Consider the elements N, P, 0, S,Cl and F. The number of valence electrons present in the elements with most and least metallic character from the above list is respectively.

To solve this problem, we need to identify the elements with the most and least metallic character from the list: Nitrogen (N), Phosphorus (P), Oxygen (O), Sulfur (S), Chlorine (Cl), and Fluorine (F). Then, we determine their valence electrons.

First, recall that metallic character decreases from left to right across a period and increases down a group in the periodic table. Valence electrons are the electrons in the outermost shell.

List the elements with their groups and valence electrons:

• N (Nitrogen): Group 15, valence electrons = 5
• P (Phosphorus): Group 15, valence electrons = 5
• O (Oxygen): Group 16, valence electrons = 6
• S (Sulfur): Group 16, valence electrons = 6
• Cl (Chlorine): Group 17, valence electrons = 7
• F (Fluorine): Group 17, valence electrons = 7

Now, compare metallic character:

• Within Group 15: P is below N, so P has higher metallic character than N.
• Within Group 16: S is below O, so S has higher metallic character than O.
• Within Group 17: Cl is below F, so Cl has higher metallic character than F.

Compare across groups and periods:

• In Period 2: Metallic character decreases left to right, so N > O > F.
• In Period 3: Metallic character decreases left to right, so P > S > Cl.

Now, compare all elements:

• P (Period 3, Group 15) has higher metallic character than S and Cl in the same period.
• P also has higher metallic character than N (same group, but P is below N).
• Thus, P has the highest metallic character in the list.

For the least metallic character:

• F is the most non-metallic element due to high electronegativity and top-right position.
• F has lower metallic character than O (same period, but F is to the right) and Cl (same group, but F is above).
• Thus, F has the least metallic character in the list.

Therefore:

• Element with most metallic character: P, valence electrons = 5
• Element with least metallic character: F, valence electrons = 7

The numbers are 5 and 7, corresponding to option B.