To draw the Lewis structure for NI3 (Nitrogen Triiodide), we start by calculating the total number of valence electrons. Nitrogen has 5 valence electrons, and each iodine atom has 7 valence electrons. Since there are three iodine atoms, we have:
- Valence electrons from nitrogen: 5
- Valence electrons from iodine: 3 × 7 = 21
Therefore, the total number of valence electrons is 5 + 21 = 26.
Next, we place nitrogen as the central atom and surround it with the three iodine atoms. We then connect each iodine to the nitrogen with a single bond, which uses up 6 electrons (3 bonds × 2 electrons each). This leaves us with 20 electrons remaining.
Now, we distribute the remaining electrons to the iodine atoms to ensure that each iodine has an octet. Each iodine will get 6 more electrons (as 3 lone pairs), accounting for all 20 electrons:
- Initial 6 from bonds + 6 from lone pairs = 12 electrons for each iodine.
- Hence, iodine has a complete octet.
Now, our Lewis structure can be represented as:
I
|
I -- N -- I
In this structure, nitrogen is in the center with three iodine atoms surrounding it, each bonded by a single bond.
Molecular Shape:
The molecular shape of NI3 is trigonal pyramidal. This is due to the three bonding pairs and one lone pair of electrons on the nitrogen atom, which repels the bonding pairs, creating a pyramid-like shape.
Electron Pair Geometry:
The electron pair geometry around the central nitrogen atom is tetrahedral because there are four areas of electron density (three bonding pairs and one lone pair).
Hybridization:
The hybridization of the central atom (nitrogen) in NI3 is sp3. This is derived from the four areas of electron density around nitrogen, indicating that it undergoes sp3 hybridization.