Question:
If n, l, m and s represent the symbols of quantum numbers, the impossible quantum number set for the electron in terms of n, l, m and s respectively is
If n, l, m and s represent the symbols of quantum numbers, the impossible quantum number set for the electron in terms of n, l, m and s respectively is
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The most common errors in quantum number sets involve the 'l' and 'm' values. Always check them sequentially: first, is 'l' valid for the given 'n'? Second, is 'm' valid for the given 'l'? This structured check helps to quickly spot the error.
Updated On: Jul 1, 2026
- 2, 0, -1, +1/2
- 3, 0, 0, -1/2
- 4, 1, +1, +1/2
- 3, 2, -1, -1/2
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
We are given four sets of quantum numbers (n, l, m, s) and we need to identify which set violates the rules governing these numbers.
Step 2: Key Formula or Approach:
The rules for the quantum numbers are:
1.
Principal quantum number (n): Can be any positive integer (1, 2, 3, ...).
2.
Azimuthal quantum number (l): Can be any integer from 0 to n-1.
3.
Magnetic quantum number (m): Can be any integer from -l to +l, including 0.
4.
Spin quantum number (s): Can be +1/2 or -1/2.
Step 3: Detailed Explanation:
Let's check each set against the rules:
-
(A) n=2, l=0, m=-1, s=+1/2:
- n=2 is valid.
- l=0 is valid (since \(0 \le 0 \le 2-1\)).
- m=-1 is
invalid. For l=0, the only possible value for m is 0. Since the rule is violated, this set is impossible.
-
(B) n=3, l=0, m=0, s=-1/2:
- n=3 is valid.
- l=0 is valid (since \(0 \le 0 \le 3-1\)).
- m=0 is valid (since for l=0, m must be 0).
- s=-1/2 is valid. This set is possible (it describes an electron in the 3s orbital).
-
(C) n=4, l=1, m=+1, s=+1/2:
- n=4 is valid.
- l=1 is valid (since \(0 \le 1 \le 4-1\)).
- m=+1 is valid (since for l=1, m can be -1, 0, +1).
- s=+1/2 is valid. This set is possible (it describes an electron in a 4p orbital).
-
(D) n=3, l=2, m=-1, s=-1/2:
- n=3 is valid.
- l=2 is valid (since \(0 \le 2 \le 3-1\)).
- m=-1 is valid (since for l=2, m can be -2, -1, 0, +1, +2).
- s=-1/2 is valid. This set is possible (it describes an electron in a 3d orbital).
Step 4: Final Answer:
The impossible quantum number set is (2, 0, -1, +1/2).
We are given four sets of quantum numbers (n, l, m, s) and we need to identify which set violates the rules governing these numbers.
Step 2: Key Formula or Approach:
The rules for the quantum numbers are:
1.
Principal quantum number (n): Can be any positive integer (1, 2, 3, ...).
2.
Azimuthal quantum number (l): Can be any integer from 0 to n-1.
3.
Magnetic quantum number (m): Can be any integer from -l to +l, including 0.
4.
Spin quantum number (s): Can be +1/2 or -1/2.
Step 3: Detailed Explanation:
Let's check each set against the rules:
-
(A) n=2, l=0, m=-1, s=+1/2:
- n=2 is valid.
- l=0 is valid (since \(0 \le 0 \le 2-1\)).
- m=-1 is
invalid. For l=0, the only possible value for m is 0. Since the rule is violated, this set is impossible.
-
(B) n=3, l=0, m=0, s=-1/2:
- n=3 is valid.
- l=0 is valid (since \(0 \le 0 \le 3-1\)).
- m=0 is valid (since for l=0, m must be 0).
- s=-1/2 is valid. This set is possible (it describes an electron in the 3s orbital).
-
(C) n=4, l=1, m=+1, s=+1/2:
- n=4 is valid.
- l=1 is valid (since \(0 \le 1 \le 4-1\)).
- m=+1 is valid (since for l=1, m can be -1, 0, +1).
- s=+1/2 is valid. This set is possible (it describes an electron in a 4p orbital).
-
(D) n=3, l=2, m=-1, s=-1/2:
- n=3 is valid.
- l=2 is valid (since \(0 \le 2 \le 3-1\)).
- m=-1 is valid (since for l=2, m can be -2, -1, 0, +1, +2).
- s=-1/2 is valid. This set is possible (it describes an electron in a 3d orbital).
Step 4: Final Answer:
The impossible quantum number set is (2, 0, -1, +1/2).
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