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Unformatted text preview: © Prep101 Chem110 Practice Exam Solutions
1. E
Solution: ⎛1
1⎞
⎟ ; R = 2.179×10−18 J.
−
∆E = Einitial − Efinal = R H ⎜
H
⎜ n2 n2 ⎟
i⎠
⎝f
⎛1
1⎞
⎟ = 4.58 × 1019 J
−
∆E = (2.179×10−18 J) ⎜
⎜ (2) 2 (5) 2 ⎟
⎝
⎠
E= hc λ , therefore λ = hc (6.626 × 10 −34 Js)(3.0 × 108 m / s )
= 434 nm
=
E
4.58 × 10 −19 J 2. E.
Solution:
Shortest λ = highest energy because Ephoton = hν = hc/λ
Energy levels get closer together as n increases, so the highest energy transition is n =
2 to n = 1.
3. C
Solution: This gives the probability of finding an electron in a region of space.
4. C
Solution:
n = 1, 2, 3,…
l = 0, 1, 2, …n1
ml = l…0…+l
ms = 1/2 or +1/2
Therefore taking into account the above criteria for the quantum numbers set c) is not
possible because l = n, which is not allowed.
5. D
Solution: Titanium is a group 4 metal with 4 valence electrons. The 4s energy level is
lower in energy than the 3d energy level, therefore the 4s will fill first with 2 valence
electrons and the remaining 2 electrons will go in the 3d energy level.
6. A
Solution: Cu2+ is atomic number 29, therefore there are 27 electrons and 9 valence
electrons. The electron configuration is: [Ar] 3d9. The electrons to create the 2+
charge are removed from the 4s orbital first, as it is of lower energy. © Prep101
7. C
Solution: Since electrons in a s orbital has a higher electron density around the
nucleus, they are more effective at screening outer electrons than p or d orbitals. This
is reflected in the Zeff.
8. C
Solution: Look at the electron configuration
a) Fe: [Ar] 4s2 3d6 Fe2+: [Ar] 3d6
b) Fe: [Ar] 4s2 3d6 Fe3+: [Ar] 3d5
c) Sc: [Ar] 4s2 3d1 Sc3+: [Ar]
d) Co: [Ar] 4s2 3d7 Co3+: [Ar] 4s2 3d5
9. C
Solution: Electron affinity becomes more negative up a group and across a period.
Since F is the furthest to the right, it has the highest electron affinity.
10. C
Solution: Atomic radii increases across a period from right to left and down a group
from top to bottom.
11. B
Solution: a) atomic radii increasing, c) is arrange in increasing radii, and d) atomic
radii of transition metals are generally the same size across a period.
12. B
Solution: The more unpaired electrons present, the stronger the paramagnetism.
a)
b)
c)
d)
e) O
O+
FN2P3+ 1s22s2p4
1s22s2p3
1s22s2p6
1s22s2p5
1s22s2p63s2 contains 2 unpaired electrons
contains 3 unpaired electrons
does not contain any unpaired electrons
contains 1 unpaired electron
does not contain any unpaired electrons paramagentic
paramagnetic
diamagnetic
paramagnetic
diamagnetic 13. C
Solution: Ionization energy is the energy required to remove an electron from the
atom. It increases from left to right along a period decreases as go down a group.
14. A
Solution: An ionic bond is formed between the most metallic element with the most
nonmetalllic element. Since Li and F are on opposite ends of the periodic table, they
form an ionic bond. KI is an ionic bond, but the electronegativity is not as high as the
LiF. The other are not ionic bonds. © Prep101
15. C
Solution:
Given Rate = kN Thus, kN
N
3.1 counts/min gram
=
=
= 0.23
13.6 counts/min gram
kN 0
N0 Now we use the integrated firstorder rate law:
⎛N⎞
ln ⎜
⎟ = − kt
⎝ N0 ⎠ and k= 0.693
0.693
=
t1/ 2
5730 years Now we can solve for t, the age of the skeleton
⎛N⎞
⎛ 0.693 ⎞
ln ⎜
⎟ = ln(0.23) = − ⎜
⎟t
⎝ 5730 years ⎠
⎝ N0 ⎠ Solving the equation gives t = 12,000 years (2 Sig. Figs.), the age of the skeleton.
16. C
Solution: Alpha particles results in a –2 decrease in Z, beta particles has a +1 increase
in Z, gamma emission does not increase Z.
17. B
Solution:
Statement i) is true
Statement ii) is false – gamma rays are emitted after alpha and beta particles, but they
are high in energy
Statement iii) is true
Statement iv) is false – positron emission does not affect the atomic mass, it affects
the atomic number
18. D
Solution: Alpha and beta particles are charged, while the gamma particle is not.
Therefore, they get deflected in a magnetic field.
19. E
Solution: The law is defined as the rate of disintegration of a radioactive material –
called activity, or the decay rate, which is directly proportional to the number of
atoms present. None of them have the correct definition. © Prep101
20. D
Solution:
This structure satisfies the rules and each atom has an octet
(except H), which eliminates 1. Calculation of the formal
charges will eliminate 2, 3 and 5.
Formal charge of
O = 6 valence e – 4 lone pair e  ½ (4 bonding e)
= 642 = 0
C = 4 valence e – 0 lone pair e – ½ (8 bonding e)
= 404  0
21. A
Solution: Bond length decreases with multiplicity – single > double > triple
22. D
Solution: The carbons are sp2 hybridized. The 2 N with the H’s are sp2. The N=N is
considered to be sp hybridized.
23. C
Solution: They are all sp3 hybrids, but only FClO3 has a tetrahedral geometry. 24. B
Solution:
Bond order = ½ (# of bonding electrons – number of antibonding electrons)
= ½ (8 – 6)
=1
25. C
Solution: Germanium is in Group 4A, so it has 4 valence electrons. Aluminium is in
Group 3A and has 3 valence electrons. Al can only form 3 bonds with Ge, so there
will be a missing bond. This will be a hole, which corresponds to a ptype
semiconductor. © Prep101
26. B
Solution: All the carbon atoms have =C, which makes them sp2 hybridized.
27. D
Solution: The groups which connect to the central C=C atoms are on opposite sides,
making them a trans configuration. The dipoles associated with the N and O atoms
are cancelled out in this configuration, thus the molecule is nonpolar.
28. C
Solution: There are actually 33σ and 9π bonds. There are 24 σ bonds, but each π
bond is made up of a σ and a π bond. Since there are 9π bonds, there are also 9σ
bonds. Therefore, there is the total number of σ bonds is 24 + 9 = 33.
29. C
Solution: 1) Name the ligands first – 2 en = ethylenediamine = bisethylenediamine;
H2O = aquo, CN = cyano. Arrange them in order: aquocyanobisethylenediamine.
2. Name the metal and oxidation state: cobalt (III)
3. Name the anion: 2 Cl = dichloride
4. Put it together: aquocyanobisethylenediaminecobalt(III) chloride
5. you do NOT need to put dichloride
30. E
Solution: Cations are written before the anion, and the metal atom is written before
the ligands. Therefore, Pt goes first, (en)2, Cl2 and NO3. There is a total of –4
charges, so, Pt has an oxidation state of IV.
31. C
Solution: Cobalt has 6 ligands and Cl contributes –2 charge. Therefore the oxidation
state of Co is (II).
32. E
Solution: A low spin complex has less unpaired electrons, so they are strong field
ligands which are weakly paramagnetic
33. D
Solution: The ligands approach the central atom from the x and y plane, thus the
dx2 – dy2 and dxy orbitals are higher in energy. The dz2 is higher in energy than the
degenerate dxz and dyz orbitals.
34. E
Solution: NH3 is a stronger field ligand than H2O, so it has a larger crystal field
splitting value. Also, NH3 absorbs yellowgreen, which is the complementary color
of purple.
35. F © Prep101
36. D
Solution:
∆ = hν = hc/λ
ν = c/λ
λ = c/ν Since 3.00 x108 m / s
Therefore: λ =
= 6.00 x 107m = 600nm, which corresponds to
14
5 x10 / s
orange. Since is absorbed, the blue colour is observed.
Note: The following wavelengths (nm) are for colours absorbed: 400 = violet, 450 =
blue, 490 = green , 570 = yellowgreen, 580 = yellow, 600 = orange, 650 = red. 37. D
Solution: Greater stability is obtained if the electron is paired in the lower level,
giving it a low spin. A highspin complex has greater stability if the unpaired
electron unpaired goes in the upper level.
38. B
Solution: The most ionic bond will have the most electronegative species and the
least electronegative species. Electronegativity increases as you move across and up
the periodic table. F is the most electronegative and Li, here, is the least E.N.
39. C
Solution: Molecules that are more branched will have less intermolecular attraction.
40. D
(sand)
41. A
Solution: other compound can form intramolecular bonds
O O
N H
O 42. D
Solution: Polarizability describes the tendency for charge separation to occur in a
molecule and dispersion forces becomes more stronger which results in molecule
attracted to each other more strongly. © Prep101
43. D
Solution: Both molecules have dispersion forces, and are similar in size. However,
the polar aldehyde group in acetaldehyde allows for dipoledipole interactions. Since,
dipoledipole interactions are stronger than dispersion forces, acetaldehyde has a
higher boiling point.
44. C
At room temp: I2is a solid, Br2 is a liquid, and the rest (Cl2, F2) are gases
Solution: London dispersion forces (induced dipoledipole) exist in all these
halogen. The greater the size of the halogen, the large its electron cloud density and
the greater the dipole induced. The greater the intermolecular forces within the
halogen, the higher the melting point.
45. E
Solution: The O atom contributes to the dipole interaction, London dispersion forces
are always present.
46. E
Solution: Graphite and diamond occurs naturally in the earth. Nanotubes are
synthesized from decomposition of graphite under high temperatures.
47. D
Solution: They increase from top to bottom (from the lightest atom to the heaviest).
As they descend down a group, the atomic radius increases, as does the number of
electrons, which results in a higher melting point.
48. E
Solution: Alkali metals are the most reactive with water. Cs is the only element in that
group from here.
49. C
Solution: The cement structure is made up of CaCO3 and SiO2 molecules which
combine in the presence of water and CO2 and link up together.
50. A
Solution: B is a nonmetal and is the least metallic in nature. It is dopant in
semiconductors. ...
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This note was uploaded on 09/29/2010 for the course SCIENCE 120 taught by Professor Frenser during the Spring '10 term at McGill.
 Spring '10
 Frenser

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