1bonding test w. solutions

8/21/2019 1Bonding Test w. Solutions

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GENERAL CHEMISTRY TOPICAL:

BondingTest 1

Time: 22 Minutes*Number of Questions: 17

* The timing restrictions for the science topical tests are optional. Ifyou are using this test for the sole purpose of contentreinforcement, you may want to disregard the time limit.


2/12

MCAT

2 as developed by

DIRECTIONS: Most of the questions in the followingtest are organized into groups, with a descriptivepassage preceding each group of questions. Study thepassage, then select the single best answer to eachquestion in the group. Some of the questions are notbased on a descriptive passage; you must also select the

best answer to these questions. If you are unsure of thebest answer, eliminate the choices that you know areincorrect, then select an answer from the choices thatremain. Indicate your selection by blackening thecorresponding circle on your answer sheet. A periodictable is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS

1

H

1.0

2

He

4.0

3Li

6.9

4Be

9.0

5B

10.8

6C

12.0

7N

14.0

8O

16.0

9F

19.0

10Ne

20.2

11

Na

23.0

12

Mg

24.3

13

Al

27.0

14

Si

28.1

15

P

31.0

16

S

32.1

17

Cl

35.5

18

Ar

39.9

19

K

39.1

20

Ca

40.1

21

Sc

45.0

22

Ti

47.9

23

V

50.9

24

Cr

52.0

25

Mn

54.9

26

Fe

55.8

27

Co

58.9

28

Ni

58.7

29

Cu

63.5

30

Zn

65.4

31

Ga

69.7

32

Ge

72.6

33

As

74.9

34

Se

79.0

35

Br

79.9

36

Kr

83.8

37

Rb

85.5

38

Sr

87.6

39

Y

88.9

40

Zr

91.2

41

Nb

92.9

42

Mo

95.9

43

Tc

(98)

44

Ru

101.1

45

Rh

102.9

46

Pd

106.4

47

Ag

107.9

48

Cd

112.4

49

In

114.8

50

Sn

118.7

51

Sb

121.8

52

Te

127.6

53

I

126.9

54

Xe

131.3

55

Cs

132.9

56

Ba

137.3

57

La *

138.9

72

Hf

178.5

73

Ta

180.9

74

W

183.9

75

Re

186.2

76

Os

190.2

77

Ir

192.2

78

Pt

195.1

79

Au

197.0

80

Hg

200.6

81

Tl

204.4

82

Pb

207.2

83

Bi

209.0

84

Po

(209)

85

At

(210)

86

Rn

(222)

87

Fr

(223)

88

Ra

226.0

89

Ac

227.0

104

Rf

(261)

105

Ha

(262)

106

Unh

(263)

107

Uns

(262)

108

Uno

(265)

109

Une

(267)

*

58

Ce

140.1

59

Pr

140.9

60

Nd

144.2

61

Pm

(145)

62

Sm

150.4

63

Eu

152.0

64

Gd

157.3

65

Tb

158.9

66

Dy

162.5

67

Ho

164.9

68

Er

167.3

69

Tm

168.9

70

Yb

173.0

71

Lu

175.0

90

Th

232.0

91

Pa

(231)

92

U

238.0

93

Np

(237)

94

Pu

(244)

95

Am

(243)

96

Cm

(247)

97

Bk

(247)

98

Cf

(251)

99

Es

(252)

100

Fm

(257)

101

Md

(258)

102

No

(259)

103

Lr

(260)

GO ON TO THE NEXT PAGE.


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Bonding Tes

KAPLAN

Passage I (Questions 17)

If in the formation of a chemical bond, theelectronegativities of the combining atoms areapproximately the same, the electron density will bedistributed evenly between the two atoms forming acovalent bond. If the combining atoms have differentelectronegativities, then the bonding electrons will be

located around the more electronegative atom; the bondthus formed is known as a polar covalent bond.

When the difference in electronegativity of the

combining atoms is very large, a complete transfer of

charge may occur, forming ionic species. The energy

change that takes place in the formation of the ionic

species can be attributed to the energy change when the

electron is transferred and to the electrostatic energy

between the ions. This energy change, the energy of the

ionic bond, can be estimated with Equation 1, where D is

the dissociation energy of the bond, I(A) is the ionization

energy of atom A, Ea(B) is the electron affinity of atom

B,andZAZBe2/40R is the Coulombic potential energyof the two ions at a distance R with ZAandZBbeing the

quantized charge on ions A and B, respectively. (+1, -2,

etc.)

D= I(A) +Ea(B) ZAZBe2/40R

Equation 1

Tables 1 and 2 contain first ionization potentials andelectron affinities of selected elements, respectively.

Table 1

Element I/eV

KNaLiBrClArF

4.345.145.39

11.8112.9715.7617.42

Table 2

Element Ea/eV

ArLiNaKBr

FCl

0.300.580.780.823.40

3.453.61

According to the valence-shell electron-pairepulsion(VSEPR) theory, molecules have definite shapewhich result from a tendency of the atoms and nonbondedelectron pairs attached to the central atom to get as faaway from each other as possible. Table 3 contains theformulas, molecular structures, and bond angles of selectedmolecules.

Formula Structure Bond Angle

CaI 2

K2O

NH

H

H

Trigonalpyramidal

NH3 107

OHH

BentH2O

SOOSO2

Trigonal

planarBF3

Cl

P Cl

ClCl

ClPCl5

Table 3

GO ON TO THE NEXT PAGE


4/12

MCAT

4 as developed by

1 . According to Equation 1, at what point is itenergetically favorable for Li and Cl to form the ioniccompound Li+Cl?

A . Ea(Cl) ZLiZCle2/40RI(Li)

B . ZLiZCle2/40R I(Li) +D+Ea(Cl)

C . Ea(Cl) I(Li) +ZLiZCle2/40R

D . I(Li) +DEa

(Cl) ZLi

ZCl

e2/40

R

2 . The bond distance in NaCl(g) has been determined tobe 250 1012m. According to Equation 1, what isthe predicted bond energy of one molecule ofNaCl(g)?

[Note: e2/40R = 5.74 eV at anRof 250 pm.]

A . 6.45 eVB . 4.21 eVC . 4.21 eVD . 6.45 eV

3 . In which of the following does ionic bondingpredominate?

I. NH3II. CaCl2

III. PCl5

IV. K2O

A . I and III onlyB . II and III onlyC . I and IV onlyD . II and IV only

4 . Based on information presented in the passage, whatwould be the predicted bond angle in borontrifluoride?

A . 120B . 109C . 90D . 108

5 . Which compound in Table 3 has two or moreresonance forms?

A . NH3B . PCl5C . SO2D . CaCl2

6 . What is the molecular structure of phosphoruspentachloride?

A . Trigonal pyramidalB . LinearC . TetrahedralD . Trigonal bipyramidal

7 . Which of the following compounds has the strongestelectrostatic interactions between the constituent

particles?

A . NH3B . PCl5C . NaCl

D . NO2



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Bonding Tes

KAPLAN

Questions 8 through 11 are NOTbased on a descriptive passage.

8 . Which of the following is a Lewis base?

A . NH4+

B . CH4C . PH

3D . CH3CH3

9 . What is the formal charge on the nitrogen atom inHNO3?

A . 1B . +1C . 0D . +2

1 0 . Which of the following is NOT a resonance structureof the others?

A .

CH2 CH CH C

O

H

B .

CH2

CH CH C

O

H

C .

CH2

C C

O

H

CH3

D .

CH2

CH CH C

O

H

1 1 . Which of the following is NOT a correct Lewis dodiagram?

A . B .

O N O

C .

O

S

O O

D .

Cl

B

ClCl

C HCH

GO ON TO THE NEXT PAGE


6/12

MCAT

6 as developed by

Passage II (Questions 1217)

Some atoms have more power to draw electronstoward themselves than others. The stronger this tendencyis, the more electronegative an element is said to be.Electronegativity can be quantified by using either of twoscales: the Mulliken scale or the Pauling scale. As shownin Equation 1, the Mulliken electronegativity of an

element is defined as the mean of the ionization potential,I, and the electron affinity,Ea, of that element.

XM =(I + Ea)

2

Equation 1

The actual values of the ionization potential and

electron affinity that are of relevance, however, are not

necessarily equal to those of the ground state element.

This is because electronegativity is a characteristic of the

atom when it is bonded to others in a molecule. The

ionization energy and electron affinity to be used shouldbe the ones for the atom in its correct hybridized state in a

molecule. The ionization energy of carbon, for example,

is the energy needed to remove an electron from its

valence 2p orbital. In the calculation of its

electronegativity in the molecule methane, however, what

is meant by the ionization energy is the energy needed to

remove an electron from the now sp3 hybridized orbital.

Because this orbital contains 25% s character, it is at a

lower energy than a p orbital and thus the ionization

energy would be greater. This also implies that the

electronegativity of an element is dependent upon its

hybridization in the molecule.

The Pauling scale is approximately proportional to

the Mulliken scale. The Mulliken and Pauling

electronegativities of elements A and B are related with

Equation 2, where XAM and XA

P are the Mulliken and

Pauling electronegativities of element A, respectively.

XAM XB

M= 2.78(XAP XB

P)

Equation 2

1 2 . Using Equation 2, if the Pauling electronegativity ofphosphorus is 2.1, predict its Mullikenelectronegativity.

A . 2.1/2.78B . 2.78/2.1C . 0.68D . The Mulliken electronegativity of phosphorus

cannot be determined without more information.

1 3 . Which of the following is a true statement about amolecule of SO2?

A . It is a polar molecule with polar bonds.B . It is a polar molecule with nonpolar bonds.C . It is a nonpolar molecule with polar bonds.D . It is a nonpolar molecule with nonpolar bonds.

1 4 . In the molecule shown below, which arrow indicatesthe direction of the dipole moment?

N

H

H

H

3

2

1

A . 1B . 2C . 3D . The molecule has no overall dipole moment.

1 5 . Which structure below has the partial chargescorrectly represented?

A .

H C N

+

C .

H C HC

+

B .

H

C

H

O

+

D .

O

H H

+



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Bonding Tes

KAPLAN

1 6 . Which of the following has the most polar bond?

A . NOB . CHC . CCD . HF

1 7 . In which of the following compounds will the

nitrogen atom have the highest electronegativity onthe Mulliken scale?

A . H3N:

B . (H3C)3N:

C .

N

D . HCN:

END OF TES


8/12

MCAT

8 as developed by

ANSWER KEY:1. A 6. D 11. B 16. D2. B 7. C 12. D 17. D3. D 8. C 13. A4. A 9. B 14. A5. C 10. C 15. A


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Bonding Tes

KAPLAN

EXPLANATIONS

Passage I (Questions 17)

1 . AThe quickest way to obtain the answer is to recognize that bond dissociation is an endothermic process, and that th

bond dissociation energy is the amount of energy one needs to supply to break the bond. A positive dissociation energy wouldindicate that the ionically-bonded state is more stable than the state of separated neutral atoms. An ionic bond forms

therefore, if D 0. Substituting in the terms on the left hand side of equation 1 for D and rearranging, one obtains choice A.To explain this more fully: the point at which bond formation becomes possible is that point where the energy

released upon formation of the ionic bond is greater than or equal to the energy one needs to supply to generate the ionicspecies. The energy required to remove an electron from a neutral atom to form a cation is positive--in other words, thiprocess costs energy. So, you may ask, why does an ionic bond form? It forms because energy is released in two ways: firstthe electron affinity of a neutral atom is positive: i.e. energy is released as it accepts one more electron to form an anionsecond, the Coulombic attraction between the two oppositely-charged ions means that the system can lower its energy bycoming close together (forming the bond). In order to make the process of forming an ionically-bonded species worthwhilethe energy one needs to invest in, the ionization energy, has to be less than or at most equal to the payback, that is, thenergy released in the form of electron affinity and lower Coulomb potential energy. Choice A describes this criterion:

energy that needs to be supplied = energy to ionize Li =I(Li)energy that one gets back = energy released from adding electron to Cl + Coulombic energy

released

= electron affinity of Cl + Coulombic energy released

What is the Coulombic energy released? When the two ions are infinitely separated, the Coulombic potential energyis zero, since R = . When the two are a distance R apart, this electrostatic potential energy is:

U =ZLiZCle

2

40R

as given in the passage. Note that ZLiandZClwill have opposite signs, and so this term is negative. This makes sense as we

would expect the energy to be lower when the two are closer than when they are infinitely far apart. (The ions are attracted tonot repelled by, each other.) The amount of Coulombic potential energy released as the ionic bond forms, then, is

U =

ZLiZCle2

40R

which is positive since the opposite signs of ZLi andZCl cancel the negative sign in front. The energy one gets back i

therefore

Ea(Cl) ZLiZCle

2

40R

and this needs to be greater than or equal toI(Li).

2 . BThis question requires you to use Equation 1 together with Tables 1 and 2. Looking at Equation 1, you can see tha

the missing values that you need to provide are the electron affinity and the ionization potential; the energy of interactionterm--the 4part--has been given to you in the question stem as 5.74 eV. (ZaandZbare +1 and -1 respectively and so do

not affect the magnitude of the interaction term). Sodium chloride is Na+and Cl, and so we want the ionization potential osodium and the electron affinity of chlorine. These values are 5.14 and 3.61, respectively. This gives:

D = I(Na) +Ea(Cl) + 5.74 = 5.14 + 3.61 + 5.74 = 4.21 eV

Remember, if the bond forms, then the bond energy (or bond dissociation energy) would have to be positive: bond-formation is exothermic; bond-breaking is endothermic.


10/12

MCAT

10 as developed by

3 . DIonic bonding predominates in those compounds which contain elements with large differences in electronegativity.

The larger the difference, the more polar the bond. The easiest, and quickest, way to answer this question is to consult theperiodic table. You should know that the most electronegative elements are located in the upper-right hand corner of the table(excluding the noble gases) and those elements that are the least electronegative are located in the lower left-hand corner.Looking at the answer choices, you can see that they are all two-item choices--in other words, look for the two compoundswhose elements are the furthest apart in the periodic table. Compound I is ammonia, which you probably should know to be

a covalent compound already. Since answer choices A and C both contain roman numeral I, they can be eliminated. Romannumeral II, calcium chloride: calcium is located in the second column, fourth row; chlorine is located in the 17th column,third row--pretty far apart from each other---mostly ionic. You may also know that calcium tends to form the +2 cation, whilechlorine forms the 1 anion to complete their octet. Oppositely-charged ions would form ionic bonds. Roman numeral III,phosphorus pentachloride: in the periodic table, phosphorus and chlorine are only separated from each other by sulfur, makingit mostly covalent. By elimination, choice D is the correct response.

4 . AThe easiest way to answer this question correctly is to go to Table 3 and see that the shape of BF 3 is listed as

trigonal planar. If there wasn't any information in the passage, you could still answer this question correctly. From theperiodic table you should be able to determine that boron has two nonvalence electrons in the 1 s subshell and three valenceelectrons that are available for bonding; in other words, boron completes its valence by forming three bonds, and only threebonds. When boron forms these three bonds, there will be six bonding electrons surrounding it: there are no nonbondingelectrons. Since there are no nonbonding electrons to worry about, the bonding electrons arrange themselves to be as far apart

from each other as possible. As a result, a trigonal arrangement is adopted which has bond angles of 120.

5 . CIn order for resonance structures to be possible in a molecule or polyatomic ion, it needs to have double and single

bonds which can be interchanged. In resonance structures, bonds and nonbonding electrons may move, but atoms do not. Intable 3, SO2is the only one of these which has a double bond. The resonance structures are therefore:

S

O O

S

O O

This implies that the 2 electrons are actually delocalized over all three atoms. Each sulfur-oxygen bond hascharacteristics intermediate between a single and a double bond.

6 . DThis question is testing you on your understanding of the valence-shell electron-pair repulsion theory (VSEPR). The

MCAT requires you to make predictions of molecular shapes using the VSEPR theory. PCl 5has a phosphorus atom at the

center surrounded by 5 chlorine atoms; there are no lone pairs, thus giving it the formula AX 5--trigonal bipyramidal, choice

D.

7 . CIn order to answer this question correctly, you need to remember that ionic bonding results from the attractions

between opposite charges. Particles in ionic compounds therefore exhibit stronger electrostatic or Coulombic interactions thanatoms in covalent bonds. Looking at the answer choices, the only choice that is predominately ionic is choice C, NaCl.

Discrete Questions

8 . CA Lewis base is a substance that can donate a pair of electrons. Among the answer choices, only PH3, choice C, has

a pair of nonbonding electrons, making it the correct response. The easiest way to determine if a substance is a Lewis base isto draw its Lewis dot structure and see if a lone electron pair exists:

P

H H

H


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MCAT

11 as developed

9 . BTo answer this question, we need to write the Lewis dot diagrams and then use any one of several formulas to find

the formal charge. One such formula is Formal charge = Valence electrons - [number of bonds + number of nonbondingelectrons]. The Lewis dot structure of HNO3is as follows:

N OH

O

O

N OH

O

O

Formal charge is equal to the valence electrons of nitrogen, which is 5, minus the sum of the number of bonds andthe number of nonbonding electrons, which in this case is 4. So, the formal charge on the nitrogen is equal to 54 which is+1.

1 0 . CRemember that in resonance forms, only electrons (bonds and nonbonding electron pairs) can move. Examining th

answer choices, you can see that the atomic linkages are all the same except for answer choice C; the carbon second from theleft in choice C now has a methyl group bonded to it, whereas the other choices have hydrogens in this position. Below arethe resonance structures showing the actual way that the electrons can be shifted around:

H2C CH

CH

C

O

H

H2 C CH

CH

C

O

H

H2C CH

CH

C

O

H

It is always important to keep in mind, however, that shifting or moving electrons is just a figure of speech: theelectrons are really delocalized over all the bonds.

1 1 . BThe first thing you should do to verify a Lewis structure of a molecule is to make sure that all the valence electron

are accounted for. For choice A, acetylene, there are two carbons, each having 4 valence electrons, and 2 hydrogens, eachhaving 1 valence electron. So, choice A needs to have 10 valence electrons, and indeed it has. For choice B, nitrogen dioxidethere is 1 nitrogen, which has 5 valence electrons, and 2 oxygens, each having six valence electrons--there should be a total o17 electrons accounted for. Counting the electrons in choice B you can see that it has only 16 electrons--choice B is th

correct answer. Choice C, sulfur trioxide, should have a total of 24 valence electrons, 6 electrons from each elementCounting the valence electrons, you'll see that it has the required 24. Choice D, boron trichloride, should have and does hav24 electrons.

Passage II (Qu estions 1217)

1 2 . DThe relationship between the Pauling and Mulliken scales of electronegativity given in Equation 2 has four

unknowns, though some may look similar at first glance. To solve such an equation we must be given, or be able todetermine, three of these unknowns. In this question, only one of the unknowns (the Pauling electronegativity of phosphorusis given--without more information there is no way to figure out the other two. To determine the equivalent Pauling oMulliken electronegativity for an element A we must know both the Pauling and Mulliken electronegativities of the otherelement B.

Likely mathematical errors might include the attempted algebraic cancellation of XB

Mwith XB

P from both sides o

the equation. This is a common mistake and there are two reasons why this approach is incorrect. Firstly, the unknowns arenot identical (though they look similar) and cannot be cancel each other out. Secondly, even if the two variables were identica(i.e. both XB

Mand XBP= XB), cancellation would not work: the variables on the right side of the equation would have to b

first multiplied by 2.78 (converting XBto 2.78XB) before XB is added to both sides of the equation (leaving 1.78X Bon

the right side of the equation). The variable would still remain and the answer still could not be determined.

1 3 . AOnce again a question concerning VSEPR theory and vector addition. The correct VSEPR class for SO2is AX2E; it

VSEPR shape is angular. As can be seen from the drawing of SO 2 in Table 3 of the first passage, the individual bonds are


12/12

MCAT

12 as developed by

polar and the vector sum of the bonds is nonzero, choice A is the correct response. If you had forgotten the shape of SO2, or

it is not made available to you, you could have realized that it contains polar bonds, only homonuclear bonds are nonpolar,and eliminated choices B and D, giving you a 50/50 chance of selecting the correct answer.

1 4 . AAnother vector addition question--the vector sum of the three H--N dipole moments is an arrow which goes straight

up the middle of the molecule. You should know that a dipole vector is represented as an arrow, with the head of the arrowpointing to the partial negative charge. In this case, nitrogen would be the center of partial negative charge since it's more

electronegative than hydrogen. Adding the three bond vectors together gives a dipole moment having the direction shown byarrow one:

N

H H

H

1 5 . AChoice A is the only one where the more electronegative atom has a partial negative charge. In this example, the

nitrogen, being the more electronegative, has a small negative charge due to increased electron density and the electropositivecarbon has a small partial positive charge.

1 6 . DThe most polar bond is the one that has the greatest difference in the electronegativities of the two elements.(Remember, electronegativity increases in going from the lower left to the upper right of the periodic table.) Among thesechoices, H-F is the most polar since hydrogen and fluorine are the furthest apart in the periodic table.

1 7 . DThe key to this question lies in understanding the point made in the second paragraph. Depending on the

hybridization of the atom, the ionization potential and the electron affinity values change and thus so does the Mullikenelectronegativity. In particular, the higher the percentage of s character, the higher the electronegativity. Choice D has thenitrogen in an sphybridized state. This means that each sigma orbital has 50% s character. This is the highest among the fourshown and thus the nitrogen atom there has the highest electronegativity.

Choices A and B both have the nitrogen in the sp3hybridized state.The nitrogen in choice C is sp2hybridized. The sigma orbital thus has 33% scharacter.

1bonding test w. solutions

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