Question

In a multielectron atom, which of the following orbitals described by three quantum numbers will have the same energy in absence of electric and magnetic fields?
A. \( n = 1, l = 0, m_l = 0 \)
B. \( n = 2, l = 0, m_l = 0 \)
C. \( n = 2, l = 1, m_l = 1 \)
D. \( n = 3, l = 2, m_l = 1 \)
E. \( n = 3, l = 2, m_l = 0 \)
Choose the correct answer from the options given below:

• A. D and E Only
• B. C and D Only
• C. B and C Only
• D. A and B Only

Hint:

In multielectron atoms, degeneracy occurs for orbitals with the same principal quantum number \(n\).

Answer 1

The Correct Option is A

In a multi-electron atom, in the absence of electric and magnetic fields, the energy of an orbital depends only on the principal quantum number $ n $ and the azimuthal quantum number $ l $. The magnetic quantum number $ m_l $ has no effect on energy under these conditions (only affects energy in presence of magnetic field — Zeeman effect).

So, we need to find pairs of orbitals that have the same $ n $ and $ l $ values:

• A. $ n = 1, l = 0, m_l = 0 $ → 1s orbital
• B. $ n = 2, l = 0, m_l = 0 $ → 2s orbital
• C. $ n = 2, l = 1, m_l = 1 $ → 2p orbital
• D. $ n = 3, l = 2, m_l = 1 $ → 3d orbital
• E. $ n = 3, l = 2, m_l = 0 $ → 3d orbital

D and E have the same $ n = 3 $ and $ l = 2 $, so they belong to the same energy level in the absence of external fields.

Final Answer:
The final answer is $ D \text{ and } E \text{ Only} $.