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 the absence of electric and magnetic fields, the energy of orbitals depends only on the principal quantum number \(n\) for a hydrogen-like atom. 
Step 1: For multielectron atoms, the energy levels are further split due to electron-electron interactions and quantum numbers \(l\) and \(m_l\).
Step 2: Orbitals with the same \(n\) but different \(l\) and \(m_l\) have the same energy due to the degeneracy of the \(n\) level. 

Final Conclusion: The correct answer is Option (1), where D and E have the same energy.