To determine the Lewis structure for dinitrogen tetroxide ([tex]\(N_2O_4\)[/tex]), let's go through the steps methodically.
1. Count the Total Number of Valence Electrons
- Nitrogen ([tex]\(N\)[/tex]) has 5 valence electrons.
- Oxygen ([tex]\(O\)[/tex]) has 6 valence electrons.
- For [tex]\(N_2O_4\)[/tex]:
- 2 Nitrogen atoms: [tex]\(2 \times 5 = 10\)[/tex] electrons
- 4 Oxygen atoms: [tex]\(4 \times 6 = 24\)[/tex] electrons
- Total valence electrons: [tex]\(10 + 24 = 34\)[/tex] electrons
2. Arrange the Atoms
- Typically, oxygen atoms tend to form peripheral positions, while nitrogen atoms often bond to each other or form central positions.
- Given the formula [tex]\(N_2O_4\)[/tex], arrange the nitrogen atoms in the center, and the oxygen atoms around them.
3. Draw Single Bonds Between Atoms
- Draw a single bond between the two nitrogen atoms.
- Draw single bonds from each nitrogen atom to two oxygen atoms.
4. Distribute Remaining Electrons to Complete Octets
- First, subtract the electrons used for bonds from the total count: [tex]\(4 \text{ N-O bonds} + 1 \text{ N-N bond} = 5 \text{ bonds} \times 2 = 10\)[/tex] electrons.
- [tex]\(34 - 10 = 24\)[/tex] electrons left.
- Distribute these 24 electrons to fulfill the octets of the oxygen atoms first.
5. Check for Octet Fulfillment and Adjust Bonds If Necessary
- Arrange the oxygen atoms around the nitrogen atoms with single bonds initially.
- Each oxygen connected through a single bond should initially have 3 lone pairs (6 electrons each) around them.
- Then, check whether each nitrogen atom has an octet. If not, consider forming double bonds between nitrogen and adjacent oxygen atoms to fulfill the octet rule.
Here is a common Lewis structure for dinitrogen tetroxide ([tex]\(N_2O_4\)[/tex]):
```
O O
|| ||
N = N
|| ||
O O
```
- Each nitrogen atom has a double bond to an oxygen atom and another single bond to an oxygen atom. This structure maintains the octet rule for both nitrogen and oxygen.
- The lone pairs on the oxygen atoms are not specifically shown here, but each oxygen atom not double-bonded to nitrogen will have three lone pairs, and each oxygen double-bonded to nitrogen will have two lone pairs.
Therefore, the correct Lewis structure for [tex]\(N_2O_4\)[/tex] involves double bonding between nitrogen and one oxygen atom each, and then having single bonds between nitrogen and the other oxygen atoms, satisfying all octet requirements.
