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Chapter 9: Problem 31

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, NH鈧勨伜, has the largest H-N-H bond angle of 109.5掳 due to its tetrahedral molecular shape. (b) The amide ion, NH鈧傗伝, has the smallest H-N-H bond angle, which is typically smaller than 107掳, 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 (NH鈧�): 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 (NH鈧傗伝): 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 (NH鈧勨伜): 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 (NH鈧�): It has one lone pair and three bonding pairs. The molecular shape is trigonal pyramidal. 2. Amide ion (NH鈧傗伝): It has two lone pairs and two bonding pairs. The molecular shape is bent or V-shaped. 3. Ammonium ion (NH鈧勨伜): 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 (NH鈧�): In a trigonal pyramidal molecular shape, the H-N-H bond angles are typically 107掳. 2. Amide ion (NH鈧傗伝): In a bent or V-shaped molecular shape with two lone pairs, the H-N-H bond angles are typically smaller than 107掳, ranging from 104.5掳 to close to 109.5掳 for different molecules. 3. Ammonium ion (NH鈧勨伜): In a tetrahedral molecular shape, the H-N-H bond angles are precisely 109.5掳. Now we can answer the original questions: (a) Ammonium ion, NH鈧勨伜, has the largest H-N-H bond angle (109.5掳). (b) Amide ion, NH鈧傗伝, has the smallest H-N-H bond angle (typically smaller than 107掳).

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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掳.
    • 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掳.
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掳 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掳 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

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(I_{2}\) ? Explain your answer.

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

(a) Predict the electron-domain geometry around the central \(\mathrm{S}\) atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). ( \(\mathbf{b}\) ) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the 1 s orbital of hydrogen with the 2 s orbital of fluorine. The 1 s orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a 1 sorbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a 1 s orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HE. These are called "nonbonding orbitals." Sketch the energy- level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HE. Where are the nonbonding electrons?

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4} \mathrm{Cl}\). How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?
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