Answer:
See figure 1
Explanation:
For the azide ion [tex]N_3^-^1[/tex] we will 5 electrons for each nitrogen and 1 electron for the negative charge, so:
[tex]5~electrons*~3~N=~15~e^-\\\\15~e^-~+~1e^-~=~16e^-[/tex]
We have to organize the 16 electrons. For example in figure 1 we will have the 3 nitrogens in line, if we check the formal charge for each atom we will get:
[tex]FC=~valence~electrons~-(lone~pairs~e^-+\frac{1}{2}~bound~electrons)[/tex]
For figure 1 we will have:
[tex]A:~FC=~5~-(4+\frac{1}{2}4)=~-1[/tex]
[tex]B:~FC=~5~-(0+\frac{1}{2}8)=~+1[/tex]
[tex]C:~FC=~5~-(4+\frac{1}{2}4)=~-1[/tex]
In total we will have a charge of => -1+1-1= -1
