What is the hybridization about the nitrogen center in `"NCl"_2^(+)`?

`"N"`, from the periodic table, has `5` valence electrons, since its electron configuration ends in `2s^2 2p^3`. Each `"Cl"` has `7` valence electrons. The charge of `+1` then means that there is one less electron than in `"NCl"_2`. That accounts for a total of `5 + 2 xx 7 - 1 = 18` valence electrons.

• Distributing `3` lone pairs on each chlorine atom, we use `12` of these electrons.
• Making two bonds off of nitrogen uses `4` more.
• The remaining `2` become a lone pair.

But then nitrogen only has 6 valence electrons. So, instinctively, the next best thing is to make a second `"N"-"Cl"` bond...

Since `"Cl"` is slightly more electronegative, this is not extremely favorable. So, it is a bit better to delocalize like this, using one valence electron from each chlorine instead of two from one:

This gives for formal charge:

• `"7 valence" - "6.5 owned" = +0.5` on each chlorine atom
• `"5 valence" - "5 owned" = 0` on the nitrogen atom

which adds up to a total charge of `+1` as expected. As a check, each atom has `8` valence electrons:

• `2+2+1+2+1/2(2) = 8` total around each chlorine atom
• `2+2+2+1/2(2)+1/2(2) = 8` total around the nitrogen atom

Assuming this is the correct structure, this indicates a hybridization of `color(blue)(sp^2)` since it has three electron groups.