Drawing Lewis structures
F
FCF
F
Lewis structures help us to see precisely
how the atoms in a molecule are bonded
together.
They also help us predict the shapes and
the properties of the molecules.
As it turns out, Lewis structures are
relatively easy to draw, as long as you
follow the five steps...
Step 1: Write down the central atom.
Usually the central atom will simply be the atom that
there’s just one of. For example in CH4, the C is the
central atom and in NF3, the N is the central atom. In
BrF5, the central atom is the Br, and in SO2, the central
atom is the S.
Sometimes, there is no central atom, for instance in HBr,
the H and the Br are just side by side. Likewise for the
molecule F2.
Step 2: Arrange the other atoms around
the central atom.
In CH4, after the central C is written down…
the four H’s can simply be placed around it…
H
HCH
H
Step 2: Arrange the other atoms around
the central atom.
In NF3, after the central N is written down…
the three F’s can simply be placed around it…
F
FNF
By the way, it really doesn’t matter whether these surrounding
atoms go on top, bottom, left or right.
Step 2: Arrange the other atoms around
the central atom.
In BrF5, after the central Br is written down…
the five F’s can simply be placed around it…
F
F Br F
F F
Note that when there are more than four atoms surrounding the
central one, you just scoot them around a bit to make room for
them all.
Step 2: Arrange the other atoms around
the central atom.
And in SO2, after the central S is written down…
the two O’s can simply be placed around it…
OSO
Step 3: Make bonds between the central
atom and each of the surrounding ones.
A bond is simply a pair of electrons being shared
between two atoms.
These are represented by a pair of dots.
So for CH4…
We simply add electron
pairs between the C and
each H.
H
HCH
H
Step 3: Make bonds between the central
atom and each of the surrounding ones.
For NF3.
We simply add electron
pairs between the N and
each F.
F
F NF
Step 3: Make bonds between the central
atom and each of the surrounding ones.
For BrF5.
We simply add electron
pairs between the Br and
each F.
F
F Br F
F F
Step 3: Make bonds between the central
atom and each of the surrounding ones.
And for SO2.
We simply add electron
pairs between the S and
each O.
O S O
Step 4: Complete the octets on every atom
that needs it.
An octet is simply a group of eight electrons – that’s
what it takes to fill an atom’s valence level.
With two above, two below, two to the left and two to the right,
it’s easy to tell in an instant whether an atom’s octet is complete or
not:
X
X
Not complete
Complete
Step 4: Complete the octets on every atom
that needs it.
Which atoms need an octet? All of the ones in group IV
on the periodic table and up.
(in other words, all those from C’s column to the right)
I
VIII
II
III IV V VI VII
C
Step 4: Complete the octets on every atom
that needs it.
So for CH4…
H is in group I;
I it does not need an
octet, so leave all four H’s the way
they are.
C is in group IV
IV; it does need an
octet. But
H
since it already
has one, just
leave it alone
too!
I
II
H
HCH
H
III
IV
C
V
VI
VII
VIII
Step 4: Complete the octets on every atom
that needs it.
For NF3…
F is in group VII
VII; so it needs an
octet. Just add electron pairs until
all the F’s are complete:
N is in group V;
V so it also needs an
octet. Just
add one more
pair right here.
F
F NF
I
II
VIII
III
IV
V
N
VI
VII
F
Step 4: Complete the octets on every atom
that needs it.
For BrF5…
F is in group VII; so it needs an
octet. Just add electron pairs until
all the F’s are complete:
Br is also in group VII
but since it
already has 10
drawn around it…
I
there’s no need
to add any
more.
F
F Br F
F F
VIII
II
III
IV
V
VI
VII
Br
Step 4: Complete the octets on every atom
that needs it.
And for SO2…
O is in group VI
VI; so it needs an octet.
Just add electron pairs until both the
O’s are complete:
OS O
S is also in group VI.
It also needs an
octet. Just add
two more pairs –
top and bottom.
I
VIII
II
III
IV
V
VI
O
S
VII
Let’s review what we’ve done so far.
Step 1: We wrote down the central atom.
Step 2: We arranged the other atoms around the central
atom.
Step 3: We made bonds between the central atom and each
surrounding atom.
Step 4: We completed the octets on each atom that needed
it (group IV and up).
These first four steps give us a “rough draft” of
our Lewis structure.
In step 5, we need to edit that rough draft
to make sure it contains the appropriate
number of dots (electrons).
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
So for CH4…
We have drawn 8 electrons.
And we now go back and see what
groups the atoms are in:
C is in group IV, so it contributes 4
valence electrons.
H is in group I, so each of the four H’s
contributes 1 valence electron.
4 + 4(1) = 4 + 4 = 8
H
HCH
H
This means our Lewis structure is supposed to have 8 electrons.
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
Since we have drawn 8 electrons and
we were supposed to draw 8
electrons, there is no need to edit the
rough draft at all. It is correct the way
it is:
H
HCH
H
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
For NF3…
8 +8
We have drawn 26 electrons.
And we now go back and see what
groups the atoms are in:
N is in group V, so it contributes 5
valence electrons.
F is in group VII, so each of the three
F’s contributes 7 valence electrons.
5 + 3(7) = 5 + 21 = 26
+ 8 + 2 = 26
F
F NF
This means our Lewis structure is supposed to have 26 electrons.
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
Since we have drawn 26 electrons and
we were supposed to draw 26
electrons, there is no need to edit the
rough draft at all. It is correct the way
it is:
F
F NF
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
For BrF5…
8 +8+8
We have drawn 40 electrons.
And we now go back and see what
groups the atoms are in:
Br is in group VII, so it has
7 valence electrons.
F is in group VII, so each of the five
F’s has 7 valence electrons.
7 + 5(7) = 7 + 35 = 42
+ 8 + 8 = 40
F
F Br F
F F
This means our Lewis structure is supposed to have 42 electrons.
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
Since we have drawn 40 electrons but we were supposed
to draw 42 electrons, we must add one more pair.
Whenever electron pairs must be
added, simply draw them in on
the central atom:
Since it’s a little crowded down
there, a little loop helps to show
that this pair of electrons belongs
on the Br.
F
F Br F
F F
Step 5: Compare how many electrons we’ve
drawn to how many we should have drawn.
Since we now have 42 electrons – the exact number we are
supposed to have – the Lewis structure is finished.
Notice that the Br now has a
total of 12 electrons:
That’s way more than an octet,
but sometimes it’s necessary to
go beyond a stable octet when
it comes to drawing Lewis
structures.
F
F Br F
F F
What if you need to add two pairs of
electrons?
Again just add them both to the central atom:
Y X Y
24
22
Total electrons drawn: 20
And what if your rough draft has too many dots
and you need to delete a pair or two?
Let’s try SO2…
8
We have drawn 20 electrons.
And we now go back and see what
groups the atoms are in:
S is in group VI, so it has 6 valence
electrons.
O is in group VI, so each of the two
O’s has 6 valence electrons.
6 + 2(6) = 6 + 12 = 18
+ 8 + 2 + 2 = 20
OS O
This means our Lewis structure is supposed to have 18 electrons.
And what if your rough draft has too many dots
and you need to delete a pair or two?
Since we have drawn 20 electrons but we were supposed to
draw only 18 electrons, we must delete one pair.
Whenever electron pairs must be
deleted, erase a nonbonding pair
on the central atom
First, we need to define some terms:
“Bonding pairs” are the electrons drawn
between two atoms, like these two…
“Nonbonding pairs” are all the others,
like these…
Then erase a nonbonding pair on one
of the surrounding atoms
OS O
And then add in a pair of bonding electrons between those
same two atoms
And what if your rough draft has too many dots
and you need to delete a pair or two?
Since we now have 18 electrons – the exact number we are
supposed to have – the Lewis structure is finished.
This group of four electrons
is known as a “double bond.”
OS O
So whenever you are
instructed to “make a double
bond,” remember to erase one
nonbonding pair on the central atom and one nonbonding
pair on a surrounding atom, then add in a bonding pair
between those same two atoms.
And what if your rough draft has too many dots
and you need to delete a pair or two?
Because you have erased four electrons and drawn in two,
you have effectively deleted two electrons from your rough
draft.
Notice how each atom still has a
stable octet.
8
8
8
OS O
Because this last concept is kind of tricky, let’s
try one more like it:
Let’s say we wanted to draw a Lewis structure for SeO3.
Step 1: Write down the central atom:
Step 2: Arrange the three O’s around it:
Step 3: Make bonds between the Se and
each of the three O’s:
Step 4: Complete the octects on Se
(group VI) and each O (group VI)
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
SeO3 = 6 + 3(6) = 6 + 18 = 24
8 + 8 + 8 + 2 = 26
O
O Se O
Since we have 26 dots, but are only supposed to have 24, we erase
and make a double bond…
two nonbonding pairs
What if you need to delete two pairs?
Sometimes this
requires making two
double bonds:
8
8
8
Y X Y
16
18
Total electrons drawn: 20
Notice how each atom still has a stable octet.
Other times it
requires making a
triple bond:
8
8
X Y
10
12
Total electrons drawn: 14
Again note how each atom still has a stable octet.
The final question to answer: what about
drawing Lewis structures for ions?
Answer: you draw them the exact same way as you do
molecules – following the same 5 steps described above.
But in step 5, when it comes to determining how many electrons are
supposed to be in the Lewis structure, you must incorporate the
charge of the ion into your total.
If it’s a positive ion, then electrons must have been lost, so the
charge tells us how many to subtract from the total.
And if it’s a negative ion, then electrons must have been gained, so
the charge tells us how many to add to the total.
For example, without the charge, NF41+ should have
5 + 4(7) = 5 + 28 = 33 electrons.
But the charge of 1+ tells us that one electron has
been lost, so that would bring the total down to 33
– 1 = 32.
Or consider the SO3 2- ion. Without the charge it
should have 6 + 3(6) = 6 + 18 = 24 electrons.
But the charge of 2- tells us that two electrons have
been gained, so that would bring the total up to 24
+ 2 = 26.
So let’s see what this looks like:
Let’s say we wanted to draw a Lewis structure for CO32-.
Step 1: Write down the central atom:
8 + 8 + 8 + 2 = 26
Step 2: Arrange the three O’s around it:
Step 3: Make bonds between the C and
each of the three O’s:
Step 4: Complete the octects on C (group
2IV) and each O (group VI)
Step 5: Compare the number of
electrons we have drawn…
O
[O C O ]
with the number we should have drawn:
CO32- = 4 + 3(6) = 4 + 18 = 22
+ 2 = 24 (because of the 2- charge)
Since we have 26 dots, but are only supposed to have 24, we erase
and make a double bond…
two nonbonding pairs
Brackets and charge should be
added every time you write a
And don’t forget to add brackets and a charge:
Lewis structure for an ion!
Now try problems 1-10 on the Lewis structures
worksheet, using the left-hand boxes for practice,
then copying your final answers into the right-hand
boxes.
Use the next few slides if you get stuck, or to check
your final answers.
1. NH3
Step 1: Write down the central atom:
Step 2: Arrange the three H’s around it:
Step 3: Make bonds between the N and
each of the three H’s:
Step 4: Complete the octect on N (group V)
but NOT on the H’s (group I)
Step 5: Compare the number of
electrons we have drawn…
8
with the number we should have drawn:
NH3 = 5 + 3(1) = 8
H
HNH
Since we have 8 dots, and we are supposed to have 8, we are all
finished!
2. BF3
Step 1: Write down the central atom:
Step 2: Arrange the three F’s around it:
8 + 8 + 8 = 24
Step 3: Make bonds between the B and
each of the three F’s:
Step 4: Complete the octects on the F’s (group
VII) but NOT on the B (group III)
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
BeF3 = 3 + 3(7) = 3 + 21 = 24
F
FBF
Since we have 24 dots, and we are supposed to have 24, we are
finished.
3. H2O
Step 1: Write down the central atom:
Step 2: Arrange the two H’s around it:
Step 3: Make bonds between the O and
each of the two H’s:
Step 4: Complete the octect on O (group VI)
but NOT on the H’s (group I)
Step 5: Compare the number of
electrons we have drawn…
8
with the number we should have drawn:
H2O = 2(1) + 6 = 8
HOH
Since we have 8 dots, and we are supposed to have 8, we are all
finished.
4. BrF3
Step 1: Write down the central atom:
Step 2: Arrange the three F’s around it:
Step 3: Make bonds between the Br and
each of the three F’s:
Step 4: Complete the octects on Br
(group VII) and each F (group VII)
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
BrF3 = 7 + 3(7) = 7 + 21 = 28
8 + 8 + 8 + 2 = 26
F
F Br F
Since we have 26 dots, but are supposed to have 28, we simply add
an extra pair on the central atom.
5. SF6
8 + 8 + 8 + 8 + 8 + 8 = 48
Step 1: Write down the central atom:
Step 2: Arrange the six F’s around it:
Step 3: Make bonds between the S and
each of the six F’s:
Step 4: Complete the octects on S (group
VI) and each F (group VII)
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
SF6 = 6 + 6(7) = 6 + 42 = 48
FF
FS
FFF
Since we have 48 dots, and we are supposed to have 48, we are all
finished.
6. CO2
Step 1: Write down the central atom:
Step 2: Arrange the two O’s around it:
Step 3: Make bonds between the C and
each of the two O’s:
Step 4: Complete the octects on C (group
IV) and each O (group VI)
Step 5: Compare the number of
electrons we have drawn…
8 + 2 + 8 + 2 = 20
OC O
with the number we should have drawn: CO2 = 4 + 2(6) = 4 + 12 = 16
Since we have 20 dots, but are only supposed to have 16, we must
delete two pairs.
Erase these nonbonding pairs…
Then make a double bond…
Then erase these pairs…
And make a second double bond.
That’s it.
7. HF
Step 1: Write down the central atom: (actually, since there are only two
atoms, just write them down side-by-side).
Step 2: (With no central atom, this step doesn’t apply.)
Step 3: Make a bond between the H and
the F:
Step 4: Complete the octect on F (group
VII) but NOT on H (group I)
Step 5: Compare the number of
electrons we have drawn…
8
with the number we should have drawn:
HF = 1 + 7 = 8
Since we have 8 dots, and are supposed to have 8, we are finished.
HF
8. Br2
Step 1: Write down the central atom: (actually, since there are only two
atoms, just write them down side-by-side).
Step 2: (With no central atom, this step doesn’t apply.)
Step 3: Make a bond between the two
Br’s:
Step 4: Complete the octects on both
Br’s (group VII)
8 + 6 =14
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
Br2 = 2(7) = 14
Since we have 14 dots, and are supposed to have 14, we are finished.
Br Br
9. CO
Step 1: Write down the central atom: (actually, since there are only two
atoms, just write them down side-by-side).
Step 2: (With no central atom, this step doesn’t apply.)
Step 3: Make a bond between the C and the O:
8 + 6 =14
Step 4: Complete the octects on both C
(group IV) and O (group VI)
Step 5: Compare the number of
electrons we have drawn…
C O
with the number we should have drawn: CO = 4 + 6 = 10
Since we have 14 dots, but are only supposed to have 10, we must
delete two pairs.
Erase these nonbonding pairs…
Then make a double bond…
Then erase these pairs…
And make a triple bond.
That’s it.
10. SO3
2-
Step 1: Write down the central atom:
Step 2: Arrange the three O’s around it:
Step 3: Make bonds between the S and
each of the three O’s:
Step 4: Complete the octects on S (group
VI) and each O (group VI)
8 + 8 + 8 + 2 = 26
O
[O S O ]
Step 5: Compare the number of
electrons we have drawn…
with the number we should have drawn:
SO32- = 6 + 3(6) = 6 + 18 = 24
+ 2 = 26 (because of the 2- charge)
Since we have 26 dots, and are supposed to have 26, we are
finished… But don’t forget to add the brackets and the charge
2-
Now that you have learned how to
draw Lewis structures, try the rest
of the worksheet.