Which of the following sets of quantum numbers is not allowed?

🧠 Spotting the Illegal Address Just like building a house requires a valid city $\to$ street $\to$ number, quantum numbers must follow $n \to l \to m_l$. If at any step the rule $l < n$ or $|m_l| \leq l$ is broken, the entire set is "not allowed."

🗺️ Evaluating the Options

• Option (1): $n=3, l=2, m_l=0$. Valid ($l < 3, 0 \leq 2$).
• Option (2): $n=3, l=2, m_l=-2$. Valid ($l < 3, |-2| \leq 2$).
• Option (3): $n=3, l=3, m_l=-3$. Not allowed. $l$ must be strictly less than $n$. (The highest $l$ for $n=3$ is $2$).
• Option (4): $n=3, l=0, m_l=0$. Valid ($l < 3, 0 \leq 0$).

⚡ The $n-1$ Filter For any $n$, the maximum value of $l$ is $n-1$. So for $n=3$, $l$ can only be 0, 1, or 2. Seeing $l=3$ in Option (3) is an immediate disqualification. No 3f subshell exists.

⚠️ Common Traps Sometimes students think $l$ can go up to $n$ because it goes from $0$ to $n-1$ (total $n$ values). But if $n$ is 3, the values are 0, 1, 2. The number of values is 3, but the maximum value is 2.

$\boxed{\text{Answer: Option (c)}}$