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

See explanation

Step-by-step 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!

Answer 2

The central nitrogen atoms in N2O4 are both sp2 hybridized.

The Lewis structure shows the number of electron pairs that surround the atoms in a molecule as dots. It is quite easy to determine the number of valence electrons in a molecule simply by observing its Lewis dot structure.

The molecule N2O4 has 34 valence electrons as shown in its dot electron structure. The central nitrogen atoms in N2O4 are both sp2 hybridized as shown. The formal charges on each atom in N2O4 are also shown.

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