Question

Given the following sets of electron quantum numbers, indicate those that could not occur, and explain your answer. (a) \(1,0,0,-\frac{1}{2}\) (b) \(1,1,0,+\frac{1}{2}\) (c) \(3,2,-2,+\frac{1}{2}\) (d) \(2,1,2,+\frac{1}{2}\) (e) \(4,0,2,+\frac{1}{2}\)

Short answer

Answer: Set (a): \(1,0,0,-\frac{1}{2}\) - Valid Set (b): \(1,1,0,+\frac{1}{2}\) - Invalid Set (c): \(3,2,-2,+\frac{1}{2}\) - Valid Set (d): \(2,1,2,+\frac{1}{2}\) - Invalid Set (e): \(4,0,2,+\frac{1}{2}\) - Invalid

Step-by-step solution

Set (a): \(1,0,0,-\frac{1}{2}\)

For this set, the quantum numbers are n=1, l=0, m=0, and s=-1/2. This combination satisfies all the rules mentioned above. Therefore, this set of quantum numbers could occur.

Set (b): \(1,1,0,+\frac{1}{2}\)

Here, the quantum numbers are n=1, l=1, m=0, and s=+1/2. Notice that since n=1, the highest possible value for l is 0. Therefore, l=1 is invalid for this set of quantum numbers, and it could not occur.

Set (c): \(3,2,-2,+\frac{1}{2}\)

For this set, the quantum numbers are n=3, l=2, m=-2, and s=+1/2. Based on the rules, this set of quantum numbers satisfies all the requirements and could occur.

Set (d): \(2,1,2,+\frac{1}{2}\)

Here, the quantum numbers are n=2, l=1, m=2, and s=+1/2. Since l=1, the range of possible m values is -1, 0, and +1. As a result, m=2 is invalid, and this set of quantum numbers could not occur.

Set (e): \(4,0,2,+\frac{1}{2}\)

In this set, the quantum numbers are n=4, l=0, m=2, and s=+1/2. The value for l is valid but having m=2 when l=0 is not. The set is invalid because m should be in the range of -l to +l, which is only 0 in this case. Therefore, this set of quantum numbers could not occur.