Question

which of the following is the correct lewis structure for the compound pbr₃?

Explanation:

• In a Lewis structure, atoms should have their valence electrons represented. Bromine (\(Br\)) has 7 valence electrons and forms single bonds. Phosphorus (\(P\)) has 5 valence electrons. In \(PBr_3\), \(P\) forms three single bonds with \(Br\) atoms. Each \(Br\) atom should have 3 lone pairs (6 electrons) around it to complete its octet.
• The first option shows \(Br\) atoms with lone pairs (dots representing non - bonding electrons) and single bonds between \(P\) and \(Br\), which is correct.
• The second option is wrong because \(Br\) atoms need to have their non - bonding electrons (lone pairs) shown in a Lewis structure.
• The third option is wrong as \(PBr_3\) has single bonds, not double or triple bonds.
• The fourth option is wrong because the bonding and non - bonding electrons are not correctly represented (incorrect bond type and incomplete lone - pair representation for \(Br\)).

Answer:

The first option (the one with \(Br\) atoms having lone pairs and single bonds) is the correct Lewis structure for \(PBr_3\).