The maximum number of 3d electrons having spin quantum number `s = +(1)/(2)` are

Options

A. 10
B. 14
C. 5
D. any number from 1 to 10

Explanation

To solve the question regarding the maximum number of 3d electrons having a spin quantum number \( s = +\frac{1}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the 3d Subshell**:
- The 3d subshell can hold a maximum of 10 electrons. This is because there are 5 d orbitals (dxy, dyz, dzx, dx²-y², dz²), and each orbital can hold 2 electrons.

2. **Electron Configuration**:
- Each of the 5 d orbitals can accommodate 2 electrons, one with spin \( +\frac{1}{2} \) and the other with spin \( -\frac{1}{2} \).

3. **Distribution of Electrons**:
- When filling the d orbitals, we follow Hund's rule, which states that electrons will fill degenerate orbitals singly first, with parallel spins, before pairing up.

4. **Counting Electrons with Spin \( +\frac{1}{2} \)**:
- Since there are 5 d orbitals, the maximum number of electrons that can have a spin quantum number \( s = +\frac{1}{2} \) is equal to the number of orbitals, which is 5. This is because each of the 5 orbitals can have one electron with spin \( +\frac{1}{2} \).

5. **Conclusion**:
- Therefore, the maximum number of 3d electrons with spin \( +\frac{1}{2} \) is 5.

### Final Answer:
The maximum number of 3d electrons having spin quantum number \( s = +\frac{1}{2} \) is **5**.

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