Using VSEPR what is the shape of ammonia (NH3) and water (H2O) when bonded to a hydrogen bond?
To determine the shape of a molecule using VSEPR (Valence Shell Electron Pair Repulsion) theory, you need to follow these steps:
1. Determine the Lewis structure of the molecule:
For ammonia (NH3), we have one nitrogen atom (N) bonded to three hydrogen atoms (H). The Lewis structure would look like this:
H
|
H-N-H
|
H
For water (H2O), we have one oxygen atom (O) bonded to two hydrogen atoms (H). The Lewis structure would look like this:
H
|
H-O-H
|
2. Count the number of electron pairs around the central atom:
In ammonia (NH3), nitrogen (N) is the central atom. It has three single bonds (with hydrogen atoms) and one lone pair of electrons. So, there are four electron pairs around the central atom.
In water (H2O), oxygen (O) is the central atom. It has two single bonds (with hydrogen atoms) and two lone pairs of electrons. So, there are four electron pairs around the central atom.
3. Determine the electron pair geometry:
For both ammonia and water, the electron pair geometry is tetrahedral because there are four electron pairs around the central atom.
4. Determine the molecular geometry:
To get the molecular geometry, you need to consider both the bonded pairs and lone pairs of electrons. Lone pairs tend to occupy more space compared to bonded pairs.
For ammonia (NH3), the three hydrogen atoms and one lone pair of electrons give the molecule a trigonal pyramidal shape.
For water (H2O), the two hydrogen atoms and two lone pairs of electrons give the molecule a bent shape.
In summary, the shape of ammonia (NH3) when bonded to a hydrogen atom is trigonal pyramidal, and the shape of water (H2O) when bonded to a hydrogen atom is bent.