Question

1. which molecule would have a larger bond angle, bcl₃ or cf₄? support your answer with lewis structures below

Explanation:

Step1: Determine the electron - pair geometry of \(BCl_3\)

Boron in \(BCl_3\) has 3 valence electrons. Each \(Cl\) atom contributes 1 electron for bonding. So, \(BCl_3\) has 3 bonding pairs and 0 lone pairs around the central \(B\) atom. According to VSEPR theory, the electron - pair geometry (and molecular geometry) is trigonal planar. The bond angles in a trigonal planar geometry are \(120^{\circ}\).
The Lewis structure of \(BCl_3\) is:

      Cl
      |
Cl - B - Cl

Step2: Determine the electron - pair geometry of \(CF_4\)

Carbon in \(CF_4\) has 4 valence electrons. Each \(F\) atom contributes 1 electron for bonding. So, \(CF_4\) has 4 bonding pairs and 0 lone pairs around the central \(C\) atom. According to VSEPR theory, the electron - pair geometry (and molecular geometry) is tetrahedral. The bond angles in a tetrahedral geometry are \(109.5^{\circ}\).
The Lewis structure of \(CF_4\) is:

      F
      |
F - C - F
      |
      F

Answer:

\(BCl_3\) has a larger bond angle. The bond angle of \(BCl_3\) is \(120^{\circ}\) (trigonal planar geometry) and the bond angle of \(CF_4\) is \(109.5^{\circ}\) (tetrahedral geometry).