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Ammonia, \(\mathrm{NH}_{3}\), reacts with incredibly strong bases to produce the amide ion, \(\mathrm{NH}_{2}\) -. Ammonia can also react with acids to produce the ammonium ion, \(\mathrm{NH}_{4}^{+}\). (a) Which species (amide ion, ammonia, or ammonium ion) has the largest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle? (b) Which species has the smallest \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angle?

Short Answer

Expert verified
(a) The ammonium ion, NH₄⁺, has the largest H-N-H bond angle of 109.5° due to its tetrahedral molecular shape. (b) The amide ion, NH₂⁻, has the smallest H-N-H bond angle, which is typically smaller than 107°, because of its bent or V-shaped molecular shape.

Step by step solution

01

Determine the electron pair geometry of each species

Let's start determining the number of electron pairs around the nitrogen atom for each species: 1. Ammonia (NH₃): Nitrogen has 5 valence electrons, and it is bonded to 3 hydrogen atoms. So, there is one lone pair and three bonding pairs, which gives a total of four electron pairs. 2. Amide ion (NH₂⁻): Nitrogen has 5 valence electrons like in ammonia, but it gains one extra electron due to the negative charge. In this species, nitrogen has two lone pairs and two bonding pairs, totaling four electron pairs. 3. Ammonium ion (NH₄⁺): Nitrogen has 5 valence electrons but loses one electron as it has a positive charge. It is bonded to 4 hydrogen atoms, and thus, it has 0 lone pairs and four bonding pairs, totaling four electron pairs. For each of the species mentioned above, there are four electron pairs around the nitrogen atom. Therefore, the electron pair geometry for each one of them is tetrahedral.
02

Determine the molecular shape of each species

Using the VSEPR theory, we can determine the molecular shape of the species: 1. Ammonia (NH₃): It has one lone pair and three bonding pairs. The molecular shape is trigonal pyramidal. 2. Amide ion (NH₂⁻): It has two lone pairs and two bonding pairs. The molecular shape is bent or V-shaped. 3. Ammonium ion (NH₄⁺): It has no lone pair and four bonding pairs, and that gives it a tetrahedral molecular shape with four equivalent H-N-H bond angles.
03

Compare H-N-H bond angles in each species

Now, based on the molecular shapes, we can predict the H-N-H bond angles: 1. Ammonia (NH₃): In a trigonal pyramidal molecular shape, the H-N-H bond angles are typically 107°. 2. Amide ion (NH₂⁻): In a bent or V-shaped molecular shape with two lone pairs, the H-N-H bond angles are typically smaller than 107°, ranging from 104.5° to close to 109.5° for different molecules. 3. Ammonium ion (NH₄⁺): In a tetrahedral molecular shape, the H-N-H bond angles are precisely 109.5°. Now we can answer the original questions: (a) Ammonium ion, NH₄⁺, has the largest H-N-H bond angle (109.5°). (b) Amide ion, NH₂⁻, has the smallest H-N-H bond angle (typically smaller than 107°).

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Key Concepts

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

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. The shape of a molecule is determined by the number and arrangement of bonded atoms and lone pairs around a central atom. In the context of the exercise:
  • Ammonia (\(\text{NH}_3\)): It has one lone pair and three hydrogen atoms bonded to the nitrogen, resulting in a trigonal pyramidal shape. This geometry is not completely symmetrical due to the influence of the lone pair.
  • Amide ion (\(\text{NH}_2^-\)): Consisting of two lone pairs and two hydrogen atoms bonded to the nitrogen, it shapes into a bent or V-shaped molecular geometry. The presence of two lone pairs causes the bonds to bend even more compared to ammonia, which directly affects the bond angles.
  • Ammonium ion (\(\text{NH}_4^+\)): With no lone pairs and four hydrogens bonded to the nitrogen, the ammonium ion adopts a tetrahedral molecular geometry. This shape is a perfect example of a symmetrical geometry.
Understanding these shapes helps predict the bond angles, an essential aspect in the study of molecular structure.
Electron Pair Geometry
Electron pair geometry accounts for both bonding pairs and lone pairs of electrons around a central atom. VSEPR (Valence Shell Electron Pair Repulsion) theory guides us in determining these configurations by considering the repulsion between electron pairs:
  • Tetrahedral Electron Pair Geometry: In molecules like ammonia, amide ion, and ammonium ion, the nitrogen atom is surrounded by four pairs of electrons, including bonds with hydrogen atoms and any lone pairs. Each configuration in this problem starts with a tetrahedral electron pair geometry as they all have a total of four electron pairs.
  • Even though all three species start from a tetrahedral arrangement:
    • Ammonia: The lack of symmetry due to one lone pair reduces the ideal bond angles to approximately 107°.
    • Amide Ion: The presence of two lone pairs bends the shape further, typically resulting in angles smaller than ammonia's.
    • Ammonium Ion: Without lone pairs, this maintains the classic tetrahedral angle of 109.5°.
This concept underscores the importance of both bonded atoms and lone pairs in influencing the shape and angles in a molecule.
Bond Angles
Bond angles are the angles between adjacent bonds in a molecule. They tell us a lot about the molecule's shape, and changes in these angles indicate alterations in the molecule's geometry due to the presence of lone pairs.
  • Influence of Lone Pairs: Lone pairs occupy more space than bonding pairs, resulting in greater repulsion that compresses bond angles.
    • In ammonia, the bond angles are approximately 107° due to one lone pair pushing the hydrogen atoms closer together. This angle is slightly less than the ideal tetrahedral angle because of this repulsive force.
    • The amide ion further reduces the bond angle due to two lone pairs, typically making these angles even smaller than those in ammonia.
  • Symmetrical Tetrahedral Shape: Ammonium ion maintains a stable tetrahedral bond angle of precisely 109.5° due to the absence of lone pairs, allowing for symmetric distribution of hydrogen atoms.
Understanding bond angles provides key insights into how electron pair repulsion shapes the physical configuration of molecules in chemistry.

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Most popular questions from this chapter

The following is part of a molecular orbital energy-level diagram for MOs constructed from 1 s atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: $$ \mathrm{H}_{2} \mathrm{He}_{2}, \mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{+}, \mathrm{or} \mathrm{H}_{2}^{-} ? $$ (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7\(]\)

What is the hybridization of the central atom in (a) \(\mathrm{PBr}_{5}\), (b) \(\mathrm{CH}_{2} \mathrm{O},\) (c) \(\mathrm{O}_{3},(\mathbf{d}) \mathrm{NO}_{2} ?\)

There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}:\) The compound on the right is called cisplatin, and the compound on the left is called transplatin. (a) Which compound has a nonzero dipole moment? (b) One of these compounds is an anticancer drug, and one is inactive. The anticancer drug works by its chloride ions undergoing a substitution reaction with nitrogen atoms in DNA that are close together, forming a \(\mathrm{N}-\mathrm{Pt}-\mathrm{N}\) angle of about \(90^{*} .\) Which compound would you predict to be the anticancer drug?

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Vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), is a gas that is used to form the important polymer called polyvinyl chloride (PVC). Its Lewis structure is (a) What is the total number of valence electrons in the vinyl chloride molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

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