Question

which of the following is not a valid set of quantum numbers?

\\(n = 2, l = 1, m = 0\\)
\\(n = 1, l = 0, m = 0\\)
\\(n = 3, l = 3, m = 3\\)

Explanation:

State the rules for quantum numbers

Quantum numbers describe atomic orbitals.
The principal quantum number \(n\) must be a positive integer:

$$n \in \{1, 2, 3, \dots\}$$

The angular momentum quantum number \(l\) depends on \(n\):

$$0 \le l \le n - 1$$

The magnetic quantum number \(m\) (or \(m_l\)) depends on \(l\):

$$-l \le m \le +l$$

Analyze the first option

Evaluate \(n = 2\), \(l = 1\), \(m = 0\).
Since \(n = 2\), \(l\) can be \(0\) or \(1\). Here, \(l = 1\), which is valid.
Since \(l = 1\), \(m\) can be \(-1, 0, 1\). Here, \(m = 0\), which is valid.

Analyze the second option

Evaluate \(n = 1\), \(l = 0\), \(m = 0\).
Since \(n = 1\), \(l\) can only be \(0\). Here, \(l = 0\), which is valid.
Since \(l = 0\), \(m\) can only be \(0\). Here, \(m = 0\), which is valid.

Analyze the third option

Evaluate \(n = 3\), \(l = 3\), \(m = 3\).
Since \(n = 3\), the allowed values for \(l\) are \(0, 1, 2\).
The value \(l = 3\) is invalid because \(l\) must be strictly less than \(n\).
Consequently, \(m = 3\) is also invalid because \(m\) cannot exceed \(l\).

Answer:

• \(n = 2, l = 1, m = 0\)
• \(n = 1, l = 0, m = 0\)
• \(n = 3, l = 3, m = 3\) (Correct answer)