Chemistry · Structure Of Atom · NEET
There are four simple rules. (1) n (principal) is a whole number: 1, 2, 3, ... — it can never be 0 or negative. (2) l (azimuthal) goes from 0 up to n-1. So if n=3, l can be 0, 1, or 2 only. (3) m (magnetic) goes from -l to +l, including 0. So if l=2, m can be -2, -1, 0, +1, +2 (that is 2l+1 = 5 values). (4) s (spin) is only +1/2 or -1/2. If all four values follow these rules, the set is allowed. If even one breaks, it is forbidden.
Because l can only reach up to n-1, never n. Think of it as: the highest subshell in a shell is always one below the shell number. If n=3, the biggest l allowed is 2 (the d subshell). A set like n=3, l=3 is forbidden because there is no 3f subshell. Quick check: whenever you see l equal to or bigger than n, mark the set as impossible.
Check in a fixed order: n, then l, then m, then s. First make sure n is a positive whole number. Then check l is between 0 and n-1. Then check m is between -l and +l (and includes 0 if a full list is given). Then check s is +1/2 or -1/2. Stop at the first rule that breaks — that set is your answer. Doing it in this order stops you from missing the trap.
Count them and check the range. For a given l, the m values must be exactly -l to +l, which is 2l+1 values, and must include 0. NEET loves to give a list that skips 0. For example, for l=2 the correct list is -2, -1, 0, +1, +2 (5 values). If a set writes -2, -1, +1, +2 (only 4 values, missing 0), that set is forbidden. Always look for a missing 0.
No. The principal quantum number n starts at 1. There is no zero shell and no negative shell. So any set with n=0 or a negative n is automatically forbidden before you even check the other numbers. This is why you always test n first.
NEET almost every year gives a question like 'which set of quantum numbers is not possible' or 'the incorrect set is'. It is a fast, sure mark if you know the four rules. You do not need to calculate anything — just apply the rules in order. Students lose this easy mark only because they forget that l stops at n-1 or that m must include 0.
The incorrect set of quantum numbers from the following is:
The relation between n_m (n_m = the number of permissible values of the magnetic quantum number m) for a given value of the azimuthal quantum number l is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
An allowed set has every value (n, l, m, s) following its rule, so a real electron can have it. A forbidden set breaks at least one rule, so no electron in an atom can ever have that combination.
No. l must be from 0 to n-1. So l can never equal n or be larger than n. If you see l bigger than or equal to n, the set is forbidden.
Exactly 2l+1 values, running from -l to +l and always including 0. For l=0 there is 1 value, for l=1 there are 3, for l=2 there are 5, and for l=3 there are 7.
Only two: +1/2 and -1/2. Any other value, like 0 or 1, makes the set forbidden.
Check n first (must be a positive whole number), then l (0 to n-1), then m (-l to +l, includes 0), then s (+1/2 or -1/2). Stop at the first rule that fails.
