1

AP Chemistry: 2017-18 Semester Review: MULTIPLE CHOICE Section

Chapters 6 thru 9, 11.2-11.3

Chp. 6 1) Which one of the following is correct?

A) ν ÷ λ = c

B) ν = cλ

C) λ = c ν

D) νλ = c

E) ν + λ = c Page Ref: Sec. 6.1

2) In the Bohr model of the atom, _____.

A) electrons travel in circular paths called orbitals

B) electron paths are controlled by probability

C) electrons can have any energy

D) electron energies are quantized

E) both A and C Page Ref: Sec. 6.3

3) Which one of the following is an incorrect orbital notation?

A) 4dxy

B) 3py

C) 4s

D) 2s

E) 3f Page Ref: Sec. 6.5

4)

Which electron configuration represents a violation of the Pauli exclusion principle?

A)

B)

C)

D)

E)

Page Ref: Sec. 6.8

2

5) The ground state electron configuration of Ga is __________.

A)

B)

C)

D)

E) [Ar]

Page Ref: Sec. 6.8

6) Which electron configuration represents a violation of Hund's rule for an atom in its ground state?

A)

B)

C)

D)

E)

Page Ref: Sec. 6.8

7) Which two elements have the same ground-state electron configuration?

A) Pd and Pt

B) Fe and Cu

C) Cu and Ag

D) Cl and Ar

E) No two elements have the same ground-state electron configuration. Page Ref: Sec. 6.8

Chp. 7

1)

In which set of elements would all members be expected to have very similar chemical properties?

A) O, S, Se

B) S, Se, Si

C) N, O, F

D) Ne, Na, Mg

E) Na, Mg, K Page Ref: Sec. 7.1

2) Electrons in the 1s subshell are much closer to the nucleus in Ar than in He due to the larger _____ in Ar.

A) diamagnetism

B) Hund's rule

C) azimuthal quantum number

D) nuclear charge

E) paramagnetism Page Ref: Sec. 7.2

3

3) Atomic radius generally increases as we move _____.

A) down a group; the period position has no effect

B) down a group and from right to left across a period

C) down a group and from left to right across a period

D) up a group and from right to left across a period

E) up a group and from left to right across a period Page Ref: Sec. 7.2

4) The atomic radius of main-group elements generally increases down a group because _____.

A) effective nuclear charge decreases down a group

B) effective nuclear charge increases down a group *the effective nuclear charge is the same within a group,

C) effective nuclear charge zigzags down a group and increases s across the period.

D) the principal quantum number of the valence orbitals increases

E) both effective nuclear charge increases down a group and the principal quantum number of the valence orbitals

increases Page Ref: Sec. 7.2

5) Which of the following correctly lists the five atoms in order of increasing size (smallest to largest)?

A) O < F < S < Ba < Mg

B) F < S < O < Mg < Ba

C) F < O < S < Mg < Ba

D) O < F < S < Mg < Ba

E) F < O < S < Ba < Mg Page Ref: Sec. 7.2

5) Of the following elements, which has the largest first ionization energy?

A) Na

B) Al

C) Se

D) Br

E) Cl Page Ref: Sec. 7.4

6) Which equation correctly represents the first ionization of phosphorus?

A) P –

(g) → P (g) + e –

B) P (g) + e –

→ P –

(g)

C) P (g) → P –

(g) + e –

D) P (g) → P+

(g) + e –

E) P+ (g) + e

– → P (g)

Page Ref: Sec. 7.4

7) The ion with the smallest diameter is ____.

A) Why? It has a greater nuclear charge than the oxide ion with 9 protons and the valence electrons are in the

second energy level while the rest of the halides increase the energy levels by one through the group.

B)

C)

D)

E)

Page Ref: Sec. 7.4

8) ____ is isoelectronic with argon and _____ is isoelectronic with neon.

A) ,

B) ,

C) ,

D) ,

E) F+, F

–

Page Ref: Sec. 7.3

4

9) Of the following elements, ____ has the most negative electron affinity. (The one that will pay out the largest energy for

the electron)

A) I

B) S

C) Se

D) Cl

E) Br Page Ref: Sec. 7.4

10)

10) Sodium is much more apt to exist as a cation than is chlorine. This is because ____.

A) chlorine is bigger than sodium

B) chlorine is a gas and sodium is a solid

C) chlorine has a greater ionization energy than sodium does

D) chlorine is more metallic than sodium

E) chlorine has a greater electron affinity than sodium does Page Ref: Sec. 7.4

11) Which of the following correctly represents the electron affinity of bromine?

A) Br+

(g) + e –

→ Br (g)

B) Br2 (g) + 2 e –

→ 2 Br –

(g)

C) Br (g) + e –

→ Br –

(g)

D) Br2(g) + e –

→ Br –

(g)

E) Br (g) → Br+ (g) + e

–

Page Ref: Sec. 7.4

Consider the following electron configurations to answer the questions that follow:

(i) 1 2 2 3

(ii) 1 2 2 3 (because it’s almost stable)

(iii) 1 2 2 3 3

(iv) 1 2 2 3 3

(v) 1 2 2 3 3

12) The electron configuration that belongs to the atom with the lowest second ionization energy is _____.

A) (i)

B) (ii)

C) (iii)

D) (iv)

E) (v) Page Ref: Sec. 7.4

13) Of the elements below, _____ is the most metallic.

A) calcium

B) sodium

C) barium

D) magnesium

E) cesium Page Ref: Sec. 7.5

14) Which one of the following compounds would produce an acidic solution when dissolved in water?

A) Na2O

B) CaO

C) SrO

D) MgO

E) CO2 Page Ref: Sec. 7.5

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15) Of the following metals, __________ exhibits multiple oxidation states.

A) Na

B) Cs

C) Al

D) Ca

E) V Page Ref: Sec. 7.6

16) Which of these oxides is most basic?

A) Al2O3

B) Na2O

C) CO2

D) MgO

E) K2OPage Ref: Sec. 7.6

17) Which of the following is not a characteristic of metals?

A) malleability

B) acidic oxides

C) ductility

D) low ionization energies

E) These are all characteristics of metals. Page Ref: Sec. 7.6

18) Consider the following properties of an element:

(i) It is solid at room temperature.

(ii) It easily forms an oxide when exposed to air.

(iii) When it reacts with water, hydrogen gas evolves.

(iv) It must be stored submerged in oil.

Which element fits the above description the best?

A) sulfur

B) magnesium

C) mercury

D) sodium

E) copper Page Ref: Sec. 7.7

19) Which of the following generalizations cannot be made with regard to reactions of alkali metals? (The symbol M

represents any one of the alkali metals.)

A) M (s) + O2 (g) → MO2 (s)

B) 2M (s) + S (s) → M2S (s)

C) 2M (s) + 2 H2O (l) → 2MOH (aq) + H2 (g)

D) 2M (s) + H2 (g) → 2MH (s)

E) 2M (s) + Cl2 (g) → 2MCl (s) Page Ref: Sec. 7.7

20) All of the following are ionic compounds except __________.

A) SiO2 network covalent

B) NaCl

C) Li3N

D) Na2O4

E) K2OPage Ref: Sec. 7.7

21)

In nature, the noble gases exist as

A) alkali metal salts

B) the gaseous fluorides

C) monatomic gaseous atoms

D) the sulfides

E) solids in rocks and in minerals Page Ref: Sec. 7.8

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Chp. 8

1) Of the ions below, only __________ has a noble gas electron configuration.

A)

B)

C)

D)

E) Cl –

Page Ref: Sec. 8.1

2) The chloride of which of the following metals should have the greatest lattice energy?

A) sodium

B) lithium

C) rubidium

D) cesium

E) potassium Page Ref: Sec. 8.2

The diagram below is the Born-huber cycle for the formation of crystalline potassium fluoride.

3)

Which energy change corresponds to the first ionization energy of potassium?

A) 3

B) 2

C) 4

D) 6

E) 5 Page Ref: Sec. 8.2

4) Using the Born-Haber cycle, the ΔHfo of KBr is equal to ___.

A) ΔHfo K (g)] + ΔHf

o Br (g)] - I1 - E(Br) + ΔHlattice

B) ΔHfo K (g)] - ΔHf

o [Br (g)] + I1 (K) - E(Br) + ΔHlattice

C) ΔHfo K (g)] - ΔHf

o [Br (g)] - I1 (K) - E(Br) - ΔHlattice

D) ΔHfo K (g)] + ΔHf

o [Br (g)] + I1 (K) + E(Br) + ΔHlattice

E) ΔHfo K (g)] + ΔHf

o [Br (g)] + I1 (K) + E(Br) - ΔHlattice

Page Ref: Sec. 8.2

5) Of the atoms below, ____ is the most electronegative.

A) Cl

B) S

C) Ca

D) Si

E) Rb Page Ref: Sec. 8.4

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6) Which of the following has the bonds correctly arranged in order of increasing polarity?

A) Mg-F, Be-F, N-F, O-F

B) Be-F, Mg-F, N-F, O-F

C) O-F, N-F, Be-F, Mg-F

D) N-F, Be-F, Mg-F, O-F

E) O-F, Be-F, Mg-F, N-F Page Ref: Sec. 8.4

7) Resonance structures differ by ___.

A) number of atoms only

B) number of electrons only

C) placement of electrons only

D) number and placement of electrons

E) placement of atoms only Page Ref: Sec. 8.6

For the questions that follow, consider the BEST Lewis structures of the following ox_yanions:

(i) (ii) (iii) (iv) (v)

8) There can be three equivalent best resonance structures of _____. (no structure has four that I can think of)

A) (i)

B) (ii)

C) (iii)

D) (iv)

E) (v) Page Ref: Sec. 8.5-8.7

9) Of the following, ____ cannot accommodate more than an octet of electrons.

A) As

B) I

C) O (cannot expand into a d orbital)

D) S

E) P Page Ref: Sec. 8.7

10) The central atom in ____ does not violate the octet rule.

A) KrF2

B) CF4

C) SF4

D) ICl4 –

E) XeF4 Page Ref: Sec. 8.7

11) A valid Lewis structure of _____ cannot be drawn without violating the octet rule.

A) BeH2

B) SO32-

C)

D)

E)

Page Ref: Sec. 8.7

12) Bond enthalpy is _____. (the energy needed to break a bond)

A) always positive

B) always negative

C) always zero

D) sometimes positive, sometimes negative

E) unpredictable Page Ref: Sec. 8.8

8

13) As the number of covalent bonds between two atoms increases, the distance between the atoms _____ and the strength of

the bond between them _____.

A) increases, increases

B) decreases, decreases

C) increases, decreases

D) decreases, increases

E) is unpredictable Page Ref: Sec. 8.8

Chp. 9 1) According to VSEPR theory, if there are three electron domains in the valence shell of an atom, they will be arranged in

a(n) ____ geometry.

A) tetrahedral

B) trigonal planar

C) linear

D) trigonal bipyramidal

E) octahedral Page Ref: Sec. 9.2

2) The Cl-C-Cl bond angle in the CCl2O molecule (C is the central atom) is slightly _____.

A) greater than 109.5°

B) less than 109.5°

C) greater than 90°

D) less than 120°

E) greater than 120° Page Ref: Sec. 9.2

Explanation: The Lewis structure of the CCl2O molecule shows that there is a double bond between the O and the

central C atom. A double bond represents a slightly larger electron domain which disrupts the ideal 120° Cl-C-Cl

bond angle, making it smaller.

3) The molecular geometry of the BrO3 –

ion is _____.

A) trigonal planar

B) trigonal pyramidal

C) bent

D) T-shaped

E) tetrahedral Page Ref: Sec. 9.2

4) The bond angles marked a, b, and c in the molecule below are about __________, __________, and __________,

respectively.

A) 120°, 120°, 90°

B) 90°, 90°, 90°

C) 120°, 120°, 109.5°

D) 109.5°, 90°, 120°

E) 109.5°, 120°, 109.5° Page Ref: Sec. 9.2

5) The central iodine atom in the ICl4 –

ion has _____ nonbonded electron pairs and _____ bonded electron pairs in its

valence shell.

A) 1, 3

B) 2, 2

C) 2, 4

D) 3, 4

E) 3, 2 Page Ref: Sec. 9.2

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6) According to VSEPR theory, if there are two electron domains on a central atom, they will be arranged such that the

angles between the domains are ____.

A) 120°

B) 109.5°

C) 90°

D) 360°

E) 180° Page Ref: Sec. 9.2

Consider the following species when answering the following questions:

(i) (ii) (iii) (iv) (v)

7) For which of the molecules is the molecular geometry (shape) the same as the VSEPR electron domain arrangement

(electron domain geometry)?

A) (i) and (ii)

B) (i) and (iii)

C) (ii) and (v)

D) (iv) and (v)

E) (v) only Page Ref: Sec. 9.2

8) Of the molecules below, only ____ is nonpolar.

A) NF3

B) IF3

C) PBr3

D) BF3

E) BrCl3 Page Ref: Sec. 9.3

9) For molecules with only one central atom, how many lone pairs on the central atom guarantees molecular polarity?

A) 1

B) 2

C) 1 or 2

D) 1 or 3

E) 3 Page Ref: Sec. 9.3

10) The electron-domain geometry of a carbon-centered compound is tetrahedral. The hybridization of the central carbon

atom is ____.

A) d

B) sp3

C)

D)

E) sp Page Ref: Sec. 9.5

11) The atomic hybrid orbital set accommodates ____ electron domains.

A) 2

B) 3

C) 4

D) 5

E) 6 Page Ref: Sec. 9.5

10

12) The hybridizations of iodine in IF3 and IF5 are ____ and ____, respectively.

A) d,

B) d,

C) ,

D) , d

E) , d

Page Ref: Sec. 9.5

13) The hybrid orbitals used for bonding by Xe in the unstable XeF2 molecule are _____ orbitals.

A) sp

B)

C) sp3d

D)

E)

Page Ref: Sec. 9.5

14) The electron-domain geometry of the AsF5 molecule is trigonal bipyramidal. The hybrid orbitals used by the As atom

for bonding are _____ orbitals.

A)

B) sp3

C) d

D)

E)

Page Ref: Sec. 9.5

Consider the following species when answering the following questions:

(i) (ii) (iii) (iv) (v)

15) In which of the molecules does the central atom utilize d orbitals to form hybrid orbitals?

A) (i) and (ii)

B) (iii) only

C) (i) and (v)

D) (iii), (iv), and (v)

E) (v) only Page Ref: Sec. 9.5

16) When three atomic orbitals are mixed to form hybrid orbitals, how many hybrid orbitals are formed?

A) one

B) four

C) six

D) five

E) three Page Ref: Sec. 9.5

17) A typical double bond __________.

A) consists of two shared electron pairs

B) imparts rigidity to a molecule

C) consists of one σ bond and one π bond

D) is stronger and shorter than a single bond

E) All of the above answers are correct. Page Ref: Sec. 9.6

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18) Which of the following molecules or ions will exhibit delocalized bonding?

SO2 SO3 SO32 –

A) SO2 and SO3

B) SO32 –

only

C) SO3 and SO32 –

D) SO2, SO3, and SO32 –

E) None of the above will exhibit delocalized bonding. Page Ref: Sec. 9.6

19)

The carbon-carbon σ bond in ethylene, CH2CH2, results from the overlap of _____.

A) p atomic orbitals

B) hybrid orbitals

C) s atomic orbitals

D) sp hybrid orbitals

E) hybrid orbitals

Page Ref: Sec. 9.6

20) The N-N bond in HNNH consists of _____.

A) one σ bond and two π bonds

B) two σ bonds and one π bond

C) two σ bonds and two π bonds

D) one σ bond and one π bond

E) one σ bond and no π bonds Page Ref: Sec. 9.6

21) The hybridization of the carbon atom labeled x in the molecule below is _____.

A) sp

B)

C)

D) d

E)

Page Ref: Sec. 9.6

22) Valence bond theory does not address the issue of ____.

A) excited states of molecules

B) covalent bonding

C) molecular shape

D) multiple bonds

E) hybridization Page Ref: Sec. 9.5

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11.2 - 11.3

1) Which one of the following exhibits dipole-dipole attraction between molecules?

A) CO2

B) XeF4

C)

D) AsH3

E) BCl3 Page Ref: Sec. 11.2

2)

Which one of the following derivatives of ethane has the highest boiling point?

A)

B)

C)

D)

E) C2I6 Page Ref: Sec. 11.2

3) Hydrogen bonding is a special case of __________.

A) dipole-dipole attractions

B) ion-dipole attraction

C) London-dispersion forces

D) none of the above

E) ion-ion interactions Page Ref: Sec. 11.2

4) The predominant intermolecular force in (CH3)2NH is ___________

A) London dispersion forces

B) ion-dipole forces

C) hydrogen bonding

D) ionic bonding

E) dipole-dipole forces Page Ref: Sec. 11.2

5) Which one of the following substances will not have hydrogen bonding as one of its intermolecular forces?

A)

B)

C)

13

D)

E)

Page Ref: Sec. 11.2

6) _________ is the energy required to expand the surface area of a liquid by a unit amount of area.

A) Meniscus

B) Surface tension

C) Volatility

D) Capillary action

E) Viscosity Page Ref: Sec. 11.3

7) Which statements about viscosity are true?

(i) Viscosity increases as temperature decreases.

(ii) Viscosity increases as molecular weight increases.

(iii) Viscosity increases as intermolecular forces increase.

A) (i) only

B) (ii) and (iii)

C) (i) and (iii)

D) none

E) all Page Ref: Sec. 11.3

8)

The shape of a liquid's meniscus is determined by __________.

A) the amount of hydrogen bonding in the liquid

B) the relative magnitudes of cohesive forces in the liquid and adhesive forces between

the liquid and its container

C) the type of material the container is made of

D) the viscosity of the liquid

E) the volume of the liquid Page Ref: Sec. 11.3

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AP Chemistry: 2016-17 Semester Review:

Extended Response Section

CHEMICAL BONDING: Chp. 8 1. (a) Draw the ALL possible Lewis structures for sulfur trioxide, SO3.

(b) Determine the formal charge on all atoms of all structures. Sulfur is 2 single bonded O is -1 the

double bonded oxygen is 0

(c) Identify the most important Lewis structure for SO3. Justify your answer.

The Lewis structures are all identical and represent the delocalization of the electrons. The pi electrons

are shared among all the atoms. The three structures are needed as a culmination of the structure, no one is

more important than the other.

CHEMICAL BONDING / MOLECULAR GEOMETRY: Chp. 8 - 9 3. (a) Xenon can react with oxygen and fluorine to form compounds such as XeO3 and XeF4. In the boxes

provided, draw the complete Lewis electron-dot diagram for each of the molecules represented

below.

XeO3 XeF4

(b) On the basis of the Lewis electron-dot diagrams you drew for part (a), predict the following:

(i) The geometric shape of the XeO3 molecule trigonal pyramidal

(ii) The hybridization of the valence orbitals of xenon in XeF4 d2sp

3

(c) Predict whether the XeO3 molecule is polar or nonpolar. Justify your prediction. Polarity of a

molecule is based upon two factors: the bond polarity and the arrangement of the atoms around the center atom.

Oxygen is more electronegative than Xe so the bonds are polar. The arrangement of oxygen around the center

atom is such a way that the center of positive and negative charge do not coincide, therefore, the dipoles do not

cancel creating a polar molecule.

15

INTERMOLECULAR FORCES: 11.1 - 11.3

4. Use appropriate chemical principles to account for each of the following observations. In each part, your

response must include specific information about both substances.

(a) At 25oC and 1 atm, F2 is a gas, whereas I2 is a solid. Both molecular fluorine and molecular iodine have

London Dispersion forces. Iodine has more electrons than fluorine allowing for more temporary dipoles.

Since iodine is more polarizable, it has stronger dispersion forces than fluorine. The stronger dispersion

forces is the cause for the difference in the state of matter.

(b) The melting point of NaF is 993oC, whereas the melting point of CsCl is 645

oC. The energy required to

melt an ionic substance is based upon Coulomb’s Law. Coulomb’s Law states the forces that hold the

ionic compound is directly related to the charge and indirectly related to the radius of the ions bonding.

NaF and CsCl both have the same ionic charges of +1 and -1. The factor that is dominate is the size of

the ions bonding. Na ion is smaller than the Cs ion and F ion is smaller than the Cl ion. Therefore, the

forces are stronger in NaF causing it to have the higher melting point.

(c) Ammonia, NH3, is very soluble in water, whereas phosphine, PH3, is only moderately soluble in water.

Both NH3 and H2O exhibit hydrogen bonding whereas, PH3 has dipole-dipole bonding. The difference

in the intermolecular forces allows the ammonia to be more soluble in water than phosphine.

PERIODIC PROPERTIES: Chp. 7 5. Using principles of atomic and molecular structure and the information in the table below, answer the

following questions about atomic fluorine, oxygen, and xenon, as well as some of their compounds.

(a) Write the equation for the ionization energy of atomic fluorine that requires 1,681.01 kJ mol – 1

.

F F+ + e

- I1 = 1,681.01 kJ mol

-1

(b) Account for the fact that the first ionization energy of atomic fluorine is greater than that of atomic

oxygen. (You must discuss both atoms in your response.) Atomic fluorine has a greater nuclear charge

with 9 protons. Oxygen has 8 protons. The greater the nuclear charge, the greater the force of attraction

for the electrons so the more energy required to remove the electron from the atom.

(c) Predict whether the first ionization energy of atomic xenon is greater than, less than, or equal to the first

ionization energy of atomic fluorine. Justify your prediction. Xe has valence electrons that are in the

fifth energy level – 5p6. Xe has more quantized energy levels than fluorine which only has valence

electrons in the second energy level – 2p5. The further the electrons are from the nucleus the lower the

ionization energy due to the smaller force of attraction to the nucleus.

Atom

First Ionization Energy (kJ

mol – 1

)

F

1,681.0

O

1,313.9

Xe

?

16

6. The table below shows the first three ionization energies for atoms of four elements from the third period of

the periodic table. The elements are numbered randomly. Use the information in the table to answer the

following questions.

(a) Which element is most metallic in character? Explain your reasoning.

Element 2 is the most metallic. This element is an alkali metal which is in Group I of the

periodic table. Metallic character increases to the left and down on the periodic table.

(b) Identify element 3. Explain your reasoning. Element 3 would be magnesium. The third ionization

energy of element 3 is much greater than the second ionization energy indicating the element has only 2 valence

electrons. This element would be in group II A of period 3 which is Mg.

(c) Write the complete electron configuration for an atom of element 3. 1s22s

22p

63s

2

(d) What is the expected oxidation state for the most common ion of element 2? +1

(e) What is the chemical symbol for element 2? Na

(f) A neutral atom of which of the four elements has the smallest radius? #1 will have the smallest

radius with the highest first ionization energy.

Element

First

Ionization Energy

(kJ mol-1

)

Second

Ionization Energy

(kJ mol-1

)

third

Ionization Energy

(kJ mol-1

)

1 1,251 2,300 3,820

2 496 4,560 6,910

3 738 1,450 7,730

4 1,000 2,250 3,360