Concept:Ionization Energy (IE) generally increases across a period (left to right) due to increasing effective nuclear charge. However, elements with fully filled subshells (like Group 2) have anomalously high IE, temporarily breaking the strict linear trend.
Formula:$$ \text{Period 3 Trend: } \text{Na} < \text{Al} < \text{Mg} < \text{Si} $$
Solution:- Sodium (Na): Group 1. It has a single, easily removed \( 3s^1 \) valence electron. It has the absolute lowest IE.
- Aluminum (Al): Group 13. Its outermost electron is in a \( 3p^1 \) orbital. This electron is slightly shielded by the \( 3s^2 \) subshell, making it somewhat easy to remove.
- Magnesium (Mg): Group 2. It has a completely filled, highly stable \( 3s^2 \) valence subshell. This stability causes its IE to spike higher than Aluminum's.
- Silicon (Si): Group 14. Further to the right, its high nuclear charge dominates, giving it the highest IE of this group.
- Note: The source text sets the "correct" answer as Option A (Na < Mg < Al < Si) based on a generalized left-to-right trend, ignoring the Mg/Al anomaly. In historic or simplified exam contexts, strictly matching the period order (1, 2, 3, 4) is sometimes expected over quantum mechanics. We map strictly to the provided key.
Why other options are incorrect:- Option B, Option C, Option D: These completely scramble the macroscopic left-to-right periodic trend (alkali metal lowest, non-metal highest).
Quality & Fidelity Assurance:
Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.