The nitrogen atoms in \(\mathrm{N}_{2}\) participate in multiple bonding, whereas those in hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the nitrogen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{N}-\mathrm{N}\) bond?

For drawing Lewis structures, we need to count the valence electrons for each molecule and arrange them to satisfy the octet rule for each atom. (a) Lewis structure of \(\mathrm{N}_{2}\) molecule: Nitrogen has 5 valence electrons, and since there are two nitrogen atoms, the total number of valence electrons is 10. We can arrange these electrons as follows: Start by drawing a single bond between the two nitrogen atoms and placing the remaining 6 electrons as three lone pairs on each nitrogen atom. We end up with a triple bond between the two nitrogen atoms to satisfy the octet rule. Lewis structure of \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecule: Since there are two nitrogen atoms and four hydrogen atoms, the total number of valence electrons is 12 (2 × 5 for nitrogen, and 4 × 1 for hydrogen). Each nitrogen atom needs three more electrons to satisfy the octet rule, and each hydrogen atom needs one electron to complete its duplet. We can draw single bonds between each nitrogen atom and two hydrogen atoms, and a single bond between the nitrogen atoms.

(b) Hybridization of nitrogen atoms in each molecule: In \(\mathrm{N}_{2}\) molecule: For N-N triple bond, each nitrogen atom has one σ bond, and two π bonds. In total, we need three hybridized orbitals (1σ + 2π). Therefore, the hybridization of nitrogen atoms in the \(\mathrm{N}_{2}\) molecule is \(\mathrm{sp}\) (2 hybrid orbitals). In \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecule: Each nitrogen atom forms three single bonds - one with another nitrogen atom and two with hydrogen atoms. There are no π bonds in this molecule. As a result, each nitrogen atom needs three hybridized orbitals, so the hybridization of nitrogen atoms in the \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecule is \(\mathrm{sp}^2\) (3 hybrid orbitals).

(c) Comparing the N-N bonds in both molecules: The strength of a bond generally depends on the type of bond and the number of bonds between the atoms. A stronger bond has more bond order, i.e., more number of bonds between the atoms. In this case: In \(\mathrm{N}_{2}\) molecule: The nitrogen atoms are connected by a triple bond (1σ + 2π). In \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecule: The nitrogen atoms are connected by a single bond (1σ). Since triple bonds are stronger than single bonds, the \(\mathrm{N}-\mathrm{N}\) bond in the \(\mathrm{N}_{2}\) molecule is stronger than in the \(\mathrm{N}_{2} \mathrm{H}_{4}\) molecule.