The configuration is `1s^(2) 2s^(2) 2p^(5)` `3s^(1)` shows :

Options

A. ground state of the fluorine atom
B. excited state of the fluorine atom
C. excited state of the neon atom
D. excited state of `O_(2)^(-)` ion

Answer

To solve the question regarding the electronic configuration `1s^(2) 2s^(2) 2p^(5) 3s^(1)`, we will follow these steps:

### Step 1: Identify the total number of electrons
The given electronic configuration can be broken down as follows:
- `1s^(2)` contributes 2 electrons
- `2s^(2)` contributes 2 electrons
- `2p^(5)` contributes 5 electrons
- `3s^(1)` contributes 1 electron

Now, we add these electrons together:
\[ 2 + 2 + 5 + 1 = 10 \text{ electrons} \]

### Step 2: Determine the element based on the total number of electrons
The total number of electrons (10) corresponds to the atomic number of an element. The element with atomic number 10 is Neon (Ne).

### Step 3: Analyze the electronic configuration
In its ground state, Neon has the electronic configuration `1s^(2) 2s^(2) 2p^(6)`. However, the given configuration is `1s^(2) 2s^(2) 2p^(5) 3s^(1)`, which indicates that one of the electrons from the `2p` subshell has been excited to the `3s` subshell.

### Step 4: Conclude the state of the element
Since the configuration shows an electron in the `3s` subshell instead of filling the `2p` subshell completely, we can conclude that this configuration represents an excited state of the Neon atom.

### Final Answer
The configuration `1s^(2) 2s^(2) 2p^(5) 3s^(1)` shows the excited state of the Neon atom.

---