Complete the VSEPR table.

\begin{tabular}{|c|c|c|}
\hline
Geometry & Domains & Angle \\
\hline
linear & 2 bonding & [tex]$180^{\circ}$[/tex] \\
\hline
trigonal planar & 3 bonding & [tex]$120^{\circ}$[/tex] \\
\hline
bent & 2 bonding / 1 lone pair & [tex]$\ \textless \ 120^{\circ}$[/tex] \\
\hline
tetrahedral & 4 bonding & [tex]$109.5^{\circ}$[/tex] \\
\hline
trigonal pyramidal & 3 bonding / 1 lone pair & [tex]$\ \textless \ 109.5^{\circ}$[/tex] \\
\hline
bent & 2 bonding / 2 lone pairs & [tex]$\ \textless \ 109.5^{\circ}$[/tex] \\
\hline
\end{tabular}



Answer :

Sure, let's complete the VSEPR table step-by-step by matching geometries with their corresponding domains and angles.

1. For the first entry, we know that the geometry corresponds to an angle of [tex]\(180^\circ\)[/tex]. The only geometry that results in this bond angle is linear. Typically, a linear geometry involves 2 domains (either both bonding or one bonding and one lone pair), which gives the bond angle of [tex]\(180^\circ\)[/tex].

2. For the second entry, we know that the geometry involves 3 bonding domains. The geometry that corresponds to 3 bonding domains is trigonal planar. The bond angles in a trigonal planar geometry are typically [tex]\(120^\circ\)[/tex].

3. For the third entry with a bent geometry and 2 bonding domains / 1 lone pair, the bond angle is typically less than [tex]\(120^\circ\)[/tex] because the lone pair repulsion pushes the bonding pairs closer together. This geometry is also derived from a trigonal planar arrangement but with one lone pair.

4. For the fourth entry, we know the geometry involves 4 bonding domains. The geometry that corresponds to 4 bonding domains is tetrahedral. The bond angles in a tetrahedral geometry are typically [tex]\(109.5^\circ\)[/tex].

5. For the fifth entry, we need a geometry that matches trigonal pyramidal. A trigonal pyramidal geometry typically involves 3 bonding domains and 1 lone pair. The bond angles are slightly less than [tex]\(109.5^\circ\)[/tex] due to the lone pair repulsion.

6. For the sixth entry, with a bent geometry and bond angle less than [tex]\(109.5^\circ\)[/tex], this results from a tetrahedral arrangement but with 2 bonding domains and 2 lone pairs.

Now we can fill in the complete table:

[tex]\[ \begin{array}{|c|c|c|} \hline \text{Geometry} & \text{Domains} & \text{Angle} \\ \hline \text{linear} & 2 & 180^\circ \\ \hline \text{trigonal planar} & 3 \text{ bonding} & 120^\circ \\ \hline \text{bent} & 2 \text{ bonding / 1 lone pair} & <120^\circ \\ \hline \text{tetrahedral} & 4 \text{ bonding} & 109.5^\circ \\ \hline \text{trigonal pyramidal} & 3 \text{ bonding / 1 lone pair} & <109.5^\circ \\ \hline \text{bent} & 2 \text{ bonding / 2 lone pairs} & <109.5^\circ \\ \hline \end{array} \][/tex]