Compare the energies of following sets of quantum numbers for a multielectron system: (A) n=4, l=1 (4p) (B) n=4, l=2…

🧠 The Shielding Effect Hierarchy In a lone-proton Hydrogen atom, all subshells in the same shell ($3s, 3p, 3d$) have equal energy. But in multielectron atoms, electrons shield each other from the nucleus. This breaks the degeneracy and forces us to use the n + l rule (Aufbau Principle).

🗺️ The (n+l) Audit Let's rank the given sets:

• (C) 3p: $n=3, l=1 \implies n+l = 4$.
• (E) 4s: $n=4, l=0 \implies n+l = 4$.
  • Conflict! Same $n+l$. The one with lower $n$ has lower energy. So 3p < 4s.
• (D) 3d: $n=3, l=2 \implies n+l = 5$.
• (A) 4p: $n=4, l=1 \implies n+l = 5$.
  • Conflict! Same $n+l$. Lower $n$ wins. So 3d < 4p.
• (B) 4d: $n=4, l=2 \implies n+l = 6$. Highest value.

Final Order: (C) 3p < (E) 4s < (D) 3d < (A) 4p < (B) 4d.

⚡ The Diagonals of Aufbau If you visualize the Aufbau diagram, the diagonal lines represent constant $(n+l)$. You just read them left-to-right: $3s \to 3p \to 4s \to 3d \to 4p \to 5s \to 4d \dots$. Scanning this mental list makes comparison instant.

⚠️ Common Traps The $n+l$ tie-breaker! Students often forget that if $n+l$ is equal, the energy is determined by the principal quantum number $n$. $4s$ feels "higher up" than $3p$, and indeed it is, despite having the same total quantum sum!

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