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Chapter 9: Problem 20

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. (a) \(\mathrm{H}_{2} \mathrm{~S},\) (b) \(\mathrm{BCl}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CBr}_{4}\) (e) TeBr \(_{4}\)

Short Answer

Expert verified
We can confidently predict the bond angles for (b) BCl3, (c) CH3I, and (d) CBr4, as they have symmetric molecular geometries (trigonal planar and tetrahedral) with no lone pairs on the central atoms. The bond angles are 120掳 for BCl3, and 109.5掳 for both CH3I and CBr4. However, we are uncertain about the bond angles in (a) H2S and (e) TeBr4, as they both contain lone pairs on their central atoms (S and Te) affecting their bond angles.

Step by step solution

01

1. Determine the Lewis Structure

Draw the Lewis structure锛坋lectron-dot structure锛� for each molecule to understand the distribution of electrons around the central atom. (a) H2S: S is the central atom, surrounded by two H atoms and two lone pairs of electrons. (b) BCl3: B is the central atom, surrounded by three Cl atoms with no lone pairs. (c) CH3I: C is the central atom, surrounded by three H atoms and one I atom with no lone pairs. (d) CBr4: C is the central atom, surrounded by four Br atoms with no lone pairs. (e) TeBr4: Te is the central atom, surrounded by four Br atoms and one lone pair of electrons.
02

2. Determine the Electronic Geometry and Molecular Geometry

Use the VSEPR theory to identify the electronic geometry and molecular geometry for each molecule. The notation AXmEn is used, where A stands for the central atom, X represents surrounding atoms, m represents the number of surrounding atoms, and n stands for the number of lone pairs on the central atom. (a) H2S: - Electronic geometry: AX2E2 (Tetrahedral) - Molecular geometry: Bent (b) BCl3: - Electronic geometry: AX3 (Trigonal planar) - Molecular geometry: Trigonal planar (c) CH3I: - Electronic geometry: AX4 (Tetrahedral) - Molecular geometry: Tetrahedral (d) CBr4: - Electronic geometry: AX4 (Tetrahedral) - Molecular geometry: Tetrahedral (e) TeBr4: - Electronic geometry: AX4E (Trigonal bipyramidal) - Molecular geometry: See-saw
03

3. Predict Bond Angles and Level of Confidence

Based on the molecular geometry, predict the bond angles, and then determine the level of confidence in the predictions. (a) H2S (Bent): - Bond angle: Approximately 104.5掳 - Confidence: Uncertain, due to two lone pairs on the central atom (S) that can cause variations in bond angle. (b) BCl3 (Trigonal Planar): - Bond angle: 120掳 - Confidence: Confident, as there are no lone pairs on the central atom (B) and the molecular geometry is symmetric. (c) CH3I (Tetrahedral): - Bond angle: 109.5掳 - Confidence: Confident, as there are no lone pairs on the central atom (C) and the molecular geometry is symmetric. (d) CBr4 (Tetrahedral): - Bond angle: 109.5掳 - Confidence: Confident, as there are no lone pairs on the central atom (C) and the molecular geometry is symmetric. (e) TeBr4 (See-saw): - Bond angle: <120掳 for Br-Te-Br (equatorial) and <90掳 for Br-Te-Br (axial) - Confidence: Uncertain, due to the presence of a lone pair on the central atom (Te) that can cause variations in bond angle.

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Key Concepts

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

Lewis Structure
To fully understand molecules, we start by drawing the Lewis Structure. This representation shows how atoms are connected and illustrates the distribution of electrons, such as bonding pairs and lone pairs, around the central atom. For example, in the molecule \(\text{H}_2\text{S}\), sulfur acts as the central atom, which is bonded to two hydrogen atoms and has two lone pairs on it.
Similarly, \(\text{BCl}_3\) displays boron as the central atom bonded to three chlorine atoms without any lone pairs.
  • \(\text{CH}_3\text{I}\) highlights carbon at the center surrounded by hydrogens and an iodine.
  • In \(\text{CBr}_4\), carbon is again central with four bromine atoms around it.
  • For \(\text{TeBr}_4\), tellurium is the centerpiece with four bromines and one lone pair.
Understanding the Lewis Structures is crucial because it helps predict molecular shapes and how these will affect molecular properties.
Molecular Geometry
Once the Lewis Structure is known, we can determine the Molecular Geometry using the VSEPR (Valence Shell Electron Pair Repulsion) theory. This theory suggests that electron pairs will arrange themselves to minimize repulsion between them. This affects the shape of the molecule significantly.
For each molecule, the notation \(AX_mE_n\) is used to denote geometry. Here, "A" is the central atom, "X" is the number of surrounding atoms, and "E" is the number of lone pairs.
  • For \(\text{H}_2\text{S}\), the geometry is bent due to the two lone pairs (AX2E2).
  • In \(\text{BCl}_3\), it forms a trigonal planar structure as it is simply \(AX3\), with no lone pairs affecting the shape.
  • Both \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) display a tetrahedral arrangement (AX4).
  • For \(\text{TeBr}_4\), the geometry is see-saw due to the lone pair, marked as AX4E.
This analysis is key for step further into understanding how the molecule might interact or react.
Bond Angles
Bond angles are an essential feature of molecular geometry. They determine the angle formed between three atoms across at least two bonds. These angles can vary depending, mainly, upon the presence of lone electron pairs, which exert repulsion on bonding pairs, thereby altering the idealized angles.
Here are basic expectations for each molecule:
  • In \(\text{H}_2\text{S}\), the bending caused by lone pairs compresses the angle to about 104.5掳.
  • \(\text{BCl}_3\)鈥檚 planar symmetry allows for perfect 120掳 angles.
  • Compounds like \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) both maintain angles of approximately 109.5掳 due to their tetrahedral shape.
  • For \(\text{TeBr}_4\), angles vary due to its asymmetric see-saw shape, with Br-Te-Br typically less than 120掳 for equatorial and less than 90掳 for axial bonds.
Understanding these angles helps in predicting the molecule's chemical behavior and interaction tendencies.
Electron Pair Repulsion
According to VSEPR theory, Electron Pair Repulsion is a central concept that explains molecular shapes. This premise indicates that electron pairs surrounding a central atom repel each other.
Thus, to minimize repulsion and achieve stability, they space themselves as far apart as possible.
  • Take \(\text{H}_2\text{S}\); due to significant repulsion from the two lone pairs, the molecule bends.
  • \(\text{BCl}_3\) lacks lone pairs, allowing for equal distribution and maintains a trigonal planar shape.
  • The tetrahedral shape in both \(\text{CH}_3\text{I}\) and \(\text{CBr}_4\) arises due to the equal repulsion of four bonds.
  • Lastly, the see-saw shape of \(\text{TeBr}_4\) is because of lone pair influence which distorts the relative symmetry.
Recognizing the role of electron pair repulsion in shaping molecules is crucial to understanding molecular structures and their potential reactive behaviors.

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Most popular questions from this chapter

The structure of borazine, \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\) is a six-membered ring of alternating \(\mathrm{B}\) and \(\mathrm{N}\) atoms. There is one \(\mathrm{H}\) atom bonded to each \(B\) and to each \(\mathrm{N}\) atom. The molecule is planar. (a) Write a Lewis structure for borazine in which the formal charge on every atom is zero. (b) Write a Lewis structure for borazine in which the octet rule is satisfied for every atom. (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? (d) Do either of the Lewis structures in parts (a) and (b) have multiple resonance structures? (e) What are the hybridizations at the \(\mathrm{B}\) and \(\mathrm{N}\) atoms in the Lewis structures from parts (a) and (b)? Would you expect the molecule to be planar for both Lewis structures? (f) The six \(\mathrm{B}-\mathrm{N}\) bonds in the borazine molecule are all identical in length at \(144 \mathrm{pm} .\) Typical values for the bond lengths of \(\mathrm{B}-\mathrm{N}\) single and double bonds are \(151 \mathrm{pm}\) and \(131 \mathrm{pm},\) respectively. Does the value of the \(\mathrm{B}-\mathrm{N}\) bond length seem to favor one Lewis structure over the other? (g) How many electrons are in the \(\pi\) system of botazine?

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure \(9.25 .\) In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}{ }^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

In the sulphate ion, \(\mathrm{SO}_{4}^{2-}\), the sulphur atom is the central atom with the other 4 oxygen atoms attached to it. (a) Draw a Lewis structure for the sulphate ion. (b) What hybridization is exhibited by the \(\mathrm{S}\) atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

(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?
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