Блог Анкара

In This Chapter

^ Hitting the high points

^ Shaping up your skill with some practice questions

^ Discovering the answers and explanations to your questions

Chapter 7 was packed with ideas, starting with the way valence electrons distribute themselves around individual atoms, building up through the way atoms connect themselves into molecules with particular shapes and finishing with the way molecular shapes help determine the properties of different compounds. In other words, Chapter 7 was a lot to take in one bite. Here is where you chew. In the following sections we review the highlights of Chapter 7, give you practice problems to help sear important concepts into your brain, and give you full explanations for the answers to help you swallow what you’ve chewed.

Shaping Up the High Points

Before you attempt any questions in this chapter, you should review the following sections of important points from Chapter 7.

Electron dot structures and Lewis structures

Electron dot structures And Lewis structures Use dots and/or lines to represent valence electrons and/or lines to represent valence electron pairs.

Basic steps to follow when drawing a Lewis structure are

1. Add up all the valence electrons for all the atoms in the molecule.

2. Pick a "central" atom to serve as the anchor of your Lewis structure.

3. Connect the other, "outer" atoms to your central atom using single bonds only.

4. Fill the valence shells of your outer atoms. Then put any remaining electrons on the central atom.

5. Check whether the central atom now has a full valence shell.

Compounds that can engage in multiple, valid Lewis structures participate in Resonance, Based on an old idea that the molecules might actually be in equilibrium between the multiple structures. Although we now know that not to be the case, the term has stuck. Individual Resonance structures Contribute to an overall, averaged structure called a Resonance hybrid. Adjacent p-orbital electrons are commonly involved in resonance, in which they participate within delocalized N Bonds.

Valence bond theory

Valence bond theory Describes covalent bonds in terms of overlap between atomic orbitals:

IU Orbitals that overlap in a manner completely symmetric to the bond axis form Sigma bonds.

I Orbitals that overlap in a manner that is symmetric to the bond axis in only one plane form Pi bonds.

VSEPR theory

VSEPR theory Explains molecular shapes in terms of mutual repulsion between electron pairs (both bonding and nonbonding pairs) around a central atom:

IU Lone (nonbonding) pairs repel more strongly than bonding pairs.

IU Valence electrons in different kinds orbitals (like S And P Orbitals) Hybridize Into mixed, equivalent orbitals. The total number of hybridized orbitals is the same as the original number of unmixed orbitals.

I One S Orbital mixes with one P Orbital to create two identical Sp Hybrids.

Sp Hybridized centers will have linear geometry. IU One S Orbital mixes with two P Orbitals to create three identical Sp2 Hybrids.

Sp2 Hybridized centers will have trigonal planar geometry. IU One S Orbital mixes with three P Orbitals to create four identical Sp3 Hybrids.

Sp3 Hybridized centers will have tetrahedral geometry.

Bond dipoles

Individual Bond dipoles Arise from differences in electronegativity in bonded atoms. Except for homonuclear diatomic molecules (like O2, H2, and so on), all atom-atom bonds will have a at least a small bond dipole. Within a molecule, bond dipoles may or may not contribute to an overall Molecular dipole Depending on geometry:

I Bond dipoles add as vectors within molecules, building on each other or canceling each other out depending on geometry.

IU Dipole moment (□,) = amount of separated charge (Q) x distance separated (d); the equation looks like this: u, = Qd

Isomers

Isomers Are compounds made up of the same atoms, but are put together in the following different arrangements:

IU Structural isomers Have different bonds.

IU Stereoisomers Have the same bonds, but different three-dimensional arrangements of atoms.

• Geometric isomers Can occur when atoms are restricted from rotating freely about their bonds.

• Optical isomers (also called Enantiomers) Are nonsuperimposable mirror images of one another and are therefore Chiral.

Testing Your Knowledge

Now that you know your sp2′s from your sp3′s, and have your VSEPR in shape, give these practice questions a try.

1. Which of the following correctly lists the molecules in order of increasing polarity?

(A) H2O, H2S, HF, H2

(B) H2, HF, H2S, H2O

(C) H2, H2S, HF, H2O

(D) H2, H2S, H2O, HF

(E) H2S, H2O, HF, H2

2. Which of the following molecules has bond angles closest to 109.5°?

(A) BF3

(B) SiS2

(C) CCl4

(D) H2O

(E) CO2

3. Identify the incorrect set of resonance structures.

A.

-1 • • +1 . • -1 IO — N — O!

-1 • • +1 • •

! O — N — O • • | • •

I O I

"-1

• • +1 . • -1

O — N — O!

• • | • •

I O I -1 • •

B.

H

C.

H

H H H

H — C — O I

M m

H

H

HH H

‘.OL I..

HCO

D.

+1

‘, O — O — O!

+1

I O — O — O I

-1 • • • •

I O — C = O

• • I • •

‘.OL

* *-1

-1 • • • • • .-1

I O — C — O I

OCO

Mm | • •

‘.OL

* *-1

4. Group IVA elements tend to hybridize in which of the following ways?

5. Within formaldehyde, CH2O, carbon possesses which of the following traits?

I. Sp2 Hybridization

II. Three equivalent sigma bonds

III. Trigonal planar bonding geometry

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) I, II and III

O

H

H

-1

O

-1

-1

E.

-1

O

Questions 6 through 10 refer to the following list of geometries:

6. Characteristic of four electron pairs, two bonding and two nonbonding

7. Typical of Sp Hybridization

8. Accounts for the nonpolarity of SiF4

9. Nitrate anion 10. PCl5

Questions 11 through 14 refer to the following molecule, and the accompanying set of choices:

O

11. Exhibit(s) Sp2 Hybridization

12. Exhibit(s) Sp3 Hybridization

13. Has a lone pair

14. Has a pi bond

Checking Your Work

Don’t play the angles. Clear up any lingering confusion by checking your answers, making sure you understand the explanations for any questions you got wrong — or the ones that you got right by guessing.

1. (D). H2 is the least polar because its sole covalent bond occurs between atoms that are identical, and which therefore have identical electronegativity. H2S and H2O are close, but H2O is slightly more polar because oxygen is slightly more electronegative than sulfur. The "bent" geometry of both H2S and H2O is key to the presence of a molecular dipole in each

Case — a linear shape would lead to canceling of the bond dipoles, resulting in nonpolar molecules. The sole covalent (nearly ionic) bond in HF is the most polar of them all.

2. (C). Bond angles of 109.5° signify a perfect tetrahedron. Many molecules approach this ideal, but perfect tetrahedral geometry typically requires that the central atom (carbon, in this case) possess four identical bonding partners (chlorines, in this case).

3. (E). The last set of resonance structures, for CO32- anion, are incorrect. Each of the structures includes an oxygen atom with an excess of valence electrons. The oxygen atom double-bonded to carbon in each structure should have only two lone pairs, not three.

4. (C). Group IVA atoms tend to engage in four covalent bonds. Because of hybridization, these bonds involve four identical Sp3 Orbitals.

5. (D). In formaldehyde, the central carbon indeed exhibits Sp2 Hybridization. This hybridization leads to a trigonal planar (though not perfectly so) molecular shape. However, the three sigma bonds are not identical, for two reasons. First, C-O bonds differ in polarity from C-H bonds. Second, the C-O bond axis includes a pi bond; double-bonding between carbon and oxygen brings those two atoms closer together than does a single bond.

6. (B). In water, for example, the two lone pairs of oxygen and the two O-H bonds lead to just such a bent shape.

7. (A). In an Sp Hybridized atom, there are only two bond axes. The electrons in these axes mutually repel to achieve the maximum separation afforded by linear geometry.

8. (D). Because of its perfect tetrahedral geometry, silicon tetrafluoride is nonpolar, despite the fact that each silicon-fluorine bond is polar.

9. (C). Nitrate displays resonance between three Sp2 Hybridized resonance structures. The hybrid structure therefore has a trigonal planar shape consistent with Sp2 Hybridization.

10. (E). This compound may have thrown you. The central phosphorous of PCl5 is "hyperva-lent," meaning that it engages in more than the three covalent bonds you’d expect from its position on the periodic table. But even if the hypervalency freaked you out, take heart: the basic principles of VSEPR still apply. Whatever the reasons for pentavalent phosphorous, the fact remains that the electrons within the five bond axes mutually repel, seeking the shape with maximum separation. That shape is trigonal bipyramidal.

11. (C). Carbon III engages in a double bond (with carbon) and in two single bonds (with hydrogen and nitrogen). This arrangement leads to three bonding axes and Sp2 Hybridization.

12. (D). Carbon I engages in four separate single bonds, one with another carbon and three with hydrogens. Four separate single bonds around a central carbon are a hallmark of Sp3 Hybridization. Nitrogen II may at first appear to be Sp2 Hybridized, because it engages in only three bonds — but the nitrogen also contains a lone pair, and is thus Sp3 Hybridized.

13. (B). Carbons I and III keep their four valence electrons occupied within various combinations of covalent bonds. Because it has one more valence electron than carbon, Nitrogen II engages in only three covalent bonds, reserving two electrons as a nonbonding "lone pair."

14. (C). Double bonds, like the carbon-carbon double bond in which Carbon III participates, include both sigma and pi bonds.