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