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