To draw the Lewis structure for the I3– ion (triodide ion), we start by calculating the total number of valence electrons. Each iodine (I) atom has 7 valence electrons, and there are three iodine atoms. Since it’s an ion with a -1 charge, we add one extra electron. Therefore:
- 3 Iodine atoms: 3 x 7 = 21 electrons
- Add 1 electron for the -1 charge: 21 + 1 = 22 electrons total
Next, we place the iodine atoms in the structure, with one iodine atom as the central atom (since it’s less electronegative than the others) and the other two as terminal atoms. We arrange the electrons around the atoms to complete their octets:
- Central Iodine atom: 2 lone pairs and 2 bond pairs (one bond pair with each terminal iodine)
- Each terminal iodine: 3 lone pairs and 1 bond pair
This gives us the structure:
The molecular geometry of I3– can be classified using the VSEPR theory. The central iodine atom has two bonded pairs and two lone pairs. Given this arrangement, the geometry is described as linear because the lone pairs are positioned to minimize repulsion, and the bond pairs will align in a straight line.
The electron domain geometry is also classified as tetrahedral since we have four regions of electron density (2 bonding pairs and 2 lone pairs).
Regarding bond angles, the ideal angle between the bonded pairs in a tetrahedral geometry is 109.5 degrees. However, due to the repulsion caused by the lone pairs, the actual bond angle in I3– is expected to be around 180 degrees, as it aligns linearly in the molecule.
Finally, the hybridization of the central iodine atom can be determined. In the case of I3–, the central iodine atom uses sp3 hybridization to form the bonds with the two terminal iodine atoms, accounting for two bond pairs and two lone pairs.
In conclusion, for the triodide ion (I3–):
- Molecular Geometry: Linear
- Electron Domain Geometry: Tetrahedral
- Bond Angles: ~180 degrees
- Hybridization: sp3