To solve the problem of calculating the minimum and maximum number of electrons that may have a magnetic quantum number \( m = +1 \) and a spin quantum number \( s = -\frac{1}{2} \) in chromium (Cr), we will follow these steps:
### Step 1: Determine the electronic configuration of chromium.
Chromium has an atomic number of 24. Its electronic configuration is:
\[ \text{Cr: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \, 3d^5 \]
### Step 2: Identify the relevant orbitals for the magnetic quantum number \( m = +1 \).
The magnetic quantum number \( m \) can take values from \(-l\) to \(+l\), where \( l \) is the azimuthal quantum number.
- For \( p \) orbitals (\( l = 1 \)), the possible \( m \) values are \(-1, 0, +1\).
- For \( d \) orbitals (\( l = 2 \)), the possible \( m \) values are \(-2, -1, 0, +1, +2\).
Thus, \( m = +1 \) can be found in both \( p \) and \( d \) orbitals.
### Step 3: Count the electrons in the relevant orbitals.
1. **For the \( 2p \) orbital**:
- The \( 2p \) subshell can hold a maximum of 6 electrons (3 orbitals, 2 electrons each).
- In the case of chromium, the \( 2p \) orbital is completely filled with 6 electrons.
- The distribution of spins in \( 2p \) is:
- \( m = -1 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = 0 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = +1 \): 1 electron (spin \( -\frac{1}{2} \))
- Therefore, there is **1 electron** with \( m = +1 \) and \( s = -\frac{1}{2} \).
2. **For the \( 3p \) orbital**:
- The \( 3p \) subshell is also completely filled with 6 electrons.
- The distribution of spins in \( 3p \) is:
- \( m = -1 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = 0 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = +1 \): 1 electron (spin \( -\frac{1}{2} \))
- Therefore, there is **1 electron** with \( m = +1 \) and \( s = -\frac{1}{2} \).
3. **For the \( 3d \) orbital**:
- The \( 3d \) subshell has 5 electrons (half-filled).
- The distribution of spins in \( 3d \) is:
- \( m = -2 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = -1 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = 0 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = +1 \): 1 electron (spin \( +\frac{1}{2} \))
- \( m = +2 \): 1 electron (spin \( +\frac{1}{2} \))
- Therefore, there is **1 electron** with \( m = +1 \) that can have either spin \( +\frac{1}{2} \) or \( -\frac{1}{2} \).
### Step 4: Calculate the minimum and maximum number of electrons.
- **Minimum number of electrons with \( m = +1 \) and \( s = -\frac{1}{2} \)**:
- From \( 2p \): 1 electron
- From \( 3p \): 1 electron
- From \( 3d \): 1 electron
- Total minimum = 2 electrons (1 from \( 2p \) and 1 from \( 3p \)).
- **Maximum number of electrons with \( m = +1 \) and \( s = -\frac{1}{2} \)**:
- From \( 2p \): 1 electron
- From \( 3p \): 1 electron
- From \( 3d \): 1 electron
- Total maximum = 3 electrons (1 from \( 2p \), 1 from \( 3p \), and 1 from \( 3d \)).
### Final Answer:
- **Minimum number of electrons**: 2
- **Maximum number of electrons**: 3
