Given below are the quantum numbers for 4 electrons: A. n=3, l=2, ml=1, ms=+12 B. n=4, l=1, ml=0, ms=+12 C. n=4, l=2,…

🧠 The (n+l) Rule Challenge We need to rank four electron states (A, B, C, D) by energy. In multielectron systems, energy is primarily governed by the sum $(n+l)$, with the principal quantum number $n$ acting as the tie-breaker.

🗺️ The Quantum Audit

• D: $n=3, l=1 \implies n+l = 4$ (3p). Lowest.
• A: $n=3, l=2 \implies n+l = 5$ (3d).
• B: $n=4, l=1 \implies n+l = 5$ (4p).
  • Tie-breaker: Since both A and B have $n+l=5$, the one with lower $n$ ($A$) has lower energy. So A < B.
• C: $n=4, l=2 \implies n+l = 6$ (4d). Highest.

Final increasing sequence: D < A < B < C.

⚡ The Periodic Fill Trace Mental trace through the Periodic Table: $3p$ (Period 3) $\to$ $3d$ (Period 4, Transition) $\to$ $4p$ (Period 4, Halogen group) $\to$ $4d$ (Period 5, Transition). Following the physical rows of the table naturally gives you the energy order!

⚠️ Common Traps Ignoring the tie-breaker! Many students think that $n=4$ must always be higher energy than $n=3$. While often true, look at $3d$ vs $4s$; $4s$ ($4+0=4$) is actually lower than $3d$ ($3+2=5$). In this problem, $3d$ and $4p$ are very close, and the $n$ rule is vital.

$\boxed{\text{Answer: (b)}}$