review questions - gordon state collegeptfaculty.gordonstate.edu/lgoodroad/summer 2011/chem...

Chemical Bonding II:Molecular Shapes,Valence Bond Theory,and MolecularOrbital Theory

Review Questions

10.1 J The properties of molecules are directly related to their shape. The sensation of taste, immuneresponse, the sense of smell, and many types of drug action all depend on shape-specificinteractions between molecules and proteins.

According to VSEPR theory, the repulsion between electron groups on interior atoms of amolecule determines the geometry of the molecule.

The five basic electron geometries are(1) Linear, which has two electron groups.(2) Trigonal planar, which has three electron groups.(3) Tetrahedral, which has four electron groups.(4) Trigonal bipyramid, which has five electron groups.(5) Octahedral, which has six electron groups.An electron group is defined as a lone pair of electrons, a single bond, a multiple bond, oreven a single electron.

ijj^^jl

H—C—H

109.5=

(a) Linear geometry \ \ (b) Trigonal planar geometry I Tetrahedral geometry I

Axial chlorineEquatorial chlorine

"P—Cl:

\Trigonal bipyramidal geometry 1 I Octahedral geometry I

369

370 Chapter 10 Chemical Bonding II

The electron geometry is the geometrical arrangement of the electron groups around the central atom.

The molecular geometry is the geometrical arrangement of the atoms around the central atom.

The electron geometry and the molecular geometry are the same when every electron group bonds twoatoms together. The presence of unbonded lone-pair electrons gives a different molecular geometry andelectron geometry.

(a) Four electron groups give tetrahedral electron geometry, while three bonding groups and one lonepair give a trigonal pyramidal molecular geometry.

(b) Four electron groups give a tetrahedral electron geometry, while two bonding groups and two lonepairs give a bent molecular geometry.

(c) Five electron groups give a trigonal bipyramidal electron geometry, while four bonding groups andone lone pair give a seesaw molecular geometry.

(d) Five electron groups give a trigonal bipyramidal electron geometry, while three bonding groups andtwo lone pairs give a T-shaped molecular geometry.

(e) Five electron groups gives a trigonal bipyramidal electron geometry, while two bonding groups andthree lone pair give a linear geometry.

(f) Six electron groups give an octahedral electron geometry, while five bonding groups and one lonepair give a square pyramidal molecular geometry.

(g) Six electron groups give an octahedral electron geometry, while four bonding groups and two lonepairs gives a square planar molecular geometry.

Larger molecules may have two or more interior atoms. When predicting the shapes of these molecules,determine the geometry about each interior atom and use these geometries to determine the entire three-dimensional shape of the molecules.

To determine if a molecule is polar, do the following:

1. Draw the Lewis structure for the molecule and determine the molecular geometry.

2. Determine whether the molecule contains polar bonds.

3. Determine whether the polar bonds add together to form a net dipole moment.

Polarity is important because polar and nonpolar molecules have different properties. Polar molecules inter-act strongly with other polar molecules, but do not interact with nonpolar molecules, and vice versa.

According to valence bond theory a chemical bond results from the overlap of two half-filled orbitals withspin-pairing of the two valence electrons.

According to valence bond theory, the shape of the molecule is determined by the geometry of the overlap-ping orbitals.

In valence bond theory, the interaction energy is usually negative (or stabilizing) when the interactingatomic orbitals contain a total of two electrons that can spin-pair.

Hybridization is a mathematical procedure in which the standard atomic orbitals are combined to formnew atomic orbitals called hybrid orbitals. Hybrid orbitals are still localized on individual atoms, but theyhave different shapes and energies from those of standard atomic orbitals. They are necessary in valencebond theory because they correspond more closely to the actual distribution of electrons in chemically-bonded atoms.

10.8

10.9

10.10

10.11

10.12 Hybrid orbitals minimize the energy of the molecule by maximizing the orbital overlap in a bond.

Chapter 10 Chemical Bonding II 373

10.29 Nonbonding orbitals are atomic orbitals not involved in a bond and will remain localized on the atom.

10.30 In Lewis theory, a chemical bond is the transfer or sharing of electrons represented as dots. Lewis theory allowsus to predict the combination of atoms that form stable molecules, and the general shape of a molecule.

Lewis theory is a quick way to predict the stability and shapes of molecules based on the number ofvalence electrons. However, it does not deal at all with how the bonds that we make are formed. Valencebond theory is a more advanced bonding theory that treats electrons in a quantum-mechanical manner. Aquantitative approach is extremely complicated but a qualitative approach allows an understanding ofhow the bonds are formed. In valence bond theory, electrons reside in quantum-mechanical orbitals local-ized on individual atoms. When two atoms approach each other, the electrons and nucleus of one atominteract with the electron and nucleus of the other atom. If the energy of the system is lowered, a chemi-cal bond forms. So, valence bond theory portrays a chemical bond as the overlap of two half-filled atomicorbitals. The shape of the molecule can be predicted from the geometry of the overlapping orbitals. Also,valence bond theory explains the rigidity of the double bond. However, valence bond theory falls shortin explaining certain phenomenon such as magnetism and certain bond properties. Valence bond theorytreats the electrons as if they reside in the quantum-mechanical orbitals that we calculate for an atom. Thisis an oversimplification that is partially compensated for by introducing the concept of hybridization. Aneven more complex quantum-mechanical model is molecular orbital theory. In molecular orbital theory, achemical bond occurs when the electrons in the atoms can lower their energy by occupying the molecu-lar orbitals of the resultant molecule. The chemical bonds in MO theory are not localized between atoms,but spread throughout the entire molecule. Molecular orbital theory uses trial functions to solve theSchrodinger equation for the molecules. In order to determine how well the trial function works, you cal-culate the energy, trying to minimize the energy. However, no matter how "good" your guess, you cannever do better than nature at minimizing energy. These minimum-energy calculations for orbitals mustbe done by computer.

All three of these models have strengths and weaknesses, none is "correct." What information you need,depends on which approach you use.

Problems by Topic

VSEPR Theory and Molecular Geometry

10.31 Four electron groups: A trigonal pyramidal molecular geometry has three bonding groups and one lone pairof electrons, so there are four electron pairs on atom A.

10.32 Three electron groups: A trigonal planar molecular geometry has three bonding groups and no lone pairs of~N. electrons so there are three electron pairs on atom A.

/ • 110.33 I (a) 4 total electron groups, 4 bonding groups, 0 lone pairs/ A tetrahedral molecular geometry has four bonding groups and no lone pairs. So, there are four total

V / electron groups, four bonding groups, and pairs.

(b) 5 total electron groups, 3 bonding groups, 2 lone pairsA T-shaped molecular geometry has three bonding groups and two lone pairs. So, there are five totalelectron groups, three bonding groups, and two lone pairs.

(c) 6 total electron groups, 5 bonding groups, 1 lone pairsA square pyramidal molecular geometry has five bonding groups and one lone pair. So, there are sixtotal electron groups, five bonding groups, and one lone pairs.

10.34 (a) 6 total electron groups, 6 bonding groups, 0 lone pairsAn octahedral molecular geometry has six bonding groups and no lone pairs. So, there are six totalelectron groups, six bonding groups, and no lone pairs.

(b) 6 electron groups, 4 bonding groups, 2 lone pairsA square planar molecular geometry has four bonding groups and two lone pairs. So, there are sixtotal electron groups, four bonding groups, and two lone pairs.


(c) 5 electron groups, 4 bonding groups, 1 lone pairA seesaw molecular geometry has four bonding groups and one lone pair. So, there are five total elec-tron groups, four bonding groups, and one lone pair.

PF3:

(b) SBr2:

Electron geometry-tetrahedral; molecular geometry-trigonal pyramidal; bond angle = 109.5°Because of the lone pair, the bond angle will be less than 109.5°.Draw a Lewis structure for the molecule:PF3 has 26 valence electrons.

• c •• r •

-p•• -F• •

Determine the total number of electron groups around the central atom:There are four electron groups on P.Determine the number of bonding groups and the number of lone pairs around the central atom:There are three bonding groups and one lone pair.Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:Four electron groups give a tetrahedral electron geometry; three bonding groups and one lonepair give a trigonal pyramidal molecular geometry; the idealized bond angles for tetrahedralgeometry are 109.5°. The lone pair will make the bond angle less than idealized.

Electron geometry-tetrahedral; molecular geometry-bent; bond angle = 109.5°Because of the lone pairs, the bond angle will be less than 109.5°.Draw a Lewis structure for the molecule:

has 20 valence electrons.

(0

Determine the total number of electron groups around the central atom:There are four electron groups on S.Determine the number of bonding groups and the number of lone pairs around the central atom:There are two bonding groups and two lone pairs.Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:Four electron groups give a tetrahedral electron geometry; two bonding groups and two lonepair give a bent molecular geometry; the idealized bond angles for tetrahedral geometry are109.5°. The lone pairs will make the bond angle less than idealized.

CHC13: Electron geometry-tetrahedral; molecular geometry-tetrahedral; bond angle = 109.5°Because there are no lone pairs, the bond angle will be 109.5°.Draw a Lewis structure for the molecule:CHC13 has 26 valence electrons.

• •Cl-

/~i •Cl •

: ci:

Determine the total number of electron groups around the central atom:There are four electron groups on C.Determine the number of bonding groups and the number of lone pairs around the central atom:There are four bonding groups and no lone pairs.


Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:Four electron groups give a tetrahedral electron geometry; four bonding groups and no lonepairs give a tetrahedral molecular geometry; the idealized bond angles for tetrahedral geom-etry are 109.5°; however, because the attached atoms have different electronegativities thebond angles are less than idealized.

(d) CS2: Electron geometry-linear; molecular geometry-linear; bond angle = 180°Because there are no lone pairs, the bond angle will be 180°.Draw a Lewis structure for the molecule:C$2 has 16 valence electrons.

Determine the total number of electron groups around the central atom:There are two electron groups on C.Determine the number of bonding groups and the number of lone pairs around the central atom:There are two bonding groups and no lone pairs.Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:Two electron groups give a linear geometry; two bonding groups and no lone pairs give a linearmolecular geometry; the idealized bond angle is 180°. The molecule will not deviate from this.

10.36 (a) CF4: Electron geometry-tetrahedral; molecular geometry-tetrahedral; bond angle = 109.5°Draw a Lewis structure for the molecule:

has 32 valence electrons.

Determine the total number of electron groups around the central atom:There are four electron groups on C.Determine the number of bonding groups and the number of lone pairs around the central atom:There are four bonding groups and no lone pairs.Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:Four electron groups give a tetrahedral electron geometry; four bonding groups and no lonepairs give a tetrahedral molecular geometry; idealized tetrahedral bond angles for tetrahedralgeometry are 109.5°.

(b) NFj. Electron geometry-tetrahedral; molecular geometry-trigonal pyramidal; bond angle = 109.5°Because of the lone pair, the bond angle will be less than 109.5°.Draw a Lewis structure for the molecule:NF3 has 26 valence electrons.

N

Determine the total number of electron groups around the central atom:There are four electron groups on N.Determine the number of bonding groups and the number of lone pairs around the central atom:There are three bonding groups and one lone pair.Use Table 10.1 to determine the electron geometry, molecular geometry, and bond angles:


10.38 ClOa will have the smaller bond angle because lone pair-bonding pair repulsions are greater than bond-ing pair-bonding pair repulsions.

Draw the Lewis structures for both structures:has 26 valence electrons. C1O4 ~ has 32 valence electrons.

• • ••

o i• i •• o •

There are three bonding groups and There are four bonding groups andone lone pair. no lone pairs.

Both have four electron groups, but the lone pair in QO^ ~ will cause the bond angle to be smaller becauseof the lone pair-bonding pair repulsions.

S?4 Draw a Lewis structure for the molecule:SF4 has 34 valence electrons.

••: F :

. •• " •:F—- s—F .

• F •• r .

Determine the total number of electron groups around the central atom:There are five electron groups on S.Determine the number of bonding groups and the number of lone pairs around the central atom:There are four bonding groups and one lone pair.Use Table 10.1 to determine the electron geometry and molecular geometry:The electron geometry is trigonal bipyramidal so the molecular geometry is seesaw.Sketch the molecule:

l\(b) C1F3 Draw a Lewis structure for the molecule:

C1F3 has 28 valence electrons.

••: F :

: F ci F •

Determine the total number of electron groups around the central atom:There are five electron groups on CI.


Determine the number of bonding groups and the number of lone pairs around the central atom:There are three bonding groups and two lone pairs.Use Table 10.1 to determine the electron geometry and molecular geometry:The electron geometry is trigonal bipyramidal so the molecular geometry is T-shaped.Sketch the molecule:

-Cl

(c) IF2 Draw a Lewis structure for the ion:IF-> ~ has 22 valence electrons.

: F••i - F

••

Determine the total number of electron groups around the central atom:There are five electron groups on I.Determine the number of bonding groups and the number of lone pairs around the central atom:There are two bonding groups and three lone pairs.Use Table 10.1 to determine the electron geometry and molecular geometry:The electron geometry is trigonal bipyramidal so the molecular geometry is linear.Sketch the ion:

[F I

(d) IBr4 Draw a Lewis structure for the ion:IBr4 " has 36 valence electrons.

• •

Br-

Br«

•Br :

Determine the total number of electron groups around the central atom:There are six electron groups on I.Determine the number of bonding groups and the number of lone pairs around the central atom:There are four bonding groups and two lone pairs.Use Table 10.1 to determine the electron geometry and molecular geometry:The electron geometry is octahedral so the molecular geometry is square planar.Sketch the ion:

Br Br

Br Br

Chapter 10 Chemical Bonding II

(a) N2 Draw the Lewis structure:

(b)

Atom

LeftN

Right N

Number ofElectron Groups

2

2

Number ofLone Pairs

1

1

MolecularGeometry

Linear

Linear

Sketch the molecule:

N==N

Draw the Lewis structure:

N N •H

Atom

LeftN

Right N


3

3

Number ofLone Pairs

1

1

MolecularGeometry

Bent

Bent


N/ \H

(c) N2H4 Draw the Lewis structure:

•• ••

Atom

LeftN

Right C


4

4

Number ofLone Pairs

1

1

MolecularGeometry

Trigonal pyramidal

Trigonal pyramidal


H

10.43 (a) Four pairs of electrons give a tetrahedral electron geometry. The lone pair would cause lonepair-bonded pair repulsions and would have a trigonal pyramidal molecular geometry.


Atom

LeftC

Center C

0

Right C


4

3

4

4

Number ofLone Pairs

0

0

2

0

MolecularGeometry

Tetrahedral

Trigonal Planar

Bent

Tetrahedral


H

(c) NH2CO2H Draw the Lewis structure and determine the geometry about each interior atom:

H N.••

-o• •

Atom

N

C

O


4

3

4

Number ofLone Pairs

1

0

2

MolecularGeometry

Trigonal Pyramidal

Trigonal Planar

Bent


O

H \

cular Shape and Polarity

Draw the Lewis structure for CC>2 and CCLj determine the molecular geometry and then the polarity.

Number of electron groups on CNumber of lone pairsMolecular geometry

0

2

0linear

•PJ •

• PI •• u •

40tetrahedral


Even though each molecule contains polar bonds, the sum of the bond dipoles gives a net dipole of zero foreach molecule.The linear molecular geometry of CC>2 will have bond vectors that are equal and opposite. * ^The tetrahedral molecular geometry of CCLj will have bond vectors that are equal and have a net dipole of zero.

10.48 Draw the Lewis structure of CHsF determine the molecular geometry and then the polarity.

• •• •' F •

IH C H

H

Number of electron groups on C 4Number of lone pairs 0Molecular geometry tetrahedralThe molecule is tetrahedral but is polar because the C - H bond dipoles are different from the C - F bond dipoles.Because the bond dipoles are different, the sum of the bond dipoles is NOT zero. Therefore, the molecule is polar.The tetrahedral molecular geometry of Cti^F will have unequal bond vectors so the molecule will have anet dipole.

10.49 (a) PF3 - polarDraw the Lewis structure and determine the molecular geometry:The molecular geometry from Exercise 35 is trigonal pyramidal.

Determine if the molecule contains polar bonds:The electronegativities of P = 2.1 and F = 4. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecule is trigonal pyramidal, the three dipole moments sum to a nonzero netdipole moment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole momentsadd to determine polarity.

(b) SBr2 - nonpolarDraw the Lewis structure and determine the molecular geometry:The molecular geometry from Exercise 35 is bent.

Determine if the molecule contains polar bonds:The electronegativities of S = 2.5 and Br = 2.8. Therefore the bonds are nonpolar.

Even though the molecule is bent, since the bonds are nonpolar, the molecule is nonpolar.


Determine if the molecule contains polar bonds:The electronegativities of H = 2.1 and S = 2.5. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is bent, the two dipole moments sum to a nonzero net dipolemoment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole moments addto determine polarity.

(a) C1O3 " - polarDraw the Lewis structure and determine the molecular geometry:

Four electron pairs, with one lone pair give a trigonal pyramidal molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of Cl = 3.0 and O = 3.5. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is trigonal pyramidal, the three dipole moments sum to anonzero net dipole moment. The molecule is polar. See Table 10.2 p. 415 in text to see howdipole moments add to determine polarity.

(b) SC12- polarDraw the Lewis structure and determine the molecular geometry:

Four electron pairs with two lone pairs give a bent molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of S = 2.5 and Cl = 3.0. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is bent, the two dipole moments sum to a nonzero net dipolemoment. The molecule is polar. See Table 10.2 p. 415 in text to see how dipole moments addto determine polarity.

(c) SC14 - polarDraw the Lewis structure and determine the molecular geometry:

••: ci :

: ci — s —

ici :

Five electron pairs with one lone pair give a seesaw molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of S = 2.5 and Cl = 3.0. Therefore the bonds are polar.


Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is seesaw, the four equal dipole moments sum to a nonzeronet dipole moment. The molecule is polar.The seesaw molecular geometry will not have offsetting bond vectors.

A(d) BrCls - nonpolar

Draw the Lewis structure and determine the molecular geometry.

.. !ciCL ci:

: a-

Six electron pairs with one lone pair gives square pyramidal molecular geometry.

Determine if the molecule contains polar bonds:The electronegativity of Br = 2.8 and Cl = 3.0. The difference is only 0.2, therefore the bondsare nonpolar. Even though the molecular geometry is square pyramidal, the five bonds arenonpolar so there is no net dipole. The molecule is nonpolar.

10.52 (a) SiCl4 - nonpolarDraw the Lewis structure and determine the molecular geometry:

: ci:

: c"—si — ci:•• ••i101

Four electron pairs with no lone pairs give a tetrahedral molecular geometry..•

Determine if the molecule contains polar bonds:The electronegativities of Cl = 3.0 and Si = 1.8. Therefore the bonds are polar.

iDetermine whether the polar bonds add together to form a net dipole:Because the molecular geometry is tetrahedral, the four equal dipole moments sum to a zeronet dipole moment. The molecule is nonpolar. See Table 10.2 p. 415 in text to see how dipolemoments add to determine polarity.

(b) CF2C12 - polarDraw the Lewis structure and determine the molecular geometry:

: ci•• I ••

: :••

Four electron pairs with no lone pairs give a tetrahedral molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of C = 2.5, F = 4.0, and Cl = 3.0. Therefore the bonds are polar.


Determine whether the polar bonds add together to form a net dipole:Even though the molecular geometry is tetrahedral, which normally yields a nonpolar mole-cule, the four dipole moments sum to a nonzero net dipole moment because of the differentelectronegativities of Cl and F. The molecule is polar. See Table 10.2 p. 415 in text to see howdipole moments add to determine polarity.

(c) SeF6 - nonpolarDraw the Lewis structure and determine the molecular geometry:

: F

: F

• •

-/.'••

(d)

Six electron pairs with no lone pairs give an octahedral molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of Se = 3.0 and F = 4.0. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is octahedral, the six equal dipole moments sum to a zero netdipole moment. The molecule is nonpolar. See Table 10.2 p. 415 in text to see how dipolemoments add to determine polarity.

- polarDraw the Lewis structure and determine the molecular geometry:

• •• c- •

•• , ••: F F :..\ I S»• • s ' ••

Six electron pairs with one lone pair give square pyramidal molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of I = 2.0 and F = 4.0. Therefore the bonds are polar.

Determine whether the polar bonds add together to form a net dipole:Because the molecular geometry is square pyramidal, the five dipole moments sum to anonzero net dipole moment. The molecule is polar.The square pyramid structure has offsetting bond vectors in the equatorial plane, but not inthe axial positions.

/N

y*.<•" '*',.

ce Bond Theory

(a) Be 2s2 No bonds can form. Beryllium contains no unpaired electrons, so no bonds can form with-out hybridization.


(b) P 3s23p3 Three bonds can form. Phosphorus contains three unpaired electrons, so three bonds canform without hybridization.

(c) F 2s22p5 One bond can form. Fluorine contains one unpaired electron, so one bond can form with-out hybridization.

10.54 (a) B 2s22p1 One bond can form. Boron contains one unpaired electron, so one bond can form withouthybridization.

(b) N 2s22p3 Three bonds can form. Nitrogen contains three unpaired electrons, so three bonds can formwithout hybridization.

(c) O 2s22p4 Two bonds can form. Oxygen contains two unpaired electrons, so two bonds can form with-out hybridization.

10.55 PH3 P[ E D l i t ) I ( T ) | ( T ) |3s 3p

The unhybridized bond angles should be 90°. So, without hybridization, there is good agreement betweenvalence bond theory and the actual bond angle of 93.3°.

10.56 SF2 sTTJ3sFITT]2s

F2

TT]2s

l l t i m i m i3p

U m t i m i2p

U T U T i m i2p

The unhybridized bond angles should be 90°. So, without hybridization, there is not very good agreementbetween valence bond theory and the actual bond angle of 98.2°.


10.57 2s22p2

C

TTTT

T T T I T I T

10.58

10.60

2s22p2

C

2sT i l l I

sp2 Only sp2 hybridization of this set of orbitals has a remaining p orbital to form a -IT bond.sp3 hybridization utilizes all 3 p orbitals.sp3d2 hybridization utilizes all 3 p orbitals and 2 d orbitals.

sp3d sp3d hybridization utilizes an s orbital, 3 p orbitals, and d orbital. Since 5 orbitals are used, 5 hybridorbitals form and 5 bonds can form.

sp3 Hybridization utilizes an s orbital and 3 p orbitals. Four orbitals are used, so 4 hybrid orbitals formand 4 bonds can form.

sp2 Hybridization utilizes an s orbital and 2 p orbitals. Three orbitals are used, so 3 hybrid orbitals form.This allows 3 cr and 1 TT bond to form for a total of 4 bonds formed.

(a) CC14 Write the Lewis structure for the molecule:

-8 •

Use VSEPR to predict the electron geometry:Four electron groups around the central atom give a tetrahedral electron geometry.

Select the correct hybridization for the central atom based on the electron geometry:Tetrahedral electron geometry has sp3 hybridization.

Sketch the molecule and label the bonds:

o-C(sp3)-Cl(/>)


(b) NH3 Write the Lewis structure for the molecule:

H

H N H• •




I Lone pair in N spi

(c) OF2 Write the Lewis structure for the molecule:

• • •• ••! F O F •

•• •• ••




I Lone pair in O s/>3

(d) CC>2 Write the Lewis structure for the molecule:

Use VSEPR to predict the electron geometry:Two electron groups around the central atom give a linear electron geometry.

Select the correct hybridization for the central atom based on the electron geometry:Linear electron geometry has sp hybridization.



10.62

<rC<sp)-0(p)

(a) CH2Br2 Write the Lewis structure for the molecule:

H

••: Br-

••H




(b) SO2 Write the Lewis structure for the molecule:

• • ••o= s — o :

Use VSEPR to predict the electron geometry:Three electron groups around the central atom give a trigonal planar electron geometry.

Select the correct hybridization for the central atom based on the electron geometry:Trigonal planar electron geometry has sp2 hybridization.

Sketch the molecule and label the bonds:irS(p)-O(p)

Lone pairin S(sp2) A

- O(p)


(c) JSTF-3 Write the Lewis structure for the molecule:

• •F-

• •N-

" i• i •• r •




o-N(sp3)-F(p) [B Lone pair in N sp3

(d) Bp3 Write the Lewis structure for the molecule:

••: F :

••: F- F :

Use VSEPR to predict the electron geometry:Three electron groups around the central atom give a trigonal planar electron geometry.


Sketch the molecule and label the bonds:Empty p orbital

o-B(5p2)-F(p)

COC12 Write the Lewis structure for the molecule:

:0:•••_ I—"•

Use VSEPR to predict the electron geometry:

Three electron groups around the central atom give a trigonal planar electron geometry.




o-C(s/)-0(p)

(b) BrF5 Write the Lewis structure for the molecule:

F :

• •: F"

IBr

F• •

Use VSEPR to predict the electron geometry:Six electron pairs around the central atoms gives an octahedral electron geometry.

Select the correct hybridization for the central atom based on the electron geometry:Octahedral electron geometry has sp3d2 hybridization.


Lone pair in lAr(

(c) Xep2 Write the Lewis structure for the molecule:

••: F-••

••Xe • F

• •

Use VSEPR to predict the electron geometry:Five electron groups around the central atom give a trigonal bipyramidal geometry.

Select the correct hybridization for the central atom based on the electron geometry:Trigonal bipyramidal geometry has sp3d hybridization.



3 lone pairs in Xe5p3t/ equatorial

(d) 13 Write the Lewis structure for the molecule:

••- 1 :: i -••

Use VSEPR to predict the electron geometry:Five electron groups around the central atom give a trigonal bipyramidal geometry.

Select the correct hybridization for the central atom based on the electron geometry:Trigonal bipyramidal geometry has sp*d hybridization.


3 lone pairs in 1sp3d equatorial

10.64 (a) SO32~ Write the Lewis structure for the ion:

_. 2-

o-••••-o:




10.68

N has four electron pairs around the atom, which is tetrahedral electron pair geometry. Tetrahedral electronpair geometry is sp3 hybridization.

C - 1 and C - 4 each have three electron groups around the atom, which is trigonal planar electron pairgeometry. Trigonal planar electron pair geometry is sp2 hybridization.C - 2 and C - 3 each have four electron pairs around the atom, which is tetrahedral electron pair geometry.Tetrahedral electron pair geometry is sp3 hybridization.O - 1 and O - 2 each have four electron pairs around the atom, which is tetrahedral electron pair geometry.Tetrahedral electron pair geometry is sp3 hybridization.N has four electron pairs around the atom, which is tetrahedral electron pair geometry. Tetrahedral electronpair geometry is sp3 hybridization.

Orbital Theory

Is + Is constructive interference results in a bonding orbital:

Bondingmolecular

orbital

Isl Is

10.70 Is - Is destructive interference results in an antibonding orbital:

Destructive

interference

1s -is

has seven electrons.

t <r

tl t

I.

AO MO AO

Antibondingmolecularorbital

has nine electrons.

T »>

n_^t

isjfl. Jti_l-

41I T &\s

AOAO = Atomic Orbital; MO = Molecular Orbital

MO AO

5 — 4 14 — 3 1Bond order = - = - stable Bond order = — - — = - stable



irC(p)-C(p) o-C(spz)-H(s)

C(sp')-C(sp3)

<rC(5p2)-H(j)

<rC(sp3)-H(s)

(c) CH3SH Write the Lewis structure for the molecule:

H

H- -S-••

H

Use VSEPR to predict the electron geometry:Four electron groups around the C atom and the S atom give a tetrahedral electron geometry.Four bonding pairs of electrons around the C give a tetrahedral molecular geometry and twobonding groups and two lone pairs around the S give a bent molecular geometry.

Determine if the molecule contains polar bonds:The electronegativities of C = 2.5, S = 2.5, and H = 2.1. The C - H bonds and the S - H bondwill be slightly polar, and the C - S bond will be nonpolar.

Determine whether the polar bonds add together to form a net dipole:In both molecular geometries the sum of the dipole moments is not zero. The molecule is polar.

Select the correct hybridization for the central atom based on the electron geometry:Tetrahedral electron geometry has sp3 hybridization on both C and S.


2 lone paii in spon S

<7C(*p3)-H(s)

o-S(sp3)-H(s)

S(sp3)-C(jp3)

crC(sp})-H«

senne

H N

O .

O : H

. o-• • '

C - 1 and C - 3 each have four electron groups around the atom. Four electron pairs give a tetrahe-dral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding pairs andzero lone pairs give a tetrahedral molecular geometry.


C - 2 has three electron groups around the atom. Three electron pairs give a trigonal planar geome-try; trigonal planar geometry has sp2 hybridization. Three bonding pairs and zero lone pairs give atrigonal planar molecular geometry.N has four electron groups around the atom. Four electron pairs give a tetrahedral electron geome-try; tetrahedral electron geometry has sp3 hybridization. Three bonding pairs and one lone pair givea trigonal pyramidal molecular geometry.O -1 and O - 2 each have four electron groups around the atom. Four electron pairs give a tetrahe-dral electron geometry; tetrahedral electron geometry has sp3 hybridization. Two bonding pairs andtwo lone pairs give a bent molecular geometry.

H

H

(b) asparagine

H H IQ:

H N d C. O H1 1 ••

H C3 H

c = o4 ••

C - 1 and C - 3 each have four electron groups around the atom. Four electron groups give a tetrahe-dral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding groupsand zero lone pairs give a tetrahedral molecular geometry.C - 2 and C - 4 each have three electron groups around the atom. Three electron groups give a trigo-nal planar geometry; trigonal planar geometry has sp2 hybridization. Three bonding pairs and zerolone groups give a trigonal planar molecular geometry.N - 1 and N - 2 each have four electron groups around the atom. Four electron groups give a tetra-hedral electron geometry; tetrahedral electron geometry has sp3 hybridization. Three bonding groupsand one lone pair give a trigonal pyramidal molecular geometry.O has four electron groups around the atom. Four electron groups give a tetrahedral electron geom-etry; tetrahedral electron geometry has sp3 hybridization. Two bonding groups and two lone pairsgive a bent molecular geometry.

H


(c) cysteine

H H

H N d CL O H** I **

H <L H

C -I and C - 3 each have four electron groups around the atom. Four electron pairs give a tetrahe-dral electron geometry; tetrahedral electron geometry has sp3 hybridization. Four bonding pairs andzero lone pairs give a tetrahedral molecular geometry.C - 2 has three electron groups around the atom. Three electron groups give a trigonal planar geom-etry; trigonal planar geometry has sp2 hybridization. Three bonding groups and zero lone pairs givea trigonal planar molecular geometry.N has four electron groups around the atom. Four electron groups give a tetrahedral electron geom-etry; tetrahedral electron geometry has sp3 hybridization. Three bonding groups and one lone pairgive a trigonal pyramidal molecular geometry.O and S have four electron groups around the atom. Four electron groups give a tetrahedral electrongeometry; tetrahedral electron geometry has sp3 hybridization. Two bonding groups and two lonepairs gives bent molecular geometry.

H

(a) cytosine

H

H

N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedralelectron geometry and sp3 hybridization. Three bonding pairs of electrons and one lone pair give atrigonal pyramidal molecular geometry.C - 2 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planar molec-ular geometry.N - 3 has two bonding groups of electrons and one lone pair; three electron groups give a trigonalplanar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bentmolecular geometry.C - 4 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonalplanar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.C - 5 has three bonding groups of electrons and zero lone pairs; three electron groups give a trigonalplanar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.


C - 6 has three bonding pairs of electrons and zero lone pairs; three electron groups give a trigonalplanar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.N outside the ring has three bonding groups of electrons and one lone pair; four electron groups givea tetrahedral electron geometry and sp3 hybridization. Three bonding groups of electrons give a trig-onal pyramidal molecular geometry.

(b) adenineNH2

( b )N - 1 has two bonding groups of electrons and one lone pair; three electron groups give a trigonalplanar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bentmolecular geometry.C - 2 has three bonding groups of electrons and zero lone groups; three electron groups give a trigo-nal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.N - 3 has two bonding groups of electrons and one lone pair; three electron groups give a trigonalplanar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bentmolecular geometry.C - 4 has three bonding groups of electrons and zero lone groups; three electron groups give a trigo-nal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.C - 5 has three bonding groups of electrons and zero lone groups; three electron groups give a trigo-nal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.C - 6 has three bonding groups of electrons and zero lone groups; three electron groups give a trigo-nal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.N - 7 has two bonding groups of electrons and one lone pair; three electron groups give a trigonalplanar electron geometry and sp2 hybridization. Two bonding groups and one lone pair give a bentmolecular geometry.C - 8 has three bonding groups of electrons and zero lone groups; three electron groups give a trigo-nal planar geometry and sp2 hybridization. Three bonding groups of electrons give a trigonal planarmolecular geometry.N - 9 has three bonding groups of electrons and one lone pair; four electron groups give a tetrahedralelectron geometry and sp3 hybridization. Three bonding groups of electrons give a trigonal pyrami-dal molecular geometry.N outside the ring has three bonding groups of electrons and one lone pair; four electron groups givea tetrahedral electron geometry and sp3 hybridization. Three bonding groups of electrons give a trig-onal pyramidal molecular geometry.

(c) thymine


N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedralelectron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidalmolecular geometry.C - 2 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.N - 3 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedral elec-tron geometry and sp3 hybridization. Three bonding pairs and one lone pair give a trigonal pyramidalmolecular geometry.C - 4 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.C - 5 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.C - 6 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.C outside the ring has four bonding pairs of electrons and zero lone pairs; four electron pairs givea tetrahedral electron geometry and sp3 hybridization. Four bonding pairs of electrons give atetrahedral molecular geometry.

(d) guanine

5 4 3 9/^ l M

N - 1 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedralelectron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidalmolecular geometry.C - 2 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.N - 3 has two bonding pairs of electrons and one lone pair; three electron pairs give a trigonal planarelectron geometry and sp2 hybridization. Two bonding pairs and one lone pair give a bent moleculargeometry.C - 4 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.C - 5 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.C - 6 has three bonding pairs of electrons and zero lone pairs; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.N - 7 has two bonding pairs of electrons and one lone pair; three electron pairs give a trigonal planarelectron geometry and sp2 hybridization. Two bonding pairs and one lone pair give a bent moleculargeometry.C - 8 has three bonding pair of electrons and zero lone pair; three electron pairs give a trigonalplanar geometry and sp2 hybridization. Three bonding pairs of electrons give a trigonal planarmolecular geometry.

416 Chapter 10 Chemical Bonding H

N - 9 has three bonding pairs of electrons and one lone pair; four electron pairs give a tetrahedralelectron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonal pyramidalmolecular geometry.N outside the ring has three bonding pairs of electrons and one lone pair; four electron pairs give atetrahedral electron geometry and sp3 hybridization. Three bonding pairs of electrons give a trigonalpyramidal molecular geometry.

10.87 4 TT bonds; 25 a bonds; the lone pair on the Os and N - 2 occupy sp2 orbitals; the lone pairs on N - 1, N - 3,and N - 4 occupy sp3 orbitals.caffeine

10.88 5 TT bonds; 21 a bondsaspirin

'V

There is rotation around the bond from C -1 to the ring and from C -1 to OH bond. There is rotation aroundthe O - 2 to the ring bond and around the O - 2 to C - 2 bond. There is rotation around the C - 2 to C - 3 bond.The C -1 to O - 1 bond is rigid, the ring structure is rigid, and the C - 2 to O - 3 bond is rigid.

10.89 (a) Water soluble: The 4 C - OH bonds, the C = O bond, and the C - O bonds in the ring, make the mol-ecule polar. Because of the large electronegativity difference between the C and O, each of the bondswill have a dipole moment. The sum of the dipole moments does NOT give a net zero dipole moment,so the molecule is polar. Since it is polar, it will be water soluble.

(b) Fat soluble: There is only one C - O bond in the molecule. The dipole moment from this bond is notenough to make the molecule polar because of all of the nonpolar components of the molecule. TheC - H bonds in the structure lead to a net dipole of zero for most of the sites in the molecule. Sincethe molecule is nonpolar, it is fat soluble.

(c) Water soluble: The carboxylic acid function (COOH group) along with the N atom in the ring makethe molecule polar. Because of the electronegativity difference between the C and O and the C andN atoms, the bonds will have a dipole moment and the net dipole moment of the molecule is NOTzero, so the molecule is polar. Since the molecule is polar, it is water soluble.

(d) Fat soluble: The two O atoms in the structure contribute a very small amount to the net dipolemoment of this molecule. The majority of the molecule is nonpolar because there is no net dipolemoment at the interior C atoms. Because the molecule is nonpolar it is fat soluble.

review questions - gordon state collegeptfaculty.gordonstate.edu/lgoodroad/summer 2011/chem...

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