Q.58. Match List-I with List-II.
List-I (Electronic configuration of neutral atom (where n = 2))
A.

Q: What is the correct answer for JEE Main Chemistry Q58?

The correct answer is A. Sol. The electronic configurations correspond to elements in the second period (n=2): A. $ns^2$ (for n=2, $2s^2$) corresponds to Beryllium (Be). B. $ns^2np^1$ (for n=2, $2s^22p^1$) corresponds to Boron (B). C. $ns^2np^3$ (for n=2, $2s^22p^3$) corresponds to Nitrogen (N). D. $ns^2np^6$ (for n=2, $2s^22p^6$) corresponds to Neon (Ne). General trend for ionization energy across a period is increasing.

Q: Q.58. Match List-I with List-II. List-I (Electronic configuration of neutral atom (where n = 2)) A. B. C. D. List-II (1st Ionization Energy (kJ mol⁻¹)) I. 2080 II. 899 III. 800 IV. 1402 Choose…

The correct answer is A. Sol. The electronic configurations correspond to elements in the second period (n=2): A. $ns^2$ (for n=2, $2s^2$) corresponds to Beryllium (Be). B. $ns^2np^1$ (for n=2, $2s^22p^1$) corresponds to Boron (B). C. $ns^2np^3$ (for n=2, $2s^22p^3$) corresponds to Nitrogen (N). D. $ns^2np^6$ (for n=2, $2s^22p^6$) corresponds to Neon (Ne). General trend for ionization energy across a period is increasing.

Q: What are the options for this JEE Main Chemistry question?

A. A-II, B-III, C-IV, D-I · B. A-IV, B-III, C-II, D-I · C. A-III, B-II, C-IV, D-I · D. A-III, B-II, C-I, D-IV

Question

Q.58. Match List-I with List-II.
List-I (Electronic configuration of neutral atom (where n = 2))
A. 
B. 
C. 
D. 
List-II (1st Ionization Energy (kJ mol⁻¹))
I. 2080
II. 899
III. 800
IV. 1402
Choose the correct answer from the options given below:

FirstInTest · Accepted Answer

Sol. The electronic configurations correspond to elements in the second period (n=2):
A. $ns^2$ (for n=2, $2s^2$) corresponds to Beryllium (Be).
B. $ns^2np^1$ (for n=2, $2s^22p^1$) corresponds to Boron (B).
C. $ns^2np^3$ (for n=2, $2s^22p^3$) corresponds to Nitrogen (N).
D. $ns^2np^6$ (for n=2, $2s^22p^6$) corresponds to Neon (Ne).
General trend for ionization energy across a period is increasing. However, there are exceptions.
Expected order: B < Be < N < O < F < Ne.
Actual ionization energies (kJ/mol):
Be ($2s^2$): 899
B ($2s^22p^1$): 800 (exception: B < Be due to $2p$ electron being easier to remove than $2s$)
N ($2s^22p^3$): 1402 (exception: N > O due to half-filled $p$-orbital stability)
Ne ($2s^22p^6$): 2080 (highest due to full octet)
Matching:
A. Be (899 kJ/mol) → II
B. B (800 kJ/mol) → III
C. N (1402 kJ/mol) → IV
D. Ne (2080 kJ/mol) → I
So, the correct match is A-II, B-III, C-IV, D-I.

Answer

A-II, B-III, C-IV, D-I

Answer

A-IV, B-III, C-II, D-I

Answer

A-III, B-II, C-IV, D-I

Answer

A-III, B-II, C-I, D-IV

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