The maximum number of electrons with s = +1/2 in an orbital for which l=2 is

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the question regarding the maximum number of electrons with a spin of \( s = +\frac{1}{2} \) in an orbital for which \( l = 2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Quantum Numbers**:
- The quantum number \( l \) represents the angular momentum quantum number, which determines the shape of the orbital. For \( l = 2 \), the corresponding type of orbital is the d-orbital.

2. **Identify the Number of Orbitals**:
- The number of orbitals for a given \( l \) value can be calculated using the formula \( 2l + 1 \).
- For \( l = 2 \):
\[ \text{Number of orbitals} = 2(2) + 1 = 5 \]
- Therefore, there are 5 d-orbitals.

3. **Determine the Electron Capacity of Each Orbital**:
- Each orbital can hold a maximum of 2 electrons, which can have different spins: \( +\frac{1}{2} \) and \( -\frac{1}{2} \).

4. **Calculate the Maximum Number of Electrons with \( s = +\frac{1}{2} \)**:
- Since there are 5 orbitals and each orbital can accommodate one electron with \( s = +\frac{1}{2} \), the total number of electrons with this spin in all orbitals is:
\[ \text{Maximum electrons with } s = +\frac{1}{2} = 5 \]

5. **Conclusion**:
- The maximum number of electrons with \( s = +\frac{1}{2} \) in an orbital for which \( l = 2 \) is **5**.

### Final Answer:
The maximum number of electrons with \( s = +\frac{1}{2} \) in an orbital for which \( l = 2 \) is **5**.

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