The number of valence electrons in an atom with the configuration `1s^(2)2s^(2)2p^(6)3s^(2)3p^(2)` is:

Options

A. 6
B. 5
C. 4
D. 2

Answer

To determine the number of valence electrons in the atom with the electronic configuration `1s^(2)2s^(2)2p^(6)3s^(2)3p^(2)`, we can follow these steps:

### Step 1: Identify the electronic configuration
The given electronic configuration is `1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(2)`.

### Step 2: Determine the highest energy level
In the electronic configuration, the highest energy level is the one with the largest principal quantum number (n). Here, the highest energy level is n = 3, which includes the orbitals `3s` and `3p`.

### Step 3: Count the electrons in the highest energy level
Now, we need to count the electrons in the highest energy level (n = 3):
- From `3s^(2)`, there are 2 electrons.
- From `3p^(2)`, there are 2 electrons.

### Step 4: Calculate the total number of valence electrons
Adding the electrons from both orbitals in the highest energy level:
- Total valence electrons = 2 (from `3s`) + 2 (from `3p`) = 4.

### Conclusion
The number of valence electrons in the atom with the given electronic configuration is **4**.

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