Question

In which of the following elements are correctly arranged in the increasing order of their electronegativity values?

• A. $\text{Li} \lt \text{Be} \lt \text{Na} \lt \text{Mg}$
• B. $\text{P} \lt \text{Si} \lt \text{C} \lt \text{N}$
• C. $\text{Cl} \lt \text{S} \lt \text{N} \lt \text{O}$
• D. $\text{P} \lt \text{S} \lt \text{N} \lt \text{O}$

Hint:

F, O, N, and Cl are the most electronegativity elements in the periodic table.
Second-period elements are always more electronegative than third-period elements of the same or neighboring groups due to their very small atomic size.

Answer 1

The Correct Option is D

Step 1: Recall the periodic trend without needing exact numbers.
Electronegativity rises across a period, left to right, and falls down a group, so the elements can be ranked purely from their positions in the periodic table.
Step 2: Place P, S, N and O on the table.
Phosphorus and sulfur sit in period 3, groups 15 and 16, while nitrogen and oxygen sit directly above them in period 2.
Step 3: Compare pairwise and chain the results.
Within period 3, sulfur beats phosphorus, so $P \lt S$. Moving up a group always raises electronegativity, so nitrogen beats phosphorus and oxygen beats sulfur, and within period 2, oxygen beats nitrogen. Chaining these gives $P \lt S \lt N \lt O$.
Step 4: Match to the options.
This increasing sequence corresponds to exactly one listed order.
Final answer: Option 4, $P \lt S \lt N \lt O$.