Question

The structure of borazine, \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(\mathrm{B}\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable? (d)Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at 1.44 A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are 1.51 \(\mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of borazine?

Short answer

The Lewis structures for borazine are as follows: (a) The structure with formal charges of zero: ``` H H | | B-N-B-N-B-N ! | H H ``` (b) The structure satisfying the octet rule, with three resonance structures: ``` H H | | B=N-B=N-B=N ! | H H ``` All atoms in both structures are sp² hybridized, making the molecule planar. The average of typical B-N single and double bond lengths is \(1.41 \mathrm{A}\), closer to borazine's bond length of \(1.44 \mathrm{A}\), suggesting it to be a resonance hybrid of the structure in (b). The molecule has 6 electrons in its \(\pi\) system.

Step-by-step solution

Draw the atoms and their valence electrons

To draw the Lewis structure, we first place the \(\mathrm{B}\) and \(\mathrm{N}\) atoms in an alternating pattern, and then add the lone electrons (3 for each \(\mathrm{B}\) atom and 5 for each \(\mathrm{N}\) atom) around each atom. Add \(\mathrm{H}\) atoms bonded to each \(\mathrm{B}\) and \(\mathrm{N}\) atom.

Connect the atoms with shared lone pairs

Now, we will connect each \(\mathrm{B}\) atom to an \(\mathrm{N}\) atom with a single bond and also connect \(\mathrm{H}\) atoms to each \(\mathrm{B}\) and \(\mathrm{N}\) atom with bonds. Here is the Lewis structure: ``` H H | | B-N-B-N-B-N ! | H H ``` (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom.

Modify the previous Lewis structure

Since the previous Lewis structure did not fulfill the octet rule for \(\mathrm{B}\) and \(\mathrm{N}\) atoms, use a double bond between \(\mathrm{B}\) and \(\mathrm{N}\) atoms instead of a single bond. This results in the following Lewis structure: ``` H H | | B=N-B=N-B=N ! | H H ``` (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of \(\mathrm{B}\) and \(\mathrm{N},\) do the formal charges seem favorable or unfavorable?

Calculate formal charges

Formal charge = Valence e⁻ - nonbonding e⁻ - (number of bonds) \(\mathrm{B}\): Formal charge = 3 - 0 - 4 = -1 \(\mathrm{N}\): Formal charge = 5 - 2 - 3 = 0 Since \(\mathrm{B}\) is less electronegative than \(\mathrm{N}\), having a negative formal charge might not be favorable. The structure in part (a) has no formal charges, making it more favorable. (d) Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures?

Check for resonance structures

For the structure in (a), there are no resonance structures, as there is no shifting of electrons to create another valid structure. For the structure in (b), considering the double bonds between \(\mathrm{B}\) and \(\mathrm{N}\) atoms, we can see that there are resonance structures as the double bonds could be located at different positions. Thus, there are 3 resonance structures for the Lewis structure in part (b). (e) What are the hybridizations at the B and N atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures?

Determine hybridizations

For part (a): \(\mathrm{B}\): 3 \(\sigma\) bonds (1 with an \(\mathrm{N}\) atom and 2 with \(\mathrm{H}\) atoms) - sp² hybridization \(\mathrm{N}\): 3 \(\sigma\) bonds (1 with a \(\mathrm{B}\) atom and 2 with \(\mathrm{H}\) atoms) - sp² hybridization For part (b): \(\mathrm{B}\): 3 \(\sigma\) bonds and 1 \(\pi\) bond (with an \(\mathrm{N}\) atom) - sp² hybridization \(\mathrm{N}\): 2 \(\sigma\) bonds and 1 \(\pi\) bond (with a \(\mathrm{B}\) atom) - sp² hybridization In both cases, all the atoms are sp² hybridized, and we would expect the molecule to be planar. (f) The six \(B-N\) bonds in the borazine molecule are all identical in length at \(1.44\) A. Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(1.51 \mathrm{A}\) and \(1.31 \mathrm{A},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other?

Compare bond lengths

The given \(B-N\) bond length in the borazine molecule is \(1.44 \mathrm{A}\). The average of the typical single and double bond lengths is \((1.51 + 1.31) \div 2 = 1.41\), which is closer to the actual bond length. This suggests that the real structure of borazine is a resonance hybrid of the structure in (b) rather than the structure in (a). (g) How many electrons are in the \(\pi\) system of borazine?

Count \(\pi\) electrons

In the resonance hybrid of borazine (structure b), there are 3 double bonds (B=N) involving 2 \(\pi\) electrons each. Thus, the total number of electrons in the \(\pi\) system of borazine is 3 x 2 = 6.

Key concepts

Formal Charge

The concept of formal charge is crucial in understanding molecular structure. It's the hypothetical charge an atom would have if all bonding electrons were shared equally between atoms. To calculate it, we use the formula:
Formal charge = (Valence electrons) - (Nonbonding electrons) - 0.5 * (Bonding electrons)
For instance, in the Lewis structure of borazine where the octet rule is satisfied, the boron atoms have a negative formal charge. This is less than ideal since boron is less electronegative than nitrogen and typically wouldn't harbor negative charge. Therefore, the structure with no formal charges, which corresponds to the one with single bonds between boron and nitrogen, is usually considered more favorable.

Octet Rule

The octet rule is the principle that atoms tend to form compounds in ways that give them eight valence electrons, corresponding to a full s and p orbital block — akin to the electron configuration of the noble gases. When creating a Lewis structure for borazine, satisfying the octet rule for all atoms can lead to different representations. Borazine features boron and nitrogen atoms, which means boron has three valence electrons and nitrogen has five. In some Lewis structures, double bonds are necessary to make sure both types of atoms adhere to the octet rule. However, whether or not this depiction is the most accurate can be determined by looking into resonance, hybridization, and actual bond lengths.

Resonance Structures

Resonance structures represent the delocalization of electrons within a molecule, where two or more valid Lewis structures can depict the molecule. These structures are not actual transitions between forms but rather a conceptual blend to understand electron distributions. For borazine's octet-satisfying Lewis structure (part b), there exist multiple resonance structures indicating that the double bonds may be spread out among different boron and nitrogen pairs. This concept of resonance helps to illustrate a more accurate representation of the molecule's actual electronic structure, which usually lies in an averaged state of these possible structures.

Hybridization

Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which can accommodate the pairing of electrons to form chemical bonds. In borazine, both the boron and nitrogen atoms undergo sp² hybridization. This results in a trigonal planar geometry as per VSEPR theory, agreeing with borazine's planar nature. Both in the case of single bonds and double bonds between boron and nitrogen, the sp² hybridization is maintained. Understanding hybridization not only informs us about the shape of the molecule but also explains the types of bonds — \( \sigma \) and \( \pi \) bonds — that are present in the molecule.

Pi System

In chemistry, a pi system is a set of electrons found in the overlapping p-orbitals across adjacent atoms, which span multiple atoms in a molecule. These systems are characterized by \( \pi \) bonds, which typically arise in double bonding scenarios. In borazine's octet rule-following structure, there are \( \pi \) electrons as part of the double bonds between the boron and nitrogen atoms. Because there are three such double bonds, each containing two \( \pi \) electrons, borazine has a total of six \( \pi \) electrons. The presence of these \( \pi \) electrons contributes to the resonance stabilization and the overall electron delocalization within borazine.

Bond Length

The concept of bond length is directly related to the type of bonding between atoms. Single bonds are generally longer than double bonds due to the stronger electron-sharing double bonds exhibit. In borazine, the observation that all B-N bonds are the same length and intermediate between typical single and double B-N bond lengths implies delocalization of electrons. This hints at the presence of resonance structures and a resonance hybrid form, further pointing towards the actual structure of borazine being a composite, rather than a molecule with distinct single and double bonds as might be simplistically depicted in some Lewis structures.