Q.The electronic configuration of the outer most shell of the most electronegative element is

Concept: Periodic Trend of Electronegativity

Electronegativity measures an atom's ability to attract shared electrons in a bond. It increases across a period (left to right) as nuclear charge rises and atomic size decreases, and decreases down a group as atomic size increases and valence electrons move farther from the nucleus.

The most electronegative element is fluorine (F), located in Group 17 (halogens), Period 2. Its electronic configuration is $1s^2 \, 2s^2 \, 2p^5$, giving an outermost shell configuration of $2s^2 \, 2p^5$. …

Understanding Electronegativity

Electronegativity measures an atom's ability to attract shared electrons in a chemical bond. The most electronegative element is the one that pulls bonding electrons toward itself most strongly.

Two periodic trends govern electronegativity:

Across a period (left to right): Electronegativity increases because nuclear charge rises while shielding remains roughly constant. The nucleus pulls more strongly on valence electrons.

Down a group (top to bottom): Electronegativity decreases because atomic size increases. Valence electrons sit farther from the nucleus, weakening the attractive force despite the higher nuclear charge.

These trends converge on a single winner: the element in the upper right corner of the periodic table (excluding noble gases, which don't form typical bonds).

Finding the Most Electronegative Element

Let's examine each configuration:

1. Option (A): $2s^2 2p^5$

   This is the second period, Group 17 (halogens). With 7 valence electrons, this element needs just one more to complete its octet. The small atomic radius (second period) means the nucleus is very close to the valence shell. This configuration belongs to fluorine (F).

2. Option (B): $3s^2 3p^5$

   Third period, Group 17. This is chlorine (Cl). While still highly electronegative, it's larger than fluorine, so the nucleus exerts less pull on bonding electrons.

3. Option (C): $4s^2 4p^5$

   Fourth period, Group 17. This is bromine (Br). Even larger atomic radius further reduces electronegativity.

4. Option (D): $5s^2 5p^5$

   Fifth period, Group 17. This is iodine (I). The largest of these halogens, hence the least electronegative among the options. …