Question

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?

Short answer

(a) Lewis structures: \(\mathrm{N}_{2}\) has a triple bond between the nitrogen atoms, while \(\mathrm{N}_{2} \mathrm{H}_{4}\) has a single bond between the nitrogen atoms and single bonds between the nitrogen and hydrogen atoms. (b) Hybridization: In \(\mathrm{N}_{2}\), the nitrogen atoms are \(\mathrm{sp}\) hybridized, while in \(\mathrm{N}_{2} \mathrm{H}_{4}\), they are \(\mathrm{sp}^2\) hybridized. (c) The \(\mathrm{N}-\mathrm{N}\) bond in \(\mathrm{N}_{2}\) is stronger than in \(\mathrm{N}_{2} \mathrm{H}_{4}\) due to its triple bond.

Step-by-step solution

Drawing Lewis Structures

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.

Identifying the Hybridization of 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).

Determining the Stronger N-N Bond

(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.

Key concepts

Lewis Structures

Lewis structures are a fantastic way to understand how atoms are bonded in a molecule. They show how valence electrons are distributed and help us visualize the molecule's shape. To draw a Lewis structure, it's essential to know the total number of valence electrons in the molecule. For instance, in the \(\mathrm{N}_2\) molecule, we have a total of 10 valence electrons. Each nitrogen atom has five electrons, making ten for the molecule. You start by connecting the two nitrogens with a single bond, then adjust by adding electrons to form a stable structure, finally resulting in a triple bond with each nitrogen atom satisfying the octet rule, having eight electrons around it.

• Count all valence electrons.
• Connect atoms using single bonds first.
• Distribute remaining electrons to satisfy the octet or duplet (for hydrogen) rule.
• Form multiple bonds if necessary to satisfy these rules.

For \(\mathrm{N}_2\mathrm{H}_4\), or hydrazine, the process is slightly different because hydrogens only need two electrons (a duplet) to fulfill their needs. Here, we count a total of 12 valence electrons from both the nitrogen and hydrogen atoms. Each nitrogen bonds to two hydrogens and to another nitrogen atom, forming single bonds that stabilize the structure. Thus, for hydrazine, the nitrogen atoms achieve the octet configuration through these bonds.

Hybridization

Hybridization explains how atoms' orbitals mix to form new hybrid orbitals used in bonding. This concept is crucial to understanding the structure and bonding nature of molecules. In the \(\mathrm{N}_2\) molecule, hybridization plays a key role because each nitrogen atom is involved in a triple bond, consisting of one sigma (\(\sigma\)) and two pi (\(\pi\)) bonds.The \(\mathrm{sp}\) hybridization occurs because each nitrogen atom forms one \(\sigma\) bond and overlaps its p orbitals to form two \(\pi\) bonds. This kind of bonding and hybridization lower energy consumption, stabilizing the molecule.In hydrazine, \(\mathrm{N}_2\mathrm{H}_4\), the nitrogen atoms are \(\mathrm{sp}^3\) hybridized since they form three sigma bonds – two with hydrogen atoms and one with the other nitrogen atom. The \(\mathrm{sp}^3\) hybridization helps in explaining the bond angles and the shape of the molecule, which is more pronounced and distinct in cases with single bonds.

Bond Strength

The strength of a bond is determined by the bond order. A higher bond order means a stronger bond, and this can often be gauged by the number of bonds between atoms. For instance, in \(\mathrm{N}_2\), the bond strength is high thanks to the triple bond present between the nitrogen atoms. This triple bond (one \(\sigma\) and two \(\pi\) bonds) is immensely stronger compared to a single bond due to higher electron sharing and overlap, making \(\mathrm{N}_2\) one of the most stable diatomic molecules.Conversely, in hydrazine \(\mathrm{N}_2\mathrm{H}_4\), the nitrogen atoms are connected by a single bond (one \(\sigma\) bond), which is inherently weaker than a triple bond. This principle helps in understanding why materials with higher bond orders tend to be more rigid and possess higher energies, impacting their chemical behavior and reactivity in chemical reactions.