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 [tex]N_2O_4[/tex], we will have in total:
[tex](5*2)+(6*4)=34~electrons[/tex]
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:
[tex]Sp^3~=~4[/tex]
[tex]Sp^2~=~3[/tex]
[tex]Sp~=~2[/tex]
With this in mind, the hybridization of nitrogen is [tex]Sp^2[/tex].
See figure 1
I hope it helps!
