Chem 121
Answers to Paper HW#7
Spring 2013
DUE Tues 4/30/13
Jespersen Chap.9 p.404 #89, 91(A&C only), 95 (A,B,C), 99, 103, 111
ALL Lewis structures must include FORMAL CHARGES. If resonance exists, all resonance
structures must be included, then circle the most stable form and give the resonance hybrid.
9.89) Draw Lewis structures
..
: F : ..
.. Te ..
:F
F
..
..:
: F..:
a) TeF4
..
:
:
:F
.. F :
.. Cl ..
:F
F
..
..:
: F..:
..
b) ClF5
_
c) PF6
d) XeF4
..
..
:F :
..
:F
..
..
:F :
# valence e = 6 + 4x7 = 34 = 17 pairs
# valence e = 7 + 7x5 = 42 = 21 prs
# valence e = 5 + 7x6 + 1 = 48 = 24 prs
..
_ : F:
P ..
F
..:
: F..:
..
: F : .. : F :
..
.. Xe
.. F
:F
..
..:
# valence e = 8 + 7x4 = 36 = 18 prs
9.91A) HIO3
# valence e = 1 + 7 + 3x6 = 26 = 13 prs
.. _
:
O
: .. _
..
H O I O:
.. .. ..
.. _
:
O
: ..
..
H O I O:
+ .. ..
..
:
O
..
..
H O I O:
.. ..
2+
no stable
because of
separation of
charges
not stable because
oxygen would not want
a positive charge
most stable
resonance form
9.91C) HCO3
# valence e = 1 + 4 + 3x6 +1 = 12 prs
_
.. _
:
O
: ..
..
H O C O:
..
.. : O : .. _
H O C O:
..
..
9.95) Draw Lewis structures:
a) NO+
+
: N O:
# valence e = 5 + 6 − 1 = 10 = 5 prs
_
b) NO2
_
_
.. .. ..
O
.. :
.. N O
.. .. ..
:O
..
.. N O
..
:O
_
..
..
N
O:
_
# valence e = 5 + 6x2 + 1 = 18 = 9 prs
resonance hybrid
_
c) SbCl6
..
: Cl :
..
: Cl
..
..
: Cl:
..
:
_ Cl :
Sb ..
Cl
.. :
:
: Cl
..
_
# valence e = 5 + 7x6 = 1 = 48 = 24 prs
9.99) Assign formal charges
: O:
a)
_
..
: Cl
..
b)
..
N O:
+ ..
..
:F:
..
:F
..
_
+ ..
N O:
..
:F :
..
c)
.. _
:O:
.. 2+ ..
:F
.. S F.. :
: O :_
..
Note that the net charge always adds up to zero, consistent
with the formula. You should always check to see that this is
so. It is a good way to catch your own mistakes.
9.103) The following are two Lewis structures that can be drawn for phosgene, a substance that has
been used as a war gas. Which is a better Lewis structure? Why?
.. _
: O:
: O:
..
: Cl
..
C
..
.. +
Cl
..
: Cl
..
C
..
Cl :
..
The structure on the right is a better structure because it does not have any formal charges. The
structure on the left has “separation of charges” which makes it less stable.
By the way, phosgene is a poisonous gas used combat in World War I. It is said to smell of freshly
mown hay. Some said it smelled like May blossoms, the blossoms of hawthorn trees.
As you notice, it does not contain phosphorus, and is named “phosgene” because it was first
synthesized by exposing a mixture of carbon monoxide (CO) and chlorine (Cl2) to sunlight. “Phos”
is the Greek word for light and “gene” is the Greek word for “born”: born of light.
9.111) Use formal charges to establish the preferred Lewis structures for ClO3– and ClO4– ions. Draw
resonance structures for both ions and determine the average Cl–O bond order in each. Which of these
ions would be expected to have the shorter Cl–O bond length?
ClO3– has 13 pairs of valence electrons
_
2+
..
:O
..
.. _
Cl O :
..
:O : _
..
..
This is too unstable to be considered because
of the separation of charges. The 2+ on the central
atom tells you that you need to move 2 pairs of electrons
in from the O. The question is from which 2 oxygen atoms.
This is how we tell there are resonance structures.
.. _
: O Cl O
.. :
:O :
..
..
..
:O
Cl
..
_ ..
:O
..
O:
..
Cl
..
O:
:O :
:O : _
..
_
..
resonance hybrid = : O
..
_
..
..
O:
Cl
Cl-O bond order = 1 2/3
:O :
_
The bond order is determined this way. There is ALWAYS one bond between Cl and O. The second
bond is there only 2 times out of 3, so that counts as 2/3 of a bond. Thus the total is 1 2/3.
(continued on the next page)
ClO4– has 16 pairs of valence electrons.
.. _
:O:
.. _
Cl O :
..
..
:O : _
..
_ ..
:O
3+
:O:
..
O
..
..
..
_
O:
Cl
This is structure is too unstable to be considered
because of the separation of charges. The 3+ on
the central atom Cl, tells us that we need to move
3 pairs of electrons in to be shared. Again, the question
is from which 3 oxygen atoms? This is how we know
there are resonance structures.
:O:
:O:
..
O
..
:O :
Cl
..
:O
_ ..
..
..
O
:O
.. : _
Cl
:O:
..
..
O
.. _
:O:
..
..
O Cl O
..
..
:O :
_
_
..
resonance hybrid = : O
: O:
_
..
O:
Cl
Cl-O bond order = 1 3/4 or 1.75
:O :
_
The bond order for ClO4– is determined as follows. There is always one bond between each Cl and O.
The second bond is there only 3 times out of 4, so its bond order is ¾. Thus, the total bond order
between Cl and O is 1 3/4 or 1.75.
The average bond order for ClO3– is 1 2/3 or 1.67.
The average bond order for ClO4– is 1 ¾ or 1.75.
ClO4– has a higher bond order (1.75 instead of 1.67). This means it is closer to a double bond (bond
order of 2) than ClO3–. A double bond has a shorter bond than a single bond when all else is kept
constant. Therefore, ClO4– is expected to be shorter.