Problem 55
Question
Which of the following has the lowest bond angle? (a) \(\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CH}_{4}\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) (d) \(\mathrm{NH}_{3}\)
Step-by-Step Solution
Verified Answer
Hydrogen sulfide (\(\text{H}_2\text{S}\)) has the lowest bond angle.
1Step 1: Understanding Bond Angles in Tetrahedral Structures
Typically, in a perfect tetrahedral structure, such as methane ( \(\text{CH}_4\)), the bond angle is 109.5 degrees. This angle is determined by the repulsions between electron pairs around the central atom, which aim to spread out as evenly as possible.
2Step 2: Identifying Molecular Shapes and Electron Pair Effects
Water (\(\text{H}_2\text{O}\)), hydrogen sulfide (\(\text{H}_2\text{S}\)), and ammonia (\(\text{NH}_3\)) all have deviations from tetrahedral geometry due to the presence of lone pairs, which repel more strongly than bonding pairs. Water generally has a bond angle of about 104.5 degrees, ammonia about 107 degrees, while the bond angles in \(\text{H}_2\text{S}\) are smaller because sulfur is less electronegative and its lone pair-bond pair repulsion is weaker than in water.
3Step 3: Determining the Smallest Bond Angle
Among the given molecules, the molecule with the most lone pairs and least effective central atom electronegativity for lone pair-bond pair repulsion would have the smallest bond angle. Hydrogen sulfide (\(\text{H}_2\text{S}\)) has angles around 92 degrees, making it the smallest among them.
Key Concepts
Tetrahedral GeometryElectron Pair RepulsionMolecular Shapes
Tetrahedral Geometry
In molecular chemistry, tetrahedral geometry is a common and important shape that many molecules exhibit. This structure is named after a tetrahedron, a three-dimensional shape with four triangular faces. Typically, when a central atom bonds with four other atoms, such as in methane (\(\text{CH}_4\)), they adopt a tetrahedral geometry.
The unique shape emerges because in three-dimensional space, the four bonds arrange themselves as far apart as possible. This minimizes repulsion between electron pairs, resulting in a bond angle of approximately 109.5 degrees.
The unique shape emerges because in three-dimensional space, the four bonds arrange themselves as far apart as possible. This minimizes repulsion between electron pairs, resulting in a bond angle of approximately 109.5 degrees.
- This setup is symmetrical and maximizes the distance between electron pairs, which are negatively charged and thus repulsive.
- The symmetrical distribution helps stabilize the molecule.
Electron Pair Repulsion
Electron pair repulsion is a crucial concept in understanding molecular shapes. It is the foundational theory behind the VSEPR (Valence Shell Electron Pair Repulsion) model, which predicts the geometry of a molecule.
According to VSEPR theory, electron pairs, both bonding and lone pairs, repel each other. They tend to position themselves as far apart as possible around the central atom. This behavior explains the shape and bond angles in molecules.
According to VSEPR theory, electron pairs, both bonding and lone pairs, repel each other. They tend to position themselves as far apart as possible around the central atom. This behavior explains the shape and bond angles in molecules.
- Bonding pairs are those involved in forming bonds between atoms.
- Lone pairs are pairs of electrons that are not involved in bonding.
Molecular Shapes
Molecular shapes are determined by the arrangement of electron pairs around a central atom. These shapes influence the molecule's physical and chemical properties.
The presence of lone pairs is a key factor in altering these shapes. Take water (\(\text{H}_2\text{O}\)) as an example. Though it has a tetrahedral electron pair geometry, the two lone pairs on the oxygen atom compress the bond angle between the hydrogen atoms to about 104.5 degrees, giving it a bent shape.
The presence of lone pairs is a key factor in altering these shapes. Take water (\(\text{H}_2\text{O}\)) as an example. Though it has a tetrahedral electron pair geometry, the two lone pairs on the oxygen atom compress the bond angle between the hydrogen atoms to about 104.5 degrees, giving it a bent shape.
- Ammonia (\(\text{NH}_3\)) has one lone pair, resulting in a trigonal pyramidal shape with a bond angle of about 107 degrees.
- Hydrogen sulfide (\(\text{H}_2\text{S}\)) exhibits the most deviation, with bond angles around 92 degrees due to weaker repulsion by its lone pairs.
Other exercises in this chapter
Problem 53
Which of the following has the most acidiccharacter? (a) HF (b) \(\mathrm{HCl}\) (c) \(\mathrm{HBr}\) (d) HI
View solution Problem 54
Which one of the sixth group elements has the highest catenation power? (a) \(\mathrm{O}\) (b) \(\mathrm{S}\) (c) \(\mathrm{Se}\) (d) \(\mathrm{Te}\)
View solution Problem 56
Which one of the following elements does not form the compound, \(\mathrm{M}_{4} \mathrm{O}_{10}(\mathrm{M}=\) element \() ?\) (a) As (b) Bi (c) \(\mathrm{Sb}\)
View solution Problem 58
Which of the following sets has the strongest tendency to form anions? (a) \(\mathrm{V}, \mathrm{Cr}, \mathrm{Mn}\) (b) \(\mathrm{N}, \mathrm{O}, \mathrm{F}\) (
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