Chemistry 1000 A and V, Midterm Test #1, October 17, 2008
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1. State the Pauli Exclusion Principle.
No two electrons in the same atoms can have the same four quantum numbers.
2. 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
The electron can behave as a wave. Waves have nodes, i.e. places where the amplitude is zero. Thus the
electron can cross the node if its amplitude as a wave is zero at the node.
3. The Ca+2 ion is isoelectronic with the Ar atom. Which one has the smaller radius and why?
The Ca+2 ion has the smaller radius because it has two more protons than the Ar atom.
4. AUFBAU would predict the electronic configuration of Silver (Ag) to be [Kr]5s24d9. Why is it actually
[Kr]5s14d10?
The filled 4d subshell and the half-filled 5s subshell are collectively at a lower energy than the filled 5s
and 9/10ths filled 4d subshells.
5. What type of hybrid orbitals does the O atom use in a molecule of OCl2? (The O atom is in the centre of
this molecule.)
This molecule has 6 + (2 x 7) = 20 valence electrons. Placing eight of these around eacl Cl atom leaves
two lone pairs to be placed on the O atom. The O atom therefore has four electron groups around it, and
so must be using sp3 hybrid orbitals.
6. Why is a molecule of oxygen paramagnetic?
Because it has two unpaired electrons in its molecular orbitals.
Part B. Answer each question. Best two answers will count. (20 marks each)
B1.
Lines in the Lyman series of the hydrogen spectrum come from transitions to the m = 1 shell from higher
shells.
(a) [8 marks] Calculate the wavelength (in nm) of the highest energy transition in this series.
The Balmer Rydberg equation is 1/ = R [1/m2 - 1/n2], where n is the upper level and m is the lower
level,  is the wavelength of the light emitted in nm, and R is the Rydberg constant (0.01097 nm-1). The
highest energy transition must start at n = and end at m=1. Thus,
1/ = 0.01097 nm-1 [1/12 - 1/2]
= 0.01097 nm-1 [1-0]
= 0.01097 nm-1
Thus,  = 1/0.01097 nm-1 = 91.2 nm
(b) [8 marks] Calculate the energy of the resulting photons in kJ/mol.
E = h= hc/ = 6.63 x 10-34 Js (3.00 x 108 m/s)/(91.2 x 10-9 m) = 2.18 x 10-18 J (per photon)
x 6.02 x 1023 mol-1 = 1.31 x 106 J/mol = 1313 kJ/mol
(c) [4 marks] In what portion of the electromagnetic spectrum are these photons (UV, Visible or IR)?
91.2 nm is in the UV portion of the electromagnetic spectrum.
B2.
Use VSEPR theory to predict the shapes of the following four species.
Wrong name of shape = zero marks! (5 marks each)
(a) XeF4
Valence electrons = 8 + (4 x 7) = 36.
Arranging the four F atoms around the Xe and joining with single bonds uses 8 electrons.
Completing the octets on the F atoms uses another 24, for a total of 32.
This leaves 4 electrons, which are placed as two lone pairs on the Xe.
The structure is therefore AX4E2, which is square planar.
(b) I3¯
Valence electrons = 3 x 7 + 1 = 22.
Arranging two of the I atoms around the third, and joining with single bonds uses 4 electrons.
Completing the octets on the terminal I atoms uses another 12, for a total of 16.
This leaves 6 electrons, which are placed as three lone pairs on the central I atom.
The structure is therefore AX2E3, which is linear.
(c) ClO3¯
Valence electrons = 7 + 3 x 6 + 1 = 26.
Arranging the three O atoms around the Cl atom, and joining with single bonds uses 6 electrons.
Completing the octets on the terminal O atoms uses another 18, for a total of 24.
This leaves 2 electrons, which are placed as a lone pair on the central Cl atom.
The structure is therefore AX3E, which is trigonal pyramidal.
(d) SO2
Valence electrons = 6 + (2 x 6) = 18.
Arranging the two O atoms around theS atom, andnd joining with single bonds uses 4 electrons.
Completing the octets on the terminal O atoms uses another 12, for a total of 16.
This leaves 2 electrons, which are placed as a lone pair on the central S atom.
The structure is therefore AX2E, which is bent.
B3.
[20 marks] Calculate the lattice energy (in kJ mol-1) of the hypothetical compound MgCl(s) by using the
following data:
Energy of sublimation of Mg(s):
First ionization potential of Mg:
Energy of dissociation of Cl2(g):
Electron affinity of Cl:
Energy of formation of MgCl(s) from the elements
+150 kJ mol-1
+738 kJ mol-1
+243 kJ mol-1
-349 kJ mol-1
-15 kJ mol-1
Mg(s) → Mg(g)
+150 kJ/mol
+
Mg(g) → Mg (g)
+738
½ Cl2(g) → Cl
½ (+243)
Cl(g) + e → Cl (g)
-349
+
Mg (g) + Cl (g) → MgCl(s)
-U (where +U is the lattice energy)
____________________________________________________
Mg(s) + ½ Cl2(g) → MgCl(s) -15 kJ/mol
Thus, -U = -15 – (150+738+121.5-349) = -675.5
Thus, the lattice energy = U = +675.5 kJ/mol