Consider the following pairs of electrons
(A) (a) `n=3 , l =1 , m_1=1, m_s=+1/2`
(b) `n=3 , l =2 , m_1=1, m_s=+1/2`
(B) (a) `n=3 , l =2 , m_1=-2, m_s=-1/2`
(b) `n=3 , l =2 , m_1=-1, m_s=-1/2`
(C) (a) `n=4 , l =2 , m_1=-2, m_s=+1/2`
(b) `n=3 , l =2 , m_1=2, m_s=+1/2`
The pairs of electrons present in degenerate orbitals is/are :

Options

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

Answer

To determine which pairs of electrons are present in degenerate orbitals, we need to analyze each pair based on their quantum numbers. Degenerate orbitals are those that have the same energy level.

### Step-by-Step Solution:

1. **Understanding Quantum Numbers**:
- Each electron is described by four quantum numbers:
- Principal quantum number (n)
- Azimuthal quantum number (l)
- Magnetic quantum number (m_l)
- Spin quantum number (m_s)

2. **Analyzing Pair (A)**:
- (a) `n=3, l=1, m_l=1, m_s=+1/2` corresponds to a 3p orbital.
- (b) `n=3, l=2, m_l=1, m_s=+1/2` corresponds to a 3d orbital.
- Since the azimuthal quantum number (l) is different (1 for p and 2 for d), these orbitals are not degenerate.

3. **Analyzing Pair (B)**:
- (a) `n=3, l=2, m_l=-2, m_s=-1/2` corresponds to a 3d orbital.
- (b) `n=3, l=2, m_l=-1, m_s=-1/2` also corresponds to a 3d orbital.
- Both electrons are in the same subshell (3d) and have the same azimuthal quantum number (l=2). Therefore, they are degenerate orbitals.

4. **Analyzing Pair (C)**:
- (a) `n=4, l=2, m_l=-2, m_s=+1/2` corresponds to a 4d orbital.
- (b) `n=3, l=2, m_l=2, m_s=+1/2` corresponds to a 3d orbital.
- Here, the principal quantum number (n) is different (4 for the first and 3 for the second), which means they are not degenerate.

5. **Conclusion**:
- The only pair of electrons that are present in degenerate orbitals is from Pair (B).

### Final Answer:
The pairs of electrons present in degenerate orbitals is: **(B)**.

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