Question

The correct order of ionisation energy of Cl, S, P, Al, Si is:

• A. \( \text{Cl}>\text{P}>\text{S}>\text{Si}>\text{Al} \)
• B. \( \text{P}>\text{Cl}>\text{S}>\text{Al}>\text{Si} \)
• C. \( \text{Cl}>\text{S}>\text{P}>\text{Si}>\text{Al} \)
• D. \( \text{Cl}>\text{Al}>\text{Si}>\text{P}>\text{S} \)

Hint:

Half-filled and fully filled subshells show higher stability, leading to higher ionisation energy than expected.

Answer 1

The Correct Option is A

Concept:
Ionisation energy is the minimum energy required to remove the most loosely bound electron from a gaseous atom. General trends:
Ionisation energy increases across a period from left to right.
It decreases down a group.
Half-filled and fully filled orbitals are more stable and have higher ionisation energy.
Step 1: All given elements belong to the 3rd period of the periodic table. \[ \text{Al} \rightarrow \text{Si} \rightarrow \text{P} \rightarrow \text{S} \rightarrow \text{Cl} \] Across a period, ionisation energy generally increases due to:
Increasing nuclear charge
Decreasing atomic size
Step 2: Exception between phosphorus and sulfur. Electronic configurations: \[ \text{P}: [Ne]\,3s^2\,3p^3 \quad (\text{half-filled}) \] \[ \text{S}: [Ne]\,3s^2\,3p^4 \] Phosphorus has a half-filled \(3p\)-subshell, which is extra stable. Sulfur has one paired electron in the \(3p\)-orbital, causing increased electron–electron repulsion. \[ \therefore \ \text{Ionisation energy of P}>\text{Ionisation energy of S} \]
Step 3: Final order of ionisation energy: \[ \text{Cl}>\text{P}>\text{S}>\text{Si}>\text{Al} \]