Exam Sample Problems by Topic – Answers in Red or Highlighted
electron configurations
Write the electron configuration for C. = 1s22s22p2
Write the electron configuration for Cr. = 1s22s22p63s23p64s13d5
Write the shorthand notation for Bi. = [Xe]6s24f145d106p3
Use these answers for questions 4 - 7.
(A) 1s2 2s22p6 3s23p5
(B) 1s2 2s22p6 3s23p6
(C) 1s2 2s22p62d10 3s23p6
(D) 1s2 2s22p6 3s23p63d5
(E) 1s2 2s22p6 3s23p63d3 4s2
4. An impossible electronic configuration = (C)
5. The ground-state configuration for the atoms of a transition element = (D)
6. The ground-state configuration of a negative ion of a halogen = (B)
7. The ground-state configuration of a common ion of an alkaline earth element = (B)
Which of the following conclusions can be drawn from J. J. Thomson's cathode ray experiments?
(A) Atoms contain electrons.
(B) Practically all the mass of an atom is contained in its nucleus.
(C) Atoms contain protons, neutrons, and electrons.
(D) Atoms have a positively charged nucleus surrounded by an electron cloud.
(E) No two electrons in one atom can have the same four quantum numbers.
Lewis structures & Molecular Geometry
Draw the Lewis structure and state the molecular geometry for H 2O, nitrate, and carbon dioxide.
H2O = bent
nitrate (NO3-) = trigonal planar (by the way, this one has resonance. The double bond could be on the left or right, too. In
reality, each bond is an average of both double and single).
Carbon dioxide (CO2) = linear
The Lewis dot structure of which of the following molecules shows only one unshared pair of valence electron?
(A) Cl2
(B) N2
(C) NH3
(D) CCl4
(E) H2O2
CCl4, CO2, PCl3, PCl5, SF6 Which of the following does not describe any of the preceding molecules?
(A) Linear
(B) Octahedral
(C) Square planar
(D) Tetrahedral
(E) Trigonal pyramidal
Consider the molecules PF3 and PF5.
a)
Draw the Lewis electron-dot structures for PF3 and PF5 and predict the molecular geometry of each.
Phosphorus trichloride (PF3) = trigonal pyramidal
Phosphorus pentafluoride (PF5) = trigonal bipyramidal
b) Is the PF3 molecular polar, or is it nonpolar? Explain. The PF3 molecule is polar because it has an electronegativity
difference of 1.8 (well above the 0.45 cut off point) and the molecule is not symmetrical due to the lone pair on the phosphorus.
c) On the basis of bonding principles, predict whether each of the following compounds exists. In each case, explain your
prediction: NF5 vs. AsF5. AsF5 can exist, but NF5 cannot. Although AsF5 will have more than an octet around its central atom
(as is the case with the PF5 molecule), this is allowable because the As has d orbitals which can be used for bonding. It is
impossible for NF5 to form because the N doesn’t have any d orbitals it can use for bonding.
Use these answers for questions 11 - 14.
(A) hydrogen bonding
(B) hybridization
(C) ionic bonding
(D) resonance
(E) van der Waals forces (London dispersion forces)
11. Is used to explain why iodine molecules are held together in the solid state. = (E)
12. Is used to explain why the boiling point of HF is greater than the boiling point of HBr = (A)
14. Is used to explain the fact that the carbon-to-carbon bonds in benzene, C6H6, are identical = (D)
Explain each of the following in terms of atomic and molecular structures and/or intermolecular forces.
(a) Solid K conducts an electric current, whereas solid KNO 3 does not. Solid K is a metal, and metals conduct electricity due to
their loosely held “sea of electrons.” Solid KNO3 does not conduct electricity because it is an ionic compound with a lattice
structure of interlocking + and – ions. These ions do not have any mobility, and thus, cannot conduct electricity in the solid
state.
(b) SbCl3 has a measurable dipole moment, whereas SbCl5 does not. SbCl3 is a polar molecule because it doesn’t have
symmetry. The lone pair on the central Sb push the three chlorine atoms toward each other in the trigonal pyramidal shape.
SbCl5, however, is symmetrical and therefore, does not have any separation of charges.
(c) The normal boiling point of CCl4 is 77 °C, whereas that of CBr4 is 190 °C. Both molecules are similar in shape and both are
polar molecules which exhibit dipole-dipole IMFs. When two compounds exhibit the same intermolecular forces, there is still a
difference in the IMF strength based on the size of the molecule. Larger molecules exhibit stronger IMFs, so because CBr 4 is
larger than CCl4, its IMF strength is higher, requiring more energy to overcome the attractions and boil, and thus, a higher b.p.
(d) NaI(s) is very soluble in water whereas I2(s) has a solubility of only 0.03 gram per 100 grams of water. NaI is very soluble
in water because it is ionic. I2 has a low solubility in water because it is nonpolar covalent. When dissolving substances “like
dissolves like” and water is a polar solvent. Although NaI is ionic and not polar, it can dissolve easily in water because water
has partially positive hydrogens that attract the I- ions in the salt crystal, and water also has partially negative oxygen, which
attracts the Na+ ions in the salt crystal. These electrostatic attractions cause the salt crystal to dissociate and dissolve. Since I 2 is
nonpolar, it doesn’t have any charges that the water can be attracted to. For a substance to dissolve, it must become completely
surrounded by solvent molecules.
Gases. Solve.
Given 500. cm3 of methane gas at 2.5 atm and 20 oC. What is the volume of the gas at STP? V2 = 1200 cm3 (use
the combined gas law)
What pressure is exerted by 2.0 moles of a gas in a 500 cm 3 container at 25 oC? P = 98 atm (use the ideal gas law)
Two hundred cubic centimeters of a gas are collected by water displacement. The conditions at time are 1.1 atm
and 30 oC. The vapor pressure of water at 30 oC is 23.4 mm Hg. What would be the pressure of the dry gas? P =
812.6 mm Hg or 1.07 atm (sig figs don’t work great with these numbers, so I used enough to show the answer
without too much rounding – use Dalton’s law of partial pressures)
2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O
How many milliliters of ethane (C2H6) at STP are required to produce 22 g of carbon dioxide? 5600 mL (use molar
volume of 22.4 L/mol since it’s at STP)
N2(g) + 3 H2(g) -----> 2 NH3(g)
How many liters of NH3 can be produced at a temperature of 27 degrees Celsius and a pressure of 760 torr, if 20.
moles of N2 are consumed? 980 L (use the ideal gas law to find the new molar volume of 24.6 L/mol since it’s not at
STP, then perform the regular stoichiometry)
A gaseous mixture containing 7.0 moles of nitrogen, 2.5 moles of oxygen, and 0.50 mole of helium exerts a
total pressure of 0.90 atmosphere. What is the partial pressure of the nitrogen? Mea culpa: this involves
something called the mole fraction (which we didn’t learn this year). The partial pressure of a gas is equal to its
mole fraction times the total pressure of the sample. Mole fraction is moles of gas X/ moles of total sample.
(A) 0.13 atm
(B) 0.27 atm
(C) 0.63 atm
(D) 0.90 atm
(E) 6.3 atm
Hydrogen gas is collected over water at 24 °C. The total pressure of the sample is 755 millimeters of
mercury. At 24 °C, the vapor pressure of water is 22 millimeters of mercury. What is the partial pressure of
the hydrogen gas? (use Dalton’s law of partial pressures)
(A) 22 mm Hg
(B) 733 mm Hg
(C) 755 mm Hg
(D) 760 mm Hg
(E) 777 mm Hg
A 2.00-liter sample of nitrogen gas at 27 °C and 600. millimeters of mercury is heated until it occupies a
volume of 5.00 liters. If the pressure remains unchanged, the final temperature of the gas is (use Charles’
law)
(A) 68 °C
(B) 120 °C
(C) 477 °C
(D) 677 °C
(E) 950. °C
2 K + 2 H2O ---> 2 K+ + 2 OH¯ + H2
When 0.400 mole of potassium reacts with excess water at standard temperature and pressure as shown in
the equation above, the volume of hydrogen gas produced is
(A) 1.12 liters
(B) 2.24 liters
(C) 3.36 liters
(D) 4.48 liters
(E) 6.72 liters
Represented above are five identical balloons, each filled to the same volume at 25°C and 1.0 atmosphere
pressure with the pure gas indicated.
(a) Which balloon contains the greatest mass of gas? Explain. If each balloon is filled to the same volume,
Avogadro’s principle says that they will all contain an equal number of moles of gas. Since CO 2 has the highest
molar mass (44 g/mol), it will contain the greatest mass of gas.
(b) Compare the average kinetic energies of the gas molecules in the balloons. Explain. If all of the balloons are at
the same temperature, and temperature is a measure of average kinetic energy, then all of the average kinetic
energies of the gas molecules in the balloons are the same.
(c) Which balloon contains the gas that would be expected to deviate most from the behavior of an ideal gas?
Explain. Ideal gases don’t exhibit IMFs and they have negligible volumes. In terms of size, CO 2 is the most massive
particle, but CH4 contains the highest number of atoms; however, due to the molecular geometry of the CH4
molecule (tetrahedral) the molecule remains fairly compact whereas the linear CO 2 molecule will take up more
room. All of the molecules are nonpolar so we can’t easily eliminate anything based on the strength of the IMFs;
however, the CO2 molecule does have a dipole moment between the C and O atoms. The dipole cancels in the
molecule overall making it nonpolar, but in reality, the oxygens are still slightly more negative than the carbon and
can interact with other CO2 molecules. For all of these reasons, the balloon containing CO2 would deviate the most
from the behavior of an ideal gas.
(d) Twelve hours after being filled, all the balloons have decreased in size. Predict which balloon will be the
smallest. Explain your reasoning. Graham’s law of effusion relates the molar mass of particles to the rate at which
they effuse (leave the balloon through small holes in the material). The smaller the molar mass of the gas, the
faster the molecule will effuse. Therefore, the balloon containing the gas with the lowest molar mass will be the
smallest—He.
Which of the following is true at the triple point of a pure substance?
(A) The vapor pressure of the solid phase always equal the vapor pressure of the liquid phase.
(B) The temperature is always 0.01 K lower that the normal melting point.
(C) The liquid and gas phases of the substance always have the same density and are therefore indistinguishable.
(D) the solid phase always melts if the pressure increases at constant temperature.
(E) The liquid phase always vaporizes if the pressure increases at constant temperature.
Use the following diagram for questions 49-51.
49. The normal boiling point of the substance represented by the phase diagram above is
(A) -15 °C
(B) -10 °C
(C) 140 °C
(D) greater than 140 °C
(E) not determinable from the diagram
50. The phase diagram above provides sufficient information for determining the
(A) entropy change on vaporization
(B) conditions necessary for sublimation
(C) deviations from ideal gas behavior of the gas phase
(D) latent heat of vaporization
(E) latent heat of fusion
51. For the substance represented in the diagram, which of the phases is most dense and which is least dense at -15 °C. This
one isn’t very obvious, so let me explain. Density increases with increasing pressure, because when pressure increases, the
molecules are being compressed, lowering volume. Since density is mass/volume, as volume gets smaller the overall fraction
becomes larger. If you look at the -15°C line very carefully, you’ll see that it hits liquid at the top and gas at the bottom with
solid actually being in between (like water – ice floats!).
Most Dense
Least Dense
(A)
Solid
Gas
(B)
Solid
Liquid
(C)
Liquid
Solid
(D)
Liquid
Gas
(E)
The diagram gives no information about densities.
Name the following acids.
H3PO3 = phosphorous acid
HCl
= hydrochloric acid
H2Se = hydroselenic acid
HNO2 = nitrous acid
H2CO3 = carbonic acid
Solve.
[H+] = 5.32 X 10 -6
[OH-] = 1.88 x 10-9
pH = 5.27
pOH = 8.73
Acid or Base? acid
[H+] = 1.20 x 10-3
[OH-] = 8.32 X 10 -12
pH = 2.92
pOH = 11.08
Acid or Base? acid
Bronsted-Lowry Pairs
Label each compound in the equation using the following terms: acid, base, conjugate acid, conjugate base.
NH3
base
HNO2
acid
+
+
H2O
acid
H2O
base
<===> NH4+ +
conj. acid
<===> H3O+ +
conj. acid
OH¯
conj. base
NO2¯
conj. base
In the titration of a weak acid of unknown concentration with a standard solution of a strong base, a pH meter was used
to follow the progress of the titration. Which of the following is true for this experiment?
(A) The pH is 7 at the equivalence point.
(B) The pH at the equivalence point depends on the indicator used.
(C) The graph of pH versus volume of base added rises gradually at first and then much more rapidly.
(D) The graph of pH versus volume of base added shows no sharp rise.
(E) The [H+] at the equivalence point equals the ionization constant of the acid.
How many milliliters of 11.6-molar HCl must be diluted to obtain 1.0 liter of 3.0-molar HCl?
(A) 3.9 mL
(B) 35 mL
(C) 260 mL
(D) 1,000 mL
(E) 3,900 mL
All of the following species can function as Brönsted-Lowry bases in solution EXCEPT
(A) H2O
(B) NH3
(C) S2¯
(D) NH4+
(E) HCO3¯
In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted
by a change in the volume of the system at constant temperature?
(A) CO(g) + NO(g) <===> CO2(g) + 1/2 N2(g)
(B) N2(g) + 3 H2(g) <===> 2 NH3(g)
(C) N2(g) + 2 O2(g) <===> 2 NO2(g)
(D) N2O4(g) <===> 2 NO2(g)
(E) NO(g) + O3(g) <===> NO2(g) + O2(g)
2 SO2(g) + O2(g) <===> 2 SO3(g)
When 0.40 mole of SO2 and 0.60 mole of O2 are placed in an evacuated 1.00-liter flask, the reaction represented above
occurs. After the reactants and the product reach equilibrium and the initial temperature is restored, the flask is found
to contain 0.30 mole of SO3. Based on these results, the equilibrium constant, K c for the reaction is…To find K, you must
first find the equilibrium amounts of each substance. Since the 0.30 mol of SO 3 is an equilibrium amount, you can use
stoichiometry to find out how much SO2 and O2 was lost to make the 0.30 mol of SO3. Then you can plug your numbers into
the K expression.
(A) 20.
(B) 10.
(C) 6.7
(D) 2.0
(E) 1.2
The solubility of CuI is 2 x 10¯6 molar. What is the solubility product constant, K sp, for CuI?
(A) 1.4 x 10¯3
(B) 2 x 10¯6
(C) 4 x 10¯12
(D) 2 x 10¯12
(E) 8 x 10¯18