Question

Consider the following pairs of electrons
A. $(a)n=3,1=1,m_1=1,m_s=+\frac{1}{2}$
$(b)n=3,1=2,m_1=1,m_s=+\frac{1}{2}$
B. $(a)n=3,1=2,m_1=-2,m_s=-\frac{1}{2}$
$(b)n=3,1=2,m_1=-1,m_s=-\frac{1}{2}$
C. $(a)n=4,1=2,m_1=2,m_s=+\frac{1}{2}$
$(b)n=3,1=2,m_1=2,m_s=+\frac{1}{2}$
The pairs of electron present in degenerate orbitals is/are:

• A. Only A
• B. Only B
• C. Only C
• D. (B) and (C)

Answer 1

Answer: (B)

Based on " $n+l$ " rule only $(B)$ has pair of electron in degenerate orbitals