Total number of electrons having `n + l = 3` in `Cr(24)` atom in its ground state is.

Options

A. 8
B. 10
C. 12
D. 6

Explanation

To determine the total number of electrons having \( n + l = 3 \) in a chromium (Cr) atom in its ground state, we can follow these steps:

### Step 1: Identify the Atomic Number
The atomic number of chromium is 24. This means a neutral chromium atom has 24 electrons.

### Step 2: Write the Electronic Configuration
The electronic configuration of chromium (Cr) is:
\[ \text{Cr: } 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \, 4s^1 \]
This configuration shows how the electrons are distributed among the various orbitals.

### Step 3: Calculate \( n + l \) for Each Subshell
We need to calculate \( n + l \) for each subshell:
- For \( 1s^2 \): \( n = 1, l = 0 \) → \( n + l = 1 + 0 = 1 \)
- For \( 2s^2 \): \( n = 2, l = 0 \) → \( n + l = 2 + 0 = 2 \)
- For \( 2p^6 \): \( n = 2, l = 1 \) → \( n + l = 2 + 1 = 3 \)
- For \( 3s^2 \): \( n = 3, l = 0 \) → \( n + l = 3 + 0 = 3 \)
- For \( 3p^6 \): \( n = 3, l = 1 \) → \( n + l = 3 + 1 = 4 \)
- For \( 3d^5 \): \( n = 3, l = 2 \) → \( n + l = 3 + 2 = 5 \)
- For \( 4s^1 \): \( n = 4, l = 0 \) → \( n + l = 4 + 0 = 4 \)

### Step 4: Identify Electrons with \( n + l = 3 \)
From the calculations:
- \( 2p^6 \) has \( n + l = 3 \) → contributes 6 electrons.
- \( 3s^2 \) has \( n + l = 3 \) → contributes 2 electrons.

### Step 5: Total Number of Electrons with \( n + l = 3 \)
Now, we sum the contributions:
\[ 6 \, (\text{from } 2p) + 2 \, (\text{from } 3s) = 8 \]

### Conclusion
The total number of electrons having \( n + l = 3 \) in a chromium atom in its ground state is **8**.