To determine the most preferred and lowest energy structure for $S{O}_3$, we need to consider formal charges and the octet rule.

Concept: Formal Charge and Octet Rule
The most stable Lewis structure (lowest energy) is generally the one where atoms have formal charges closest to zero, and all atoms (especially the more electronegative ones) satisfy the octet rule. Sulfur, being in Period 3, can expand its octet.

Why Option (B) is correct:
Let's calculate the formal charges for the central sulfur atom and the oxygen atoms in structure (B).

• Sulfur (S): Valency = 6. Number of non-bonding electrons = 0. Number of bonds = 6 (three double bonds).
  Formal Charge (S) = $6-0-\frac{1}{2}(12)=6-6=0$
• Oxygen (O): Valency = 6. Number of non-bonding electrons = 4. Number of bonds = 2 (one double bond).
  Formal Charge (O) = $6-4-\frac{1}{2}(4)=6-4-2=0$

In structure (B), all atoms have a formal charge of zero, and sulfur expands its octet to 12 electrons, which is permissible for Period 3 elements. This structure minimizes formal charges and is therefore the most stable and lowest energy.

Option Analysis:

• (A) In this structure, sulfur has three single bonds and one double bond, resulting in a formal charge of +1 on sulfur and -1 on two oxygen atoms. This is less stable due to charge separation.
• (C) In this structure, sulfur has three single bonds, resulting in a formal charge of +3 on sulfur and -1 on all three oxygen atoms. This is highly unstable due to significant charge separation.
• (D) This structure shows sulfur with two double bonds and one single bond, resulting in a formal charge of +1 on sulfur and -1 on one oxygen atom. This is less stable than (B) due to charge separation.

Correct Answer: (B)