Arrange the following orbitals in decreasing order of energy: A. n=3, l=0, m=0 B. n=4, l=0, m=0 C. n=3, l=1, m=0 D.…

🧠 The n+l Energy Rule In multi-electron atoms, the energy rank of an orbital is determined by the sum of $n$ (principal) and $l$ (azimuthal) quantum numbers. If there's a tie, the higher shell ($n$) wins.

🗺️ The Audit

• A: $n=3, l=0 \to n+l = 3$ (3s)
• C: $n=3, l=1 \to n+l = 4$ (3p)
• B: $n=4, l=0 \to n+l = 4$ (4s)
• D: $n=3, l=2 \to n+l = 5$ (3d)

Between B and C, both have $n+l=4$. Since B has a higher $n$ ($4 > 3$), it is higher in energy than C. Order (Increasing): A < C < B < D. Order (Decreasing): D > B > C > A.

⚡ The "Tie-Breaker" Anchor For any $n+l$ tie, the orbital with the larger shell number ($n$) is always the higher energy level. $4s$ is higher than $3p$ even though they share the same $(n+l)$ score.

⚠️ Common Traps Don't ignore the $n$-shell during a tie. Also, ensure you arrange in decreasing order as requested.

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