CHEMISTRY 11 AP – VALENCE BOND THEORY WORKSHEET
1) Draw Lewis structures for each of the following species. State the shape and the central atom
hybridization for each. Indicate the total number of σ and/or π bonds in each molecule or ion.
valence
(a) SCl2
.
:
a
-
@
S
O
0
-
6+14=20
=
a
-
o
0
AX2E2
SP
g.
:
→
:
o
:
.
4=16
-
.
.
.
e-
?q
!,
:c
20
:
bonds
bent
6+35+1=42
if
?E
⇐
¥
ftp.f#eIEiiitIEfsobonas
pyramidal
valence e-
(b) TeF5-
=
E
±f
:
-10=32
:
:
AXSE
square
t÷Lsp3d2
:
.
ye
:
|
o
e-
Valence
(c) XeF4
•
.
.
.
.
.
.
.
.
:
-
.
:
-
8=28
III
IIIxe
:
:
AX4E2
.
8+28=36
=
square planar
40
bonds
Valence
(d) CO32-
2 other
they
will have the
:O
-6=18
dc]÷sp23o
[
are
,
4+18+2=24
=
:
resonance
structures
possible but
e-
.io/?o ;
same
geometry bonds &
hybridization
AX }
,
.
T
I
bonds
bond
trigonal planar
:
:O
if
\|⇐sp3d2
j=
f÷E¥¥÷¥#s±dl?IIi!¥
Valence
(e) IOF5
e-
7+6+35=48
=
-
12=36
:
:
:
1
60
:
Axio
/
\
:
f
/
°
0
a
'
'
bonds
.
.
.
:
F
:
:
octahedral
(f) IF4+
-
:D
7+28-1=34-8
AX4E
.
:
seesaw
+
.no
sp3=4sp3
bridged
2) Draw the following molecule showing all of the orbitals involved.
%
p
orbitals
←sp←spsP=£o%°y
orbitals
D-*
:
170
.pt#gIeIgotsP:Gq0oooa
3) Predict the shape, approximate bond angles and hybridization around each of the indicated atoms in
the following molecules. Indicate the total number of σ and/or π bonds in each molecule.
H
H
N
C
N
F
F
Br
3
0
2
F
Atfgegistoriggnpalpyrimidal
C
4
AXAE
N
5
F
1888800
C
F
H
C
°
C
N
N
o
H
4
IT
shown
in
blue
shown
in
red
:
O .H
bent
<
120
Exeoeasibsnpt
sp2
4) What is the hybridization
at each
3s
3pcarbon atom in the following molecule? How many σ bonds are
there in the molecule? How many π bonds are there in the molecule. Identify all the 120° bond
molecule.
angles
inInthethe
hydrocarbon
120bar dangles
are
found
the
AXS
in
around
q•
H
H
l
spy H
0080080
&
136
H
trigonalplanar
H
als, ide
the MO
is bond
planes.
•
•
C
•
C
y
.
•
C
H
C
q
C H
ofo
C
SP
q
H
geometry
(B) What is the hybridization at each
carbon atom in the
showninbwe
thespcarbons
molecule? (C) How many s bonds are there in the molecule?
( shown by(D))How many p bonds? (E)ztshowninred
Identify all the 120° bond angles
.
spa
in the molecule. [Section 9.6]
5) The nitrogen atom in N2 participate in multiple bonding, whereas those in hydrazine, N2H4, do not.
Draw
Lewis
Thestructures
drawing
shows the
overlap
of two hybrid
orbitalsatoms in each
forbelow
both molecules.
What
is the hybridization
of the nitrogen
molecule?
stronger N – N bond?
Explain. of the followto Which
form molecule
a bondhas
in the
a hydrocarbon.
(B) Which
ing types of bonds is being formed: (i) C ¬ C s, (ii) C ¬ C p,
valence
or (iii) C e¬ H s? (C) Which of theNatty
following-10+4=14
could be the
identity of the hydrocarbon: (i) CH4, (ii) C2H6, (iii) C2H4, or
H
H
(iv) C2H2? [Section 9.6]
N2
:N
3H)
19
91
10-2=8
:N=N
=
:
q
g
SP
The dia
-10=4 MOs o
same ro
of elect
the mo
the p2p
a great
atomic
atomic
1
1
-
N
:
(H
SP
The molecule shown below is called furan. It is represented in
typical shorthand way for organic molecules, with hydrogen
The triple bond found in N2 would be a much stronger bond since it is the combination of three
atoms not shown.
overlapping orbitals forming 2 π and 1 σ bonds. This combination of bonds would require an
incredible amount of energy to break. In comparison, the single overlapping orbitals form the σ
O
bonds in N2H4 would require a lot less energy to break.
6) Draw the Lewis structure for NH3.
Valence
:
e-
=
5+3=8
(B) What is the molecular formula for furan? (C) How many
valence electrons are there in the molecule? (D) What is the
hybridization at each of
the carbon atoms? (E) How many elec'
trons are in the p systemHof the molecule? (F) The C ¬ C ¬ C
bond angles in furan are much smaller than those in benzene. The
H
likely reason is which of the following: (i) The hybridization of the
carbon atoms in furan is different from that in benzene, (ii) Furan
does not have another resonance structure equivalent to the one
I
#
Molecular S
-6=2
(sections 9.
An AB
length
scribe
many n
from th
(B) Met
describ
bond a
State the shape and the central atom hybridization for the central atom.
AX3E
.
sp3
trigonal pyramidal
:
Draw the electron configuration’s orbital diagram for the ground state of nitrogen’s electrons.
Energy !
2p
2s
1s
979
:
1
How many equivalent molecular orbitals are required around the central atom? Four
Draw the electron configuration’s oribtal diagram showing the hybridization that will explain how
these equivalent molecular orbitals are achieved.
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ftp.ff
Energy !
2p
-
2s
since
needtohaue
equal energies
but Nonlyneeds
toform 3 bonds
SP
.
(We can ignore the full 1s subshell since it is core and therefore not
involved in bonds)
:3
°
H
:O
H
D
°
¥00
unpaired
orbitals
Draw the molecule of NH3 showing all of the overlapping orbitals involved.
←
4 orbitals
are
't
for
bonding
italreadyhas
present
compare
this
( lone
2
electrons
pair )
tits
picture
required
4sp3 orbitals
thisisthesporbitalthatissince
Unavailable
e-
Lewisstrocturetoseethebonds
calized in the p system of ozone?
Butadiene, C4H6, is a planar molecule that has the following
7) Butadiene, C4H6, is a planar molecule that has the following carbon – carbon bond lengths:
carbon–carbon bond lengths:
H2C
1.34 Å
CH
1.48 Å
CH
1.34 Å
B
thePredict
the lengths
bondfound
angles
around
Explain
different bond
in this
molecule.each
and sketch the molecule.
CH2
of the carbon atoms
The double bonds in the molecule are shorter than the C – C bond since the sp2 hybrid orbitals
involved
those bonds possess
more of
their s characteristics
(they are bond
33% s and
67% p) than the
C
inCompare
the bond
lengths
to the average
lengths
sp3 hybrid orbitals involved in the C – C bond (they are 25% s and 75% p). Specifically, the
listed
Tableare8.5.
Can toyou
explain
differences?
electrons found
in ain
s orbital
held closer
the nucleus
and any
as a result
cause their bonds to be
shorter than bonds formed as a result of overlapping p orbitals. If a triple bond was present it would
is a six-membered
ring
(B) shorter
The structure
of borazine,
be even
since the sp hybrid
orbital has B
the
most
(they are 50% s and
50%of
p).
3N
3Hs6,character
alternating B and N atoms. There is one H atom bonded to each
B and to each N atom. The molecule is planar. (B) Write a Lewis
structure for borazine in which the formal charges on every atom
is zero. (C) Write a Lewis structure for borazine in which the octet
rule is satisfied for every atom. (D) What are the formal charges on
the atoms in the Lewis structure from part (b)? Given the electronegativities of B and N, do the formal charges seem favorable or
unfavorable? (E) Do either of the Lewis structures in parts (a) and
(b) have multiple resonance structures? (F) What are the hybridizations at the B and N atoms in the Lewis structures from parts
(a) and (b)? Would you expect the molecule to be planar for both
Lewis structures? (G) The six B ¬ N bonds in the borazine molecule are all identical in length at 1.44 Å. Typical values for the
bond lengths of B ¬ N single and double bonds are 1.51 Å and
1.31 A° , respectively. Does the value of the B ¬ N bond length
seem to favor one Lewis structure over the other? (H) How many
electrons are in the p system of borazine?
Suppose that silicon could form molecules that are precisely
the analogs of ethane 1C2H62, ethylene 1C2H42, and acetylene
1C2H22. How would you describe the bonding about Si in
terms of hydrid orbitals? Silicon does not readily form some
of the analogous compounds containing p bonds. Why might
this be the case?
One of the molecular orbitals of the H2- ion is sketched below: