Fall 2006
Chemistry 1000 Test #2A
INSTRUCTIONS:
1.
____/ 65 marks
1) Please read over the test carefully before beginning. You should have
7 pages of questions, and a periodic table (8 pages total). If you need
extra space, please use the bottom half of the periodic table page and
clearly indicate which question is being answered.
2) If your work is not legible, it will be given a mark of zero.
3) Marks will be deducted for extraneous incorrect answers.
4) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
5) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
6) You may use a calculator.
7) You have 90 minutes to complete this test.
Acetonitrile (CH3CN) has the connectivity shown below. Answer parts (a) and (c)
directly on this diagram.
[10 marks]
H 109.5
o
180o
H C C N ..
109.5o
109.5o
H
109.5o
(a)
Draw the best Lewis electron dot structure for acetonitrile.
(b)
Indicate the molecular geometry at each central atom.
C (of CH3) = tetrahedral
net dipole
180o
C (of CN) = linear
(c)
Indicate all of the bond angles in this molecule. (The clearest way to do this is probably
to label them directly on your structure.)
(d)
List the bonds in order of increasing bond length:
i.
C-H (shortest bond)
ii.
C≡N
iii.
C-C (longest bond)
(e)
Draw the net dipole for this molecule.
(f)
What is the hybridization of the following atoms in acetonitrile:
C (of CH3) = sp3
C (of CN) = sp
(g)
How many sigma bonds are there in acetonitrile? 5
(h)
How many pi bonds are there in acetonitrile? 2
N = sp
2.
Which of the two molecules below is more polar? Justify your answer.
[5 marks]
BBr3 vs. PBr3
BBr3 is a nonpolar molecule. Its trigonal planar geometry makes all of the bond dipoles cancel
out, leaving the molecule with no net dipole.
Br
trigonal planar
B
no net dipole
Br
Br
nonpolar molecule
PBr3 is a polar molecule. Its trigonal pyramidal geometry does not allow the bond dipoles to
cancel out, leaving the molecule with a net dipole.
trigonal pyramidal
P
Br
Br
net dipole
Br
polar molecule
3.
(a)
What is the correct name for CO32-?
(b)
Draw all of the resonance structures for CO32-. Place charges on the appropriate atom(s).
..O..
.. . -1
.. O
.
.. C .. .-1
.. O
O
.. .
..
.. .-1
..O C O
..
.. .
-1
(c)
[7 marks]
carbonate
What is the average C-O bond order in CO32-?
1⅓ = 1.33
.. . -1
.. O
.
.. C ..
.. O
.O.
..
-1
4.
Complete the following table. Where charge(s) are necessary, place them on the
appropriate atom(s) in the Lewis electron dot structure.
[6 marks]
Formula
PO43-
Lewis Electron Dot Structure
..
.. O
..
...-1
.. O
.
-1
+1
P
..
.. O
..
-1
or
.. O-1..
..
..
.. Cl
..
SCl2
5.
.. -1
.
O
.. .
...-1
.. O
.
P
.. -1
.
O
.. .
Electron Pair
Geometry
Molecular
Geometry
Predicted
Bond Angle
tetrahedral
tetrahedral
109.5˚
tetrahedral
bent
..O..
.. ..
S
.. .
Cl
.. .
Complete the following table.
~104.5˚
[5 marks]
Molecular Formula
Name
Na3N
sodium nitride
Cu2O
copper(I) oxide
Al(OH)3
aluminum hydroxide
NH4BrO3
ammonium bromate
SnCl4·5H2O
tin(IV) chloride pentahydrate
6.
Use diagrams to explain why it is not possible to rotate about the double bond in
molecule A to make molecule B:
[4 marks]
F
F
C
X
C
H
H
F
H
C
C
H
F
molecule B
molecule A
A π bond has a node along the axis between the two nuclei. Rotating one of the atoms in
the π bond will rotate the orbitals on that atom and break the symmetry required for the π
bond. The bond will be broken by twisting it. Thus, twisting a π bond requires enough
energy to break the bond (and is therefore difficult to do).
H
F
H
C
C
F
molecule A
(p orbitals have correct
symmetry to overlap)
7.
rotate
X
rotate
H
C
F
C
H
F
F
X
pi bond broken!
(p orbitals do not have correct
symmetry to overlap)
H
H
C
C
F
molecule B
(p orbitals have correct
symmetry to overlap)
Use diagrams to explain why an s orbital cannot be involved in making a π bond.
(Hint: If you’re stumped, try drawing orbitals that can make a π bond and see why they
are able to do so.)
[4 marks]
A π bond has a node along the line connecting the two nuclei. An s orbital does
not have this node, so it doesn’t have the correct symmetry to make a π bond.
node (not present in s orbitals)
8.
(a)
Use molecular orbital theory to describe the bonding in fluorine.
Complete the molecular orbital correlation diagram below by:
i.
drawing and naming the atomic orbitals,
ii.
drawing and naming the molecular orbitals, and
iii.
adding the electrons
[12 marks]
σ∗2p
π∗2p
2p
2p
Energy
σ2p
π2p
σ∗2s
2s
2s
σ2s
F
(b)
F2
F
Fluorine gas reacts violently with sodium metal. Use molecular orbital theory to explain
what happens if a fluorine molecule acquires two more electrons (as it does when
reacting with sodium). In your answer, be sure to mention where those two electrons go!
The two additional electrons are added to the σ*2p orbital, decreasing the bond order to
zero. Therefore, the F-F bond is broken, and we get two fluoride ions (F-).
9.
(a)
An unknown solid is found to contain 31.74% Fe, 40.92% O and 27.34% S by mass.
[6 marks]
Calculate the empirical formula for this unknown solid.
(b)
Assuming that the empirical and molecular formulae are the same, name the unknown
solid.
(a)
100 g of unknown molecule contains 31.74 g Fe, 40.92 g O and 27.34 g S.
nFe = mFe
MFe
nO = mO
MO
=
=
nFe
(b)
(31.74 g) .
(55.847 g/mol)
= 0.5683 mol
nS = mS
MS
(40.92 g)
=
(27.34 g) .
(15.9994 g/mol)
(32.066 g/mol)
nO = 2.558 mol
nS = 0.8526 mol
Therefore,
nFe : nO : nS = 0.5683 mol Fe : 2.558 mol O : 0.8526 mol S
0.5683
0.5683
0.5683
= 1 mol Fe : 4.5 mol O : 1.5 mol S
nFe : nO : nS = 2 mol Fe : 9 mol O : 3 mol S
Therefore,
the empirical formula is Fe2S3O9
Fe2S3O9 = Fe2(SO3)3 = iron(III) sulfite
10.
Kelly needs anhydrous nickel(II) bromide for her next experiment but all she has
available is nickel(II) bromide trihydrate. Since she has a hot oven available, she decides
that she’ll make some anhydrous nickel(II) bromide by dehydrating the hydrate. What is
the minimum mass of hydrate that Kelly must heat in order to obtain 4.75 g of anhydrous
nickel(II) bromide?
[6 marks]
NiBr2 · 3H2O
272.547 g/mol
→
NiBr2
+ 3 H2O
218.501 g/mol
nNiBr2 = mNiBr2
MNiBr2
=
4.75 g
.
218.501 g/mol
nNiBr2 = 0.0217 mol
nNiBr2.3H2O = nNiBr2 = 0.0217 mol
mNiBr2.3H2O = nNiBr2.3H2O × MNiBr2.3H2O
= (0.0217 mol) × (272.547 g/mol)
mNiBr2.3H2O = 5.92 g