lecture 3: 123.101

Unit One Parts 3 & 4:
molecular bonding

Unit One
Parts 3&4
H O H3C Br O Br
H CH3

Locating electrons
Describing bonds Pages
Shape of molecules 34 & 43

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

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


Na Cl

Na Cl
here we have an
atom of sodium (Na)
and an atom of

Ionic bonds
chlorine (Cl)

Pg
34

if we take one
electron from Na and
give it to Cl...

Na Cl

Na Cl

Ionic bonds Pg
34

Na Cl

Na+ Cl-

we get 2 charged species
(cation = positive charge &
anion = negative charge)

Ionic bonds Pg
34

Na+ Cl-

NaCl
opposite charges
attract and give us an
ionic bond

Ionic bonds Pg
34

covalent bonds
H H

if we bring 2 atoms
together and they...

Pg
34

covalent bonds
H H

share their 2
electrons we have a
covalent bond Pg
34

covalent bonds
H H
this is the bond we’ll be
dealing with most often
and is represented by
the black line

H H

2 electrons
per bond
Pg
34

covalent bonds
H H

H H
please remember

2
that this line is
2 electrons

electrons
per bond
Pg
34

chemistry
apain

these are
just
extremes

where do we ﬁnd
electrons?

Aufbau Principle

THIS IS THE LONG
VERSION...NOT THE
VERSION I DO IN
THE LECTURES

lowest energy orbital

Aufbau Principle

don’t worry about
the name...just that
electrons like to have
lowest energy
possible...

lowest energy orbital

rather like many
students...

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

1 18
1
H
H He
2 13 14 15 16 17

Li Be B C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s1
energy

2s 2px 2py 2pz
just one electron

hydrogen
so in first orbital

Pg
1s 43

Pauli Exclusion Principle

no two electrons
are identical

1 18
2
H He
He
2 13 14 15 16 17

Li Be B C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s2
energy

2s 2px 2py 2pz

helium
Pg
1s 43

1 18
2
H He
He
2 13 14 15 16 17

Li Be B C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s2
energy

2s 2px 2py 2pz
one electron has spin
+½ (up) and the other
spin –½ (down)
helium
Pg
1s 43

1 18
2
H He
He
2 13 14 15 16 17

Li Be B C N O F Ne

3 4 5 6 7 8 9 10 11 12


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

1 18
2
H He
He
2 13 14 15 16 17

Li Be B C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s2
energy

2s 2px 2py 2pz
so can only ever
have two electrons
per orbital helium
Pg
1s 43

1 18

H He
2 13 14 15 16 17
3
Li Be B C N O F Ne
Li
3 4 5 6 7 8 9 10 11 12


2px 2py 2pz

1s22s1
energy

2s
lithium
Pg
1s 43

1 18

H He
2 13 14 15 16 17
3
Li Be B C N O F Ne
Li
3 4 5 6 7 8 9 10 11 12


2px 2py 2pz

1s22s1
energy

lithium obeys both
rules...fill lowest orbital
2s first (until full) then fill
next lowest)
lithium
Pg
1s 43

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

3 4 5 6 7 8 9 10 11 12


2px 2py 2pz

1s22s2
energy

2s
beryllium
Pg
1s 43

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


2px 2py 2pz

1s22s22p1
energy

2s
boron
Pg
1s 43

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

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

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!)

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)

1 18

H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s22s22p 12p 1
2px 2py 2pz
x y
energy

2s
1s22s22p2
carbon
Pg
1s 43

1 18

H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s22s22p 12p 1
2px 2py 2pz could be 2pz,
makes no
x y
difference...
energy

2s
1s22s22p2
carbon
Pg
1s 43

that's a lot of
electrons...

luckily we don’t care
about all them...

all you have to
remember is...

©jaci XIII@ﬂickr

1 18

H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s22s22p2
carbon
atomic = number of
number electrons Pg
45

Valence electrons
1 18

H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne

3 4 5 6 7 8 9 10 11 12


1s22s22p 12p 1
2px 2py 2pz
x y
energy

2s 1s22s22p2
carbon
Pg
1s
43

Valence electrons
1 18

H He
2 13 14 15 16 17
6
Li Be B C
C N O F Ne

3 4 5 6 7 8 9 10 11 12


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

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

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

44

C C

1s22s22p2 2s22p2

N N

1s22s22p3 2s22p3

absolute
rubbish...but more
group 1 2 13 14 15 16 17 18
comprehendible!

H He

41

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

41

C C

1s22s22p2 2s22p2

N N

...and ignore
1s22s22p3 2s22p3 first ‘1’

valence
electrons 1 2 3 4 5 6 7 8
H He

41

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

what do valence
electrons tell us?

H
1 2

H H
4
3 O
H
the number of
bonds

the shape of
molecules

109°

atoms are
happy if they
have a full
valence shell...

Ne

1s22s22p6

noble gas

Ne ...commonly
this means 8
electrons

1s22s22p6

noble gas

8
electrons
Ne

full
shell

C

1s22s22p2
4 bonds

N

1s22s22p3
3 bonds

O

1s22s22p4
2 bonds
Pg
36

Pg
45

C 4
valence
electrons
so for carbon to
get to 8 it needs
4 more electrons

1s22s22p2
Pg
45

Pg
46

C
or 4 new
covalent bonds

4 bonds Pg
46

C

nitrogen has 5
valence1s22s22p2
electrons...so
4 bonds

needs 3 more...

N

1s22s22p3
3 bonds

O

1s22s22p4
2 bonds
Pg
36

N 5
valence
electrons

1s22s22p3
Pg
46

so forms 3
covalent bonds

N

3 bonds Pg
46

C

1s22s22p2
4 bonds

oxygen needs 2
more electrons so
forms 2 covalent
bonds
N

1s22s22p3
3 bonds

O

1s22s22p4
2 bonds
Pg
36

O 6
valence
electrons

1s22s22p4
Pg
46

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

Pg

8
36

H
H C H
H
Octet rule: 8 valence electrons

Pg

8
37/46
H H
H C N O
H H
Octet rule: 8 valence electrons

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

Pg
Lewis structures 41
H–F easy..H = 2
electrons (full s
orbital) & F = 8...
H + F H F ≡ H F

Methanol CH3OH

H H
H H

Pg
Lewis structure shows
all valence electrons
represented by our
simple diagram H–F

H + F H F ≡ H F

Methanol CH3OH

H H
H H

Pg

H + F H F ≡
works for
more complex
H F
molecules

Methanol CH3OH

H H
H H

Pg

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
H H

Acetone CH3COCH3
3 C + O + 6H

how do we deal
with more complex
molecules?

Pg
44

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

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

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

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

–
Borohydride
anion BH4 –

what happens if
we have a negative
charge (anion)?

Pg
44

–
Borohydride
anion BH4 –

take the atoms
as normal and...

B + 3H + H

Pg
44

–
Borohydride
anion BH4 –

add electron

B + 3H + H
...add an electron

Pg
44

–
Borohydride
anion BH4 –

add electron

H H
B + 3H + H H B H ≡ H B H
H H
Pg
44

–
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

–
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

+
Ammonium
cation NH4 +

lose electron

if we have a
positive charge
(cation) we do
the opposite...

Pg
44

+
Ammonium
cation NH4 +

lose electron

start with our
normal atoms...

N + 3H + H

Pg
44

+
Ammonium
cation NH4 +

lose electron

N + 3H + H
then remove an
electron

Pg
44

+
Ammonium
cation NH4 +

lose electron

H H
N + 3H + H H N H ≡ H N H
H H
Pg
44

where is the
charge?

is it on one atom?

all over the molecule...

No, its all over the
molecule! But...

but the truth isn't useful, so...

formal charges localise
charge on an atom...

this is ‘electron book-
keeping’...we are just
assigning charge to one
atom to help explain
chemistry...

formal charges localise
charge on an atom...

formal number of number of ½ number
charge = valence – unshared – of shared
(fc) electrons electrons electrons
...on an atom

Pg
47

...according to
the atoms
position in the
periodic table

Pg
47

...in lone pairs...

Pg
47

...or the number
of bonds to that
atom

Pg
47


H H
N + 3H + H H N H ≡ H N H
H H
cation
N fc = 5-0-½(8)=+1
Pg
47


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

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


O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone

Pg
47


O
O + O + O
O O
O ≡ O
O
O3 neutral
ozone

lhs O; fc = 6-4-½(4)=0

Pg
47


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


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


O
O + O + O
O O
O ≡ O
O
≡ O O
O
ozone neutral as
+ & – 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


these charges
explain why
ozone is so
reactive!
O
O + O + O
O O
O ≡ O
O
≡ O O
O
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

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
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

a bond
O

2
H
O
is electrons

a bond

2
is electrons

quantum model
more accurate...

Atomic orbitals
it's a quantum world...

we’ve looked at a
nice simple model
so far...

Atomic orbitals
it's a quantum world...

mathematicians and
physicists have shown it’s a bit
more complicated in ‘reality’

but I don't like maths...

so...here's some pretty pictures...

90%
atomic orbital

atomic orbital is the
volume of space in which
there is a 90% chance of
finding an electron
Pg
36

2
remember: a

atomic orbital maximum of 2
electrons per
orbital

electrons
Pg
37

a 1s orbital is also
a sphere...just a
lot smaller

2s Pg
38
Pic: Dr. Jonathan Gutow

let’s ignore this
nasty little effect
of maths...

2s Pg
38
Pic: Dr. Jonathan Gutow

px z py z pz z

y y y

x x x

2p Pg
38

px z py z pz z

y y y

x x x

2p
each of the three
2p orbitals is
dumbbell
shaped... Pg
38

px z py z pz z

y y y

x x x

2p
...they are
identical in all
ways except...
Pg
38

px z py z pz z

y y y

x x x

2p
...they point in
different directions
(hence the names)
Pg
38

px z py z pz z

y y y

this is one orbital
(just has two different
x x coloured areas)
x

2p Pg
34

our simple Lewis model
helps explain a lot of
chemistry...especially
reactions... what is a
bond?

what is a
bond?
...but it fails to explain
such fundamental
concepts as shape...

...actually, it can explain
shape if we use VSEPR
theory...but anyways,
lets use those orbitals

what is a
bond?

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

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

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

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

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

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

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

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

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

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

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

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

Pg
single (σ) bond 38
σ*
this is one orbital
NOT three
energy

C• σ C•
2py C–C 2py

single (σ) bond

the blue bit is the
sigma orbital...ignore

Pg
the red orbitals for
the time being...

35

single bond

or the simple
C C
version...
THIS IS ALL YOU
NEED TO KNOW

σ (sigma) bond

single bond

σ (sigma) bond

sp3
C
an atom with 4 σ
bonds is called an sp3
atom (as 1 x s and 3 x Pg
p used in bonding)
38

sp3 1
YOU NEED TO
KNOW THIS
4 2
4
points 3

tetrahedral H
Br
C
Br
H

109°

sp3
Pg
41

tetrahedral H C
Br

Br
H
sp3 atoms are
tetrahedral in shape
(the bonds stay as far
apart as possible)

109°

sp3
Pg
41

tetrahedral sp3

maximum separation of four points
maximum separation of four valence electron pairs

Pg
double (σ + π) bonds 38
C C

C=C
π*
energy

C=C
π
two 2p orbitals can
combine side-to-side

carbon carbon
2pz C C 2pz

Pg
C C

C=C
π*
energy

the new bond
is a pi π bond
C=C
π

carbon carbon
2pz C C 2pz

Pg
C C

C=C
π*
energy

C=C
π

here we have a
C–C σ bond and
carbon a pi π bond carbon
2pz C C 2pz

Pg
C C

C=C
π*
energy

C=C
π
the pi π bond is
one orbital (with
two bits to it)
carbon carbon
2pz C C 2pz

double (π) bond

Pg
38

double (π) bond

it is called a pi π orbital as
rotation around the C–C
axis causes a change (from
red to blue) so no longer
like a cylinder Pg
38

double (π) bond

remember: this is ONE
orbital (just two different
coloured halves)

Pg
35

double (π) bond

we have an inner σ bond
(the rod) and an outer π
bond (the orbital) hence it
is a double bond

Pg
38

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

H3C CH3 CH3

CH3 CH3
this can effect
O H
shape of molecule

multistep enzyme- light isomerises
process cis-retinal

H3C CH3 CH3 CH3 O

H

CH3

Pg
38

H3C CH3 CH3

CH3 CH3

O H
we must break π
bond before
catalysed reverse complexed alkene can rotate
process cis-retinal

H3C CH3 CH3 CH3 O

H

CH3

Pg
38

H3C CH3 CH3

CH3 CH3
the change in
O H
shape initiates the
visual cascade and
multistep enzyme- light isomerises our sight
process cis-retinal

H3C CH3 CH3 CH3 O

H

CH3

Pg
38

H3C CH3 CH3

CH3 CH3

O H

process cis-retinal

H3C CH3 CH3 CH3 O why do you think
red path is easy
H but blue hard?

CH3

Pg
38

double bond

or the simple
version...
THIS IS ALL YOU
NEED TO KNOW

π (pi) bond

double bond

π (pi) bond

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

sp2 1
3
3
2
1 double bond and 2
single bonds and we

points have an sp2 atom

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

trigonal Pg
41
planar
sp 2

maximum separation of three points
maximum separation of three valence electron pairs

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

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

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

linear

an atom with two
groups on it will be

180°
sp Pg
linear (a straight line)
as the orbitals stay as
far apart as possible

40

linear
sp

maximum separation of two points
maximum separation of two valence electron pairs

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

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

...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

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

what is
oxygen?
H3C
H
CO2H
OH O
O
OCH3
H

OH O
dynemicin A

Pg
40

...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

sp, sp2 or sp3?

H
O look at a simpler
system...water, sp,

H sp2 or sp3?

sp, sp2 or sp3?

H
O H draw Lewis
structure...

sp, sp2 or sp3?

H
O H we have FOUR
groups around O,
two lone pairs &
two H atoms. So it
is...

tetrahedral
H
O
H sp3

tetrahedral
H
O
H sp3
that is why we draw
water as a bent
molecule...its shape
is based on a
tetrahedron...

tetrahedral
H
O
...any atom with
four atoms or lone
pairs around it is
sp3 with all that
entails!
H sp3

sp, sp2
or sp
O
3?
what kind of atom
is the oxygen?
C
H H

1 double
bond
O ...and two lone
pairs, so three
groups around the
C oxygen so it is...

H H

trigonal
planar
H
O
C
H
sp2

sp, sp2
or sp3?
N C H
what kind of atom
is the nitrogen?

1 triple
bond
N C H
and one lone pair
so two groups so it
is...

what have
....we learnt?

•e l e c t r o n s
where they are
•b o n d s
what they are
Courtesy: National Science Foundation
•s h a p e

lecture 3: 123.101

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lecture 3: 123.101