Question

Out of \( \mathrm{N, P, S, Cl, F} \), number of valence electrons in the {least metallic element and most metallic element respectively is: }

• A. \(5, 7\)
• B. \(7, 5\)
• C. \(6, 5\)
• D. \(5, 6\)

Hint:

To compare metallic character, always check {position in the periodic table}: down the group $\uparrow$ metallic nature, across the period $\rightarrow$ non-metallic nature.

Answer 1

The Correct Option is B

Concept:
Metallic character:

Increases down a group

Decreases from left to right
across a period
Valence electrons are determined by the group number of the element.
Step 1: Identify the Least Metallic Element
Among the given elements: \[ \mathrm{N,\ P,\ S,\ Cl,\ F} \]

Fluorine (\(\mathrm{F}\)) is the most electronegative element.
Hence, it is the least metallic
.
Fluorine belongs to Group 17: \[ \Rightarrow \text{Valence electrons} = 7 \]
Step 2: Identify the Most Metallic Element


Phosphorus (\(\mathrm{P}\)) lies below nitrogen in Group 15.
Metallic character increases down the group.
Hence, \(\mathrm{P}\) is the most metallic
among the given elements. Phosphorus belongs to Group 15: \[ \Rightarrow \text{Valence electrons} = 5 \]
Final Conclusion:
\[ \boxed{7,\ 5} \]