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CHEM 1000 A, V Midterm Test #1 Answers
October 14, 2005
Part A. Answer each question with a few sentences or equations where necessary. (5 Marks each)
1.
Referring to the periodic table on the last page of this test, are the following elements s-block, p-block,
d-block or f-block elements?
(a) Cs __s________ (b) Cd__d_______
2.
(c) U___f________ (d) I___p________ (e) H___s_______
The standard for atomic weights is an atom of 12C, which has a mass of exactly 12.000 amu. Why is the
atomic weight of carbon 12.011 g/mol?
The atomic weight is a weighted average of the atomic weights of all of the isotopes of an element.
Carbon has several isotopes heavier than 12C, so the average atomic weight is greater than 12.000 g
mol-1.
3.
The valence electrons in a Sr atom are 5s2. What are the four quantum numbers for each of these two
electrons?
Electron 1: n = __5___ l = ___0__ ml = __0___ ms = __+ ½ ___
Electron 2: n = __5___ l = ___0__ ml = __0___ ms = __ ½ ___
4.
Explain why the ionization energy increases going from left to right across a period.
Going left to right across a period, electrons are being added to the same principal shell, and so have
roughly the same energy as each other. However, protons are being added to the nucleus, increasing the
effective nuclear charge, pulling the electrons closer and reducing the radius.
5.
What is Hess’ Law?
Hess’ law states that the energy of a reaction is equal to the sum of the energies of the steps of that
reaction.
6.
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.
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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.
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) Referring to the structure of ethylene, C2H4, shown below, answer the following questions:
(i)
A is a _sigma_ bond.
It is formed by the overlap of an __sp2___ orbital of one carbon with an _sp2_ orbital of the
other.
(ii)
B is a _pi_ bond.
It is formed by the overlap of a _p_ orbital of one carbon with a __p__orbital of the other carbon.
(iii) C is a _sigma__ bond
It is formed by the overlap of an __sp2_ orbital of the carbon with a ___1s__ orbital of the
hydrogen.
(iv)
This molecule has a single / double / triple bond between the carbon atoms (circle one)
C
B
A
B2.
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(a) The Lyman lines in the spectrum of hydrogen terminate at m=1. Calculate the wavelength (in nm) of
the lowest energy line in the Lyman series. Also calculate the frequency of this radiation (s-1).
(10 marks)
1
1
 1
R 2  2

m n 
Here, m=1 since the line is in the Lyman series. The lowest energy transition will come from n=2, i.e.
the next energy level up. Thus,
1
1 1 
 0.01097 nm 1  2  2 

1 2 
 0.01097 nm 1 (0.75)
 0.0082275 nm 1
thus,  

1
 121.5 nm
0.0082275 nm 1
c 3.00 108 ms 1

 2.47 1015 s 1
 121.5 109 m
(b) NO2(g) is an important species in the formation of smog in large cities. The N-O bond in this
molecule can be broken by light having a wavelength less than 429 nm. Calculate the N-O bond
dissociation energy (in kJ mol-1). (10 marks)
Eh
c

 3.00 108 ms 1 
 6.63 1034 J s 

9
 429 10 m 
 4.64 1019 J (per photon)
 6.02 1023 photon mol 1  279109 J mol 1
 279 kJ mol1
B3. Use the following information to calculate the electron affinity of fluorine (F):
1.
2.
3.
4.
5.
The energy change upon combining Li(s) with F2(g) is 617 kJ mol-1
The bond dissociation energy of F2(g) is +79.5 kJ mol-1
The heat of sublimation of Li(s) is +161 kJ mol-1
The ionization potential of Li is +520 kJ mol-1
The lattice energy of LiF(s) is +1050 kJ mol-1
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Li(s) → Li(g)
+161 kJ mol
Li(g) → Li+(g) + e¯
+520 kJ mol-1
½ F2(g) → F(g)
½ (+79.5 kJ mol-1)
F(g) + e¯ → F¯(g)
??
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Li+(g) + F¯(g) → LiF(s)
(1050 kJ mol-1)
_________________________________________
Li(s) + ½ F2(g) → LiF(s)
617 kJ mol-1
Thus, the electron affinity of F is found by difference:
617  (+161 + 520 + ½ (79.5) – 1050) = 288 kJ mol-1