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JEE Advanced 2022 Paper 1 · Chemistry · Atomic Structure · easy

Match each orbital in List-I with the correct description of its nodes in List-II. 3p | 1. Total nodes 2, of which 1 is radial | | Q. 4d | 2. Total…

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Match each orbital in List-I with the correct description of its nodes in List-II.

| List-I | List-II |
|---|---|
| P. 3p | 1. Total nodes 2, of which 1 is radial |
| Q. 4d | 2. Total nodes 3, of which 1 is radial |
| R. 4f | 3. Total nodes 3, none radial |
| S. 5s | 4. Total nodes 4, all radial |
| | 5. Total nodes 3, of which 2 are radial |
  1. A. P→1, Q→2, R→3, S→4
  2. B. P→1, Q→5, R→3, S→4
  3. C. P→2, Q→1, R→3, S→4
  4. D. P→1, Q→2, R→5, S→4
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Answer: (A) P→1, Q→2, R→3, S→4

The number of radial nodes is n−l−1n-l-1 and the number of angular nodes is ll, so the total number of nodes is n−1n-1. P (3p): n=3,l=1n=3,l=1: radial =1=1, angular =1=1, total =2=2: entry 1. Q (4d): n=4,l=2n=4,l=2: radial =1=1, angular =2=2, total =3=3: entry 2. R (4f): n=4,l=3n=4,l=3: radial =0=0, angular =3=3, total =3=3 with none radial: entry 3. S (5s): n=5,l=0n=5,l=0: radial =4=4, angular =0=0, total =4=4, all radial: entry 4. Entry 5 (total 3, two radial) belongs to 4p (n−l−1=2n-l-1=2), which is not in List-I. Correct code: P→1, Q→2, R→3, S→4.
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