2. Unit One
Parts 3&4
H O H3C Br O Br
H CH3
Locating electrons
Describing bonds Pages
Shape of molecules 34 & 43
3. Unit One
3&4
if we know where
Parts
electrons are we can
predict reactions and
shape...they really are
key to understanding
chemistry
H O H3C Br O Br
H CH3
Locating electrons
Describing bonds Pages
Shape of molecules 34 & 43
4. Unit One
Parts 3&4
H O H3C Br O Br
H CH3
as I’ve taken the
material out of order,
I’ll give you some
Locating electrons
page numbers
Describing bonds Pages
Shape of molecules 35 & 45
21. 1 18
1
H
H He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s1
energy
2s 2px 2py 2pz
hydrogen
Pg
1s 43
22. 1 18
1
H
H He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s1
energy
2s 2px 2py 2pz
just one electron
hydrogen
so in first orbital
Pg
1s 43
24. 1 18
2
H He
He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s2
energy
2s 2px 2py 2pz
helium
Pg
1s 43
25. 1 18
2
H He
He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s2
energy
2s 2px 2py 2pz
one electron has spin
+½ (up) and the other
spin –½ (down)
helium
Pg
1s 43
26. 1 18
2
H He
He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s2
energy
2s 2px 2py 2pz doesn’t matter what it
means...just remember
helium
an electron can only be
up or down
Pg
1s 43
27. 1 18
2
H He
He
2 13 14 15 16 17
Li Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s2
energy
2s 2px 2py 2pz
so can only ever
have two electrons
per orbital helium
Pg
1s 43
28. 1 18
H He
2 13 14 15 16 17
3
Li Be B C N O F Ne
Li
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
2px 2py 2pz
1s22s1
energy
2s
lithium
Pg
1s 43
29. 1 18
H He
2 13 14 15 16 17
3
Li Be B C N O F Ne
Li
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
2px 2py 2pz
1s22s1
energy
lithium obeys both
rules...fill lowest orbital
2s first (until full) then fill
next lowest)
lithium
Pg
1s 43
30. 1 18
H He
2 13 14 15 16one more
...adding 17
4 electron is easy...
Li Be
Be B C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
2px 2py 2pz
1s22s2
energy
2s
beryllium
Pg
1s 43
31. 1 18
H He
2 13 14 15 16 17
5
Li Be B
B C N O F Ne
Na Mg ...and another... Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
2px 2py 2pz
1s22s22p1
energy
2s
boron
Pg
1s 43
32. 1 18
H He
2 13 14 15 16 17
5
Li Be B
B C N O F Ne
Na Mg it could go in any of
Al Si P S Cl Ar
3 4 5 6 7 8 9 10 x,11 y12 2pz,
2p 2p or
they’re identical...well
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn they are As Se Br Kr
energetically Ga Ge
2px 2py 2pz
1s22s22p1
energy
2s
boron
Pg
1s 43
33. 1 18
H He
2 13 14 15 16 17
5
Li Be B
B C N O F Ne
but, where does
Na Mg the next (and most Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
important as its
K Ca Sccarbon) go?? Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Ti V Cr
2px 2py 2pz
1s22s22p1
energy
2s
boron
Pg
1s 43
34. Hund's rule
electrons as far apart as
possible
(de ge n er a t e o rb i tals)
(as long as it doesn’t
violate any of the
previous rules!)
35. Hund's rule
makes sense as like
charges always
repel...
electrons as far apart as
possible
(de ge n er a t e o rb i tals)
36. 1 18
H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s22s22p 12p 1
2px 2py 2pz
x y
energy
2s
1s22s22p2
carbon
Pg
1s 43
37. 1 18
H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s22s22p 12p 1
2px 2py 2pz could be 2pz,
makes no
x y
difference...
energy
2s
1s22s22p2
carbon
Pg
1s 43
38. that's a lot of
electrons...
luckily we don’t care
about all them...
40. 1 18
H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s22s22p2
carbon
atomic = number of
number electrons Pg
45
41. Valence electrons
1 18
H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
1s22s22p 12p 1
2px 2py 2pz
x y
energy
2s 1s22s22p2
carbon
Pg
1s
43
42. Valence electrons
1 18
H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne
Na Mg Al Si P S Cl Ar
3 4 5 6 7 8 9 10 11 12
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
only need consider high
energy electrons or those
on the outside called the
1s22s22p 12p 1
2px 2py 2pz valence electrons.
x y
energy
2s 1s22s22p2
carbon
Pg
1s
44
43. C C
if we consider the Bohr
model of the atom, the
1s22s22p2 atomwhere we thinkplanet 2
one
resembling a
of an
2s22p
with moons orbiting (or
the solar system)
N N
1s22s22p3 2s22p3
group 1 2 13 14 15 16 17 18
H He
Li Be B C N O F Ne Pg
44
44. C C
1s22s22p2 2s22p2
then the valence
electrons are those on the
outer edge (like Neptune
for young-upstarts or
Pluto for us oldies) N N
1s22s22p3 2s22p3
group 1 2 13 14 15 16 17 18
H He
Li Be B C N O F Ne Pg
44
45. C C
1s22s22p2 2s22p2
then the valence
electrons are those on the
outer edge (like Neptune
for young-upstarts or
Pluto for us oldies) N N
1s22s22p3 2s22p3
group 1 2 13 14 15 16 17 18
H He
Li Be B C N O F Ne Pg
44
46. C C
1s22s22p2 2s22p2
N N
1s22s22p3 2s22p3
absolute
rubbish...but more
group 1 2 13 14 15 16 17 18
comprehendible!
H He
Li Be B C N O F Ne Pg
41
47. C C
1s22s22p2 2s22p2
N N an easy we to
remember the number
of valence electrons is
1s22s22p3 2s22p3
to take group
number...
group 1 2 13 14 15 16 17 18
H He
Li Be B C N O F Ne Pg
41
48. C C
1s22s22p2 2s22p2
N N
...and ignore
1s22s22p3 2s22p3 first ‘1’
valence
electrons 1 2 3 4 5 6 7 8
H He
Li Be B C N O F Ne Pg
41
49. C C
1s22s22p2 2s22p2
N N
1s22s22p3 2s22p3
valence
electrons 1 2 3 4 5 6 7 8 so oxygen
(group 16) has
H He 6 valence
Li Be B C N O F Ne
electrons
Pg
41
66. C
1s22s22p2
4 bonds
hopefully, you can see
3
this is where those
magic numbers in
lecture one came
N bonds from!
1s22s22p3
O
1s22s22p4
2 bonds
Pg
34
67. Pg
8
36
H
H C H
H
Octet rule: 8 valence electrons
68. Pg
8
37/46
H H
H C N O
H H
Octet rule: 8 valence electrons
69. Pg
Lewis structures 37/46
Hydrofluoric acid HF
H + F H F ≡ H F
use octet rule to draw
Methanol CH OH
3
the structure of stable
molecules...
H H
C + O + 4H H C O H ≡H C O H
H H
70. Pg
Lewis structures 41
Hydrofluoric acid HF
H–F easy..H = 2
electrons (full s
orbital) & F = 8...
H + F H F ≡ H F
Methanol CH3OH
H H
C + O + 4H H C O H ≡H C O H
H H
71. Pg
Lewis structures 37/46
Lewis structure shows
all valence electrons
Hydrofluoric acid HF
represented by our
simple diagram H–F
H + F H F ≡ H F
Methanol CH3OH
H H
C + O + 4H H C O H ≡H C O H
H H
72. Pg
Lewis structures 37/46
Hydrofluoric acid HF
H + F H F ≡
works for
more complex
H F
molecules
Methanol CH3OH
H H
C + O + 4H H C O H ≡H C O H
H H
73. Pg
Lewis structures 37/46
Hydrofluoric acid HF
H + F H F ≡ H F
Methanol CH3OH
Note: it helps to leave
lone pairs (of electrons)
on diagram...this is
where a lot of chemistry
occurs... H
H
C + O + 4H H C O H ≡H C O H
H H
74. Acetone CH3COCH3
3 C + O + 6H
how do we deal
with more complex
molecules?
Pg
44
75. Acetone CH3COCH3
3 C + O + 6H
first draw all the
atoms where you think O
they might go...
H H
C
H C C H
H H
Pg
44
76. Acetone CH3COCH3
3 C + O + 6H
now join all the atoms
together...some of the
atoms have full
valence shells so we
can draw them in as on O
the next slide...
H C H
C C
H H
H H
Pg
44
77. Acetone CH3COCH3
3 C + O + 6H
the central C and O
both have only 7
valence electrons...
O
H C H
C C
H H
H H
Pg
44
78. Acetone CH3COCH3
3 C + O + 6H
O
...but if they share 4
electrons they both have H C H
8 valence electrons...this C C
gives us a double bond
H H
(alkene)
H H
O O
C
≡ Pg
H3C CH3 44
83. –
Borohydride
anion BH4 –
add electron
does it matter
which atom we give
the electron to?
H H
B + 3H + H H B H ≡ H B H
H H
Pg
44
84. –
Borohydride
anion BH4 –
add electron
does it matter
which atom we give
the electron to?
H H
B + 3H + H H B H ≡ H B
no! (but in this case
H
H H– makes more
H
chemical sense)
Pg
44
85. +
Ammonium
cation NH4 +
lose electron
if we have a
positive charge
(cation) we do
the opposite...
Pg
44
93. this is ‘electron book-
keeping’...we are just
assigning charge to one
atom to help explain
chemistry...
formal charges localise
charge on an atom...
94. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
...on an atom
Pg
47
95. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
...according to
the atoms
position in the
periodic table
Pg
47
96. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
...in lone pairs...
Pg
47
97. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
...or the number
of bonds to that
atom
Pg
47
98. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
H H
N + 3H + H H N H ≡ H N H
H H
cation
N fc = 5-0-½(8)=+1
Pg
47
99. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
H H
N + 3H + H H N H ≡ H N H
H H
no charge on H as:
cation
H = 1-0-½(2) = 0 N fc = 5-0-½(8)=+1
Pg
47
100. formal number of number of number of
charge = valence – unshared – bonds
(fc) electrons electrons
H H
the simplified
N + 3Hformula of bonds)
+ (just use
number
H H N H ≡ H N H
H H
cation
N fc = 5-0-4=+1
Pg
47
101. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone
Pg
47
102. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone
lhs O; fc = 6-4-½(4)=0
Pg
47
103. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone
lhs O; fc = 6-4-½(4)=0
central O; fc = 6-2-½(6)=+1
rhs O; fc = 6-6-½(2)=-1 Pg
47
104. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone
lhs O; fc = 6-4-½(4)=0
central O; fc = 6-2-½(6)=+1
rhs O; fc = 6-6-½(2)=-1 Pg
47
105. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
≡ O O
O
O3 neutral atom's formal
ozone charges
lhs O; fc = 6-4-½(4)=0
central O; fc = 6-2-½(6)=+1
rhs O; fc = 6-6-½(2)=-1 Pg
47
106. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
O
O + O + O
O O
O ≡ O
O
≡ O O
O
ozone neutral as
O3 neutral atom's formal
+ & – cancel each
ozone charges
other out
lhs O; fc = 6-4-½(4)=0
central O; fc = 6-2-½(6)=+1
rhs O; fc = 6-6-½(2)=-1 Pg
47
107. formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
these charges
explain why
ozone is so
reactive!
O
O + O + O
O O
O ≡ O
O
≡ O O
O
O3 neutral atom's formal
ozone charges
lhs O; fc = 6-4-½(4)=0
central O; fc = 6-2-½(6)=+1
rhs O; fc = 6-6-½(2)=-1 Pg
47
108. formal number of number of number of
charge = valence – unshared – bonds
(fc) electrons electrons
O
O + O + O
O O
O ≡ O
O
≡ O O
O
O3 neutral atom's formal
the simplified ozone charges
formula (just use
number of bonds)
lhs O; fc = 6-4-2=0
central O; fc = 6-2-3=+1
rhs O; fc = 6-6-1=-1 Pg
47
127. our simple Lewis model
helps explain a lot of
chemistry...especially
reactions... what is a
bond?
128. what is a
bond?
...but it fails to explain
such fundamental
concepts as shape...
129. ...actually, it can explain
shape if we use VSEPR
theory...but anyways,
lets use those orbitals
what is a
bond?
130. single (σ) bond
H• + H• H H
energy
here we have 2
hydrogen atoms
(each with 1 electron
in a 1s orbital)
Pg
H•
1s
H•
1s 37
131. single (σ) bond
H• + H• H H
σ*
to form a covalent
bond they must
energy
share their
electrons...
σ Pg
H•
1s
H–H H•
1s 35
132. single (σ) bond
H• + H• H H
σ*
energy
...this is achieved
by combining the
two atomic
Pg
orbitals to give...
σ
H•
1s
H–H H•
1s 35
133. single (σ) bond
H• + H• H H
σ* ...a new molecular
orbital, a sigma σ
orbital (or bond)
energy
σ Pg
H•
1s
H–H H•
1s 35
134. single (σ) bond
H• + H• H H
...this bonding
σ* orbital is lower in
energy than the
atoms...so a bond
will form
energy
σ Pg
H•
1s
H–H H•
1s 35
135. single (σ) bond
H• + H• H H
a consequence of the
maths is we also get an
anti-bonding sigma
orbital (σ*)...2 orbitals
σ*
must give 2 new orbitals
energy
σ Pg
H•
1s
H–H H•
1s 37
136. single (σ) bond
H• + H• H H
σ*
energy
...but lets ignore this
confusing little devil for
the time being!
σ Pg
H•
1s
H–H H•
1s 37
137. single (σ) bond
it is called a σ orbital as
is symmetrical along
bond axis (you can rotate
it like a cylinder and it
doesn’t change)
Pg
H H 47
138. single (σ) bond all bonds to H are
sigma (as all are like a
cylinder)...here we
overlap 1s of H with 2p
of C and get sigma
bond)
C• + H• C H
Pg
37
139. Pg
single (σ) bond 38
σ*
energy
if we take two 2p
orbitals and combine
them head-to-head
C• σ C•
2py C–C 2py
140. Pg
single (σ) bond 38
σ*
...we get a sigma
σ bonding
orbital...it is still
energy
like a cylinder...
C• σ C•
2py C–C 2py
141. Pg
single (σ) bond 38
σ*
energy
...this is the
normal single
bond we observe
in alkanes etc.
C• σ C•
2py C–C 2py
142. Pg
single (σ) bond 38
σ*
this is one orbital
NOT three
energy
C• σ C•
2py C–C 2py
143. single (σ) bond
the blue bit is the
sigma orbital...ignore
Pg
the red orbitals for
the time being...
35
144. single bond
or the simple
C C
version...
THIS IS ALL YOU
NEED TO KNOW
σ (sigma) bond
161. H3C CH3 CH3
CH3 CH3
the p bond prevents
O H
alkenes from rotating (the
two bonds can’t twist pass
multistep enzyme- each other)...
light isomerises
catalysed reverse complexed
process cis-retinal
H3C CH3 CH3 CH3 O
H
CH3
Pg
38
162. H3C CH3 CH3
CH3 CH3
this can effect
O H
shape of molecule
multistep enzyme- light isomerises
catalysed reverse complexed
process cis-retinal
H3C CH3 CH3 CH3 O
H
CH3
Pg
38
163. H3C CH3 CH3
CH3 CH3
O H
we must break π
bond before
multistep enzyme- light isomerises
catalysed reverse complexed alkene can rotate
process cis-retinal
H3C CH3 CH3 CH3 O
H
CH3
Pg
38
164. H3C CH3 CH3
CH3 CH3
the change in
O H
shape initiates the
visual cascade and
multistep enzyme- light isomerises our sight
catalysed reverse complexed
process cis-retinal
H3C CH3 CH3 CH3 O
H
CH3
Pg
38
165. H3C CH3 CH3
CH3 CH3
O H
multistep enzyme- light isomerises
catalysed reverse complexed
process cis-retinal
H3C CH3 CH3 CH3 O why do you think
red path is easy
H but blue hard?
CH3
Pg
38
166. double bond
or the simple
version...
THIS IS ALL YOU
NEED TO KNOW
π (pi) bond
168. sp2
an atom with three σ
orbitals and one π
orbital is called an sp2
atom (we only count the
C
orbitals used in making
Pg
s orbitals)
38
169. sp2 1
3
3
2
1 double bond and 2
single bonds and we
points have an sp2 atom
170. trigonal
planar
120°
sp2 atoms are trigonal planar
sp2
(flat and pointing to the
corners of a triangle)...again,
this is because the orbitals
Pg try to be as far apart as
possible
41
171. trigonal Pg
41
planar
sp 2
maximum separation of three points
maximum separation of three valence electron pairs
172. triple (σ + 2x π) bonds
σ
H C C H
π (2pz + 2pz) σ
a triple bond (like an π (2py + 2py)
alkyne) is formed from one
σ bond and two π bonds (at
right angles to each other
due to the direct of the p
orbitals that made them) σ π
H C C
π
H Pg
39
173. triple (σ + 2x π) bonds so...two p orbitals combine
head-to-head to give a σ bond
and two pairs of p orbitals
combine side-to-side to give
the two π orbitals (& there are
only two π orbitals)
σ
H C C H
π (2pz + 2pz) σ
π (2py + 2py)
σ π
H C C
π
H Pg
39
174. sp
an atom with two σ
orbitals and two π orbitals
is called an sp atom (as
two orbitals made the
basic σ skeleton)
C
Pg
39
178. H3C
H
CO2H
OH O
O
OCH3
H
here is a real
OH O molecule...we should be
dynemicin A able to identify the types
of atoms present...
Pg
40
179. four groups attached so
it must be sp3 and as
those groups try to stay
as far apart as possible
it is tetrahedral
H3C
H
CO2H
OH O
O
OCH3
H
OH O
dynemicin A
sp3
tetrahedral
Pg
40
180. ...only three groups so
sp2 and flat, trigonal
planar
H3C
H
CO2H
OH O
O sp2
OCH3 trigonal
H
planar
OH O
dynemicin A
sp3
tetrahedral
Pg
40
181. sp
linear
straight line, two
groups must be sp
H3C
H
and linear
CO2H
OH O
O sp2
OCH3 trigonal
H
planar
OH O
dynemicin A
sp3
tetrahedral
Pg
40
182. what is
oxygen?
H3C
H
CO2H
OH O
O
OCH3
H
OH O
dynemicin A
Pg
40
183. ...is it sp as
what is attached to two
oxygen? carbon atoms?
H3C
H
CO2H
OH O
O
OCH3
H
OH O
dynemicin A
Pg
40
184. sp, sp2 or sp3?
H
O look at a simpler
system...water, sp,
H sp2 or sp3?
185. sp, sp2 or sp3?
H
O H draw Lewis
structure...
186. sp, sp2 or sp3?
H
O H we have FOUR
groups around O,
two lone pairs &
two H atoms. So it
is...