Rule for degeneracy in multielectron atoms

In a multielectron atom, the energy of an orbital depends mainly on: $n \text{ and } l$ and does not depend on the magnetic quantum number $m_l$ (in the absence of electric and magnetic fields).

Therefore, orbitals having the same $n$ and $l$ but different $m_l$ will have the same energy.

Check each given orbital

• A: $n=1,\ l=0,\ m_l=0$
  This is $1s$.

• B: $n=2,\ l=0,\ m_l=0$
  This is $2s$.

• C: $n=2,\ l=1,\ m_l=1$
  This is one of the $2p$ orbitals.

• D: $n=3,\ l=2,\ m_l=1$
  This is one of the $3d$ orbitals.

• E: $n=3,\ l=2,\ m_l=0$
  This is another one of the $3d$ orbitals.

Find orbitals with same energy

Compare their $n$ and $l$ values:

• A: $(1,0)$
• B: $(2,0)$
• C: $(2,1)$
• D: $(3,2)$
• E: $(3,2)$

Only D and E have the same $n$ and $l$; they differ only in $m_l$.

Hence, they will have the same energy.

Match with given options

The correct option is: $\boxed{\text{D: D and E Only}}$

Comparison with stored answer

Stored correct answer = D
Derived answer = D

So they agree.