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Chemistry 1000 (Fall 2012)
Problem Set #5: Lewis Structures and VSEPR
Solutions
Additional Practice Problems
1.
For each of the elements listed below, indicate the number of bonds a neutral atom will have in
any Lewis dot structure when it obeys the octet rule.
(a)
carbon
4 (c)
sulfur
2 (e)
oxygen 2
(g)
boron 3
(b)
hydrogen 1 (d)
nitrogen 3 (f)
iodine 1
(h)
silicon 4
2.
For each of the polyatomic ions named below:
(i)
draw a Lewis dot structure (including formal charges where appropriate),
(ii)
name the molecular geometry,
(iii) draw a 3-dimensional picture, and
(iv)
label the bond angles.
(a)
nitrate
(e)
.. O ..
120o
carbonate
.. O ..
120o
120o
C .. -1
-1 ..
.. O 120
.
O
.. .
..
trigonal planar
trigonal planar
o
o
(b)
120o
+1
.. .-1
-1 ..
.. O N O
.. 120 .. .
nitrite
(f)
..
N
.. ..
.
O
O
~120
.. .
..
-1
bicarbonate
.. O ..
~120o (O-C-O)
~120o
C
-1
O ~120o
o
.. ..
..
bent
.. H
O
..
~109.5o (C-O-H)
trigonal planar (around C)
bent (around O)
(c)
.. O ..
sulfate*
..
..O
(g)
ammonium
.. -1
S
O ..
.. O....
..
H
-1
109.5o
tetrahedral
(all angles are 109.5 o )
(d)
..
sulfite*
..
..O
.. .-1
O.
.. ....
O
-1 ..
H
109.5o
109.5o
+1
N
H
H
109.5o
tetrahedral
(h)
S
trigonal pyramidal
(all angles are just under 109.5o)
hydroxide
-.1 ..
.O
..
H
linear
(only 2 atoms so no angles)
3.
(a)
(b)
4.
(a)
(b)
Which of the polyatomic ions in Question 2 can have more than one resonance structure?
For each ion listed in part (a), indicate the number of possible resonance structures.
nitrate (3 resonance structures)
nitrite (2 resonance structures)
carbonate (3 resonance structures)
bicarbonate (2 resonance structures)
sulfate (6 resonance structures)
sulfite (3 resonance structures)
Consider the Lewis dot structures you drew for nitrate and nitrite (in question 2).
Which has a higher average N-O bond order?
nitrite (average bond order for nitrite = 1.5; … for nitrate = 1.33)
How will the different N-O bond lengths compare?
The N-O bonds in nitrite will be shorter because the bond order is higher.
5.
Using an argument based on formal charge and on other relevant concepts, explain why we
never see molecules containing X-F-Y, i.e. a fluorine atom bonded to two different atoms in a
molecule. (X and Y could be any two atoms.)
Fluorine is a second period element, so it cannot exceed an octet of valence electrons. As such,
a fluorine bonded to two different atoms could have no more than two lone pairs of electrons.
This would result in a fluorine atom with a charge of +1:
Qf = 7 – [½ (4) + (4)] = +1
Fluorine is the most electronegative element and therefore cannot exist in molecules which
would force it be assigned a positive formal charge.
Thus, molecules containing X-F-Y do not exist.
6.
Draw the best Lewis structure for each molecule whose skeleton is shown below.
(a)
(b)
H
H
H
H
C
H
..N
C
C
C
H
H
..
..Cl
..
C
.
Se
.. .
C
H
7.
For each of the molecules/ions below:
 draw the best Lewis structure (any non-zero formal charge must be clearly labeled)
 name the molecular geometry
 give the bond angles (use the symbol ~ to indicate “approximately”; if no ~ is
given, it will be assumed you mean the angles to be exact)
.. .. ..
(a)
SiH2Cl2
(b)
XeH2F2
H
H
H
Si
.
....
Cl .
.. .
Cl .
..
H
.. ..
..Xe ..F ..
....F ..
or
..
..
.. ..
..
tetrahedral
square planar
square planar
~109.5
~90
90