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Chapter 9: Problem 37

(a) Does \(C S_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{SO}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Short Answer

Expert verified
(a) CS₂ does not have a dipole moment due to its linear geometry that causes the dipole moments of the S-C bonds to cancel each other out. (b) SO₂ has a dipole moment due to its bent geometry. The net dipole points from the more electronegative oxygen atoms towards the less electronegative sulfur atom.

Step by step solution

01

1. Draw the Lewis structure for CS₂

First, count the number of valence electrons in the molecule: Carbon (C) has 4 valence electrons, and Sulfur (S) has 6, so we have: 4 + 2(6) = 16 total valence electrons. Draw the Lewis structure with carbon in the center and a sulfur atom on either side. Each sulfur atom forms a double bond with the central carbon atom, making the central carbon atom have an octet of electrons. The sulfur atoms also have an octet of electrons.
02

2. Determine the molecular geometry for CS₂ using VSEPR theory

VSEPR theory states that electron groups (including lone pairs and single, double, and triple bonds) will take positions in space in order to minimize repulsion. In CS₂, there are two electron groups around the central carbon atom, and both are bonding groups (double bonds to S atoms). According to VSEPR theory, with two electron groups and no lone pairs on the central atom, the molecule will have a linear geometry.
03

3. Determine if CS₂ has a net dipole moment

Since the molecule has a linear geometry, the two sulfur atoms are symmetrically placed on either side of the central carbon atom. This means that each S-C bond has a dipole moment, but they cancel each other out, so there is no net dipole moment. Consequently, CS₂ does not have a dipole moment.
04

4. Draw the Lewis structure for SO₂

First, count the number of valence electrons in the molecule: Sulfur (S) has 6 valence electrons, and Oxygen (O) has 6, so we have: 6 + 2(6) = 18 total valence electrons. Draw the Lewis structure with sulfur in the center and an oxygen atom on either side. There is a double bond between the sulfur atom and one oxygen atom, and a single bond between the sulfur atom and the other oxygen atom. The sulfur atom has a lone electron pair.
05

5. Determine the molecular geometry for SO₂ using VSEPR theory

In SO₂, there are three electron groups around the central sulfur atom (two bonding groups and one lone pair). According to VSEPR theory, with three electron groups and one lone pair on the central atom, the molecule will have a bent geometry.
06

6. Determine if SO₂ has a net dipole moment

SO₂ has a bent geometry, which results in the two oxygen atoms being placed asymmetrically around the central sulfur atom. This creates a net dipole moment in the molecule, pointing from the more electronegative oxygen atoms (with the higher electron density) to the less electronegative sulfur atom (with the lower electron density). In conclusion: (a) CS₂ does not have a dipole moment. (b) SO₂ has a dipole moment, and the net dipole points from the oxygen atoms towards the sulfur atom.

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Most popular questions from this chapter

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(96 \mathrm{pm}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? \(\mathrm{In}\) what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxvgen?

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?

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.
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