Textbook Question
Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:
(a) Predict the bond angles around each of the carbon atoms and sketch the molecule.
Ch.9 - Molecular Geometry and Bonding Theories
Chapter 9, Problem 103a
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.
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Identify the total number of valence electrons in the molecule. Boron (B) has 3 valence electrons, nitrogen (N) has 5, and hydrogen (H) has 1. Calculate the total for B3N3H6.
Arrange the B and N atoms in an alternating pattern to form a six-membered ring, as described in the problem.
Place one hydrogen atom bonded to each B and each N atom, ensuring that each atom follows the octet rule where applicable (except for hydrogen, which follows the duet rule).
Distribute the remaining valence electrons as lone pairs on the nitrogen atoms to satisfy the octet rule for nitrogen, while ensuring that boron has a complete set of bonds.
Check the formal charges on each atom to ensure they are zero. Adjust any double bonds between B and N if necessary to achieve zero formal charges on all atoms.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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, allowing for the determination of formal charges and molecular geometry. In the case of borazine, drawing the Lewis structure involves placing electrons to satisfy the octet rule for each atom while ensuring that the formal charges are minimized.
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Lewis Dot Structures: Ions
Formal Charge
Formal charge is a theoretical charge assigned to an atom in a molecule, calculated based on the number of valence electrons, the number of non-bonding electrons, and the number of bonds. It helps in assessing the stability of a molecule and determining the most favorable Lewis structure. For borazine, achieving a formal charge of zero on each atom is crucial for the stability and accuracy of the proposed structure.
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Planarity and Hybridization
Planarity in molecular structures refers to the arrangement of atoms in a single plane, which affects the molecule's reactivity and properties. In borazine, the alternating boron and nitrogen atoms create a planar six-membered ring. Understanding the hybridization of the atoms involved (sp2 for B and N in this case) is essential for predicting bond angles and the overall geometry of the molecule.
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Related Practice
Textbook Question
Butadiene, C4H6, is a planar molecule that has the following carbon–carbon bond lengths:
(b) From left to right, what is the hybridization of each carbon atom in butadiene?
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. (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 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?