Molecular Polarity Assignment This assignment involves using Lewis theory, valence bond theory, and VSEPR theory to describe the bonding and shape of two molecules

Molecular Polarity Assignment This assignment involves using Lewis theory, valence bond theory, and VSEPR theory to describe the bonding and shape of two molecules. The concepts of bond and molecular polarity will be used to predict if the assigned molecules possess a molecular dipole. For each of the assigned molecular formulas: e) Identify the hybrid state of all the central atom(s) in each molecule based on the orbital representation and electron geometry. [A2] f) Determine if bond polarity exists using electronegativity values. If so, draw dipole vectors on the shape to show the bond dipoles. [A2] g) Determine if molecular polarity exists. If so, draw a dashed vector to indicate the direction of the molecular dipole. [A2] Assigned compounds: I) carbonyl fluoride, COF2 II) diboron tetrafluoride, B2F4

The Correct Answer and Explanation is:

Here is the analysis of the two assigned compounds—carbonyl fluoride (COF₂) and diboron tetrafluoride (B₂F₄)—using Lewis theory, valence bond theory, VSEPR theory, and concepts of polarity:

I) Carbonyl Fluoride, COF₂

(e) Hybridization of the central atom (Carbon):

• The central atom is carbon, bonded to one oxygen (double bond) and two fluorine atoms (single bonds).
• There are three regions of electron density around carbon.
• According to VSEPR theory, this gives a trigonal planar geometry.
• The hybridization of carbon is sp².

(f) Bond polarity:

• Electronegativity values:
  • C = 2.55
  • O = 3.44
  • F = 3.98
• The C=O bond and each C–F bond are polar since the bonded atoms have differing electronegativities.
• Draw dipole vectors from carbon toward oxygen and fluorine, as both are more electronegative than carbon.

(g) Molecular polarity:

• The bond dipoles do not cancel out because of the differing strengths and directions (O is less electronegative than F, but the double bond adds electron density).
• Thus, COF₂ is a polar molecule.
• A dashed arrow representing the molecular dipole should point from carbon toward the more negative side, likely between the direction of the fluorines and oxygen.

II) Diboron Tetrafluoride, B₂F₄

(e) Hybridization of the central atoms (each Boron):

• Each boron is bonded to two fluorine atoms and to the other boron atom.
• That gives three regions of electron density around each boron.
• Geometry is trigonal planar around each boron.
• So, hybridization of each boron is sp².

(f) Bond polarity:

• Electronegativity:
  • B = 2.04
  • F = 3.98
• The B–F bonds are highly polar with dipole vectors pointing toward fluorine.
• The B–B bond is nonpolar (same atoms).

(g) Molecular polarity:

• The B–F bond dipoles are arranged symmetrically in a planar molecule.
• Since dipoles point in opposite directions and are equal in magnitude, they cancel out.
• Therefore, B₂F₄ is nonpolar overall.
• No molecular dipole vector is drawn.

Summary: