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

To determine the elements with the most and least metallic character among \(\mathrm{N, P, S, Cl, F}\), we must first understand the periodic trend of metallic character. Metallic character decreases across a period (from left to right on the periodic table) and increases down a group (from top to bottom on the periodic table).

• Fluorine (\( \mathrm{F} \)): It is in Group 17 (Halogens) and is the most electronegative element. This makes it the least metallic element in the given set.
• Phosphorus (\( \mathrm{P} \)): It is located in Group 15. Elements in Group 15 tend to have more metallic character than those in Group 16 or 17, and since Phosphorus is lower in the periodic table compared to Nitrogen (another element from Group 15), it is the most metallic element among the given choices.

Now, let’s find the number of valence electrons for each of these elements:

• Fluorine (\( \mathrm{F} \)): It is the least metallic element. Being in Group 17, it has 7 valence electrons.
• Phosphorus (\( \mathrm{P} \)): It is the most metallic element in the given set. Being in Group 15, it has 5 valence electrons.

Thus, the number of valence electrons in the least metallic element (\( \mathrm{F} \)) is 7, and in the most metallic element (\( \mathrm{P} \)) is 5.

Therefore, the correct option is: (7, 5).