Explain why it is not necessary to find the Lewis structure with the smallest formal charges to make a successful prediction of molecular geometry in the VSEPR theory. For example, write Lewis structures for \(S O_{2}\) having different formal charges, and predict the molecular geometry based on these structures.

Short Answer

Expert verified
Even with different Lewis structures and hence different formal charges, SO2 exhibits a 'bent' or 'V-shaped' geometry. This happens because VSEPR theory predicts molecular geometry based on the arrangement of bonded atoms and lone pairs around the central atom, not the formal charges. Therefore, it's not necessary to find the Lewis structure with the smallest formal charges to predict molecular geometry using VSEPR theory.

Step by step solution

01

Drawing Lewis Structures (SO2)

Draw Lewis structures for sulfur dioxide, SO2. First, let's consider the structure with full octets and minimal formal charges. In this structure, Sulfur (S) is the central atom and in bonding with two Oxygen (O) atoms, it forms a double bond with one O and a single bond with another. This structure will have a formal charge of 0 on all atoms.
02

Different Lewis Structures (SO2)

For a different Lewis structure with non-minimal formal charges, consider a structure where sulfur forms double bonds with both the oxygen atoms. This structure would have a formal charge of +1 on sulfur and -1 on one of the oxygen atoms.
03

VSEPR Theory and Geometry

The VSEPR theory asserts that the molecular geometry is primarily determined by the arrangement of bonds and lone pairs around the central atom. In both structures, the central sulfur atom has two bonded atoms and one lone pair. According to VSEPR theory, this corresponds to a 'bent' or 'V-shaped' geometry, regardless of the formal charge.

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