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Chemistry for Engineers
Homework 5 (part 2)
BONDING
Answer Key
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Hand in by Wednesday
Chemistry for Engineers
25 October 13:00
Homework 5 (part 2): Bonding
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Q1: Identify the correct labelling for the polarity of these molecules
e.g. “d) - Y)”. For each molecule, deduce the oxidation state of each element.
W)
X)
F
F
F
δ+
δ−
δ−
F
F
δ+
δ+
δ−
δ−
H
H
δ−
δ−
δ+
δ+
S
H
H
δ+
I
δ−
S
δ−
δ−
δ+
δ−
H
…...of the following molecules:
Si
I
S
S
H
I
F
δ+ F
F
δ+
δ+
c)
δ−
δ−
δ+
δ+
Si
I
I
F
F
I
F
S i δ+
Si
I
F
F
δ−
δ−
δ+
F
F δ−
I
b)
i) geometry of bonding (including likely bond angles)
at the atoms indicated
ii) molecule polarity
δ−
Se
Se
Se
Se
F
δ+
H
H
δ+
δ+
(9 marks)
Q2:By counting (and stating) the number of valence electrons
possessed by each of the following species, write Lewis structures
for:
IBr
SF3+
SiF5COCl2
(12 marks)
Q3:
For the following different species:
PF6
-
PF3
Q4: Draw Lewis structure then use VSEPR to describe
the…..
Z)
δ+
δ+
δ−
a)
Y)
PF4
a)
b)
c)
d)
e)
Al in AlCl3
S in SF2
C in HCCH
Kr in KrF2
I in ICl3
(20 marks)
Q5:
a) Draw the Lewis resonance structures of the phosphate
anion (PO43-)
b) What is the charge on each oxygen atom?
+
PF2
-
a) deduce the oxidation state of phosphorus
b) how many valence electron pairs are there? Draw the Lewis structure
c) What is the P-O bond order?
(11 marks)
Total: 70 marks
c) what is the geometry of bonding (e.g. linear, trigonal bipyramidal etc)
d) Hence list the four species in order of increasing F-P-F bond angle
(smallest angle to largest)
(18 marks)
1
Q1: Identify the correct labelling for the polarity of these molecules
e.g. “d) - Y)”. For each molecule, deduce the oxidation state of each
element.
W)
X)
δ+
F
F
a)
F
F
δ−
F
δ+
F
δ+
F
F
Se
δ−
Se
δ+
Se
δ+
δ−
δ−
F
δ+
δ−
I
δ−
I
Si
I
δ+
δ+
F
I
Si δ+
F δ−
δ−
δ−
H
c)
Z)
δ−
δ−
Se
I
b)
Y)
δ−
H
S
δ+
F
δ−
H
F
F
δ+
δ+
δ+
δ+
H
H
H
S is -2
H is +1
S
S
δ−
δ−
δ+
I
δ+
δ+
H
S
Si is +4
F is -1
I is -1
I
δ−
Si
I
δ−
H
δ+
δ+ F
Si
F
Se is +2
F is -1
(Q1: 3 x 3 marks)
Q2:Write Lewis structures for:
IBr
SF3+ SiF5- COCl2
IBr
Count up valence electrons
Count up valence electron pairs
= 14
=7
Determine which atoms are bond to which (struc. formula) = Br I
Form single bonds between neighbouring atoms:
(3 electron pairs remain unused)
Br
I
Use remaining electron pairs to complete octets of each atom:
Br
I
(Q2: 4 x 3 marks)
2
Q2:Write Lewis structures for:
IBr
SF3+ SiF5- COCl2
SF3+
Count up valence electrons
Count up valence electron pairs
= 26
= 13
F
Determine structural formula:
Form single bonds between neighbouring atoms:
(10 electron pairs remain unused)
F
S
F
Use remaining electron pairs to complete octets of each atom:
(F octet must be completed first)
leaves one lone pair on S
F
S
F
+
F
(Q2: 4 x 3 marks)
Q2:Write Lewis structures for:
IBr
SF3+ SiF5- COCl2
SiF5Count up valence electrons
Count up valence electron pairs
= 40
= 20
_
Determine structural formula:
Form single bonds between neighbouring atoms:
(15 electron pairs remain unused)
F
F
F
Si
F
F
Use remaining electron pairs to complete octets of each atom:
(F octet must be completed first)
leaves no lone pairs on Si
F
F _Si
F
F
F
(Q2: 4 x 3 marks)
3
Q2: COCl2
Count up valence electrons
Count up valence electron pairs
= 24
= 12
O
Determine structural formula:
Cl
Place C at centre and form single bonds between
neighbouring atoms: (9 electron pairs remain unused)
Use remaining electron pairs to complete
octets of each electronegative atom:
C
Cl
_
+
C
F
O
F
Not enough electrons to complete carbon’s octets
so C=O double bond must form
using O lone pair:
(Q2: 4 x 3 marks)
F
C
F
O
(1 mark)
Q3:
For the following species:
PF6-
PF3
PF4+
PF2-
a) deduce the oxidation state of phosphorus
b) how many valence electron pairs are there? Draw the Lewis structure
c) what is the geometry of bonding (e.g. linear, trigonal bipyramidal
etc)
d) Hence list the four species in order of increasing F-P-F bond angle
(smallest angle to largest)
(Q3: 4 + 8 + 4 + 2 marks)
4
Q3:
For the following species:
PF6+5
48
24
P ox. state
valence electrons
valence electron pairs
F
_
F
P
F
F
F
F
F-P-F bond angle
geometry
F
F
P
PF4+
+5
32
16
PF3
+3
26
13
PF2+3
20
10
F
F
F
P+
F
F
_
P
F
F
PF690º
PF2~106º
octahedral
tetrahedral
distorted by
two lone pairs
PF4+
109.5º
PF3
~108º
tetrahedral
distorted by
one lone pair
tetrahedral
(Q3: 4 + 8 + 4 + 2 marks)
Q4: Use VSEPR to describe the
i) geometry of bonding (including likely bond angles) at the atoms
indicated
ii) molecule polarity
of the following molecules:
a)
b)
c)
d)
e)
Al in AlCl3
S in SF2
C in HCCH
Kr in KrF2
I in ICl3
(Q3: 5 x 4 marks)
5
Recap: how to determine hybridisation:
Strategy: The steps for determining the hybridization of the central
atom in a molecule are:
-count total number of valence electrons
-draw
Lewis Structure to determine no. of electron pairs around central
atom
sp2
sp
sp3
-use VSEPR to
determine the bonding
geometry
-relate geometry to
hybridisation using:
sp3d
sp3d2
a) Al in AlCl3 - 24 valence electrons (12 pairs)
single bonds - 3 pairs
complete Cl octets so:
Cl lone pairs - 9 pairs
Cl
Cl
Cl
cannot complete Al octet
AlCl3 is electron
deficient
Al
sp2
sp
sp3
Al - 3 bonding pairs
trigonal geometry
(non-polar)
sp3d
sp3d2
6
b) S in SF2 - 20 valence electrons (10 pairs)
single bonds - 2 pairs
complete F octets so:
F lone pairs - 6 pairs
F
F
S
2 pairs left to complete S octet
S - 4 surrounding
electron pairs
sp3
sp2
sp
distorted tetrahedral
geometry
-polar
F-S-F bond angle < 109.5º
sp3d
sp3d2
c) C in C2H2 - 10 valence electrons (5 pairs)
single bonds - 3 pairs
reamining electrons must be put on C - 2 pairs
C octets incomplete - must form
triple bond
multiple bonds have no
affect on geometry
each C surrounded by 2
bonding pairs
H C
H C
C
H
H
sp2
sp
C
sp3
linear geometry
(non-polar)
sp3d
sp3d2
7
d) Kr in KrF2 - 22 valence electrons (11 pairs)
single bonds - 2 pairs
complete F octets so:
F lone pairs - 6 pairs
3 electron pairs remain
must be placed on Kr
(octet exceeded)
F
F Kr
F
Kr - 5 surrounding electron
pairs
-trigonal bipyramidal
geometry
-lone pairs in equator
-molecule is linear ⇒ non-polar
F
e) I in ICl3 - 28 valence electrons (14 pairs)
single bonds - 3 pairs
complete Cl octets so:
Cl lone pairs - 9 pairs
Cl
I
Cl
Cl
2 electron pairs remain
must be placed on I
(octet exceeded)
Cl
I - 5 surrounding electron pairs
Cl
-trigonal bipyramidal
geometry
-lone pairs in equator
-molecule is distorted T-shaped (angles < 90º) ⇒ polar
Cl
8
Q5:
a) Draw the Lewis resonance structures of the phosphate anion (PO43-)
valence electrons = 32 (16 pairs)
place P as central atom and use 4 pairs – single bonds
_
_
+
P
O
_O
O
O_
_
O
P can expand valency so double bond can be drawn:
now move π bond:
_
_
O
_O
P
O
O_
O
_O
P
_
_
O
O_
O
P
O
O
P
O_
_O
_
_
O
O_
O
_O
P
O
O
(Q5: 8 + 1.5 + 1.5 marks)
Total: 70 marks
b) What is the charge on each oxygen atom?
- 0.75
c) Given that the strength of a typical P-O single bond is 200
kJ/mol, estimate the P-O bond strength in the phosphate ion.
Bond order = 1.25
Bond strength expected = 250 kJ/mol.
(Q5: 8 + 1.5 + 1.5 marks)
Total: 70 marks
9