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Dartmouth College
Illustrative Chemistry 6 Credit test questions and their solutions
As the following examples illustrate, the Chemistry 6 Credit test is, in the main, not a multiple-choice examination,
but rather, it consists of a collection of short problems that provide the student an opportunity to display her or his
problem-solving abilities. On such questions, major partial credit is awarded for the development of an approach that
will lead to a successful answer. So, make sure that your answer shows clearly the approach you are using to solve the
problem.
The solutions to these questions are provided, not only for you to check your answers, but also to indicate the depth of
answer that is expected. You should not be surprised to find that the depth of understanding expected in your answers to
these questions sometimes goes beyond that expected in high school.
To be awarded credit for Chemistry 6, a student must score at least 65% on the Chemistry 6 Credit test.
You will be provided with an information sheet similar to the one included below.
Information sheet
ln
dx = ln x + C
x
dx = – 1 + C
x
x2
[X]
= –kt
[ X ]0
where [X]0 is the concentration of X at time t = 0
and [X] is the concentration of X at time t = t
1 – 1
= kt
[X]
[ X ]0
k= Ae
– Ea
RT
ln x = 2.303 log 10 x
NA = 6.022 × 1023 atoms/mol
R = 8.3145 J K–1 mol–1 = 2.0 cal K –1 mol–1
2
p2
K.E. = mv =
2
2m
c = 3 × 108 m s–1
21.8 × 10 –19 Z2
Joules
n2
ε = hν
c = νλ
En = –
h = 6.626 × 10–34 J s
1nm = 10–9 m
e = 1.60 × 10–19 Coulombs
1Å = 10–8 cm = 10–10 m
1 J = 6.24 × 1018 eV
1 Volt × 1 Coulomb = 1 Joule
m(electron) = 9.11 × 10–31 kg
Total Energy = K.E. + P.E.
P.E. =
Q1 Q2
αr
1 D = 3.38 × 10–30 C m
α = 1.113 × 10–10 C2 J–1 m–1
2
H
2.1
Li
1.0
Be
1.5
B
2.0
C
2.5
N
3.0
O
3.5
F
4.0
Na
0.9
Mg
1.2
Al
1.5
Si
1.8
P
2.1
S
2.5
Cl
3.0
K
0.8
Ca
1.0
Ga
1.6
Ge
1.8
As
2.0
Se
2.4
Br
2.8
Rb
0.8
Sr
1.0
In
1.7
Sn
1.8
Sb
1.9
Te
2.1
I
2.5
Cs
0.7
Ba
0.9
Tl
1.8
Pb
1.8
Bi
1.9
Po
2.0
At
2.2
Electronegativities of the representative elements
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Question 1
Measurements of initial reaction rates were performed to determine the details of the differential rate law for
the overall reaction (in aqueous solution):
BrO3– + 5Br – + 6H + → 3Br2 + 3H 2O
with the following results:
[BrO3–]
(mol L–1)
[Br – ]
(mol L–1)
[H+ ]
(mol L–1)
Initial Reaxn Rate
(mol L–1s–1)
_________________________________________________________
0.10
0.10
0.10
1.20 × 10–3
0.20
0.10
0.10
2.40 × 10–3
0.10
0.30
0.10
3.60 × 10–3
0.20
0.10
0.30
2.16 × 10–2
________________________________________________
The differential rate law may be written in the form:
–
–
d BrO3
dt
– a
= kexp BrO3
Br –
b
H+
c
(i). Determine the values of a, b and c.
(ii). Using the data given in the above table and the result of part (i) of this question, calculate a value for
the rate constant, kexp (state your units clearly).
Question 2
The decomposition of species A was studied at 576K by recording the concentrations of A at known times
with the results shown below: A → Products
(rate constant, kexp)
t (min) [A] mol L–1
_______________________
0
8.70 × 10 –3
40
6.15 × 10 –3
80
4.35 × 10 –3
120
3.08 × 10 –3
160
2.18 × 10 –3
_______________________
Calculate a value for the rate constant kexp. Show your method clearly and be sure to state the units of
kexp.
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Question 3
Rate constants k measured at different temperatures for the reaction
CH3N2CH3 → 2 CH3
+ N2
(1)
are given in the following table:
T (°C)
k (s–1)
________________________
250
1.8 × 10–6
268
1.5 × 10–5
287
6.0 × 10–5
303
1.6 × 10–4
320
9.5 × 10–4
________________________
Calculate the activation energy, Ea for the reaction described in equation (1). (Show your method clearly
and state your units clearly). (Results which are inaccurate by more than 5% will not receive full credit).
Question 4
The maximum wavelength (λ) of electromagnetic radiation required to eject electrons from the surface of
tungsten metal is 272 nm. Calculate the maximum kinetic energy observed amongst the electrons ejected
from tungsten metal by electromagnetic radiation with wavelength λ = 2000 Å. Express your answer in
kJ mol–1 and show your method clearly.
Question 5
The energy required to remove the outermost electron from a ground state sodium (Na) atom is 496 kJ
mol–1. If a collection of Na atoms is heated in an electric discharge, two prominent emissions occur, one at
a wavelength of 589 nm, and one at a wavelength of 820 nm. The former is emission resulting from an
excited state with the electron configuration [Ne]3p1, while the latter is emission resulting from an excited
state with the electron configuration [Ne]3d1.
(i)
Write down the electron configuration for the ground state of a Na atom.
(ii)
We take the zero of energy to correspond to Na+ and a free electron, e–, at infinite separation
and both at rest. Calculate the energy (relative to the zero of energy defined above) of the 3p
level for a Na atom. Express your answer in kJ mol–1 and show your method clearly.
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Question 6
(A) The ground state electron configuration of the gas phase O atom is [He]2s2 2p4. Using this notation
write down
(i) the ground state electron configuration
and
(ii) the number of unpaired electrons
for each of the following gas phase atoms or ions.
(a) Si
(b) Al3+
(c) As
(B) Explain why the energy spacing between the 2s and the 2p levels in the gas phase nitrogen atom, N, is
greater than the energy spacing between the 2s and the 2p levels in the gas phase boron atom, B.
Question 7
(i) The H atom 3py orbital is given by ψ3py = constant × r (6 – r) exp (–r/3) sinθ sinφ where the distance r
is expressed in atomic units. On the graph below, sketch a plot of the Radial Probability Distribution
function ( r2 R2(r)) vs. r for a H atom 3p orbital. Mark off the horizontal axis in atomic units.
Your plot should identify the positions of nodes, relative positions of maxima, and which maximum is the
largest. Give a brief explanation of your method.
(ii) Using the plot you made in part (i), explain how you would calculate the probability of finding an
electron in a 3p orbital within a distance of 3 Å from the nucleus.
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Question 8
Predict the molecular geometries of each of the following molecules or ions. Show your method clearly
and represent your prediction with a simple sketch and a descriptive name or phrase. Include a
qualitative statement about deviations of bond angles from their idealized values (i.e. <180, <120,
etc.).
(b) (BiCl5)2–
(a) GaI3
Question 9
Consider the molecule HSCN (H–S–C–N) and the ion SCN – (S–C–N) – (the skeletal structures in
parentheses are only meant to indicate which atoms are connected). From a consideration of Lewis electron
dot structures for such species, predict
(i) which species has the shorter bond between C and N;
and,
(ii) which species has the shorter bond between C and S.
(Full justification of your prediction will be required for full credit).
Question 10
(i) Which of the following molecules possesses a permanent electric dipole moment?
(a) CS2
(b) BF3
(c) PCl3
(e) CCl4
(ii) Arrange the following substances in order of increasing radius : Br –, Kr, Sr2+ , K + . Explain your
method.
Question 11
(a) Identify the hybrid orbitals used on all the atoms in the most important resonance structure for the
following molecules; AND
(b) identify the idealized local geometry around all atoms except the terminal atoms.
(Skeletal molecular structures are given in parentheses)
(i) H 2 CNN
(H
C
N)
N
H
H
(ii) (CH 3CO) 2O
(H
H
C
C
H
O
O
C
C
O
H
H
)
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Solutions to Sample Questions
Question 1
(i) From data sets 1 and 2, doubling [BrO3–] increases the initial rate by a factor of 2. ∴ a = 1.
From data sets 1 and 3, increasing [Br–] by a factor of 3 increases the initial rate by a factor of 3. ∴ b = 1.
From data sets 2 and 4, increasing [H+ ] by a factor of 3 increases the initial rate by a factor of 9. ∴ c = 2.
(ii) From part(i), the differential rate law is given by Rate = k [BrO3–] [Br–] [H+ ]2
Use any data set to evaluate the rate constant k. For example, using data set 1,
1.20 × 10–3 mol L–1 s–1 = k (0.10 mol L–1) (0.10 mol L–1) (0.10 mol L–1)2
∴ k = 1.20 × 101 mol–3 L3 s–1
Question 2
The first step is to determine the order of the reaction -- note that explicit determination of the reaction order
was required for major credit. There are two ways to determine the order of the reaction in this question.
(i) Plot ln [A] vs. t. The resulting linear plot indicates that the reaction is first-order , where the slope of
this plot = – k. Also, this could have been shown by substituting at least two different pairs of data in the
[A]
expression ln
= – k t (see the information sheet).
[ A ]0
6.15 × 10–3
For example, ln
= – k 40 min ⇒ k = 0.3469 = 8.67 × 10–3 min–1
–3
40 min
8.70 × 10
or
ln
4.35 × 10–3
8.70 ×
10–3
= – k 80 min ⇒ k = 0.6931 = 8.66 × 10–3 min–1
80 min
The constancy of k confirms that the reaction order = 1.
(ii) An alternate method is to note that [A]80 = [A]0/2 and [A]120 = [A]40/2. That is, whether we start with
[A]0 or [A]40, it takes 80 minutes to reduce [A] to one-half of its initial value. Thus, the reaction half-life,
t1/2, is independent of the initial concentration indicating that the reaction is first-order.
[ A ]0 /2
When t = t1/2, [A] = [A]0/2, and
ln
= – k t1/2
[ A ]0
Thus,
t1/2 = ln 2
k
or
k = ln 2 = 0.6931 = 8.66 × 10–3 min–1
t1/2
80 min
(Note that using the differential rate law here is incorrect since the concentrations and hence the reaction
rates vary very considerably over the time intervals given).
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Question 3
– Ea
RT
From the information sheet, k = A e
. Taking natural logarithms of both sides of this equation gives:
E
a
ln k = ln A –
, where both A and Ea are independent of temperature. Using this equation at any two
RT
temperatures T1 and T2 yields:
ln k1 = ln A – Ea (1) and k2 = ln A – Ea (2)
R T1
R T2
Subtracting equation (2) from equation (1) gives:
ln k1
k2
= – Ea
R
1 – 1 .
T1
T2
We may now use any pair of data points to evaluate Ea. For example:
ln
i.e. ln 0.1852
∴ Ea =
2.13 × 10–6 s–1
1.15 ×
10–5
s–1
= – Ea
R
Ea
8.3145 (J mol –1 K–1 )
= –
1.6862
1 – 1
523
541
1.9120 × 10–3 – 1.8484 × 10–3
8.3145 J mol–1 K–1
6.357 ×
10–5
K–1
K–1
= 220 × 103 J mol–1 = 220 kJ mol–1 .
Question 4
Applying the conservation of energy to the photoelectric effect experiment
h ν = Φ + Kinetic Energy of ejected electron
Here, h ν is the energy of the incident photon (wavelength λ) and Φ is the binding energy of the electron to
the metal.
Thus, the first step in answering this question is to determine the binding energy Φ. The maximum
wavelength photon has an energy that is just sufficient to overcome the binding energy Φ.
That is,
h c = h c = Φ and Φ = 7.303 × 10–19 J
272 nm
λ max
With the value of Φ in hand, Kinetic Energy of ejected electron = h ν – 7.303 × 10–19 J
=
h c – 7.303 × 10–19 J = ( 9.932 × 10–19 – 7.303 × 10–19 ) = 2.63 × 10–19 J/electron
200 nm
This is easily converted to units of kJ/mol, by multiplying by Avogadro's number. Thus,
Kinetic Energy = 2.63 × 10–19 J/electron
6.02 × 1023 electrons/mol = 158 kJ/mol
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Question 5
(i) Na has 11 electrons; electron configuration is 1s2 2s2 2p6 3s1 or [Ne] 3s1.
(ii) The ionization energy of the ground state Na atom = Efinal – Einitial where Efinal is the energy of Na+
and a free electron at infinite separation and both at rest, and Einitial is the ground state energy of Na. From
the information given:
Efinal – Einitial = 0 – E([Ne] 3s1) = 496 kJ mol–1
The emission line at 589 nm results from an electronic transition that originates at the Na excited state [Ne]
3p1 and terminates at the ground state [Ne] 3s1. This is summarised in the following figure:
Energy
E ([Ne] 3p1 )
E ([Ne] 3s1 )
Thus, for the transition
∆E = E ([Ne] 3s1 ) – E ([Ne] 3p1 ) =
h c = 3.373 × 10–19 J/atom = 203 kJ/mol
589 nm
That is, the excited state [Ne] 3p1 is 203 kJ mol–1 higher in energy than the ground state
[Ne] 3s1. Relative to the zero of energy given (i.e. the ionization limit), the Na ground state has an energy
of – 496 kJ mol–1. Thus, relative to this same zero of energy the energy of the Na excited state [Ne] 3p1 is
(–496 + 203) = – 293 kJ mol–1.
Question 6
(A) (a) Si: 1s2 2s2 2p6 3s2 3p2 ≡ [Ne] 3s2 3p2: there are 2 unpaired electrons.
(b) Al3+ : 1s2 2s2 2p6 ≡ [Ne]: there are 0 unpaired electrons
(c) As: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 ≡ [Ar] 4s2 4p3: there are 3 unpaired electrons
(B) An examination of the radial distribution plots for 2s and 2p atomic orbitals shows that a 2s electron
penetrates to the nucleus more strongly than does a 2p electron. Thus, in a many-electron atom, the
effective nuclear charge Zeff experienced by a 2s electron is greater than that experienced by a 2p electron.
That is, Zeff (2s) > Zeff(2p) and for the orbital energies E(2s) < E(2p) -- recall that the orbital energies are
negative.
In boron, when the 2s electron penetrates to the nucleus it "feels" a nuclear charge of +5e, whereas in
nitrogen it will "feel" a nuclear charge of +7e. Since the 2s electron is screened predominantly by the 1s
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electrons, increasing Z from 5 to 7 produces an increase in Zeff. Thus, Zeff(2sN) is greater than Zeff(2sB)
and E(2sN) < E(2sB). Using similar arguments for the 2p electron, Zeff(2pN) > Zeff(2pB) and E(2pN) <
E(2pB). However, since the 2p electron penetrates to the nucleus to a smaller extent that does a 2s electron,
the effect of increasing Z from 5 to 7 produces a smaller increase in Zeff(2p) than in Zeff(2s), and E(2p)
decreases to a smaller extent than E(2s). Thus, the 2s – 2p orbital energy spacing will be greater in
nitrogen than in boron. (Note that all of these points had to be considered for full credit)
Question 7
(i) From the wavefunction given, the radial part of the wavefunction for the 3py orbital,
R3p ∝ r (6 – r) exp (–r/3). Thus, R3p will vanish at r = 0 au, at r = 6 au, and as r → ∞. Thus, r2 R3p2 will
also vanish at r = 0, 6 au and as r → ∞. However, only values of r, other than r = 0 and r → ∞, which
make R3p vanish represent radial nodes. Thus, the 3py orbital has one radial node at r = 6 au. The plot of
r2 R3p2 vs. r has two maxima, with the principal (i.e. the largest) maximum lying at a larger value of r.
r 2 R 23p (r)
r (atomic units)
(ii) The importance of the r2 R3p2 vs. r plot is that the area under the curve between two values of r, r1 and
r2, for example, defines the probability of finding the 3py electron at a distance between r1 and r2 from the
nucleus. The first step in this part is to convert r = 3 Å to atomic units:
3 × 10–10 m
r (atomic units) =
= 5.67 au.
0.529 × 10–10 m
Thus, the required probability is the area under the r2 R3p2 vs. r plot between r = 0 and r = 5.67 au.
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Question 8
(a) GaI3 has 3 + 3 (7) = 24 valence electrons and the Lewis electron dot structure shown:
Because there are three electron groups around the central Ga atom the idealized electron group geometry is
trigonal planar. All the electron groups are bonding electron groups and the idealized molecular geometry is
also trigonal planar.
I
Trigonal Planar
Ga
I
I
Because all the bonding pairs around Ga are equivalent, there are no deviations from ideality
and ∠ I – Ga – I = 120°.
(b) (BiCl5)2– has 5 + 5(7) + 2 = 42 valence electrons and the Lewis electron dot structure shown:
There are six electron groups around the central Bi atom and the idealized electron group geometry is
octahedral. Of the six electron groups, five are bonding, each representing a BiCl bonding electron pair,
and the sixth is a nonbonding electron pair on Bi. Thus, the idealized molecular geometry is square-based
pyramidal.
Cl a
Cl b
Cl
Cl c
Bi
Square-based Pyramidal
Cl
Because nonbonding pair ↔ bonding pair repulsion > bonding pair ↔ bonding pair repulsion there will be
deviations from this idealized structure. Thus, ∠ Clb – Bi – Cla and analogous angles will be less than 90°
and ∠ Clb – Bi – Clc and analogous angles will be less than 90°.
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Question 9
The number of valence electrons in HSCN = 1 + 6 + 4 + 5 = 16. There are three Lewis electron dot
structures ((a), (b) and (c) shown below) that satisfy the octet rule:
0
[ H
+2
S
0
C
–2
..
N: ]
..
0
0
0
0
C
N:]
..
S
..
[H
0
+1
0
[H
S
C
..
–1
..
N]
..
(a)
(b)
(c)
A consideration of the formal charges shown in the above structures, indicates that the relative importance
of these structures is (b) >> (c) >> (a). In structure (b) there is no separation of formal charge, while
structure (c) separates smaller opposite formal charges than does structure (a). In view of the relative
importance of these three structures, the CN bond in HSCN is almost a pure triple bond, and the CS bond
in HSCN is almost a pure single bond.
The number of valence electrons in SCN– is 6 + 4 + 5 + 1 = 16. There are three Lewis electron dot
structures ((a), (b) and (c) shown below) that satisfy the octet rule:
+1
[:S
0
C
(a)
–2
.. –
N: ]
..
–1
..
[ :S
..
0
C
(b)
0
N:]
0
–
..
[ S
..
0
–1
C
N]
.. –
..
(c)
Neither (b) nor (c) involves a separation of formal charge. However, since nitrogen is more electronegative
than S, structure (c) is of greater relative importance than structure (b). The overall order of relative
importance is (c) > (b) >> (a). Thus, the CN bond in SCN–, although predominantly a double bond has
some triple bond character and the CS bond in SCN–, although predominantly a double bond has some
single bond character.
From this analysis, we conclude that HSCN has the shorter CN bondlength and that SCN– has the shorter
CS bondlength.
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Question 10
(i) From Lewis structure considerations, the most important resonance structure is that shown for each
molecule:
In CS2 there are two electron groups around C and the molecule is linear; thus, the two C – S bond dipole
moment vectors cancel -- i.e.
S
C
S
and CS2 does not possess a permanent dipole moment.
The molecular geometry of BF3 is trigonal planar, the three B – F bond dipole moment vectors cancel and
BF3 does not possess a permanent dipole moment.
The molecular geometry of PCl3 is pyramidal. In this case, the three P – Cl bond dipole moment vectors
will not cancel and PCl3 possesses a permanent dipole moment.
The molecular geometry of CCl4 is tetrahedral, the C – Cl bond dipole moment vectors cancel and CCl4
does not possess a permanent dipole moment.
(ii)
K+ < Sr 2+ < Kr < Br –
In K+, the outermost electrons are in the n = 3 shell, whereas for all the other species the outermost
electrons are in the n = 4 shell. Because the species Kr, Br– and Sr2+ are isoelectronic with the
configuration [Ar] 3d10 4s2 4p6, and because Z(Sr2+ ) > Z(Kr) >
Z(Br–), the effective nuclear charges have the same relative order -- i.e. Zeff(Sr2+ ) > Zeff(Kr) > Zeff(Br–).
Thus, the outermost electrons in Sr2+ are held more tightly than those in Kr and Sr2+ has a smaller radius
than Kr. Similarly, the outermost electrons in Kr are held more tightly than those in Br– and Kr has a
smaller radius than Br–.
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Question 11
(i)
The geometry around the C atom is trigonal planar and the geometry around the central N is linear.
(ii)
The geometry around the terminal C atoms is tetrahedral, that around the C = O carbon atoms is trigonal
planar, and the geometry around the central O atom is bent.