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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 charges 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 \(B\) 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 \(\mathrm{B}\) and \(\mathrm{N}\) 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 \(1.44 \AA\). Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(1.51 \AA \mathrm{A}\) 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?

Step by step solution

01

(a) Lewis Structure with Formal Charges of Zero

1. Recall the rules for creating Lewis structures of molecules. Each atom should satisfy the octet rule if possible - meaning that they should have 8 valence electrons in total from their own atom and any shared electrons within bonds. 2. Count the total number of valence electrons in the molecule: 3B (3 valence electrons each) + 3N (5 valence electrons each) + 6H (1 valence electron each) = 9 + 15 + 6 = 30 valence electrons. 3. Arrange the atoms in an alternating fashion, placing Hydrogen atoms outside of the ring, and connect them with single bonds, and place the remaining electrons as lone pairs: B / \\ N - B - N \\ / N 4. Each N atom has 6 electrons, and each B atom has only 6 electrons as well. To make the formal charge of each atom zero, we can make double bonds between adjacent N and B atoms, which will create a resonance structure: B / \\ =N - B - N= \\ / N Now, all the atoms have formal charges of zero and adhere to the octet rule.

02

(b) Lewis Structure Satisfying the Octet Rule

The Lewis structure obtained in part (a) already satisfies the octet rule for all the atoms.

03

(c) Formal Charges of the Octet Rule Lewis Structure and Electronegativity

Formal charges in the structure from part (a) are already zero. Since the electronegativities of B and N differ, the transfer of electrons from B to N would lead to unfavorable charge distributions. However, the presence of resonance ensures the stability of the structure.

04

(d) Resonance Structures

Yes, the Lewis structure in part (a) has multiple resonance structures. There are three identical resonance structures possible that arise due to the movement of electrons in the double bonds.

05

(e) Hybridizations and Planarity

The hybridization at B and N atoms in the Lewis structures: 1. B atom: Three \(\sigma\) bonds and no lone pair, so hybridization is \(\mathrm{sp^2}\). 2. N atom: Two \(\sigma\) bonds and one lone pair, so hybridization is also \(\mathrm{sp^2}\). Both \(\mathrm{/sp^2}\) hybridizations have a planar geometry, so the molecule is expected to be planar for both Lewis structures.

06

(f) Bond Lengths Comparison

The bond length of the B-N bonds in borazine is \(1.44 \ \mathrm{A}\). This value lies between the typical bond lengths of B-N single (\(1.51 \ \mathrm{A}\)) and double bonds (\(1.31 \ \mathrm{A}\)). This bond length suggests that there is resonance among the B-N bonds, as the bond length is indicative of a partial double bond character in the molecule. Therefore, the structure from part (a) with resonance structures seems more favorable.

07

(g) Electrons in the π System

The π system consists of the electrons involved in the double bonds within the molecule. In the Lewis structure from part (a), we have three double bonds in the alternate B-N pairs, each contributing two π electrons. Hence, there are 6 electrons in the π system of borazine.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Formal Charge Calculation

Formal charges are key to understanding the distribution of electrons in a molecular structure and are crucial for predicting the stability of molecules. To calculate the formal charge on an atom, we use the formula:
\[\text{Formal Charge} = \text{(Valence electrons)} - \text{(Non-bonding electrons)} - \frac{1}{2} \times \text{(Bonding electrons)}\]
The sum of the formal charges in a stable molecule must be equal to the overall charge on the molecule. For borazine, we aim for a formal charge of zero on each atom to achieve the most stable structure, reflective of the actual charge distribution within the molecule.
In borazine, applying the calculation steps shows that each boron (B) and nitrogen (N) atom can achieve a formal charge of zero when the B and N atoms are connected by alternating single and double bonds. This arrangement satisfies both formal charge requirements and octet rule for all the atoms.

Octet Rule

The octet rule is a fundamental concept in chemistry, which suggests that atoms are most stable when they have eight electrons in their valence shell, akin to the electron configuration of noble gases. Hydrogen is an exception, which is most stable with two electrons.
When we draw the Lewis structure of borazine, we aim for an electronic configuration where each boron and nitrogen atom adheres to the octet rule. In the solution provided, we demonstrate that the bonding arrangement with alternating single and double bonds between B and N atoms fulfills the octet rule for all atoms in the ring, and each hydrogen atom is also satisfied with its duet.

Resonance Structures

Resonance structures are a set of two or more Lewis Structures that collectively describe the real distribution of electrons across a molecule. None of the individual resonance structures are the real structure, but rather, the actual distribution of electrons is a hybrid of all the structures.
For borazine, the double bonds between the boron and nitrogen atoms can be distributed in different ways, leading to several equivalent structures that differ only in the placement of these bonds. This is illustrative of the resonance phenomenon and indicates that the actual electron distribution is delocalized over the entire ring, thus stabilizing the molecule.

Molecular Hybridization

Molecular hybridization refers to the mixing of atomic orbitals to form new hybrid orbitals that can better explain molecular bonding and geometry. In borazine, both boron and nitrogen atoms undergo sp² hybridization. This means that one s and two p orbitals mix to create three equivalent sp² hybrid orbitals.
Since sp² hybrid orbitals form a trigonal planar shape, this hybridization leads to the planar geometry of borazine. All B and N atoms are in the same plane, reinforcing that the Lewis structures conform with the molecule's geometry as being planar.

Pi Electron System

The pi (π) electron system comprises electrons found in π bonds, which are side-by-side overlaps of p orbitals. In the case of borazine, the pi system includes the electrons shared in the double bonds of the ring.
Since each double bond has two pi electrons and there are three such bonds in borazine, the π electron system consists of six electrons in total. These electrons are delocalized over the ring, allowing for the resonance stabilization of the molecule and contributing to its aromatic character, akin to the π electron system found in benzene.