Question

Draw the Lewis structure of N₂O₄ and then choose the appropriate pair of hybridization states for the two central atoms. Your answer choice is independent of the orientation of your drawn structure.

Answer 1

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

If the formula is $N_2O_4$, we will have in total:

$(5*2)+(6*4)=34~electrons$

Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).

To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

$Sp^3~=~4$

$Sp^2~=~3$

$Sp~=~2$

With this in mind, the hybridization of nitrogen is $Sp^2$.

See figure 1

I hope it helps!