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CHEM 1000 A, V Midterm Test #1
October 18, 2002
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.
If one atom of 12C weights exactly 12 amu, why is the atomic mass of carbon equal to
12.011 g?
This is because carbon consists of isotopes other than 12C, notably 13C, which increase the
average atomic mass to 12.011 g/mol
2.
Why do atomic light sources have only discreet frequencies?
The light from an atomic source is produced when electrons fall from a higher energy shell to a
lower energy one. The energy, hence the wavelength of the photon, corresponds to the
difference in energies of the upper and lower shell. Since these energies are quantized, so is the
energy difference, and therefore so is the energy or wavelength of the photon.
3.
A p-orbital is shown below. An electron in this orbital occupies both lobes. How can the
electron get from one lobe to the other if the node is a place where the electron can never exist?
Node
Electrons have wavelike properties. Since waves can propagate through a node, so can
electrons.
4.
The ground electronic configuration for copper (Cu) is predicted by AUFBAU to be
[Ar] 4s23d9. However, Cu actually has a different configuration. State what you think the
actual configuration is, and state why it is different from what AUFBAU predicts.
It is likely [Ar]4s13d10. In this configuration, there is one half-filled subshell, and a filled
subshell, which are more energetically stable configuration that a filled 4s subshell and a
9/10ths filled 3d subshell.
5.
Explain how the ionization energy changes going from left to right across a period, and briefly
explain why this happens.
The ionization energy (or potential) increases going left to right across a period. This is because
electrons are being added to the same principal shell, and so have approximately the same
energies. However, protons are being added to the nucleus at the same time. The effective
nuclear charge therefore increases going left to right, creating a larger energy of attraction for
the electrons, thus requiring a larger energy to remove one.
6.
SF6 is an octahedral molecule. What type of hybrid orbitals does the S atom use to bond with
the F atoms?
Since there are six charge clouds around the central S atom, it must be using sp3d2 hybrid
orbitals.
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Part B. Answer any two questions. If you answer all three, the best two answers will count.
(20 marks each)
B1.
(a) Use VSEPR theory to predict the shape of the SF3+ ion.
Valence electrons = 6 + (3 x 7) – 1 = 26.
Arranging the F atoms around the central S atom and completing the octets on the F atoms uses
24 electrons. The other two are a lone pair on the S atom. The molecule is therefore of the form
AX3E, and is trigonal pyramidal.
(b) Use VSEPR theory to predict the shape of AsCl5.
Valence electrons = 5 + (5 x 7) = 40.
Arranging the Cl atoms around the central As atom and completing the octets on the Cl atoms
uses all 40 electrons. The molecule is therefore of the form AX5, and is trigonal bipyramidal.
(c) A molecular orbital (MO) diagram for oxygen (O2) is shown below. What will happen to the
bond length and bond energy if we ionize O2 to O2+ and why?
The electron that is lost during ionization will come from the highest energy MO, i.e. the *2p.
Since this is an antibonding orbital, the bond order will increase upon ionization. Thus, the
bond will be shorter and stronger in the ion.
(d) How can you deduce from the MO diagram that O2 has a double bond?
The bond order is BO = ((8 – 4)/2) = 2, thus the molecule has a double bond.
2
p
2
p
2
p
2
s
2
s
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B2.
(a) We have used the Balmer-Rydberg equation to calculate the wavelength of photons emitted
by atoms falling from an upper to a lower energy state. However, if we shine photons onto a sample of
hydrogen, we can actually cause the electrons to jump to higher levels. The same equation applies in
such a case. Suppose the electron in a ground state (m=1) hydrogen atom absorbs a photon having a
wavelength of 97.20 nm. To what upper level does the electron move?
1/= R[1/m2 – 1/n2]
thus, 1/97.2nm = 0.01097 nm-1 [1/12 – 1/n2]
0.01029 = 0.01097 [1 – 1/n2]
0.9378 = 1 – 1/n2
solving, n = 4
(b) What is the energy of the photons having a wavelength of 97.20 nm? Express in kJ mol-1.


E = h = hc/

= 6.63 x 10-34 J s (3.00 x 108 m s-1) / (97.2 x 10-9 m)
= 2.05 x 10-18 J (per photon)
x 6.02 x 1023 mol-1 = 1231870 J (mol photons)-1
= 1232 kJ mol-1
B3. Born Haber cycles were used to make the first reliable prediction of an electron affinity. Calculate
the electron affinity of fluorine (F) given the following data:
Heat of sublimation of potassium:
+90 kJ mol-1
Ionization potential of potassium:
+419 kJ mol-1
Bond dissociation energy of fluorine (F2):
+159 kJ mol-1
Lattice energy of KF(s):
+821 kJ mol-1
Net energy of formation of KF(s) from K(s) and F2(g): -569 kJ mol-1
K(s)  K(g)
K(g)  K+(g) + e½ F2(g)  F(h)
F(g) + e-  F-(g)
K+(g) + F-(g)  KF(s)
+90 kJ mol-1 (vaporization of K(s))
+419 kj mol-1 (ionization of K(g))
+159/2 = +79.5 kJ mol-1 (dissociation of F2(g), per mole of F)
?? (electron affinity of F)
-821 (opposite of lattice energy of KF(s))
Adding the above equations results in the equation for the formation of KF(s) from the elements:
K(s) + ½ F2(g)  KF(s)
-569 kJ mol-1
Thus, the electron affinity of fluorine is –569 – (90 + 419 + 79.5 – 821) = -335.6 kJ mol-1