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Q. No.The formation of the oxide ion, O²⁻(g), from an oxygen atom occurs in two successive steps:
O(g) + e⁻ → O⁻(g) ΔH = −141 kJ mol⁻¹
O⁻(g) + e⁻ → O²⁻(g) ΔH = +780 kJ mol⁻¹
Although O²⁻ is isoelectronic with neon, its formation in the gaseous state is energetically unfavourable. What is the main reason for this?
| 1. | Electron repulsion outweighs the stability gained by achieving a noble gas configuration. |
| 2. | O– ion has a comparatively smaller size than the oxygen atom. |
| 3. | Oxygen is more electronegative. |
| 4. | Addition of electrons in oxygen results in a large size of the ion. |
1. Al < Ca < O < C < F
2. Al < O < C < Ca < F
3. C < F < O < Al < Ca
4. Ca < Al < C < O < F
The first ionization enthalpy of sodium (Na) is 5.1eV. Using this information, determine the value of the electron gain enthalpy of \(\mathrm{Na^+}\):
1. +10.2 eV
2. –5.1 eV
3. –10.2 eV
4. +2.55 eV
The correct order of increasing electron affinity for the elements, O, S, F and Cl is:
| 1. | Cl < F < O < S | 2. | O < S < F < Cl |
| 3. | F < S < O < Cl | 4. | S < O < Cl < F |
Which one of the following arrangements does not give the correct picture of the trends indicated against it?
1. F2> Cl2> Br2> I2 : Oxidising power
2. F<Cl>Br>I : Electron gain enthalpy
3. F2> Cl2> Br2> I2 : Bond dissociation energy
4. F> Cl > Br > I : Electronegativity
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Q. No.
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