Question

question 4 (2 points)
which of following structures is the best lewis structure for pf₅?

Explanation:

Step1: Count valence electrons

Phosphorus has 5 valence electrons and each fluorine has 7 valence electrons. So total valence electrons for \(PF_5\) is \(5+(5\times7)=40\).

Step2: Analyze each structure

• Structure 1: Each F has 8 electrons (satisfied), P has 10 electrons (expanded octet, allowed for P as it has d - orbitals). Total electrons \(= (5\times8)+ 10=50\) (incorrect, as total should be 40).
• Structure 2: Each F has 8 electrons (satisfied), P has 10 electrons (expanded octet). Total electrons \(=(5\times8)+10 = 50\) (incorrect). But wait, actually, in Lewis structure, each single bond contributes 2 electrons. For \(PF_5\), there are 5 single bonds (\(P - F\)). Each F has 6 non - bonding electrons. So total electrons \(=5\times2+(5\times6)=40\).
• Structure 3: Has a \(P = F\) double bond. Fluorine is the most electronegative element and does not form double bonds in \(PF_5\) (unstable).
• Structure 4: Is not a valid octet - like structure for all F atoms (some F atoms do not have 8 electrons in a proper sense as per Lewis structure rules).

Answer:

2