Camosun College
Chemistry 230
Assignment # 1
Review and Preview
1. For each of the following compounds, tell whether its bonds are ionic, covalent, or of both kind:
MgF2 ,
NaOCH3 ,
CH3Na ,
BrCN ,
OF2
2. For each of the following compounds, draw the Lewis structures, and predict which ones will be
polar:
CH3COCH3
(CH3)4C
CH2=CH2
NF3
(CH3)2CHCHO
CH3CH=CHCH3
AlCl3
CH2=CHBr
3. Draw Lewis structures for the following species including all resonance structures. Assign formal
charges where relevant:
CH3SCH3
ClO3-
SO42-
CH3CHClCH3
CH3CO2H
OH-
H2O2
CH2NO2-
CH2CHNO2
4. Explain what orbitals are involved in the formation of covalent bonds in:
a.
CH3CH2CH3
b.
CH3C≡CH
c.
CH2=CHCH3
d.
(CH3)3C+
5. What is the hybridization of the indicated atom in each of the following compounds?
O
CH3C
CH 3CH2OH
O
C
N
H 3C
S
CH3
H 3C
CH 3
6. Write a condensed structural formula for each compound given below:
O
O
Cl
OH
OH
7. Write resonance structures for the azide ion, N3- . Explain how these resonance structures account for
the fact that both bonds of the azide ion have the same length (1.15 Å).
8. Based on your understanding of bonding theories, explain the following facts:
a. NF3 has a smaller dipole moment than does NH3
b. Carbon monoxide has a much smaller dipole moment than expected.
c. The carbon-carbon single bond in propyne is shorter than the carbon-carbon single bond in
propane.
1.46 Å
H3C
C
CH
1.53 Å
H 3C CH 2 CH 3
9. Using hybridized orbitals, show how σ- and π-bonds are formed in the following structures:
H
a.
H
C
H
C
H
b.
H
O
C
H
10. Explain what is meant by the terms: bonding molecular orbital and antibonding molecular orbital.
Show how two p orbitals combine to form two π molecular orbitals.
Chemistry 230-Answers to Assignment #1
1. MgF2
NaOCH3
CH3Na
BrCN
OF2
ionic
both. There are covalent bonds in the anion CH3Oboth. There are covalent bonds in the anion CH3covalent
(Br-C≡N)
covalent
2. CH3COCH3
there are 24 valence electrons. Of these, 20 are used in 10 bonding pairs and 4 are
used in two nonbonding pairs. Molecule is polar
O
H
CH2=CH2
C
C
H H
H
H
molecule is nonpolar
H
H
C
C
H
NF3
H
C
H
molecule is polar
N
F
F
F
CH3CH=CHCH3
two possible structures (cis- and tran- stereoisomers. The trans-isomer is nonploar,
whereas the cis-isomer is relatively nonploar)
H
H
C
C
H 3C
AlCl3
H 3C
H
C
CH 3
C
H
CH 3
molecule is nonploar
Cl
Al
Cl
(CH3)4C
Cl
molecule is nonploar
H 3C
H 3C
C
CH3
H3C
(CH3)2CHCHO
molecule is polar
H3C
H
H3C
CH2=CHBr
O
C
C
H
molecule is polar
H
Br
C
C
H
H
3. CH3SCH3
S
H 3C
CH3
CH3CHClCH3
H
H 3C
SO4
Cl
H 3C
2-
O
O
C
S
O
O
O
O
S
O
O
O
O
S
O
O
O
O
O
S
O
O
O
S
2+
O
O
OHO
H
H2O2
H
O
O
ClO3
H
-
O
O
O
O
Cl
O
O
O
Cl
O
O
Cl
O
O
2+
Cl
O
(What is the difference between the last resonance structure and the first three structures of ClO3-?)
CH3CO2H
O
H
O
C
H
C
H
C
O
H
C
H
O
H
H
H
CH2CHNO2
O
H
O
N
C
O
H
N
C
H
C
H
O
O
H
C
H
N
C
H
C
H
H
H
O
CH2NO2H
O
H
O
N
H
4. a.
N
O
H
N
O
H
O
CH3CH2CH3
H
H
H
H
C
C
C
H
H
H
sp 3-s bond
H
sp 3-sp 3 bond
b.
CH3C≡CH (the two π-bonds are the result of p-p orbitals overlap)
sp 3-s bond
H
H
C
C
C
sp-s bond
H
H
sp-sp bond
sp 3-sp bond
c.
CH2=CHCH3 (the π-bond is the result of p-p orbitals overlap)
H
sp2-s bond
H
C
C
H
sp2-sp2 bond
d.
sp2-sp3 bond
H
sp3-s bond
C
H
H
(CH3)3C+
sp2-sp3 bond
H
H 3C
C
H 3C
C
H
sp3-s bond
H
O
5. The simplest way to predict the hybridization of a particular atom is to think of the electron pair
geometry about that atom.
sp3 hybridization as the electron pair geometry around the oxygen is tetrahedral
CH 3CH2OH
O
CH 3CH2
CH3C
H
sp hybridization as the electron pair geometry around the nitrogen is linear
N
O
C
H 3C
CH3
sp2 hybridization as the electron pair geometry around the oxygen is trigonal
planar
O
sp3 hybridization as the electron pair geometry around the sulf ur is tetrahedral
S
H 3C
CH 3
6.
(CH3)2CHCH2OH
OH
O
(CH3)2CHCOCH(CH3)2
OH
O
(CH3)2CHCH2CH2OH
Cl
CH3CHClCH2CH2COCH3
7. The azide ion has the chemical formula N3- (a total of 16 electrons are available for the Lewis dot
structure of the ion):
2-
N
N
N
2-
N
N
N
N
N
N
None of the resonance structures drawn above can represent the azide ion structure. The actual
structure is a “hybrid” of all of them. The two bonds between the nitrogens are indistinguishable and
are of the same length. They are neither single, double, nor triple, but somewhere in between (1.15 Å).
Keep in mind that the nitrogen-nitrogen triple bond is normally 1.10 Å and the nitrogen-nitrogen
double bond is normally about 1.20 Å.
8. a. The electronegativities of the atoms are: N = 3.0, H = 2.2, and F = 4.0. Thus, the difference in the
electronegativities of the atoms in an N-H bond is approximately the same as the difference in the
electronegativities of the atoms in an N-F bond.
F
H
N
F
F
H
N
H
However, the dipole moment of a molecule is dependent on the vector sum of all bonds in the
molecule. NF3 has a smaller dipole moment than NH3 because in NF3 all the bond dipoles point in
different directions out from the nitrogen, whereas in NH3 all the bond dipoles point toward one end of
the molecule.
b. There are 3 possible structures for CO, only the first one is an acceptable one since in the other two
structures the octet rule is not obeyed.
C
O
C
C
O
O
Since oxygen is more electronegative than carbon, we would expect polarization of the molecule
towards the O atom. However, when formal charges are assigned, it becomes obvious that there is a
counter effect to that polarization.
C
C
O
O
The C atom is assigned a formal negative charge and the O atom is assigned a formal positive charge.
Therefore, the polarization towards the O atom is going to be reduced (that is, the end where the O
atom is may not be as electronegative as one would normally expect) and the molecule will have a
much smaller dipole than expected.
c. In order to explain the bond length, it is a good idea to figure out the hybridization schemes of the
carbon atoms involved.
H3C
H 3C CH2 CH 3
C
C
H
sp 3-sp bond
sp 3-sp 3 bond
You can immediately recognize that origins of the two bonds are different. One is an sp-sp3 bond while
the other is an sp3-sp3 bond. There is a higher s-character in the carbon-carbon single covalent bond in
propyne. Since the 2s orbital is closer to the nucleus than the 2p orbital, we would expect a bond with a
higher s-character to be a shorter bond (electrons are held more tightly by the nucleus). Therefore, the
carbon-carbon single covalent bond in propyne is shorter than the carbon-carbon single covalent bond
in propane.
9. a. CH2=CH2 molecule:
The carbon atoms are sp2 hybridized. The σ-bond between the two carbons is an sp2-sp2 bond whereas
the σ-bonds between the carbons and the hydrogens are sp2-s bonds.
H
H
H
C
C
H
C
C
H
H
H
-bond skeleton of ethene
H
The π-bond is formed by an overlap between the remaining p orbitals (one on each carbon)
H
H
....
-bond
H
H
C
H
H
H
H
C
H
C
H
C
H
H
b. CH2O molecule:
Both the carbon atom and the oxygen atom are sp2 hybridized. The σ-bond between the oxygen and the
carbon atoms is sp2-sp2 bond. The bonds between the carbon and the hydrogens are sp2-s bonds.
H
H
C
O
C
O
H
H
-bond skeleton of methanal
The π-bond is formed by an overlap between the remaining p orbital on the oxygen atom and the
remaining p orbital on the carbon atom.
-bond
H
....
H
H
C
H
O
H
C
O
H
10. Bonding Molecular Orbital (BMO): is an orbital that results when two atomic orbitals with the
same sign interact (the two atomic orbitals are said to be in phase). Electrons in a bonding orbital
increase bond strength.
Anti-Bonding Molecular Orbital (ABMO): a molecular orbital that results when two atomic orbitals
with the opposite sign interact (the two are said to be out of phase). Electrons in an anti-bonding orbital
decrease bond strength.
interf erence
*-antibonding molecular orbital
(higher in energy)
re-inf orcement
-bonding molecular orbital
(lower in energy)