Draw (on paper) Lewis structures for tetrabromophosphonium ion and the nitrate ion. How many equivalent Lewis structures are necessary to describe the bonding in PBr
? How many equivalent Lewis structures are necessary to describe the bonding in NO
The Correct Answer and Explanation is:
To answer this question, we will draw the Lewis structures of the tetrabromophosphonium ion (PBr₄⁺) and the nitrate ion (NO₃⁻), and then explain how many equivalent Lewis structures are necessary for each.
1. Tetrabromophosphonium Ion (PBr₄⁺)
Structure:
- Central atom: Phosphorus (P)
- Four bonded atoms: Bromine (Br)
- Charge: +1
Drawing the structure:
Phosphorus has 5 valence electrons. Each bromine atom has 7 valence electrons. The +1 charge indicates a loss of one electron from the total count.
Total valence electrons:
- P: 5
- 4 Br atoms: 4 × 7 = 28
- Subtract 1 for the +1 charge:
Total = 5 + 28 – 1 = 32 electrons
Place P in the center and form single bonds with each Br (each bond uses 2 electrons): 4 × 2 = 8 electrons used.
Remaining: 32 – 8 = 24 electrons, used to complete octets of the Br atoms (6 each).
Final structure:
- P has four single bonds to Br.
- Br atoms each have three lone pairs.
- The structure has a +1 formal charge mostly localized on the phosphorus.
Number of equivalent Lewis structures:
Only 1 Lewis structure is needed. All Br atoms are in identical positions and single bonded; there are no resonance structures.
2. Nitrate Ion (NO₃⁻)
Structure:
- Central atom: Nitrogen (N)
- Three bonded atoms: Oxygen (O)
- Charge: –1
Total valence electrons:
- N: 5
- 3 O: 3 × 6 = 18
- Add 1 for the –1 charge:
Total = 5 + 18 + 1 = 24 electrons
Place N in the center and connect each O with a single bond: 3 × 2 = 6 electrons.
Remaining: 24 – 6 = 18 electrons
Assign 6 electrons (3 lone pairs) to each O atom: 3 × 6 = 18 electrons used.
Now check octets: Each O has 8 electrons, N only has 6. To satisfy the octet rule, create a double bond with one O by converting a lone pair into a bonding pair.
Final structure:
- One N=O double bond
- Two N–O single bonds
- Formal charges: The two single-bonded O atoms carry a –1 charge each; N has a +1 charge.
Number of equivalent Lewis structures:
3 resonance structures are needed. The double bond can be placed with any of the three O atoms. These are equivalent and collectively describe the delocalized bonding in the nitrate ion.
Conclusion:
- PBr₄⁺: 1 Lewis structure is sufficient; no resonance.
- NO₃⁻: 3 equivalent Lewis structures are required due to resonance among the oxygen atoms.
