How to Draw the Lewis Structure for ClBr3 and Determine Its Electron Geometry and Molecular Geometry?

To draw the Lewis structure for chlorine tribromide (ClBr3), we start by counting the total number of valence electrons available. Chlorine (Cl) has 7 valence electrons and each bromine (Br) atom also has 7 valence electrons. Therefore, the total is:

7 (from Cl) + 3 × 7 (from 3 Br) = 28 valence electrons

Next, we place the chlorine atom in the center since it is less electronegative than bromine. We then bond the three bromine atoms to the chlorine atom, using 6 of the 28 valence electrons:

Cl – Br, Cl – Br, Cl – Br

This uses up 6 electrons, leaving us with 22 electrons. Each bromine atom needs 6 more electrons (3 pairs) to complete its octet. That uses up the remaining 18 valence electrons:

So far we have:

  • 3 bonds to Br (6 electrons used)
  • 3 lone pairs on each Br (18 electrons used)
  • 1 Cl atom in the center with 2 lone pairs left

This leaves us with the following structure:

Br:      :Cl:      Br

         :Br:

The central chlorine atom has 1 bond with each bromine and carries 2 lone pairs. This configuration leads to an electron geometry described as octahedral because there are 6 regions of electron density (3 bonds + 3 lone pairs). However, for molecular geometry, we only consider the atoms. Since there are 3 bonded atoms and 2 lone pairs, the molecular geometry of ClBr3 is trigonal bipyramidal shaped, with the bromine atoms occupying the equatorial positions and the lone pairs occupying the axial positions.

In summary:

  • Electron Geometry: Octahedral
  • Molecular Geometry: Trigonal Bipyramidal

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