Q. \( \)lewis structure for \( \text{Na} \).\(\ \)

To draw the Lewis structure for sodium nitride, \(\text{Na}\,^{-}\), you first need to understand what “\(\text{Na}\)” is and what charge is implied.

Important clarification: If the species is written as \(\text{Na}\) only, the Lewis structure is different than if it is written as \(\text{Na}^{-}\) (a sodium anion). Your prompt says “\(\text{na}\)” but the usual Lewis-structure question for “na” is commonly actually for \(\text{Na}\) or \(\text{Na}^{+}\). I will provide the Lewis structures for the most common possibilities:

Below are all three cases.

1) Lewis structure for neutral sodium atom, \(\text{Na}\)

Step 1: Find valence electrons.

Sodium (group 1) has \(1\) valence electron.

Step 2: Draw the symbol and place the valence electron as a dot.

Neutral \(\text{Na}\) has no overall charge.

Result:

\(\text{Na}\) with one dot (one valence electron):

\(\bullet\text{Na}\) (or \(\text{Na}\) with one dot around it)

2) Lewis structure for sodium ion, \(\text{Na}^{+}\)

Step 1: Determine electrons after ion formation.

Sodium forms a \(+1\) ion by losing its one valence electron.

So the ion has \(0\) valence electrons.

Step 2: Draw the symbol with the charge.

Since there are no valence electrons, you draw only the ion with brackets and the charge.

Result:

\(\text{Na}^{+}\) in Lewis form: \([\text{Na}]^{+}\) with no dots.

3) Lewis structure for sodium anion (rare/charge-mismatched case), \(\text{Na}^{-}\)

Step 1: Count valence electrons for \(\text{Na}^{-}\).

Neutral \(\text{Na}\) has \(1\) valence electron.

If it becomes \(\text{Na}^{-}\), it gains one electron.

So it would have \(2\) valence electrons.

Step 2: Place \(2\) dots on the symbol.

The ion would be \(-1\) overall, so you show the charge.

Result:

\([\text{Na}]^{-}\) with two valence electrons as dots.

Final answer (most likely intended): If your question is the common Lewis structure for sodium ions in ionic compounds, the answer is:

\([\text{Na}]^{+}\) with no valence electrons (no dots).