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.
For N2O5, we have two nitrogen atoms (N) and five oxygen atoms (O). Each nitrogen contributes 5 valence electrons, while each oxygen contributes 6 valence electrons. So the total number of valence electrons for N2O5 is:
(2 × 5) + (5 × 6) = 10 + 30 = 40 valence electrons.
Step 2: Determine the central atom.
In this case, nitrogen (N) is the central atom since it forms the N-O-N bridge.
Step 3: Connect the atoms with single bonds.
Connect the two nitrogen atoms (N) with a single bond (N-O-N) to form the bridge.
Step 4: Distribute the remaining electrons around the atoms.
After connecting the two nitrogen atoms, we now have 38 valence electrons remaining. Distribute these electrons around the atoms, starting with the atoms that don't make multiple bonds.
Start by placing two lone pairs (four electrons) on each oxygen atom (O). This will account for 20 electrons (5 × 4 = 20).
Now, we have 18 electrons remaining. Place these electrons as lone pairs around the central nitrogen atom (N), giving it a total of 8 electrons (2 lone pairs).
Step 5: Check for octet rule satisfaction.
Check if each atom (except hydrogen) has an octet of electrons. In this case, all atoms—nitrogen (N) and oxygen (O)—have an octet, except for the oxygen atoms that are bonded directly to the nitrogen atom. These oxygens only have 6 electrons (rather than 8). So, we need to form multiple bonds using one of the lone pairs from the central nitrogen atom.
Step 6: Form multiple bonds.
Take one lone pair from the central nitrogen atom and form a double bond with each oxygen atom that doesn't satisfy the octet rule. This will give each oxygen a total of 8 electrons, satisfying the octet rule.
Finally, we have the Lewis structure of N2O5 with one oxygen bridging the two nitrogen atoms (N-O-N single bond), and two other oxygens bonded to each nitrogen atom.
To count the number of equivalent resonance structures that satisfy the octet rule and where oxygen makes at most 2 bonds, we need to examine the possible arrangements of the double bond.
In this case, since there are two oxygen atoms bonded to each nitrogen atom, there are two possible arrangements for the double bond between the nitrogen atom (N) and an oxygen atom (O). This means that we have two equivalent resonance structures for N2O5 that satisfy the octet rule and where oxygen makes at most 2 bonds.
The final resonance structures for N2O5 are:
O O
║ ║
O = N - O - N = O
║ ║
O O