Answer :
Let's break down how to determine the Lewis structure of dinitrogen tetroxide (N₂O₄) step-by-step.
1. Count the total number of valence electrons:
Each nitrogen (N) atom has 5 valence electrons, and each oxygen (O) atom has 6 valence electrons.
- 2 nitrogen atoms: [tex]\(2 \times 5 = 10\)[/tex] electrons
- 4 oxygen atoms: [tex]\(4 \times 6 = 24\)[/tex] electrons
Therefore, the total number of valence electrons is [tex]\(10 + 24 = 34\)[/tex] electrons.
2. Determine the skeletal structure:
Generally, in compounds where different atoms are present, the less electronegative atoms tend to be central atoms. Here, we arrange the two nitrogen atoms in the center with two oxygen atoms attached to each nitrogen:
O O
\ /
N - N
/ \
O O
3. Form single bonds between the atoms in the skeletal structure:
Each bond between atoms will use up 2 electrons. We have 5 bonds here:
- [tex]\( N - N \)[/tex] bond: 2 electrons
- [tex]\( N - O \)[/tex] bonds: [tex]\(4 \times 2 = 8\)[/tex] electrons
This uses up [tex]\(2 + 8 = 10\)[/tex] electrons. Subtracting these from the total:
[tex]\(34 - 10 = 24\)[/tex] electrons remaining.
4. Distribute the remaining electrons to complete the octets starting with the outer atoms (oxygen), then the central atoms (nitrogen):
Each oxygen needs 8 electrons (an octet) including the bonding electrons. Each oxygen atom already has 2 bonding electrons, needing 6 more.
- For 4 oxygen atoms, each needing 6 more electrons: [tex]\(4 \times 6 = 24\)[/tex] electrons
Using the remaining 24 electrons completes the octet for each oxygen atom. There are no additional electrons left.
5. Check the arrangement for the octet rule (full valence shells):
Check every atom for a complete octet:
- All 4 oxygen atoms now have 8 electrons each.
- The two nitrogen atoms also have 8 electrons each (contributed through N–N bond and N–O bonds).
Thus, the final Lewis structure of dinitrogen tetroxide (N₂O₄) should look like this, with each nitrogen bonded to two oxygen atoms and a single bond between the two nitrogen atoms:
O O
\ /
N - N
/ \
O O
This arrangement ensures that all atoms achieve a stable octet configuration.
1. Count the total number of valence electrons:
Each nitrogen (N) atom has 5 valence electrons, and each oxygen (O) atom has 6 valence electrons.
- 2 nitrogen atoms: [tex]\(2 \times 5 = 10\)[/tex] electrons
- 4 oxygen atoms: [tex]\(4 \times 6 = 24\)[/tex] electrons
Therefore, the total number of valence electrons is [tex]\(10 + 24 = 34\)[/tex] electrons.
2. Determine the skeletal structure:
Generally, in compounds where different atoms are present, the less electronegative atoms tend to be central atoms. Here, we arrange the two nitrogen atoms in the center with two oxygen atoms attached to each nitrogen:
O O
\ /
N - N
/ \
O O
3. Form single bonds between the atoms in the skeletal structure:
Each bond between atoms will use up 2 electrons. We have 5 bonds here:
- [tex]\( N - N \)[/tex] bond: 2 electrons
- [tex]\( N - O \)[/tex] bonds: [tex]\(4 \times 2 = 8\)[/tex] electrons
This uses up [tex]\(2 + 8 = 10\)[/tex] electrons. Subtracting these from the total:
[tex]\(34 - 10 = 24\)[/tex] electrons remaining.
4. Distribute the remaining electrons to complete the octets starting with the outer atoms (oxygen), then the central atoms (nitrogen):
Each oxygen needs 8 electrons (an octet) including the bonding electrons. Each oxygen atom already has 2 bonding electrons, needing 6 more.
- For 4 oxygen atoms, each needing 6 more electrons: [tex]\(4 \times 6 = 24\)[/tex] electrons
Using the remaining 24 electrons completes the octet for each oxygen atom. There are no additional electrons left.
5. Check the arrangement for the octet rule (full valence shells):
Check every atom for a complete octet:
- All 4 oxygen atoms now have 8 electrons each.
- The two nitrogen atoms also have 8 electrons each (contributed through N–N bond and N–O bonds).
Thus, the final Lewis structure of dinitrogen tetroxide (N₂O₄) should look like this, with each nitrogen bonded to two oxygen atoms and a single bond between the two nitrogen atoms:
O O
\ /
N - N
/ \
O O
This arrangement ensures that all atoms achieve a stable octet configuration.
