UNIT #2: Nomenclature and Chemical Bonding Review Answers
Part A: Bonding
1. a) What is a chemical bond?
A bond is an attraction between two atoms.
b) Draw the bonding continuum and identify the range of values corresponding to the various types of
bonds found in compounds.
0 COVALENT 0.5
POLAR COVALENT
1.7
IONIC
c) Explain the difference between each type of chemical bond and give an example of each.
Generally, covalent bonds involve the sharing of e- and ionic bonds involve attraction between
oppositely charged ions. “Pure “ covalent bonds form between two atoms with the same
electronegativity (e.g. H-H). These have zero dipole (charge imbalance). Bonds with ∆EN values 0 0.5 have very small dipoles and are generally called non-polar due to their low dipoles. Bonds
between 0.5-1.7 are polar covalent, meaning the sharing of electrons is unequal and there are positive
and negative bond dipoles. Bonds with ∆EN values > 1.7 are considered ionic. In this case, electrons
are usually considered to be lost or gained completely, forming negative and positive ions.
2. Determine the type of bond present in each substance using electronegativity values.
CBr4
∆EN= 2.8-2.5 = 0.3
CO2
∆EN= 3.5-2.5 = 1.0
BaF2
Bond Type: POLAR
COVALENT
∆EN= 4.0 – 0.9 = 3.1
Bond Type: COVALENT
H2S
∆EN= 2.1-1,8 = 0.3
Bond Type: COVALENT
Bond Type: IONIC
2. Use the ionization equation method to predict the formula of aluminum bromide. This
involve adding two equations (loss and gain of e-) to make a third.
Al  Al3+ + 3e3 Br + 3e-  3BrAl + 3Br  AlBr3
Part B: Nomenclature
1. a) What is the valence of the metal in VO4? Name the compound using the IUPAC system.
+8; vanadium (VIII) oxide
b) What is the valence of phosphorus in the compound PCl5? Name the compound using the Greek
prefix system.
+5; phosphorus pentachloride
c) What is the valence of carbon in the compound C2H4? Name the compound using IUPAC.
-2; ethene  unusual to have negative valence as first element
2. Correct the following mistakes: I nthe chemical formulae:
a) lead (IV) dihydrogen phosphite = Pb4H2PO3
b) dinitrogen tetroxide N4O2
Pb(H2PO3)4
N2O4
c) calcium sulfate = Ca(SO4)2
d) ammonium hydroxide = NH3HO
CaSO4
NH4OH
3. Complete the table.
Compound
Name of Compound
Formula
AsCl3 (Greek)
arsenic trichloride
barium sulfite
Pb(NO3)2∙ 4H2O
Ferric oxide
Fe2O3
sodium peroxide
Na2O2
Name of Compound
lead (II) nitrate tetrahydrate
Compound Formula
BaS
HNO2 (aq)
nitrous acid
Sn(ClO2)2
tin (II) chlorite
iron (III) iodide
Zn(OH) 2
zinc hydroxide
tin(II) hydrogen carbonate
Sn(HCO3)2
potassium sulfide
copper (II) dihydrogen
phosphate pentahydrate
Cu(H2PO4)25H2O
K2S
FeI3
HI (aq)
hydroiodic acid
nitrogen monoxide
H3PO4(aq)
phosphoric acid
carbonic acid
Be(HCO3) 2
beryllium hydrogen
carbonate
ammonium phosphate
NaClO4
sodium perchlorate
fluorine gas
NO
H2CO3
(NH4)3PO4
F2
CuNO2 (ous/ic)
cuprous nitrite
hydrobromic acid
HBr
CaO
calcium oxide
magnesium nitride
Mg3N2
1. Name the following hydrocarbons.
a)
CH3
CH3 CH CH2 CH2 CH3
2-methylpentane
b)
CH3
CH3 C CH2 CH CH2 CH2 CH2 CH3
CH3
c) 1-ethyl-3-methylcyclopentane
CH2 CH3
4-ethyl-2,2-dimethyloctane
d) 2-bromo-3-methyl-2-pentene
Br CH3
H3C
CH3 C C CH2
CH2 CH3
CH3
2. Draw the complete or condensed structural formula for the following hydrocarbons.
a) methylcyclobutane
b) 2-methyl-1-butene
c) 2,2-dimethylpropane
CH3
H2C
C
CH2
CH3
CH3
H3C
CH3
C
CH3
CH3
3. Are any of these substances isomers? Explain why or why not. [A 2]
a) and b) are isomers since they have the same formula (C5H10) but unique shapes. C) is not
an isomer of the others sicne its formula is C5H12.
Part D: Lewis Structures and Shapes of Molecules
1. VESPR theory uses two factors related to the central atom(s) of a molecule to predict the 3dimensional shape of a molecule. What are these 2 factors?
1. Number of atoms bonded to a central atom.
2. Number of lone pairs on the central atom.
2.
Draw the Lewis Structure, 3-D structure and name the shape for the following compounds.
Be sure to include all bonds and lone pair electrons.
Compound
Lewis Structure
3-D Structure of the
Compound
Shape Name for the
3-D Structure
Trigonal pyramidal
(H’s should be Cl’s)
AsCl3
Linear
C2H2
Trigonal planar
BCl3
Tetrahedral
CF2H2
H’s should be F!
SF2
Bent or V-shape
1. Draw the Lewis structure of the following compounds or polyatomic ions.
a) SF6
b) XeF2
b) hydroxide ion
c) ammonium ion
Part E: Short Answer
1.
Why is it incorrect to call sodium chloride a molecule? What is a better name for this type of
substance.
Molecules are bonded by covalent or polar covalent bonds (usually non-metals). NaCl is an
ionic compound and consists of a continuous crystal lattice of Na+ and Cl- ions.
2.
Explain the difference between intramolecular bonds and intermolecular bonds. Include a
labeled diagram of a water molecule in your answer.
Intramolecular bonds are bonds within molecules (e.g. covalent or polar covalent).
Intermolecular bonds are weak bonds between molecules. When a molecular compound
melts or boils, only intermolecular attractions break.
3.
Why are small molecular substances (e.g. chlorine, ammonia) gases at room temperature while
small ionic compounds (e.g. sodium chloride) solids?
As discussed above, non-polar molecules have weak intermolecular attractions that can be
easily broken at room temperature. They are therefore gases at 20oC. ionic compounds must
break ionic bonds to melt or boil. Since these are stronger than intermolecular bonds, these
substances remain as solids.
4.
Methane and methanol are both small molecular compounds. However they have very
different physical properties. For example, methane boils (liquid  gas) at -182oC methanol
(CH3OH) boils at 65oC.
a) Draw a Lewis structure for each molecule. Be sure to include unbonded pairs (lone pairs).
b) Draw and name the shape of each molecule. .
tetrahedral
tetrahedral & V-shape
c) Calculate the electronegativity differences. If any bond dipoles are present, indicate them on
your shape diagrams.
d) Explain the difference in boiling point between these 2 small molecular compounds.
There is a negative dipole on the oxygen of methanol and a positive dipole on the oxygenbonded hydrogen. This allows for stronger dipole-dipole intermolecular bonding, which
creates stronger attractions. This makes it more difficult to break the intermolecular bonds on
methanol.
5. A chemist tested three white powders. The following results were obtained:
1.xylitol (polar cov)
2.cesium chloride (ionic)
3. decane (non-polar)
Melting Point (oC)
6.5
645
-30
Solubility in Water
Soluble
Soluble
Insoluble
Conductivity in Water
No
Yes
N/A
Match the properties with the substances: cesium chloride (CsCl); decane (C10H22); xylitol (C5H12O5).