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

For a hydrogen-like atom in the absence of electric and magnetic fields, orbital energy is determined solely by the principal quantum number \(n\).
Step 1: In multielectron atoms, electron-electron interactions and the quantum numbers \(l\) and \(m_l\) cause further splitting of energy levels.
Step 2: Orbitals sharing the same \(n\) but differing in \(l\) and \(m_l\) exhibit identical energy due to the degeneracy of the \(n\) level. 

Final Conclusion: Option (1) is correct, stating that D and E possess the same energy.