I am asked to draw the 'dot and cross' diagram to show the ionic bonding in sodium chloride. I am asked to show only the outer electronic structure.

Sodium has 2.8.1 electronic configuration. After losing 1 electron to chlorine, it has 2.8 electronic configuration.

For sodium ion, do I draw 8 outer electrons or can it just be represented by [Na]+ to show that it has lost 1 valence electron?

Because I know that for chloride ion (2.8), I will need to show 8 electrons in the outer electronic structure.

Na element is written as Na. (that's the symbol plus the outer electron). So when it loses that 1 outside electron, it is represented by Na+. You are correct about the Cl but remember to put a negative charge on the chloride ion.

In the dot and cross diagram for ionic bonding, you represent the outer electronic structure of each ion involved in the bond.

For sodium ion (Na+), it has lost 1 electron from its original electron configuration, which was 2.8.1. So, in the dot and cross diagram, you only need to show the remaining electrons, which is 2.8. Therefore, you can represent the sodium ion as [Na]+ to indicate that it has lost 1 valence electron.

On the other hand, for chloride ion (Cl-), it has gained 1 electron to achieve a stable octet configuration, which is the electron configuration for noble gases. The electron configuration for chloride ion is 2.8.8. In the dot and cross diagram, you represent all 8 outer electrons for the chloride ion.

To summarize, in the dot and cross diagram for the ionic bonding in sodium chloride, you represent the sodium ion as [Na]+, showing 2.8, and the chloride ion as Cl-, showing 2.8.8.

I'm asked to draw the dot and cross diagram of sodium chloride

Correct 2.8.1=+1charge gaining 2.8๐Ÿ‘๐Ÿ‘