1. Identify the orbitals from the quantum numbers

For each electron, use $n$ and $l$:

• $l=0 \to s$
• $l=1 \to p$
• $l=2 \to d$

So,

• A: $n=3, l=2 \Rightarrow 3d$
• B: $n=4, l=1 \Rightarrow 4p$
• C: $n=4, l=2 \Rightarrow 4d$
• D: $n=3, l=1 \Rightarrow 3p$

The values of $m_l$ and $m_s$ do not affect the energy in a multi-electron atom for this comparison; energy mainly depends on $n$ and $l$.

2. Apply the $(n+l)$ rule

For multi-electron atoms, orbital energy increases with increasing value of $(n+l)$. If two orbitals have the same $(n+l)$ value, then the one with smaller $n$ has lower energy.

Now calculate:

• D = 3p: $n+l = 3+1=4$
• A = 3d: $n+l = 3+2=5$
• B = 4p: $n+l = 4+1=5$
• C = 4d: $n+l = 4+2=6$

3. Arrange in increasing energy

• Lowest is D (3p) since $(n+l)=4$
• Next are A (3d) and B (4p), both with $(n+l)=5$
  • Compare $n$: for $3d$, $n=3$; for $4p$, $n=4$
  • Smaller $n$ has lower energy, so $3d < 4p$
• Highest is C (4d) since $(n+l)=6$

Thus, $D < A < B < C$

4. Match with the given options

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, the derived answer agrees with the stored answer.