Draw one of the Lewis structure of the N2O5. In each case one oxygen bridges the nitrogens (N-O-N single bond) and 2 other oxygens are bonded to each N. How many equivalent resonance structures are there that satisfy the octet rule and where O makes at most 2 bonds?
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To draw the Lewis structure of N2O5, we need to follow a few steps:
Step 1: Determine the total number of valence electrons.
- Nitrogen (N) has 5 valence electrons, and there are two of them, so that's 2 × 5 = 10 valence electrons.
- Oxygen (O) has 6 valence electrons, and there are five of them, so that's 5 × 6 = 30 valence electrons.
- The total number of valence electrons for N2O5 is 10 + 30 = 40.
Step 2: Connect the atoms to form a skeleton structure.
- Place two nitrogen atoms (N) in the center and connect them with a single bond.
N ≡ N
- Place oxygen atoms (O) around the nitrogen atoms, making sure to satisfy the octet rule (except for the central nitrogen, which can exceed the octet).
O O
| |
N ≡ N
Step 3: Distribute the remaining electrons.
- We have 40 valence electrons, and 4 electrons (2 pairs) have already been used in the N≡N bond.
- Place 2 electrons on each oxygen atom.
- Distribute the remaining electrons as lone pairs to fulfill the octet rule on each atom.
O O
|| ||
N ≡ N
- At this point, we have distributed 16 electrons (8 pairs) in the structure, and we have 40 - 16 = 24 electrons remaining.
Step 4: Add remaining electrons as needed.
- Starting from the central nitrogen, arrange the remaining electrons to fulfill the octet rule.
- One oxygen bridges the nitrogens (N-O-N single bond), which requires 2 electrons.
O ...
|| ...
N ≡ N
- Place two more oxygen atoms (O) and distribute the remaining electrons, satisfying the octet rule.
O ... ... O
|| ... ... ||
N ≡ N O
- At this point, we have used all the 40 valence electrons, and the Lewis structure of N2O5 with one oxygen bridging the nitrogens (N-O-N single bond) and 2 other oxygens bonded to each N is complete.
Step 5: Determine the equivalent resonance structures that satisfy the octet rule and where oxygen makes at most 2 bonds.
- To determine the resonance structures, we can move one of the lone pairs from an oxygen atom to form a double bond with the nitrogen atom.
Resonance structure 1:
O ... ... O
||= ... = ||
N ≡ N O
Resonance structure 2:
O ... ... O
|| =... ...|=
N ≡ N O
- Both resonance structures satisfy the octet rule and fulfill the condition where oxygen makes at most 2 bonds.
- Therefore, there are two equivalent resonance structures for N2O5.