What is the VSEPR shape of NBr3?
I think its trigonal pyramidal
molecular geometry is trigonal pyramidal.
so that's not it?
I thought that was a good way of saying you were correct.
The molecular geometry is trigonal pyramidal.
The electronic geometry is tetrahedral.
To determine the VSEPR (Valence Shell Electron Pair Repulsion) shape of a molecule such as NBr3, we need to follow a step-by-step process:
Step 1: Determine the central atom.
In the molecule NBr3, the central atom is nitrogen (N) since it is less electronegative than bromine (Br).
Step 2: Count the total number of valence electrons.
Nitrogen (N) contributes 5 valence electrons, and each bromine (Br) atom contributes 7 valence electrons. So, for NBr3, we have 5 + (3 × 7) = 26 valence electrons.
Step 3: Determine the electron pair arrangement.
To determine the electron pair arrangement, we use the electron domain geometry. The electron domain geometry is based on the number of electron groups (bonded pairs and lone pairs) around the central atom. In this case, we have one nitrogen-bromine bond (N-Br) and three lone pairs on the nitrogen atom, giving us a total of 4 electron groups.
Step 4: Predict the molecular geometry.
Based on the electron domain geometry, we can predict the molecular geometry. Since we have three lone pairs and one bonding pair, the electron domain geometry is trigonal pyramidal. This means that the molecule will have a pyramid-like shape with the nitrogen atom at the apex and the three bromine atoms spread out in a trigonal planar arrangement on the base.
So, you are correct! The VSEPR shape of NBr3 is indeed trigonal pyramidal.