Test 3 Review
Information: This test will cover sections 6.5 – 6.9, 7.3 – 7.5, 8.1 – 8.5, 8.8, 9.1 – 9.6,
and 10.1 – 10.6. The test will not be multiple choice. There will be a question with a lot
of blank space for you to fill. You will be given a periodic table and you should bring
your own calculator (non-graphing only).
Information you need to know:
From Test 1 & Test 2:
• Selected prefixes to the metric system and the conversions between them, found in
• Conversion of temperature from Celsius to Kelvin
• Names of the groups, found in Table 2.3
• Diatomic molecules
• Polyatomic ions, given to you in your class guide (Periodic Table, Compounds, Ions
and Naming folder)
• Acids found in your class guide (Periodic Table, Compounds, Ions and Naming
• Avogadro’s Number
• % Yield = actual yield 100
• How to use the solubility rules (Table 4.1)
• Strong acids and strong bases (Table 4.2)
• Acid + Base H O + salt
• Oxidation Number rules found in the text on page 132
moles of solute
Liters of solution
• M 1 =1M V ;2w2ere M = molarity and V = volume
• ∆E = q + w; where ∆E = change in energy (Joules, J), q = heat (J), and w = work (J)
• How to assign signs to ∆E, q, w, ∆H
• ∆H of a reverse reaction has the opposite sign than the forward reaction
• If multiply a reaction by a value (such as 2), multiply ∆H by the same value
• q = mc∆T, where q = heat (J), c = specific heat capacity (J/g*°C), ∆T = change in
temperature = T – T (ºC)
• ∆Hºfof an element = 0, where ∆Hº = efthalpy of formation of a substance
• ∆H rxnΣn ∆Hº profucts – Σn ∆Hº reactanfs; where ∆H º = enthalprxnhange of a
reaction under standard conditions; n = number of moles, and ∆Hº = enthalpy of
formation of a substance
New for Test 3:
• Quantum numbers (n, l, m, anl m ) ans how to assign acceptable values for them • Periodic trends for size, ionization energy, electron affinity, and electronegativity,
including isoelectronic species
• How to draw Lewis symbols for atoms and Lewis Structures for molecules
• Formal charge = # valence electrons – bonds – non-bonding electrons
• How to rank strengths of ionic and covalent bonds and lengths of covalent bonds
• Bond and molecule polarity
• How to assign electron domain geometry, molecular geometry, and bond angles
• Determine hybrid orbitals, and sigma and pi bonds
• STP = Standard Temperature (273.15 K) and Pressure (1.0000 atm)
• PV = nRT, where P = pressure (atm); V = volume (L); n = moles (mol); R = constant
that will be given to yo ÷; T = temperature (K)
P1 1 P2V2
• n T = n T , where P = Pressure (atm); V = volume (L); n = moles (mol); T =
1 1 2 2
dRT g mass g
• μ= , where μ = molar mass ÷ ; d = density = ÷; R = constant
P mol volume L
that will be given to you ÷ ; T = temperature (K); P = Pressure (atm); V =
μP g g
• d= , where d = density ÷; μ = molar mass ÷ ; P = Pressure (atm); R =
RT L mol
constant that will be given to yo ÷; T = temperature (K)
• PT= P 1 P 2 P +3… + P , wxere P = pressure (atm)
P =χ P χ = # moles gas A
• A A T , where P = pressure (atm); and A total # moles gas(unitless)
1. What are the correct 4 quantum numbers for an outermost electron in Ca?
n = 4, l = 0, l = 0, s = + ½
n = 4 since the outermost shell of Ca is in period 4
l = 0 since the outermost electrons are in the s area of the periodic table
m l 0 since there is only one orientation of the s orbital
m s + ½; this could also equal – ½, both answers would be correct 2. Which of the following sets of quantum numbers cannot exist, and why?
a) n = 4, l = 4, m =l-3, m = s ½ b) n = 5, l = 2, m l +2, m = s ½
c) n = 2, l = 1, m = 0, m = + ½ d) n = 1, l = 0, m = 0, m = + ½
l s l s
Answer A cannot exist because the l value is too high. The l value should be a whole
number somewhere between 0 and n-1. The l value does not meet this criterion.
3. State which of the following corresponds to a shell, subshell, orientation, or electron
spin. a) l = 2, b) m s + ½, c) n = 2, d) m = -2l
a) corresponds to a subshell
b) corresponds to an electron spin
c) corresponds to a shell
d) corresponds to the orientation of an orbital
4. Which of the following atoms is largest and why?
a) Ca b) K c) Ar d) Cl e) S
Both Ca and K have an n level = 4, while Ar, Cl, and S have an n level = 3. Since Ca and
K have the higher n levels, one of those two must be the largest. Next, we look at
number of protons. Since Ca has a higher number of protons (20 versus 19 for K), the
protons will hold the electrons tighter, and thus Ca will be smaller than K. So K is the
5. Which of the following atoms/ions is largest and why?
2- - + 2+
a) O b) F c) Na d) Mg
This is an isoelectronic series (they all have the same number of electrons). Since the
series is isoelectronic, positive ions are smaller than neutral atoms, and neutral atoms are
smaller than negative ions. So, O is the biggest, and Mg is the smallest.
6. Define ionization energy and state which of the following has the highest ionization
a) B b) C c) N d) O e) F
Ionization energy is the energy required to remove an electron from an atom. The
smaller an atom, the harder it is to remove an electron because the attraction between the
positive nucleus and the negative electrons is strong. Since F is the smallest of the atoms,
it should have the highest ionization energy – it will be most difficult to remove an
electron from F. 7. Define electron affinity and state which of the following has the largest electron
a) Al b) Si c) P d) S e) Cl
Electron affinity is the ability of an atom to attract an electron. This is also based on size.
The smaller an atom, the easier it is to attract an electron because there is a high attraction
between the positive nucleus and the negative electron. Cl is the smallest of these atoms,
and it should have the highest electron affinity – it will be easiest to give an electron to
8. Which of the following would have the highest second ionization energy?
a) Na b) Mg c) Al d) Si e) P
The second ionization energy is the amount of energy required to remove another
electron from a positive ion. These energies are usually quite high since the attraction
between positive and negative is strong in cations. However, if removing a second
electron will ruin a noble gas configuration, the energy for this is very high, since full
orbitals (and half-full orbitals) are particularly stable. So Na will have the highest second
ionization energy since removing a second electron will destroy a full orbital.
9. Draw the Lewis symbol for P.
P has 5 valence electrons.
10. Draw the Lewis symbol for S . 2-
S has 8 valence electrons.
11. Draw the Lewis structure for CCl .
Valence electrons = 1(C) + 4 (C