SEMESTER 1
EXAM REVIEW ANSWER KEY
Unit I – Chemical Foundations
60
10
II. 1) adeh 2) a) 1 b) 3 c) 2 d) 5 3) 1.231 x 10 4) 0.00109 5) 0.000687kg 6) 8.010 x 10 um
7) 401kJ 8) $1.59/lb 9) 6.71 x 10 8 miles/hr 10) 9.5 g/mL 11) 5.4 g/cm3
Unit II – Atomic Theory & The Periodic Table
1) According to the equation E = hv, violet light has greater energy than red light, therefore the violet
photons have sufficient energy to expel the electrons from the metal, whereas the red ones do not.
More intense light is emitting more photons per second, so more electrons per second are expelled,
but they do not contain any more energy.
2) Generally, the mass is the most important term in the equation, as the mass increases, the measurable
wavelength of the object decreases (they are inversely proportional). Another way to see it is that
more massive objects have less measurable wavelengths (they are more “matter-like”), whereas
less massive objects have more measurable wavelengths (they are more “energy-like”).
3) Bohr saw the electrons in an atom occurring in specifically defined energy levels, as defined by
Planck’s Quantum Theory. As an electron passes from a more excited state (higher energy level)
to a ground state, it releases a photon of light. The energy of that photon corresponds to a certain
wavelength of light, which has a certain characteristic color.
4) Fluorine has seven valence electrons. Adding one more electron will give it a complete valence shell
(stable octet) which is a very favorable arrangement (lower in Energy). When one more electron
is added, the atom will give off energy.
5) Neon has eight valence electrons which is a very favorable arrangement. Although it is possible to add
another electron, because the electron would need to be placed in the next higher energy level, an
input of energy would be required.
6) As you go Li  Na  K, the atomic radius increases. As you go down a column on the table,
there are more energy levels containing electrons which are a greater distance from the nucleus,
therefore increasing the effective size of the atom.
7) As you go Na  Mg  Al, the atomic radius decreases. As you travel across a row on the table,
more protons are gained by the atom without any additional electron shielding. As a result, there
is a greater force of attraction between the nucleus and the valence electrons, therefore the valence
shell is drawn in closer.
8) As you go Na  Mg  Al, the 1st ionization energy increases. As you travel across a row on the
table, more protons are gained by the atom without any additional electron shielding. As a result,
there is a greater force of attraction between the nucleus and the valence electrons, therefore it
requires more energy to remove the first valence electron.
9) Once the first valence electron is removed, a sodium ion now has a stable octet, therefore making it
very unfavorable to remove a second electron. Magnesium has two valence electrons, so its
first two ionization energies are comparably small.
10) (4,3,x,x) ; 14 possibilities, (4,3,2, ½), (4,3,-1, ½); 32 electrons
11)
Potassium
Magnesium
Cadmium
Dysprosium
alkalai
metals
alkaline
earth metals
transition
metals
rare earths
(lanthanides)
1
metal
[Ar] 4s1
paramagnetic
2
metal
[Ne] 3s2
diamagnetic
XX
metal
[Kr] 5s24d10
diamagnetic
XX
metal
[Xe]6s24f10
paramagnetic
Chlorine
halogens
7
nonmetal
[Ne]3s23p5
paramagnetic
Neon
noble gases
8
nonmetal
[He] 2s22p6
diamagnetic
SEMESTER 1
EXAM REVIEW ANSWER KEY
12) a 13) c, the absolute value of mℓ must be  n 14) Pauli Excluion Principle 15) Hund’s Rule
16) a) sulfide b) sodium atoms c) K d) S e) Na + f) Cl-1
17) When a fluoride atom gains an electron, the number of protons in the nucleus remains the same, so the
effective nuclear charge remains the same, but that charge is distributed over a larger number of
electrons, making the attractive force less. As a result, the electrons move further away from the
nucleus, making the ion effectively larger.
18) a) 54 protons, 77 neutrons, 54 electrons b) 26 protons, 30 neutrons, 23 electrons
c) 53 protons, 74 neutrons, 54 electrons
Unit III – Nomenclature & Chemical Equations
1) a) H2SO4 b) Ca(OH)2 c) XeF6 d) FeCr2O7 e) HCN f) Rb2O g) Na3PO3 h) Sn3(PO4)4
i) CO j) BaSO4 . 8 H2O k) ammonia l) nickel(II) sulfide m) ammonium carbonate
n) aluminum chloride o) chromium(III) hydroxide p) carbon tetrachloride
q) iron(III) oxalate r) acetic acid s) sodium hydride t) magnesium fluoride
2) a) Ni + Cu+2  Cu + Ni+2 b) NaH + H2O  Na+ + OH- + H2 c) C3H6O + O2  CO2 + H2O
d) Pb+2 + SO4-2  PbSO4 e) NH3 + HC2H3O2  NH4+ + C2H3O2- f) SO3 + H2O  H+ + HSO4g) CN- + Ni+2  Ni(CN)4-2 (any number on CN up to 6 is OK) h) NaC2H3O2 + H+  HC2H3O2 + Na+
Unit IV – Stoichiometry
II. 1) a) 0.0193 mol HF b) 5.02 mol HF c) 0.0300 mol HF d) 0.0610 mol HF
2) a) 4,1,1,2 b) 8.65g SiF4 c) 6.50g SiF4 , 1.24g SiO2 left over d) 91.1%
3) 35.7% Ba 4) C2H6 5 ) 0.0712g Fe(OH)3 6) 6.25 x 10-5M 7) 1.17M 8) 0.0602M
Unit V – Bonding
1) NaCl. The ionic radius is smaller, therefore, according to Coulomb’s Law, the ions will exhibit a
greater attractive force for each other.
2) shortest  longest: CO2, CO3-2, CH3OH. In CO2, both C – O bonds are double bonds, which are
shorter than single bonds. In CO3-2, the resonance structure makes all three C – O bonds
effectively 1 1/3 bonds, which will be longer than double bonds, and shorter than single bonds.
In CH3OH, the C – O bond is a single bond, which is the longest of the C – O bonds.
Cl
P
O
N
O
Cl
O C
Cl
O
O
5) a)
b)
sp3, trigonal pyramid, P, 109.5o, 3σ sp, linear, NP, 180o, 2σ, 2π
c)
sp2, trigonal planar, P, 120o, 3σ, 1π
F
F
F
F
Br
F
F
F
P
F
d)
sp3d2, square pyramid, P, 90o, 5σ
F
F
e)
sp3d, trigonal bipyramid, NP, 90o/120o, 5σ
6) C2H2. This molecule has a triple C-C bond which will be much stronger than the double bond in C2H4
and the single bond in C2H6. Multiple bonds are always stronger than single bonds, therefore they
require more energy to break.
7) a) 2,4-dimethyl pentane b) 3-bromohexane c) 1-amino, 3,5-dibromobenzene
SEMESTER 1
EXAM REVIEW ANSWER KEY
Unit VI – Phases & Gas Laws
II. 1) e 2) b 3) c 4) d 5) B 6) D 7) A 8) C 9) a
III. 1) both are nonpolar, therefore both exhibit London Dispersion Forces, but because the formula mass
of argon is greater, it has more electrons, therefore the dispersion forces are stronger, making it
more difficult to separate the atoms, therefore causing the boiling point to be greater.
2) 1208 mmHg OR 1.59 atm 3) a. all the same b. C 4) 34.1 mL 5) 47.6 g/mol 6) 37.9 mL
7) 1.36 g/L 8) 92 g/mol, N2O4