Concept Explanation: Ionization Energy and Electron Affinity

Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. Electron affinity is the energy change when an electron is added to a neutral atom in its gaseous state. These properties depend on electronic configuration, stability, and effective nuclear charge.

Step-by-Step Analysis:

Step 1: Analyze Statement (I)
Statement: Ionisation energy of O(g) is less than Ionisation energy of O⁻(g)
O(g) has electron configuration: 1s²2s²2p⁴. O⁻(g) has configuration: 1s²2s²2p⁵.
Removing an electron from O⁻(g) means breaking a stable half-filled p-subshell (2p⁵), which requires more energy than removing from O(g) which has unstable configuration (2p⁴). Thus, ionization energy of O(g) is indeed less than O⁻(g).
Conclusion: True (T)

Step 2: Analyze Statement (II)
Statement: Ionisation energy of Ne(g) is greater than Ionisation energy of Ne⁺(g)
Ne(g) has stable closed-shell configuration: 1s²2s²2p⁶. High energy is needed to remove an electron.
Ne⁺(g) has configuration: 1s²2s²2p⁵. Removing an electron from this unstable configuration requires less energy.
Thus, ionization energy of Ne(g) is greater than Ne⁺(g).
Conclusion: True (T)

Step 3: Analyze Statement (III)
Statement: Electron affinity of O⁺(g) is greater than Electron affinity of O(g)
Electron affinity is energy released when adding an electron.
O⁺(g) has configuration: 1s²2s²2p³ (half-filled, stable). Adding an electron would disrupt stability, so less energy is released (lower electron affinity).
O(g) has configuration: 1s²2s²2p⁴ (unstable). Adding an electron achieves stable half-filled 2p⁵, so more energy is released (higher electron affinity).
Thus, electron affinity of O⁺(g) is less than O(g), not greater.
Conclusion: False (F)

Step 4: Analyze Statement (IV)
Statement: Ionisation energy of N(g) is less than Ionisation energy of N⁺(g)
N(g) has stable half-filled configuration: 1s²2s²2p³. High energy is needed to remove an electron.
N⁺(g) has configuration: 1s²2s²2p² (unstable). Removing an electron requires less energy.
Thus, ionization energy of N(g) is greater than N⁺(g), not less.
Conclusion: False (F)

Final Answer:

Based on the analysis: (I) T, (II) T, (III) F, (IV) F
So the correct option is TTFF. Comparing with given options, none exactly match, but closest is TTTF (which has first three same but last different). However, strictly as per analysis, it should be TTFF. But since options are given, and TTTF is listed, it might be a trick, but our reasoning gives TTFF.

Note: In the options provided, TTFF is not listed, but TTTF is. Re-check (IV): Ionisation energy of N(g) is indeed greater than N⁺(g), so statement (IV) is false. So pattern is TTFF. Since not in options, but based on theory, answer should be TTFF.

Related Topics:

• Periodic Trends: Ionization energy generally increases across a period and decreases down a group.
• Stability of Half-filled and Fully-filled Subshells: Atoms with half-filled or fully-filled orbitals have extra stability.
• Effective Nuclear Charge: The net positive charge experienced by valence electrons.
• Successive Ionization Energies: Energy required to remove electrons one after another.

Formulae and Theory:

Ionization Energy (IE): Energy required for $A\left(g\right)\to A^{+}\left(g\right)+e^{-}$

Electron Affinity (EA): Energy change for $A\left(g\right)+e^{-}\to A^{-}\left(g\right)$ (usually exothermic).

Factors affecting IE and EA:

• Atomic size: Larger atoms have lower IE.
• Shielding effect: More inner shells reduce effective nuclear charge, lowering IE.
• Electronic configuration: Stable configurations (like half-filled) have higher IE.