Practice Exam for Exam 2
1. Photodissociation of water
H2O + hν → H2 + ½ O2
has been suggested as a source of hydrogen. The ∆Hrxn° for the reaction, calculated from
thermochemical data, is +285.8 kJ/mol water. Calculate the wavelength, in nm, that would be
required to effect this reaction.
( 6.626 × 10
hc
= =
? nm
E
1 molecule H2O ×
−34
)(
J s 2.998 × 10 8 m s−1
)
3
1 mole H2O
285.8 kJ 10 J
×
×
23
6.022 × 10 molecules H2O 1 mole H2O 1 kJ
×
1 nm
10 −9 m
= 418.6 nm
2. A compound of tin and chlorine is a colorless liquid. The vapor has a density of 7.49 g/L at 151C and
1.00 atm. What is the molecular weight of the compound? Why do you think the compound is
molecular and not ionic? Write the Lewis formula for the compound.
dRT
M =
=
P
( 7.49 ) ( 0.082058
g
L
L atm
mol K
1.00 atm
K)
)( 424=
260.59 g mol −1
There are a couple of reasons to suspect that the compound is molecular and not ionic:
•
•
The compound is a liquid at room temperature. Ionic compounds tend to be solids with relatively
high melting points.
The electronegativity difference between Sn and Cl is only 1.20 which is below the threshold of
1.7 for a compound to be considered ionic.
With a molar mass of 261 g and assuming there is only 1 Sn atom there must be 4 Cl atoms
142
261
=
− 119 142 = 4
35.5
Cl
Cl
So the formula is SnCl4 which has the Lewis structure
Sn
Cl
Cl
3. For the following molecules (a) draw the Lewis Structure, (b) give the electron pair geometry, (c) give
the molecular geometry, (d) state whether or not it is polar, and (e) give the hybridization on the
central atom.
I3-
b.
trigonal bipyramidal
-
I
Cl
e.
sp2
b.
trigonal bipyramidal
Cl
Cl
d.
polar
Cl
e.
sp3d
SbCl5
Sb
c.
trigonal planar
C
d.
non-polar
I
b.
trigonal planar
O
c.
linear
I
Cl
COCl2
XeO2F2
F
c.
trigonal bipyramidal
Cl
O
d.
non-polar
Cl
b.
trigonal bipyramidal
Xe
F
e.
sp3d
c.
see-saw
O
d.
polar
e.
sp3d
4. Calculate the wavelength, in nm, of light emitted when an electron drops from the 5th energy level to
the 2nd energy level in a He+ ion.
1 Z 2 RH
=
λ
hc
 1
1
 2 − 2
n

 f ni 
( 2 ) ( 2.180 × 10 −18 J)
2
( 6.626 × 10
−34
)(
8
J s 2.998 × 10 m s
−1
)
 1
1
 2 − 2
5 
2
= 9.218339 × 106 m−1
λ= 1.084794 × 10 −7 m ×
1 nm
= 108.5 nm
10 −9 m
5. Estimate the enthalpy change for the combustion of ethane, C2H6, using bond energies. The Lewis
H
structure of ethane is:
H
H
C
C
H
H
H
2 C2H6 + 7 O2  4 CO2 + 6 H2O
∆H ≈ BE ( broken ) − BE ( formed)
= 2 ( C − C ) + 6 ( C − H ) + 7 ( O = O )  − 8 ( C = O ) + 12 ( O − H ) 
= 2 ( 346 kJ) + 6 ( 414 kJ) + 7 ( 498 kJ)  − 8 ( 799 kJ) + 12 ( 463 kJ ) 
= −5286 kJ
6. Give the molecular orbital diagram for N2, N2+, and N2-. State which is more stable and why you
determined it to be more stable.

2p*

2p*

2p

2p

2p*

2p*

2p

2p

2p*

2p*

2p

2p

2s*

2s

2s*

2s

2s*

2s
N2
N2+
N2-
B.O. = 3
B.O. = 2.5
B.O. = 2.5
B.O = ½ (bonding e - antibonding e-)
Because N2 has the highest bond order, it should be the most stable.