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Chapter 9: Problem 21

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S},\) (b) \(\mathrm{HCN},\) (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\), (d) \(\mathrm{CH}_{3} \mathrm{~F} ?\)

Short Answer

Expert verified
There are 2 nonbonding electron pairs in \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\), 1 nonbonding electron pair in \(\mathrm{HCN}\), 0 nonbonding electron pairs in \(\mathrm{H}_{2} \mathrm{C}_{2}\), and 3 nonbonding electron pairs in \(\mathrm{CH}_{3} \mathrm{~F}\).

Step by step solution

01

(a) Determine valence electrons for \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\)

First, we need to count the number of valence electrons in the molecule. Carbon (C) has 4 valence electrons, hydrogen (H) has 1, and sulfur (S) has 6. Since there are two \(\mathrm{CH}_3\) groups and one S atom in the molecule, we have a total of 2 × (4 + 3 × 1) + 6 = 20 valence electrons.
02

(a) Assign electrons to covalent bonds in \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\)

The molecule consists of two \(\mathrm{CH}_3\) groups and one S atom. In each \(\mathrm{CH}_3\) group, we need 3 single bonds (C-H), each requiring 2 electrons, and a single bond (C-S) to connect the three atoms. So, we have a total of 2 × (3 × 2 + 2) = 16 electrons used for bonding.
03

(a) Calculate nonbonding electron pairs in \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\)

Subtracting the number of bonding electrons from the total valence electrons, we get the number of nonbonding electron pairs: 20 - 16 = 4. These four nonbonding electrons are in two pairs on the sulfur atom. So, there are 2 nonbonding electron pairs in \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\).
04

(b) Determine valence electrons for \(\mathrm{HCN}\)

First, we need to count the number of valence electrons in the molecule. Hydrogen (H) has 1 valence electron, carbon (C) has 4, and nitrogen (N) has 5. So, we have a total of 1 + 4 + 5 = 10 valence electrons.
05

(b) Assign electrons to covalent bonds in \(\mathrm{HCN}\)

The molecule consists of three atoms, H, C and N. We need 1 single bond (H─C) and 1 triple bond (C≡N) to connect the three atoms. Each single bond needs 2 electrons, and each triple bond needs 6 electrons. So, we have a total of 2 + 6 = 8 electrons used for bonding.
06

(b) Calculate nonbonding electron pairs in \(\mathrm{HCN}\)

Subtracting the number of bonding electrons from the total valence electrons, we get the number of nonbonding electron pairs: 10 - 8 = 2. These two nonbonding electrons are in one pair on the nitrogen atom. So, there is 1 nonbonding electron pair in \(\mathrm{HCN}\).
07

(c) Determine valence electrons for \(\mathrm{H}_{2} \mathrm{C}_{2}\)

First, we need to count the number of valence electrons in the molecule. Carbon (C) has 4 valence electrons and hydrogen (H) has 1. Since there are two H atoms and two C atoms, we have a total of 2 × 1 + 2 × 4 = 10 valence electrons.
08

(c) Assign electrons to covalent bonds in \(\mathrm{H}_{2} \mathrm{C}_{2}\)

The molecule consists of two H atoms and two C atoms. We need 2 single bonds (H─C) and 1 triple bond (C≡C) to connect the four atoms. Each single bond needs 2 electrons, and each triple bond needs 6 electrons. So, we have a total of 2 × 2 + 6 = 10 electrons used for bonding.
09

(c) Calculate nonbonding electron pairs in \(\mathrm{H}_{2} \mathrm{C}_{2}\)

There are no remaining electrons after assigning them to the covalent bonds. Therefore, there are 0 nonbonding electron pairs in \(\mathrm{H}_{2} \mathrm{C}_{2}\).
10

(d) Determine valence electrons for \(\mathrm{CH}_{3} \mathrm{~F}\)

First, we need to count the number of valence electrons in the molecule. Carbon (C) has 4 valence electrons, hydrogen (H) has 1, and fluorine (F) has 7. Since there is one \(\mathrm{CH}_3\) group and one F atom, we have a total of (4 + 3 × 1) + 7 = 14 valence electrons.
11

(d) Assign electrons to covalent bonds in \(\mathrm{CH}_{3} \mathrm{~F}\)

The molecule consists of one \(\mathrm{CH}_3\) group and one F atom. In the \(\mathrm{CH}_3\) group, we need 3 single bonds (C-H) and 1 single bond (C-F) to connect the four atoms. So, we have a total of (3 × 2 + 2) = 8 electrons used for bonding.
12

(d) Calculate nonbonding electron pairs in \(\mathrm{CH}_{3} \mathrm{~F}\)

Subtracting the number of bonding electrons from the total valence electrons, we get the number of nonbonding electron pairs: 14 - 8 = 6. These six nonbonding electrons correspond to three pairs on the fluorine atom. So, there are 3 nonbonding electron pairs in \(\mathrm{CH}_{3} \mathrm{~F}\).

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Most popular questions from this chapter

Why are there no \(s p^{4}\) or \(s p^{5}\) hybrid orbitals?

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) \(\mathrm{BCl}_{3}3,\) (c) \(\mathrm{CH}_{3} \mathrm{I},(\mathrm{d}) \mathrm{CBr}_{4}\), (e) \(\mathrm{TeBr}_{4}\)

(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 \(\mathrm{F}\) interacting with a \(1 s\) orbital 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 HF. 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.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right) .\) (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about Si in terms of hydrid orbitals? Silicon does not readily form some of the analogous compounds containing \(\pi\) bonds. Why might this be the case?

An \(\mathrm{AB}_{3}\) molecule is described as having a trigonal-bipyramidal electron-domain geometry. How many nonbonding domains are on atom A? Explain.
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