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Answer Key
2012 Ch 1a, Problem Set Six
Side 1 of 6
Problem One – Lewis and VSEPR – 8 points each
d. PH3, trigonal pyramidal
a. CO2, linear
O
C
O
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 (be liberal with this, i.e.
wedges are not necessary if the diagram looks reasonable). 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
structures. -2 per missing resonance structure. No deductions for extra correct structures (if there are any).
Max deduction 8 points.
Answer Key
2012 Ch 1a, Problem Set Six
Side 2 of 6
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
S
O
F
F
H
The geometry must be named, and must be somewhat clear from the sketches (be liberal with this, i.e.
wedges are not necessary if the diagram looks reasonable). 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 structures. 1.5 per missing resonance structure. No deductions for extra correct structures (if there are any). Max
deduction 6 points.
Answer Key
2012 Ch 1a, Problem Set Six
Side 3 of 6
Problem Three – Ions –15 points each
C
a CF+
F
2+
(Graders: Also accept a double-bonded species for full credit.)
(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.)
O
O
O
b PO43-
P
O
O
O
P
O
+ 3 equivalent
+ 5 equivalent +
3 equivalent
structures
structures
structures
nd
(5 pts: last structure; 6 pts: 2 last structure; 4 pts: # equivalent structures for 2nd last structure)
(NOTE: The first three structures are not as reasonable because of the negative charge on less
EN phosphorous atom. Do not deduct nor award points if these three structures appear.)
O
O
O
c ClO4-
O
O
Cl
O
O
O
O
Cl
O
O
Cl 2+
O
O
O
Cl
O
O
3+
O
+ 3 equivalent
+ 5 equivalent + 3 equivalent
structures
structures
structures
(3 pts: 1st structure; 3 pts: 2nd structure; 3 pts: # of equivalent structures for 2nd structure;
3 pts: 3rd structure; 3 pts: # of equivalent structures for 3rd structure)
(NOTE : The last two structures are not as reasonable because of the large positive charge on the
EN chlorine atom. Do not deduct nor award points if these two structures appear.)
d OH3+
O
H
H
H
e ClO2-
For all parts:
Lone pairs must all be drawn, except for on F (−2 penalty if any are missing). −2 if any formal charges are missing.
Max deduction per part −15.
Answer Key
2012 Ch 1a, Problem Set Six
Side 4 of 6
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 <109.4 since the lonepair: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, VIII11), 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 bondingpair: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.
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
2012 Ch 1a, Problem Set Six
Side 5 of 6
Problem Four – Bond Angle – 8 points each
d.
NF3 and NCl3
Larger bond angle: NCl3
GRADERS:
Accept either
explanation 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 bonding-pair: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.
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
2012 Ch 1a, Problem Set Six
Side 6 of 6
Problem Five – Atmospheric Chemistry – 15 points
.. .
:Cl
..
.. - O:
.. +. ..
:O
.. - N = O:
.. . ..
:O = N - O:
..
+
-
:
.. ..
..
:Cl
O
N
O:
.. .. +1 ..-1
- 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)
-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)