The correct electronic configuration of `Na^(+)` is:

Explanation

To find the correct electronic configuration of the sodium ion \( \text{Na}^+ \), we can follow these steps:

### Step 1: Determine the Atomic Number of Sodium
Sodium (Na) has an atomic number of 11. This means that a neutral sodium atom has 11 electrons.

### Step 2: Write the Electronic Configuration of Neutral Sodium
The electronic configuration of neutral sodium (Na) can be written as:
\[ \text{Na}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^1 \]
This indicates that sodium has 2 electrons in the first shell (1s), 8 electrons in the second shell (2s and 2p), and 1 electron in the third shell (3s).

### Step 3: Remove One Electron for the Sodium Ion
Since \( \text{Na}^+ \) indicates that sodium has lost one electron, we need to remove the outermost electron, which is in the 3s subshell.

### Step 4: Write the Electronic Configuration of \( \text{Na}^+ \)
After removing one electron from the neutral sodium configuration, the electronic configuration of \( \text{Na}^+ \) becomes:
\[ \text{Na}^+: 1s^2 \, 2s^2 \, 2p^6 \]
This configuration shows that \( \text{Na}^+ \) now has a total of 10 electrons, which corresponds to the electronic configuration of neon (Ne), a noble gas.

### Final Answer
The correct electronic configuration of \( \text{Na}^+ \) is:
\[ 1s^2 \, 2s^2 \, 2p^6 \]

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