Question

In which of the following, elements are arranged in the correct order of their first electron gain enthalpy values?

• A. \(S<O<Br<I\)
• B. \(O<S<I<Br\)
• C. \(I<S<O<Br\)
• D. \(Br

Hint:

Remember the exceptional order: \[ Cl>F \quad \text{and} \quad S>O \] in terms of electron accepting tendency.

Answer 1

The Correct Option is B

Concept: Electron gain enthalpy is the enthalpy change when an isolated gaseous atom accepts an electron. \[ X(g)+e^- \rightarrow X^-(g) \] More negative electron gain enthalpy means greater tendency to accept an electron. Important periodic trends:
• Electron gain enthalpy generally becomes more negative across a period.
• It generally becomes less negative down a group.
• Oxygen has less negative value than sulphur because the small size of oxygen causes greater electron-electron repulsion.
• Chlorine has the most negative value among halogens, while bromine is more negative than iodine.

Step 1: Compare oxygen and sulphur.
Although oxygen lies above sulphur in Group 16, oxygen has a very small atomic size. The incoming electron experiences strong repulsion in the compact \(2p\)-orbital. Therefore, \[ \Delta H_{eg}(O) > \Delta H_{eg}(S) \] or equivalently, \[ O<S \] in terms of negativity.

Step 2: Compare bromine and iodine.
Bromine lies above iodine in Group 17. Since bromine is smaller in size, it attracts the incoming electron more strongly. Hence, \[ Br>I \] with respect to electron affinity tendency.

Step 3: Arrange all elements.
Using known electron gain enthalpy values: \[ O<S<I<Br \] This is the correct increasing order of negativity of electron gain enthalpy.

Step 4: Identify the correct option.
Therefore, \[ \boxed{O<S<I<Br} \] Hence the correct option is \[ \boxed{\text{(B)}} \]