionization reactions of Boron is given by Option 3: $A_{1} < A_{2} < A_{3}$. Let's explore why this is the correct order and why the other options are incorrect. ### Explanation for Option 3: $A_{1} < A_{2} < A_{3}$ 1. **Ionization Energy Concept**: Ionization energy is the energy required to remove an electron from an atom or ion in the gaseous state. As electrons are removed, the remaining electrons experience a greater effective nuclear charge because the same number of protons are attracting fewer electrons. This increased attraction makes it progressively harder to remove additional electrons. 2. **First Ionization Energy ($A_{1}$)**: This is the energy required to remove the first electron from a neutral Boron atom. Since the atom is neutral, the electron is relatively easier to remove compared to subsequent electrons. 3. **Second Ionization Energy ($A_{2}$)**: This is the energy required to remove an electron from a singly charged cation ($A^{+}$). The removal of the first electron results in a positive charge, increasing the effective nuclear charge on the remaining electrons, making it more difficult to remove the second electron compared to the first. 4. **Third Ionization Energy ($A_{3}$)**: This is the energy required to remove an electron from a doubly charged cation ($A^{2+}$). With two electrons removed, the effective nuclear charge is even higher, making it the most difficult to remove the third electron. Thus, the ionization energies increase as more electrons are removed, leading to the order $A_{1} < A_{2} < A_{3}$. ### Why Other Options are Incorrect: 1. **Option 1: $A_{1} > A_{2} > A_{3}$**: This option suggests that the first ionization energy is the highest, which contradicts the principle that removing electrons from a positively charged ion requires more energy due to increased effective nuclear charge. 2. **Option 2: $A_{1} = A_{2} = A_{3}$**: This option implies that all ionization energies are equal, which is incorrect because the effective nuclear charge increases with each electron removed, making subsequent ionizations require more energy. 3. **Option 4: $A_{3} = A_{2} < A_{1}$**: This option suggests that the first ionization energy is the lowest, which is incorrect. The first ionization energy is the lowest because it involves removing an electron from a neutral atom, whereas subsequent ionizations involve removing electrons from positively charged ions. In conclusion, Option 3 correctly reflects the increasing difficulty of removing electrons as the effective nuclear charge increases with each ionization, making $A_{1} < A_{2} < A_{3}$ the correct order of ionization energies for Boron.