Draw the Lewis structure of NCO (with minimized formal charges) and then choose the appropriate formal charges for each of the atoms.
Ô-C=N:
Question 22 of 28
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A) N=0, C = 0, O = -1
B) N = -1, C = +1, O = -1
C) N= 0, C = 0, O = 0
D) N=+1, C = -1, O = -1
E) N = -1, C = 0, O = 0
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The Correct Answer and Explanation is :
Let’s analyze the Lewis structure of the NCO molecule (also called isocyanate ion) and calculate formal charges to find the correct answer.
Step 1: Count total valence electrons
- Nitrogen (N): 5 valence electrons
- Carbon (C): 4 valence electrons
- Oxygen (O): 6 valence electrons
Total = 5 + 4 + 6 = 15 valence electrons
Because NCO is typically an ion with a negative charge (-1), add 1 electron:
Total electrons = 15 + 1 = 16 valence electrons to distribute.
Step 2: Determine bonding framework
Given the formula Ô–C=N, the connectivity is Oxygen bonded to Carbon bonded to Nitrogen.
Step 3: Draw a possible Lewis structure
Start by placing bonds and lone pairs to fulfill the octet rule and minimize formal charges:
O — C — N
Now assign electrons:
- Single bonds use 2 electrons each.
- Remaining electrons are placed as lone pairs to complete octets.
Step 4: Minimize formal charges
Try the following bonding:
- Double bond between Carbon and Nitrogen: C=N
- Double bond between Oxygen and Carbon: O=C
- Lone pairs assigned to complete octets.
This gives:
O=C=N⁻
Let’s distribute electrons for this:
- O=C bond: 2 pairs (4 electrons)
- C=N bond: 2 pairs (4 electrons)
- Lone pairs on O: 2 pairs (4 electrons)
- Lone pairs on N: 1 pair (2 electrons)
- Lone pairs on C: 0 pairs (carbon typically has no lone pairs in stable molecules)
Sum of electrons:
- Bonds: 4 + 4 = 8
- Lone pairs: 4 (O) + 2 (N) = 6
- Total: 8 + 6 = 14 electrons — We have 16 electrons total to use, so we can add one lone pair on nitrogen, making it 2 lone pairs (4 electrons) on N, total 16.
Revised structure:
- O double bonded to C, with two lone pairs (4 electrons)
- C double bonded to N
- N with 2 lone pairs (4 electrons)
Step 5: Calculate formal charges
Formal charge (FC) formula: FC=V−(L+B2)FC = V – (L + \frac{B}{2})
Where:
- VV = valence electrons of the atom
- LL = lone pair electrons on the atom
- BB = bonding electrons (shared)
Calculate for each atom:
Oxygen (O):
- Valence electrons V=6V = 6
- Lone pairs L=4L = 4 (two lone pairs)
- Bonding electrons B=4B = 4 (double bond)
FCO=6−(4+42)=6−(4+2)=6−6=0FC_O = 6 – (4 + \frac{4}{2}) = 6 – (4 + 2) = 6 – 6 = 0
Carbon (C):
- Valence electrons V=4V = 4
- Lone pairs L=0L = 0
- Bonding electrons B=8B = 8 (two double bonds: 4 electrons with O, 4 with N)
FCC=4−(0+82)=4−(0+4)=4−4=0FC_C = 4 – (0 + \frac{8}{2}) = 4 – (0 + 4) = 4 – 4 = 0
Nitrogen (N):
- Valence electrons V=5V = 5
- Lone pairs L=4L = 4 (two lone pairs)
- Bonding electrons B=4B = 4 (double bond with C)
FCN=5−(4+42)=5−(4+2)=5−6=−1FC_N = 5 – (4 + \frac{4}{2}) = 5 – (4 + 2) = 5 – 6 = -1
Step 6: Final formal charges
- O = 0
- C = 0
- N = -1
Step 7: Answer choice matching
The correct formal charges correspond to:
E) N = -1, C = 0, O = 0
Explanation (300+ words)
The molecule NCO (isocyanate ion) is composed of nitrogen, carbon, and oxygen atoms. The best Lewis structure minimizes formal charges and satisfies the octet rule for all atoms.
First, by counting total valence electrons for N (5), C (4), and O (6), plus one extra electron due to the negative charge, we get 16 valence electrons to distribute. The bonding connectivity follows the formula: oxygen bonded to carbon bonded to nitrogen (O–C–N).
A logical structure that fulfills octets and minimizes formal charges has double bonds between oxygen and carbon (O=C) and between carbon and nitrogen (C=N). Oxygen carries two lone pairs (4 electrons), nitrogen has two lone pairs as well, and carbon is fully bonded with no lone pairs.
Using the formal charge formula FC=V−(L+B2)FC = V – (L + \frac{B}{2}), oxygen has 6 valence electrons, 4 in lone pairs, and 4 shared in bonds, resulting in a formal charge of zero. Carbon has 4 valence electrons, no lone pairs, and 8 bonding electrons, also resulting in zero formal charge. Nitrogen has 5 valence electrons, 4 in lone pairs, and 4 bonding electrons, giving it a formal charge of -1.
This distribution fits the negative charge on the ion, places it on nitrogen, which is more electronegative than carbon, and keeps the formal charges minimal overall. Other structures either produce higher formal charges or do not satisfy the octet rule properly.
Hence, the formal charges for the atoms in the NCO ion with minimized formal charges are:
- Nitrogen: -1
- Carbon: 0
- Oxygen: 0
Matching answer E.