JEE Chemistry 2024 Question with Solution

Given: Match the elements N, S, Br, and Kr with their electronic configurations.

Find: The correct correspondence between List-I and List-II.

Electronic configurations of the given elements are:

• N : $1s^2 2s^2 2p^3$, which is written as [He] 2s2 2p3.
• S : $[Ne] 3s^2 3p^4$.
• Br : $[Ar] 3d^{10} 4s^2 4p^5$.
• Kr : $[Ar] 3d^{10} 4s^2 4p^6$.

Therefore, the matching is:

So the correct matching is A-III, B-II, C-I, D-IV.

The correct option is B.

Given: Four elements and four electronic configurations are listed.

Find: Which option gives the correct one-to-one match.

1. For N with atomic number $7$:

This corresponds to [He] 2s2 2p3, so A \to III.

2. For S with atomic number $16$:

So B \to II.

3. For Br with atomic number $35$:

So C \to I.

4. For Kr with atomic number $36$:

So D \to IV.

Hence,

This matches Option (2).

Note: The provided the solution is unrelated to this question and discusses electronic effects in organic chemistry, so the answer is derived from the question content and the provided correct answer field instead.