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Ch.9 - Molecular Geometry and Bonding Theories
All textbooksBrown 14th EditionCh.9 - Molecular Geometry and Bonding TheoriesProblem 103e
Chapter 9, Problem 103e

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (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? Would you expect the molecule to be planar for both Lewis structures?

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Step 1: Draw the Lewis structure of borazine, B3N3H6. Each boron atom is bonded to one nitrogen atom and one hydrogen atom. Each nitrogen atom is bonded to one boron atom and one hydrogen atom. The boron and nitrogen atoms form a six-membered ring, with the hydrogen atoms attached to the outside of the ring.
Step 2: Determine the number of electron domains around each boron and nitrogen atom. An electron domain is any area where electrons are most likely to be found, such as a single bond, double bond, triple bond, or lone pair of electrons. In borazine, each boron atom is involved in three electron domains (two single bonds and one double bond), and each nitrogen atom is involved in three electron domains (two single bonds and one lone pair).
Step 3: Use the number of electron domains to determine the hybridization of the boron and nitrogen atoms. If an atom is involved in three electron domains, it is sp2 hybridized. Therefore, both the boron and nitrogen atoms in borazine are sp2 hybridized.
Step 4: Consider the geometry of the molecule. Molecules with sp2 hybridization are planar because the three electron domains arrange themselves as far apart as possible in a plane, forming a trigonal planar shape.
Step 5: Therefore, we would expect the molecule to be planar for both Lewis structures, as both the boron and nitrogen atoms are sp2 hybridized and form a planar arrangement of electron domains.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding requirements of a molecule. In borazine, the boron (B) and nitrogen (N) atoms undergo hybridization to form sp2 hybrid orbitals, which allows for the formation of sigma bonds and the necessary geometry for the planar structure of the molecule.
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Hybridization

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They help visualize the arrangement of electrons and the connectivity of atoms, which is crucial for predicting molecular geometry and hybridization. In borazine, the Lewis structure illustrates the alternating B and N atoms and their respective hydrogen attachments.
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Planarity in Molecules

Planarity refers to the arrangement of atoms in a molecule such that they lie in a single plane. This is often a result of hybridization and the presence of double bonds or lone pairs that can influence molecular geometry. In borazine, the sp2 hybridization of B and N leads to a trigonal planar arrangement around each atom, contributing to the overall planar structure of the six-membered ring.
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Square planar complexes show the most complex splitting pattern.
Related Practice
Textbook Question

Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:

(c) The middle C¬C bond length in butadiene (1.48 Å) is a little shorter than the average C¬C single bond length (1.54 Å). Does this imply that the middle C¬C bond in butadiene is weaker or stronger than the average C¬C single bond?

Textbook Question

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero.

Textbook Question

The structure of borazine, B3N3H6, is a six-membered ring of alternating B and N atoms. There is one H atom bonded to each B and to each N atom. The molecule is planar. (c) What are the formal charges on the atoms in the Lewis structure from part (b)? Given the electronegativities of B and N, do the formal charges seem favorable or unfavorable? What are the formal charges on the atoms in the Lewis structure from part (b)?

Textbook Question

The highest occupied molecular orbital of a molecule is abbreviated as the HOMO. The lowest unoccupied molecular orbital in a molecule is called the LUMO. Experimentally, one can measure the difference in energy between the HOMO and LUMO by taking the electronic absorption (UV-visible) spectrum of the molecule. Peaks in the electronic absorption spectrum can be labeled as π2p-π2p*, σs-σ2s*, and so on, corresponding to electrons being promoted from one orbital to another. The HOMO-LUMO transition corresponds to molecules going from their ground state to their first excited state. (c) The electronic absorption spectrum of the N2 molecule has the lowest energy peak at 170 nm. To what orbital transition does this correspond?

Textbook Question

One of the molecular orbitals of the H2- ion is sketched below:

(a) Is the molecular orbital a s or p MO? Is it bonding or antibonding?