The electrons, identified by quantum numbers n and l, (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3 — Atomic Structure Chemistry Question

💡 Solution & Explanation

### Step 1 - The $(n + l)$ Rule In multi-electron atoms, the energy of an electron in an atomic orbital is determined by both the principal quantum number ($n$) and the azimuthal quantum number ($l$). The relative energies of various orbitals are governed by the **$(n + l)$ rule**: 1. **Rule 1:** An orbital with a lower value of $(n + l)$ has lower energy and is filled first. 2. **Rule 2:** If two orbitals have the same $(n + l)$ value, the orbital with the lower $n$ has lower energy. --- ### Step 2 - Calculate $(n + l)$ for Each State * **State (i):** $n = 4, l = 1$ → 4p orbital: $(n + l) = 5$ * **State (ii):** $n = 4, l = 0$ → 4s orbital: $(n + l) = 4$ * **State (iii):** $n = 3, l = 2$ → 3d orbital: $(n + l) = 5$ * **State (iv):** $n = 3, l = 1$ → 3p orbital: $(n + l) = 4$ --- ### Step 3 - Apply the Rules 1. States with $(n+l) = 4$ (ii, iv) have lower energy than states with $(n+l) = 5$ (i, iii). 2. Within $(n+l) = 4$: state (iv) has $n=3$, state (ii) has $n=4$ → iv < ii 3. Within $(n+l) = 5$: state (iii) has $n=3$, state (i) has $n=4$ → iii < i Final increasing energy order: $$\text{(iv)} < \text{(ii)} < \text{(iii)} < \text{(i)}$$ --- ### Step 4 - Evaluation of Options * **Option (A):** iv < ii < iii < i — **Correct.** * **Option (B):** ii < iv < i < iii — Incorrect. Violates Rule 2: 3p has lower $n$ than 4s. * **Option (C):** i < iii < ii < iv — Incorrect. This is the reverse (decreasing) order. * **Option (D):** iii < i < iv < ii — Incorrect. Places 3d below 3p, violating Rule 1. $$\text{Correct Option: } \boxed{\text{A}}$$