Question

Lewis dot structure for N3-

Answer 1

Answer:

See figure 1

Explanation:

For the azide ion $N_3^-^1$ we will  5 electrons for each nitrogen and 1 electron for the negative charge, so:

$5~electrons*~3~N=~15~e^-\\\\15~e^-~+~1e^-~=~16e^-$

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:

$FC=~valence~electrons~-(lone~pairs~e^-+\frac{1}{2}~bound~electrons)$

For figure 1 we will have:

$A:~FC=~5~-(4+\frac{1}{2}4)=~-1$

$B:~FC=~5~-(0+\frac{1}{2}8)=~+1$

$C:~FC=~5~-(4+\frac{1}{2}4)=~-1$

In total we will have a charge of => -1+1-1= -1

Answer 2

Answer :

The steps involved in the electron dot structure of $N^{3-}$ are :

First we have to determine the total number of valence electron in $N^{3-}$.

Number of valence electrons in N = 5

The charge on N is (-3). So, we add 3 electrons.

Total number of valence electrons = 5 + 3 = 8 electrons

The image is shown below.