Question

level: 2/2
covalent bonding and molecules
4 more questions to go
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 valence electrons. Total valence electrons \(=6 + 3\times7=27\). In the given structure, sulfur has 10 electrons (which violates octet rule as maximum should be 8 for main - group elements in period 3 when no \(d\) - orbital participation is considered in simple Lewis structures).
• For \(PF_2\) (Option B): Phosphorus has 5 valence electrons. Each fluorine has 7 valence electrons. Total valence electrons \(=5+2\times7 = 19\). In the given structure, phosphorus has 6 electrons (violates octet rule).
• For \(H_3O\) (Option C): Oxygen has 6 valence electrons. Each hydrogen has 1 valence electron. Total valence electrons \(=6 + 3\times1=9\). But a stable \(H_3O\) species would be \(H_3O^+\) (valence electrons \(=6 + 3\times1-1 = 8\)). The given structure has an extra hydrogen (violates electron - count and bonding rules).
• For \(SBr_2\) (Option D): Sulfur has 6 valence electrons. Each bromine has 7 valence electrons. Total valence electrons \(=6+2\times7 = 20\). Sulfur has 8 electrons (satisfies octet rule as \(S:6 + 2\) (from two single bonds) \(=8\)), and each bromine has \(7 + 1\) (from single bond) \(=8\) electrons.

Answer:

D.