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We had a test and and we had a question asking the number of valence electrons in an azide ion, $\ce{N3^{-}}$. I find out that the answer is 16 but I could not understand how. I saw the structure. It says that the end nitrogen have a -1 charge and the middle nitrogen has a +1 charge.
From where does the one of the end nitrogen get the electron assuming the other end nitrogen gets it from the middle nitrogen?
Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence. Therefore, a single elemental nitrogen has five valence electrons.
There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.
Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form. So, the total number of valence electrons is $15 + 1 = 16$.
Nitrogen is a 5 group element hence has valency 5 and AZIDE ion has an negative charge it means it gained an electron hence as AZIDE HAS 3 NITROGENS and a negative charge it will be 5 multiplied by 3 = 15 + one electron gained = 16
