Which of the following set of quantum numbers is permissible? — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - Rules for Permissible Quantum Numbers For an electron in an atom, its state is described by four quantum numbers $(n, l, m, s)$ satisfying: 1. $n \in \{1, 2, 3, \ldots\}$ (positive integer) 2. $0 \leq l \leq n-1$ (integer) 3. $-l \leq m \leq +l$ (integer) 4. $s \in \left\{+\frac{1}{2}, -\frac{1}{2}\right\}$ ### Step 2 - Analyzing Option (A): $(4, 1, +2, +1/2)$ * $n = 4$: Permissible * $l = 1$: Permissible ($0 \leq 1 \leq 3$) → p-subshell * $m = +2$: **NOT permissible** — for $l=1$, allowed $m \in \{-1, 0, +1\}$. Since $+2 > l$, this violates Rule 3. * **Option (A) is invalid.** ### Step 3 - Analyzing Option (B): $(4, 2, -1, +1/2)$ * $n = 4$: Permissible * $l = 2$: Permissible ($0 \leq 2 \leq 3$) → d-subshell * $m = -1$: Permissible — for $l=2$, allowed $m \in \{-2, -1, 0, +1, +2\}$ * $s = +1/2$: Permissible * **All four values satisfy the rules. Option (B) is valid.** ### Step 4 - Analyzing Options (C) and (D) * **Option (C): $(4, 0, 0, 1)$** — $s = 1$ is not a permitted spin value. Only $+1/2$ or $-1/2$ are allowed. **Invalid.** * **Option (D): $(4, 4, +2, -1/2)$** — $l = 4$ for $n = 4$ violates Rule 2: $l_{\max} = n-1 = 3$. **Invalid.** $$\text{Correct Option: } \boxed{\text{B}}$$