Answer: Determine the electron and molecular geometries of each molecule. For molecules

Step 1 of 3

(a) The following table is used to identify the electron and molecular geometry of the molecule.

\(\begin{array}{|c|c|c|c|c|c|c|} \hline \begin{array}{l} \text { Electron } \\ \text { groups } \end{array} & \begin{array}{l} \text { Bonding } \\ \text { groups } \end{array} & \begin{array}{l} \text { Lone } \\ \text { pair } \end{array} & \begin{array}{l} \text { Electron } \\ \text { geometry } \end{array} & \begin{array}{l} \text { Angle } \\ \text { between } \\ \text { electron } \\ \text { groups } \end{array} & \begin{array}{l} \text { Molecular } \\ \text { geometry } \end{array} & \text { Example } \\ \hline 2 & 2 & 0 & \text { Linear } & 180^{\circ} & \text { Linear } & \mathrm{CO}_{2} \\ \hline 3 & 3 & 0 & \text { Trigonal planar } & 120^{\circ} & \text { Trigonal planar } & \mathrm{COH}_{2} \\ \hline 3 & 2 & 1 & \text { Trigonal planar } & 109.5^{\circ} & \text { Bent } & \mathrm{SO}_{3} \\ \hline 4 & 4 & 0 & \text { Tetrahedral } & 109.5^{\circ} & \text { Tetrahedral } & \mathrm{CH}_{4} \\ \hline 4 & 3 & 1 & \text { Tetrahedral } & 109.5^{\circ} & \text { Trigonal planar } & \mathrm{NH}_{3} \\ \hline 4 & 2 & 2 & \text { Tetrahedral } & 109.5^{\circ} & \text { Bent } & \mathrm{H}_{2} \mathrm{O} \\ \hline\end{array}\)

The Lewis structure of \(N_2\) is as follows.

It is a diatomic molecule. Each nitrogen atom has one electron pair and one lone pair. But it is a homogeneous diatomic molecule.

Hence, the electron and molecular geometry of the molecule are linear.