To draw the Lewis structure for nitrogen tribromide (NBr3), we start by determining the total number of valence electrons in the molecule. Nitrogen (N) has 5 valence electrons and each bromine (Br) has 7 valence electrons. Since there are three bromine atoms, we can calculate the total as follows:
- Valence electrons from N: 5
- Valence electrons from 3 Br: 3 x 7 = 21
- Total = 5 + 21 = 26 valence electrons
Next, we place the nitrogen atom at the center, surrounded by the three bromine atoms. Each bond between nitrogen and bromine will use 2 electrons. So, we use 6 electrons for the three N-Br bonds. The structure looks like this:
N | Br-Br | Br
Now, we have used 6 of the 26 available electrons, leaving us with 20 electrons. We will distribute these remaining electrons to fulfill the octet rule for the bromine atoms, placing 6 electrons (or 3 lone pairs) around each bromine atom:
.. ..
:Br:
.. ..
|
N3
|
:Br:
.. ..
Since nitrogen only needs 8 electrons and has 3 bonding pairs (6 electrons) with each bromine, it satisfies the octet rule too. All 26 valence electrons are accounted for, and we have successfully drawn the Lewis structure for NBr3.