Figure are [WO possible Lewis structures for the sulfite ion (SO3 Using formal charges determine which is the more stable structure or B Show all your workings_ Figure Choosing from the structures given in Figure draw out the three resonance structures which will be the most important contributors t0 the hybrid structure of SO;? Explain your choice_ mark
The Correct Answer and Explanation is:
Here is a detailed solution to the questions provided.
(i) Determining the More Stable Structure using Formal Charges
To determine the more stable Lewis structure, the formal charges on each atom in both structures A and B must be calculated. The structure with formal charges closest to zero and with any negative charges on the most electronegative atoms is considered more stable.
The formula for formal charge (FC) is:
FC = (Number of Valence Electrons) – (Number of Non-bonding Electrons) – ½(Number of Bonding Electrons)
Calculations for Structure A:
- Sulfur (S): Sulfur is in Group 16, so it has 6 valence electrons. In structure A, it has 2 non-bonding electrons (one lone pair) and 8 bonding electrons (one double bond and two single bonds).
FC(S) = 6 – 2 – ½(8) = 0 - Doubly-bonded Oxygen (O=): Oxygen is in Group 16 (6 valence electrons). This oxygen has 4 non-bonding electrons and 4 bonding electrons.
FC(O=) = 6 – 4 – ½(4) = 0 - Singly-bonded Oxygens (O-): There are two of these. Each has 6 non-bonding electrons and 2 bonding electrons.
FC(O-) = 6 – 6 – ½(2) = -1
Summary for Structure A: The formal charges are S(0), O(0), O(-1), O(-1). The sum of the charges is 0 + 0 + (-1) + (-1) = -2, which matches the ion’s charge.
Calculations for Structure B:
- Sulfur (S): It has 6 valence electrons. In structure B, it has 2 non-bonding electrons and 6 bonding electrons (three single bonds).
FC(S) = 6 – 2 – ½(6) = +1 - Singly-bonded Oxygens (O-): There are three of these. Each has 6 non-bonding electrons and 2 bonding electrons.
FC(O-) = 6 – 6 – ½(2) = -1
Summary for Structure B: The formal charges are S(+1), O(-1), O(-1), O(-1). The sum of the charges is +1 + 3(-1) = -2, which also matches the ion’s charge.
Conclusion:
Structure A is the more stable structure. This is because it minimizes the formal charges on the atoms. In structure A, two atoms have a formal charge of zero, whereas in structure B, all atoms carry a formal charge. A lower magnitude of formal charges on the atoms within a molecule or ion indicates greater stability.
(ii) Most Important Resonance Structures
The most important contributors to the resonance hybrid are the most stable Lewis structures. Based on the formal charge analysis in part (i), the structures based on form A are the most significant.
The three equivalent resonance structures are formed by delocalizing the double bond among the three oxygen atoms.
The three resonance structures are:
Explanation:
These three structures are chosen as the most important contributors because they are all forms of Structure A, which was determined to be the most stable representation of the sulfite ion.
The stability of these structures arises from the minimization of formal charges. In each of these three resonance forms, the formal charge on the central sulfur atom is 0, and the formal charge on one of the oxygen atoms is also 0. The overall -2 charge of the ion is distributed over the two most electronegative atoms (oxygen), each carrying a formal charge of -1. This arrangement is more stable than Structure B, where there is a larger separation of charge (a +1 charge on sulfur and a -1 charge on all three oxygens).
Since these three structures are equivalent in terms of stability and formal charge distribution, they contribute equally to the true resonance hybrid structure of the sulfite ion (SO₃²⁻).thumb_upthumb_down
