Number of electrons having `m_(l)=0` for sodium atom is

Options

A. 2
B. 7
C. 9
D. 8

Answer

To determine the number of electrons with a magnetic quantum number \( m_l = 0 \) for a sodium atom, we will follow these steps:

### Step 1: Understand the Electron Configuration of Sodium
Sodium (Na) has an atomic number of 11, which means it has 11 electrons. The electron configuration of sodium is:
\[ \text{Na: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^1 \]

### Step 2: Identify the Quantum Numbers
The magnetic quantum number \( m_l \) can take values from \(-l\) to \(+l\), where \( l \) is the azimuthal quantum number associated with the subshell:
- For \( s \) subshell (\( l = 0 \)), \( m_l = 0 \)
- For \( p \) subshell (\( l = 1 \)), \( m_l = -1, 0, +1 \)

### Step 3: Analyze Each Subshell
Now we will analyze the subshells in the sodium atom's electron configuration to find how many electrons have \( m_l = 0 \):

1. **1s Subshell:**
- Configuration: \( 1s^2 \)
- \( l = 0 \) → \( m_l = 0 \)
- Electrons: 2 (both have \( m_l = 0 \))

2. **2s Subshell:**
- Configuration: \( 2s^2 \)
- \( l = 0 \) → \( m_l = 0 \)
- Electrons: 2 (both have \( m_l = 0 \))

3. **2p Subshell:**
- Configuration: \( 2p^6 \)
- \( l = 1 \) → \( m_l = -1, 0, +1 \)
- Electrons: 6 (1 electron has \( m_l = 0 \))

4. **3s Subshell:**
- Configuration: \( 3s^1 \)
- \( l = 0 \) → \( m_l = 0 \)
- Electrons: 1 (this electron has \( m_l = 0 \))

### Step 4: Count the Total Electrons with \( m_l = 0 \)
Now, we will sum up all the electrons that have \( m_l = 0 \):
- From \( 1s \): 2 electrons
- From \( 2s \): 2 electrons
- From \( 2p \): 1 electron
- From \( 3s \): 1 electron

Total electrons with \( m_l = 0 \):
\[ 2 + 2 + 1 + 1 = 6 \]

### Final Answer
The number of electrons having \( m_l = 0 \) for a sodium atom is **6**.

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