JEE Main 2022 · 27 June · evening · Numerical answer · +4 / −0
Consider the following set of quantum numbers.
The number of correct sets of quantum numbers is __________.
| n | 1 | m₁ | |
|---|---|---|---|
| A. | 3 | 3 | -3 |
| B. | 3 | 2 | -2 |
| C. | 2 | 1 | +1 |
| D. | 2 | 2 | +2 |
Answer and solution
Answer 2
Solution
Chemistry ma’am · 3 steps
Check l against n first; once l is allowed, m has to fit inside -l to +l.
- The two rules. For a shell n, l can only be 0, 1, …, n-1, so l is always smaller than n. For a given l, mₗ runs from -l to +l in whole steps.
Set n, l, mₗ Is l ≤ n-1? Is mₗ within ± l? Verdict A 3, 3, -3 No, l can be at most 2 not checked invalid B 3, 2, -2 Yes Yes valid (a 3d orbital) C 2, 1, +1 Yes Yes valid (a 2p orbital) D 2, 2, +2 No, l can be at most 1 not checked invalid - Missing subshells. In A, l = 3 would be a 3f orbital, and in D, l = 2 would be a 2d orbital. Neither exists, because l can never equal n.
- So. Only sets B and C are allowed, so the number of correct sets is 2.
The common slip: 4. Checking only that mₗ lies between -l and +l makes all four sets look valid and gives 4. Sets A and D fail the rule l ≤ n-1.
The idea to keep. Always test l < n first. A set like 2d or 3f is invalid even if mₗ looks fine for that l, because the subshell itself does not exist.
Answer from the NTA final answer key. Solution by PrepWiser.