QUESTION IMAGE
Question
three inequivalent lewis structures for carbonyl sulfide, sco, are shown below. use the concepts of formal charge and electronegativity to choose the structure that is the best representation by answering the questions below.
- assign formal charges to the elements in each of the structures.
- the best lewis structure for sco is
Part 1: Assign Formal Charges
The formula for formal charge is:
$$\text{Formal Charge} = \text{Valence Electrons} - \text{Non - bonding Electrons} - \frac{1}{2}\text{Bonding Electrons}$$
Structure A ($\ce{S=C=O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 4, Bonding electrons = 4 (double bond).
$$\text{Formal Charge (S)} = 6 - 4 - \frac{4}{2} = 0$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (two double bonds).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 4, Bonding electrons = 4 (double bond).
$$\text{Formal Charge (O)} = 6 - 4 - \frac{4}{2} = 0$$
Structure B ($\ce{S\equiv C - O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 2, Bonding electrons = 6 (triple bond).
$$\text{Formal Charge (S)} = 6 - 2 - \frac{6}{2} = +1$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (triple + single bond).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 6, Bonding electrons = 2 (single bond).
$$\text{Formal Charge (O)} = 6 - 6 - \frac{2}{2} = -1$$
Structure C ($\ce{S - C\equiv O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 6, Bonding electrons = 2 (single bond).
$$\text{Formal Charge (S)} = 6 - 6 - \frac{2}{2} = -1$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (single + triple bond).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 2, Bonding electrons = 6 (triple bond).
$$\text{Formal Charge (O)} = 6 - 2 - \frac{6}{2} = +1$$
Part 2: Best Lewis Structure for $\ce{SCO}$
To determine the best Lewis structure, we use two criteria:
- Formal Charges: Smaller (preferably zero) formal charges are better.
- Electronegativity: Negative formal charges should be on more electronegative atoms.
- Structure A: All formal charges are 0 (ideal).
- Structure B: $\ce{S}$ has $+1$, $\ce{O}$ has $-1$. $\ce{O}$ is more electronegative than $\ce{S}$, so the negative charge on $\ce{O}$ is favorable, but Structure A has all zero charges.
- Structure C: $\ce{S}$ has $-1$, $\ce{O}$ has $+1$. $\ce{O}$ is more electronegative than $\ce{S}$, so a positive charge on $\ce{O}$ is unfavorable.
Final Answers
1. Formal Charges
- Structure A: S = $\boldsymbol{0}$, C = $\boldsymbol{0}$, O = $\boldsymbol{0}$
- Structure B: S = $\boldsymbol{+1}$, C = $\boldsymbol{0}$, O = $\boldsymbol{-1}$
- Structure C: S = $\boldsymbol{-1}$, C = $\boldsymbol{0}$, O = $\boldsymbol{+1}$
2. Best Lewis Structure
The best Lewis structure for $\ce{SCO}$ is Structure A (since it has all zero formal charges, which is the most stable configuration).
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Part 1: Assign Formal Charges
The formula for formal charge is:
$$\text{Formal Charge} = \text{Valence Electrons} - \text{Non - bonding Electrons} - \frac{1}{2}\text{Bonding Electrons}$$
Structure A ($\ce{S=C=O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 4, Bonding electrons = 4 (double bond).
$$\text{Formal Charge (S)} = 6 - 4 - \frac{4}{2} = 0$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (two double bonds).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 4, Bonding electrons = 4 (double bond).
$$\text{Formal Charge (O)} = 6 - 4 - \frac{4}{2} = 0$$
Structure B ($\ce{S\equiv C - O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 2, Bonding electrons = 6 (triple bond).
$$\text{Formal Charge (S)} = 6 - 2 - \frac{6}{2} = +1$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (triple + single bond).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 6, Bonding electrons = 2 (single bond).
$$\text{Formal Charge (O)} = 6 - 6 - \frac{2}{2} = -1$$
Structure C ($\ce{S - C\equiv O}$)
- Sulfur (S): Valence electrons = 6, Non - bonding electrons = 6, Bonding electrons = 2 (single bond).
$$\text{Formal Charge (S)} = 6 - 6 - \frac{2}{2} = -1$$
- Carbon (C): Valence electrons = 4, Non - bonding electrons = 0, Bonding electrons = 8 (single + triple bond).
$$\text{Formal Charge (C)} = 4 - 0 - \frac{8}{2} = 0$$
- Oxygen (O): Valence electrons = 6, Non - bonding electrons = 2, Bonding electrons = 6 (triple bond).
$$\text{Formal Charge (O)} = 6 - 2 - \frac{6}{2} = +1$$
Part 2: Best Lewis Structure for $\ce{SCO}$
To determine the best Lewis structure, we use two criteria:
- Formal Charges: Smaller (preferably zero) formal charges are better.
- Electronegativity: Negative formal charges should be on more electronegative atoms.
- Structure A: All formal charges are 0 (ideal).
- Structure B: $\ce{S}$ has $+1$, $\ce{O}$ has $-1$. $\ce{O}$ is more electronegative than $\ce{S}$, so the negative charge on $\ce{O}$ is favorable, but Structure A has all zero charges.
- Structure C: $\ce{S}$ has $-1$, $\ce{O}$ has $+1$. $\ce{O}$ is more electronegative than $\ce{S}$, so a positive charge on $\ce{O}$ is unfavorable.
Final Answers
1. Formal Charges
- Structure A: S = $\boldsymbol{0}$, C = $\boldsymbol{0}$, O = $\boldsymbol{0}$
- Structure B: S = $\boldsymbol{+1}$, C = $\boldsymbol{0}$, O = $\boldsymbol{-1}$
- Structure C: S = $\boldsymbol{-1}$, C = $\boldsymbol{0}$, O = $\boldsymbol{+1}$
2. Best Lewis Structure
The best Lewis structure for $\ce{SCO}$ is Structure A (since it has all zero formal charges, which is the most stable configuration).