Name
CHM 115 EXAM #3
Practice KEY
Complete the following showing all of your work. When calculations are required, show the set-up, units,
and report answers with the correct number of significant figures. Short answers should be composed of
well-organized thoughts and written in full sentences.
(18 points for completion of BOTH parts (a) and (b))
1.
(a) In the spaces provided, do the following BrF5
Draw the BEST Lewis structure of this
interhalogen molecule and show formal
charge calculations for each type of atom
FC (X) = valence – bonds – lone pair eFC (Br) = 7 – 5 – 2 = 0
All F: FC (F) = 7 – 1 – 6 = 0
Give the names of (a) the electron pair
geometry and (b) the molecular shape,
(a) otahedral
(b) square based pyramid (or square pyramid)
Specify the hybridization of the Br AND
draw the energy diagram for the hybridized orbital
(you know - the boxes with electrons in them).
↑↓
↑↓
↑↓
↑↓
↑↓
↑↓
l.p.
σ
σ
σ
σ
σ
hybridization is sp3d2
Fluorine electrons are shown in red, l.p.
stands for lone pair, and σ is a sigma
bond.
Is this species polar or nonpolar? Explain. Polar because the five bond dipoles do not cancel.
(b) In the spaces provided, do the following for BrF3
Draw the BEST Lewis structure of this
interhalogen molecule and show formal
charge calculations for each type of atom
FC (X) = valence – bonds – lone pair eFC (Br) = 7 – 3 – 4 = 0
All F: FC (F) = 7 – 1 – 6 = 0
Give the names of (a) the electron pair
geometry and (b) the molecular shape,
(a) trigonal bipyramidal
(b) T-shaped
Specify the hybridization of the Br AND
draw the energy diagram for the hybridized orbital
(you know - the boxes with electrons in them).
↑↓
↑↓
↑↓
↑↓
↑↓
l.p.
l.p.
σ
σ
σ
hybridization is sp3d
Fluorine electrons are shown in red, l.p.
stands for lone pair, and σ is a sigma
bond.
Is this species polar or nonpolar? Explain. Polar because the three bond dipoles do not cancel.
(22 points)
2.
For the following two species complete the tasks specified below in a - d.
i. CH3COOH
ii. C2H4
a. Draw the best Lewis
structures.
b. Specify the hybridization
of both
the C atoms and the O atom
specified in (i)
and the C atoms in (ii)?
C in CH3
sp3
C in CO
sp2
c. Show the energy level
diagram for both the C atoms
and the O atom in (i) and the
C atoms in (ii), fill the
diagrams with the correct
number of valence electrons,
and identify the types of
electrons present (i.e. σ, π,
lone pair)
C in CH3
C in CH2
sp2
O between CO and H
↑
σ
sp3
C in CH2
↑ ↑
σ
σ
↑
σ
For both C
C in CO
↑
↑
σ
σ
↑
↑
↑
↑
σ
σ
↑
σ
↑
π
π
σ
O between CO and H
↑↓ ↑↓ ↑
↑
lp
lp σ σ
d. What are the bond angles
around both the C atoms and
the O atom in (i) and the C
atoms in (ii)?
C in CH3
C in CO
~109.5°
C in CH2
120°
~120°
O between CO and H
~109.5°
(8 points)
3.
Reaction of solid white phosphorus (P4) with chlorine gas (Cl2) yields phosphorus trichloride as
the sole product at 100 % yield. (a) Write a balanced chemical equation.
P4 (s) + 6 Cl2 (g) à 4 PCl3 (l)
(b) When 1.00 g of solid white phosphorus (P4) is reacted with excess chlorine gas (Cl2),
determine the maximum amount of PCl3 that can be formed?
1 mole PCl3 1 mole P
x
x1.00 g P
1 mole P
30.97 g P
mole PCl3 = 0.0323 mole PCl3
mole PCl3 =
g PCl3 =
2
137.33g PCl3
0.0323 mole PCl3
mole PCl3
4.
(20 points)
Fill in the MO energy diagrams for the species listed below. The answer the questions that appear
under the MO energy diagram of each species.
nitrogen
Ν
2
10 valence,
14 total e-
pernitride
Ν
σ∗2p
π∗2p
π2p
↑↓
↑↓
12 valence,
16 total e-
σ∗2p
π∗2p
↑↓
−2
2
↑ π∗2p
π2p
↑↓
↑ π∗2p
π2p
σ2p
↑↓
↑↓
σ2p
↑↓
σ∗2s
↑↓
σ∗2s
↑↓
σ2s
↑↓
σ2s
↑↓
σ∗1s
↑↓
σ∗1s
↑↓
σ1s
↑↓
σ1s
Show the calculation of the bond order for
nitrogen.
BO = ½ (bonding – antibonding)
BO = ½ (10 - 4) = 3
Is nitrogen diamagnetic or paramagnetic?
Explain your answer.
Molecular nitrogen, N2, is diamagnetic
because it possesses no unpaired
electrons (all electrons are paired).
Draw the Lewis structure of nitrogen.
Show the calculation of the bond order for the
above ion.
BO = ½ (bonding – antibonding)
BO = ½ (10 - 6) = 2
Is the above ion diamagnetic or paramagnetic?
Explain your answer.
The pernitride ion, N22-, is paramagnetic
because it possesses unpaired electrons.
Draw the Lewis structure for the above ion.
.. ..
[:N=N:]
-1
Does the bond order for nitrogen from the
Lewis structure match the bond order from
the MO diagram?
Yes it does!
-1
Does the bond order for the above ion from the
Lewis structure match the bond order from the
MO diagram?
Yes it does! (BUT; the Lewis structure does
not predict the paramagnetism )
Which should have the longer, weaker bond; nitrogen or the above ion? Explain your choice!
The pernitride ion definitely should have the longer and weaker bond. The bond is only a
double bond (vs. a triple bond in N2) AND the negative formal charge (-1) on each N atom
would repel one another.
3
π2p
(17 points)
5.
à 8 CO2 (g) + 10 H2O (l)
2 C4H10 (g) + 13 O2 (g)
a. Balance the reaction above.
b. If 25.0 g of C4H10 reacts with 100.0 g of O2, how much CO2 (in BOTH moles and
grams) can be formed? Show which reactant is the limiting reactant!!
1 mole C H
mole C4H10 = (25.0 g C4H10)(
4 10/58.12 g) = 0.430 mole
1 mole O
mole O2 = (100.0 g O2)(
2/32.00 g) = 3.125 mole
we only need 2.795 mole of O2 to react with 0.430 mole of C4H10, so C4H10 is the limiting
reactant
8 mole CO
mole CO2 = (0.430 mole C4H10)(
2/2 mole C4H10) = 1.72 mole CO2
g
mass CO2 = (1.72 mole CO2)*(44.02 /mole) = 75.7 g CO2
c. If the dry CO2 is collected and has a mass of 25.0 g, what is the % yield?
% Yield = 100*
(actual yield)
/(theo. yield) =
98.5 %
(8 points)
6.
Balance the following redox reaction in acidic solution.
-
-
I (aq) + MnO4 (aq)
→
I2 (aq) + Mn2+ (aq)
Balance the half reactions (using H+ and H2O as needed) & combine
2 * [8 H+(aq) + 5 e- + MnO4
5*[2I
-
→
(aq)
16 H+(aq) + 10 I
(aq)
-
Mn2+ (aq) + 4 H2O(l) ]
→
(aq)
I2 (aq) + 2 e- ]
+2 MnO4
(aq)
→
2 Mn2+ (aq) + 5 I2 (aq) + 8 H2O(l)
(7 points)
7.
A 5.00 mL sample of battery acid [H2SO4 (aq)] is completely neutralized (converted to
aqueous sodium sulfate and water) by 45.60 mL of 0.9995 M sodium hydroxide [NaOH
(aq)]. Complete and balance the reaction and calculate the molarity of the original 5.00
mL H2SO4 (aq) sample. By calculation, demonstrate that the acid in the optimal
concentration range (between 4.3 – 5.0 M)
H2SO4 (aq)
+ 2 NaOH (aq)
→
2 H2O(l) + Na2SO4 (aq)
1molH 2 SO4 0.9995molNaOH 0.04560 L NaOH sol' n
x
x
= 4.558 M
2molNaOH
1L NaOH sol' n 5.00 x10 − 3 L battery acid
Which indicates that the acid is a bit under the optimal concentration range.
4
BONUS: for up to 6 points (total)
Write the products for the reactions listed below and make sure the reactions are balanced
molecular (that is total) equations. Also, name the type of reaction (oxidation-reduction,
acid-base neutralization, or precipitation) in the blank to the right .
(2 points each)
a.
2 Ba (s)
+
O2 (g)
2 BaO (s)
b.
2 HCl (aq) + Ba(OH)2 (aq) → 2 H2O(l) + BaCl2 (aq)
c.
AlCl3 (aq) + 3 AgNO3 (aq) → Al(NO3)3 (aq) + 3 AgCl (s)
→
5
oxidation-reduction
acid-base neut.
precipitation