Minimum number of electrons having `m_(s)=(-1/2)` in Cr is "_______"

Answer

To find the minimum number of electrons having a spin magnetic quantum number \( m_s = -\frac{1}{2} \) in chromium (Cr), we can follow these steps:

### Step 1: Determine the Atomic Number and Electronic Configuration
Chromium has an atomic number of 24. The electronic configuration of chromium is:
\[ \text{Cr: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5 \]

### Step 2: Identify the Electrons in Each Orbital
- **s orbitals**: Each s orbital can hold 2 electrons, one with \( m_s = +\frac{1}{2} \) and one with \( m_s = -\frac{1}{2} \).
- **p orbitals**: Each p orbital can hold 6 electrons, with 3 having \( m_s = +\frac{1}{2} \) and 3 having \( m_s = -\frac{1}{2} \).
- **d orbitals**: Each d orbital can hold 10 electrons, but in chromium, we have 5 electrons in the 3d subshell. According to Hund's rule, each of the 5 d electrons will have the same spin first, so they will all have \( m_s = +\frac{1}{2} \) before pairing occurs.

### Step 3: Count the Electrons with \( m_s = -\frac{1}{2} \)
1. **s orbitals**:
- 1s: 1 electron with \( m_s = -\frac{1}{2} \)
- 2s: 1 electron with \( m_s = -\frac{1}{2} \)
- 3s: 1 electron with \( m_s = -\frac{1}{2} \)
- 4s: 0 electrons with \( m_s = -\frac{1}{2} \) (only 1 electron with \( m_s = +\frac{1}{2} \))

2. **p orbitals**:
- 2p: 3 electrons with \( m_s = -\frac{1}{2} \) (3 electrons total, 3 with \( m_s = +\frac{1}{2} \))
- 3p: 3 electrons with \( m_s = -\frac{1}{2} \) (3 electrons total, 3 with \( m_s = +\frac{1}{2} \))

3. **d orbitals**:
- 3d: 0 electrons with \( m_s = -\frac{1}{2} \) (all 5 electrons have \( m_s = +\frac{1}{2} \))

### Step 4: Calculate the Total
Now, we can sum the electrons with \( m_s = -\frac{1}{2} \):
- From s orbitals: \( 1 + 1 + 1 = 3 \)
- From p orbitals: \( 3 + 3 = 6 \)
- From d orbitals: \( 0 \)

Adding these together gives:
\[ 3 + 6 + 0 = 9 \]

### Final Answer
The minimum number of electrons having \( m_s = -\frac{1}{2} \) in chromium is **9**.

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