Total number of electrons having `n + l = 4` in `V(23)` atom in its ground state is

Answer

To solve the problem of finding the total number of electrons in the Vanadium (V) atom with atomic number 23 that have \( n + l = 4 \) in its ground state, we can follow these steps:

### Step 1: Write the Electronic Configuration of Vanadium
The atomic number of Vanadium is 23. The electronic configuration can be written as:
\[ \text{V: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^3 \]

### Step 2: Identify the Values of \( n \) and \( l \)
In the electronic configuration:
- The \( s \) orbital has \( l = 0 \)
- The \( p \) orbital has \( l = 1 \)
- The \( d \) orbital has \( l = 2 \)

### Step 3: Find Combinations Where \( n + l = 4 \)
We need to find the combinations of \( n \) and \( l \) such that:
\[ n + l = 4 \]

1. **For \( n = 3 \)**:
- \( l = 1 \) (from \( 3p \)): \( n + l = 3 + 1 = 4 \) → contributes 6 electrons (from \( 3p^6 \))

2. **For \( n = 2 \)**:
- \( l = 2 \) (from \( 2d \)): Not applicable since there is no \( 2d \) orbital.

3. **For \( n = 4 \)**:
- \( l = 0 \) (from \( 4s \)): \( n + l = 4 + 0 = 4 \) → contributes 2 electrons (from \( 4s^2 \))

### Step 4: Count the Total Electrons
Now we can sum the contributions from the valid combinations:
- From \( 3p^6 \): 6 electrons
- From \( 4s^2 \): 2 electrons

Total number of electrons where \( n + l = 4 \):
\[ 6 + 2 = 8 \]

### Final Answer
The total number of electrons in the Vanadium atom with \( n + l = 4 \) in its ground state is **8**.

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