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Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory
Tro - Chemistry: A Molecular Approach 5th Edition
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Tutti i libri di testoTro 5th EditionCh.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO TheoryProblema 37
Capitolo 11, Problema 37

Which species has the smaller bond angle, H3O+ or H2O? Explain.

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Identify the molecular geometry of each species: H2O is bent, and H3O+ is trigonal pyramidal.
Consider the number of lone pairs on the central atom: H2O has two lone pairs, while H3O+ has one lone pair.
Recall that lone pairs repel more strongly than bonding pairs, affecting bond angles.
Understand that more lone pairs result in smaller bond angles due to increased repulsion.
Conclude that H2O, with more lone pairs, has a smaller bond angle compared to H3O+.

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Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding pairs and lone pairs of electrons around the central atom, which influence the shape and angles between bonds. Understanding molecular geometry is crucial for predicting bond angles and the overall structure of molecules.
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Molecular Geometry with Two Electron Groups

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used to predict the geometry of individual molecules based on the repulsion between electron pairs. According to VSEPR, electron pairs will arrange themselves to minimize repulsion, which directly affects bond angles. This theory helps explain why H3O+ has a different bond angle compared to H2O.
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Molecular Shapes and VSEPR

Bond Angles

Bond angles are the angles formed between two adjacent bonds at an atom. They are influenced by the number of lone pairs and bonding pairs of electrons surrounding the atom. In the case of H3O+ and H2O, the presence of an additional hydrogen ion in H3O+ alters the bond angle compared to H2O, leading to a smaller angle due to increased electron repulsion.
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Bond Angles
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