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Chapter 9: Problem 59

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?

Short Answer

Expert verified
(a) The total number of valence electrons in the vinyl chloride molecule is 18. (b) 10 valence electrons are used to make σ bonds in the vinyl chloride molecule. (c) 2 valence electrons are used to make π bonds in the vinyl chloride molecule. (d) 6 valence electrons remain in nonbonding pairs in the vinyl chloride molecule. (e) The hybridization at each carbon atom in the vinyl chloride molecule is sp².

Step by step solution

01

Calculate the total number of valence electrons.

To find the total number of valence electrons, we add up the valence electrons of each atom within the vinyl chloride molecule, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\). Carbon has 4 valence electrons, hydrogen has 1 valence electron, and chlorine has 7 valence electrons. There are two carbon atoms, three hydrogen atoms, and one chlorine atom in the molecule. So, the total number of valence electrons is: Total valence electrons = (2 × 4) + (3 × 1) + (1 × 7) = 8 + 3 + 7 = 18 (a) The total number of valence electrons in the vinyl chloride molecule is 18.
02

Calculate the number of valence electrons used to make σ bonds.

There are single bonds between C-C, C-H, and C-Cl in the molecule. Each single bond is a σ bond. The 3 single bonds to hydrogen atoms use 2 valence electrons each (as one electron comes from hydrogen and one comes from carbon) and the single bond between carbon and chlorine also uses 2 valence electrons (one from carbon and one from chlorine). The C-C σ bond also uses 2 valence electrons. So, the total number of valence electrons used in making σ bonds is: Total σ electrons = (3 × 2) + 2 + 2 = 6 + 2 + 2 = 10 (b) 10 valence electrons are used to make σ bonds in the vinyl chloride molecule.
03

Calculate the number of valence electrons used to make π bonds.

The vinyl chloride molecule has a double bond (C=C) between the two carbon atoms. One of the bonds is a σ bond, which we already counted in Step 2, and the other bond is a π bond. Therefore, there's only one π bond in the molecule, and it uses 2 valence electrons. (c) 2 valence electrons are used to make π bonds in the vinyl chloride molecule.
04

Calculate the number of nonbonding pairs of electrons.

Subtract the total number of valence electrons used in making σ and π bonds from the total number of valence electrons in the molecule: Total nonbonding pairs of electrons = Total valence electrons - Total σ electrons - Total π electrons = 18 - 10 - 2 = 6 (d) 6 valence electrons remain in nonbonding pairs in the vinyl chloride molecule.
05

Determine the hybridization of each carbon atom.

The first carbon atom (C1) forms 3 sigma bonds (to the other carbon atom and two hydrogen atoms) and one π bond (with the other carbon atom). Since it forms 3 sigma bonds, its hybridization is sp². The second carbon atom (C2) forms 2 sigma bonds (to the chlorine atom and the first carbon atom) and one π bond (with the first carbon atom). Since it forms 2 sigma bonds, its hybridization is sp². (e) The hybridization at each carbon atom in the vinyl chloride molecule is sp².

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Most popular questions from this chapter

Molecules that are brightly colored have a small energy gap between filled and empty electronic states (the HOMOLUMO gap; see Exercise 9.104 ). Suppose you have two samples, one is lycopene which is responsible for the red color in tomato, and the other is curcumin which is responsible for the yellow color in turmeric. Which one has the larger HOMO-LUMO gap?

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Consider the following \(\mathrm{XF}_{4}\) ions: $\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\( and \)\mathrm{AlF}_{4}^{-}$ (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry? (d) Which of the ions will exhibit a see-saw molecular geometry?

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}$ ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_4) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{OSF}_{4}(g) $$ The \(O\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.3) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of OSF \(_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.
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