Questions: SiH4 and NH3 both have an identical electron domain geometry. However, SiH4 has a tetrahedral molecular geometry while NH3 has a trigonal pyramidal molecular geometry. Draw a Lewis structure of each molecule and indicate their electron domain geometry.

Solution Steps

To draw the Lewis structure of $\mathrm{SiH}_{4}$ (silane), we start by identifying the central atom, which is silicon (Si). Silicon has four valence electrons, and each hydrogen (H) atom has one valence electron. The total number of valence electrons is:

\[ 4 (\text{from Si}) + 4 \times 1 (\text{from each H}) = 8 \text{ valence electrons} \]

In the Lewis structure, silicon forms single bonds with each of the four hydrogen atoms, using all 8 valence electrons. The structure is:

\[ \text{H} - \text{Si} - \text{H} \] \[ \text{H} - \text{H} \]

The electron domain geometry considers all regions of electron density (bonds and lone pairs) around the central atom. In $\mathrm{SiH}_{4}$, there are four single bonds and no lone pairs around silicon. This results in a tetrahedral electron domain geometry.

For $\mathrm{NH}_{3}$ (ammonia), nitrogen (N) is the central atom with five valence electrons. Each hydrogen atom contributes one valence electron, giving a total of:

\[ 5 (\text{from N}) + 3 \times 1 (\text{from each H}) = 8 \text{ valence electrons} \]

In the Lewis structure, nitrogen forms three single bonds with hydrogen atoms and has one lone pair of electrons. The structure is:

\[ \begin{array}{c} \text{H} \\ | \\ \text{N} - \text{H} \\ | \\ \text{H} \end{array} \]

For $\mathrm{NH}_{3}$, the electron domain geometry includes three bonding pairs and one lone pair around nitrogen. This results in a tetrahedral electron domain geometry. However, the presence of the lone pair gives $\mathrm{NH}_{3}$ a trigonal pyramidal molecular geometry.

Final Answer