Answer Key
2015 Ch 1a, Problem Set Six
Side 1 of 7
Problem One – Lewis and VSEPR – 8 points each
a. CO2, linear
O
C
O
d. PH3, trigonal pyramidal
O
C
O
P
acceptable, not
required
b. PF5, trigonal bipyramidal
F
F
P
H
H
H
e. HCP, linear
H
C P
F
F
F
f. IF3,t-shaped
F
c. CO, linear
F
C O
I
F
The geometry must be named, and must be somewhat clear from the sketches.
4 points for listing the geometry,
3 for the Lewis diagram/charges,
1 point for the geometry in the sketch.
Lone pairs must all be drawn, except for on F (−1 penalty if any are missing).
−1 if any formal charges are missing.
–2 for each extra incorrect resonance structure.
–2 per missing resonance structure.
No deductions for extra correct structures (if there are any). Max deduction 8 points.
Answer Key
2015 Ch 1a, Problem Set Six
Side 2 of 7
Problem Two – Lewis and VSEPR – 6 points each
a. BrF5 , square pyramidal
d. BH2NH2, planar (or trigonal
planar)
F F F
Br
F
F
H
B
N
O
N
N
c. CH3CCH, tetrahedral, linear
H
H
C
H
N
H
H
note: lacks
octet at
Boron
e. SF2O, trigonal pyramidal
S
C
H
B
O
H
C
H
N
H
b. N2O, linear
N
H
O
F
F
H
The geometry must be named, and must be somewhat clear from the sketches.
3 points for the shape,
2.5 for the Lewis diagram/charges,
0.5 for the shape in the sketch.
Lone pairs must all be drawn, except for on F (−1 penalty if any are missing).
−1 if any formal charges are missing.
–1.5 for each extra incorrect resonance structure.
–1.5 per missing resonance structure.
No deductions for extra correct structures (if there are any). Max deduction 6 points.
S
O
F
F
Answer Key
2015 Ch 1a, Problem Set Six
Side 3 of 7
Problem Three – Ions –15 points each
a
C
F
2+
CF+
(Also accepted a double-bonded species for full credit, but –8 points for single-bond)
(The structure given above forms octets and maximizes bonding but places 2+ charge on
fluorine, which is highly undesirable. Despite this issue, the structure shown above is
correct but not because the Lewis diagram truly predicts it.)
bO
O
P
O
O
P
O
O
P
O
O
O
(+5 equivalent
structures)
O
O
O
(+3 equivalent
structures)
–8 points if either structure is missing
–3 points if number of equivalent structures was incorrect
PO43O
P
O
O
O
3–
P
O
O
O
2–
O
P
O
O
O
O
P
O
O
O
O
P
O
O
O
O
(+3 equivalent
(+5 equivalent
(+3 equivalent
structures)
structures)
structures)
(NOTE: These other three structures are not as reasonable because of the negative
charge on less EN phosphorous atom. No points for these structures.)
c
O
O
O
ClO4-
O
O
Cl
O
Cl
O
O
O
O
Cl
O
O
Cl 2+
O
O
O
Cl
O
O
+ 3 equivalent
+ 5 equivalent
structures
structures
5 points for each structure
–3 points if number of equivalent structures was incorrect
O
O
O
O
O
Cl 2+
O
O
O
Cl
O
O
3+
O
+ 3 equivalent
structures
(NOTE : The last two structures are not as reasonable because of the large positive
charge on the EN chlorine atom. No points for these structures.)
3+
O
Answer Key
2015 Ch 1a, Problem Set Six
Side 4 of 7
d
OH3+
O
H
H
e
O
ClO2-
H
Cl
O
O
Cl
O
(+1 equivalent
structure)
–8 points for missing one of the two structures. –3 points for incorrect number of
equivalent structures.
For all parts:
–8 if the sum of formal charges is not equal to the molecule’s total charge.
Lone pairs must all be drawn, except for on F (−2 penalty if any are missing).
−2 if any formal charges are missing.
Answer Key
2015 Ch 1a, Problem Set Six
Side 5 of 7
Problem Four – Bond Angle – 8 points each
3 points for answer, 5 points for explanation
a.
NH3 and NH4+
Larger bond angle: NH4+
Explanation: Both molecules have the same electronic domain geometry
(tetrahedral) but they have different molecular domain geometries – NH3 is
trigonal pyramidal and NH4+ is tetrahedral – because NH3 has a lone pair of
electrons. A symmetric tetrahedral molecule, like CH4 or NH4+, has bond angles
of 109.5°, but the presence of the lone pair of electrons in NH3 causes the bond
angle to get smaller. (lecture notes, VIII-10)
b.
BCl3 and PCl3
Larger bond angle: BCl3
Explanation: BCl3 is a symmetric trigonal planar molecule with 120° bond angles.
(lecture notes VIII-6) PCl3 is trigonal pyramidal with bond angles <109.5° since
the lone-pair:bonding-pair repulsion is greater than the bonding-pair:bonding-pair
repulsion. Thus, BCl3 has a larger bond angle.
c.
OF2 and SF2
Larger bond angle: OF2
Explanation: Just as for the OH2, SH2, SeH2 series presented in class (lecture
notes, VIII-11), the lone pairs get “fatter” on the central atom as the radius of the
central atom increases and the strength of the lone-pair:bonding-pair repulsion
increases. As the lone pairs get “fatter”, they push the attached atoms closer
together since the bonding-pair:bonding-pair repulsion is the same for both
molecules. Since the lone pairs on OF2 are not as “fat” as the lone pairs on SF2,
then the bond angle in OF2 is larger.
*Note: Many answers contained explanations based on differing electronegativity
of the central atom. In this context, electronegativity is not relevant, and no credit
was given for explanations that did not address the radius of the central atom.
Side note: Actual bond angles (OF2: 103.8°; SF2: 98°). source: J. Barrett,
“Structure and Bonding.” Royal Society of Chemistry, 2001 (p.114)
Answer Key
2015 Ch 1a, Problem Set Six
Side 6 of 7
d.
NF3 and NCl3
Larger bond angle: NCl3
Either explanation was
accepted for full credit.
Explanation: In this case the central atom remains the same; so the lone pair
doesn’t get any “fatter.” The bonding atoms do change though. As the bonding
atoms get larger, they repel each other more strongly. The lone-pair:bonding-pair
repulsion increases slightly as the bonding atom gets larger, but the bondingpair:bonding-pair repulsion increases more and the bond angle increases. Since
Cl is larger than F, NCl3 will have a larger bond angle caused by the increased
bonding-pair:bonding-pair interactions. Repulsion must be mentioned for full
credit, –2 points if repulsion was not mentioned. (“chlorine is bigger” by itself is
not a complete answer.)
Alternate explanation: Traditionally, we focused on electronegativities of bonding
atoms. In OGC6, p. 95 you’ll read “because Cl is more electronegative than H, it
tends to attract electrons away from the central atom, reducing the electron-pair
repulsion.” If you replace “H” with “Cl” and “Cl” with “F” in the text, the
statement is still true and explains why NCl3 has a larger bond angle than NF3.
This explanation is also consistent with the original VSEPR rules developed by
Gillespie: "The strength of the repulsion between single bonds decreases with
increasing electronegativity of the ligand and/or decreasing electronegativity of
the central atom."
Although this alternative explanation has recently fallen out of favor, we’ll still
accept it as a correct answer. But even Ron Gillespie – one of the main
developers of VSEPR – has re-evaluated his long-time position on
electronegativity vs. ligand size as the root cause of bond angle variation. In a
recent review article (July 2008), Gillespie states “Overall ligand size explains
those bond angles that are not consistent with the electronegativity rule but also
those that are consistent with the rule, so it is reasonable to replace the
electronegativity rule of the original VSEPR model with the rule that bond angles
increase with ligand size.” (R.J. Gillespie, “Fifty Years of the VSEPR Model,”
Coordination Chemistry Reviews, 252:12-14, p. 1320.) This variation of VSEPR
is sometimes called the Ligand Close Packing (LCP) model.
Side note: Actual bond angles (NF3: 102.1°; NCl3: 107.1°). source: J. Barrett,
“Structure and Bonding.” Royal Society of Chemistry, 2001 (p.135)
Answer Key
2015 Ch 1a, Problem Set Six
Side 7 of 7
Problem Five – Atmospheric Chemistry – 15 points
.. .
:Cl
..
.. - O:
+
.. . ..
:O
.. - N = O:
.. . ..
:O = N - O:
..
+
-
:
..
.. ..
:Cl
N - O:
.. - +1
..-1
.. - O
- O:
-
= O:
a) Draw the Lewis structures of ClO, NO2, and ClONO2, including any significant
resonance structures. (3 pts: ClO; 4 pts: NO2; 4 pts: ClONO2 and –1 point for every
incorrect structure.)
: :
-1
.. ..
:Cl
N = O:
.. - +1
..
.. - O
OR
.. .. .. ..
:Cl
.. - N = O:
..
.. - O
.. - O
b) Use these Lewis structures to explain why ClO and NO2 are so much more reactive
than ClONO2.
ClO and NO2 both have atoms with incomplete octets. Species such as these,
called free radicals, are very reactive because of their unpaired electrons. ClONO2,
on the other hand, has no unpaired electrons and so is much more stable (it also has
some resonance stabilization, but this is insignificant relative to the paired vs
unpaired electron effect).
(4 points: anything about incomplete octets.)