Question

which of the lewis structures in the image is correct?

Explanation:

Step1: Check valence electrons

• For \(SBr_3\) (Option A): Sulfur has 6 valence electrons, each bromine has 7. Total valence electrons \(=6 + 3\times7=27\). In the structure, sulfur forms 3 single bonds (6 electrons used) and has 1 lone pair (2 electrons), total \(6 + 2=8\) electrons around sulfur. Each bromine has 6 lone - pair electrons and 2 in the bond, total 8.
• For \(PF_2\) (Option B): Phosphorus has 5 valence electrons, each fluorine has 7. Total valence electrons \(=5+2\times7 = 19\). In the given structure, phosphorus has 2 single bonds (4 electrons) and 3 non - bonding electrons (violates octet rule as it should have 8 electrons around phosphorus in a stable Lewis structure for main - group elements).
• For \(H_3O\) (Option C): Oxygen has 6 valence electrons, each hydrogen has 1. Total valence electrons \(=6 + 3\times1=9\). But a stable \(H_3O^+\) has 8 valence electrons (\(O\) has 8 electrons around it in \(H_3O^+\) structure), and the given structure for \(H_3O\) (assuming no charge) is incorrect as it violates the octet rule for oxygen (should have 8 electrons around \(O\), here it has 7 if we count the non - bonding electron).
• For \(SBr_2\) (Option D): Sulfur has 6 valence electrons, each bromine has 7. Total valence electrons \(=6+2\times7 = 20\). In the given structure, sulfur has 2 single bonds (4 electrons) and 3 non - bonding electrons (violates octet rule as sulfur should have 8 electrons around it in a stable Lewis structure for main - group elements).

Answer:

A. \(Br -

$$\begin{array}{c} \ddot{S}\\ | \\ Br \end{array}$$

-Br\)