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

Answer

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

### Step 1: Write the Electron Configuration of Chromium
Chromium has an atomic number of 24, which means it has 24 electrons. The electron configuration of chromium is:
\[ \text{Cr: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \, 4s^1 \]

### Step 2: Identify the Spin States of Electrons
In quantum mechanics, electrons can have two possible spin states: \( m_s = +\frac{1}{2} \) (spin up) and \( m_s = -\frac{1}{2} \) (spin down). When two electrons occupy the same orbital, one will have a spin of \( +\frac{1}{2} \) and the other \( -\frac{1}{2} \).

### Step 3: Analyze the Electron Configuration
- For the filled orbitals:
- \( 1s^2 \): 2 electrons (1 spin up and 1 spin down)
- \( 2s^2 \): 2 electrons (1 spin up and 1 spin down)
- \( 2p^6 \): 6 electrons (3 pairs, so 3 spin up and 3 spin down)
- \( 3s^2 \): 2 electrons (1 spin up and 1 spin down)
- \( 3p^6 \): 6 electrons (3 pairs, so 3 spin up and 3 spin down)

- For the partially filled orbitals:
- \( 3d^5 \): 5 electrons (each with spin \( +\frac{1}{2} \), since they are unpaired)
- \( 4s^1 \): 1 electron (spin \( +\frac{1}{2} \))

### Step 4: Count the Electrons with \( m_s = -\frac{1}{2} \)
Now we will count the electrons with \( m_s = -\frac{1}{2} \):
- From \( 1s^2 \): 1 electron
- From \( 2s^2 \): 1 electron
- From \( 2p^6 \): 3 electrons
- From \( 3s^2 \): 1 electron
- From \( 3p^6 \): 3 electrons
- From \( 3d^5 \): 0 electrons (all are spin up)
- From \( 4s^1 \): 0 electrons (spin up)

Adding these together:
\[ 1 + 1 + 3 + 1 + 3 = 9 \]

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

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