Drawing Lewis Structures
1 Determine the total number of valence electrons for ALL atoms.
Don’t be concerned with which atom gave what; the SUM TOTAL is what
is important. If the entire molecule is charged (i.e. a polyatomic anion or
cation) add one valence electron for each unit of negative charge (if it is
an anion) and remove one valence electron for each unit of positive
charge (if it is a cation).
2 Write a skeleton structure for the molecule, making the least
electronegative atom the central atom.
The
order
of
electronegativity
for
the
nonmetals
is
F>O>N>Cl>Br>I>S>C>H. Metals will almost always be the central
atom if they are present. Hydrogen is never the central atom.
a If there are more than one of the least electronegative atom, your
skeletal structure should have those two attached to one another
(EXCEPTION: Hydrogen)
3 Connect each member of the skeleton structure to the central atom(s) using
a single line to represent a bond.
Each bond is comprised of two electrons so each line indicates a twoelectron bond (two and only two electrons).
4 Count the number of bonds that you made in step 3 and multiply that by 2.
Subtract that number from the number of total valence electrons from step
1. This is the number of electrons you have left to distribute.
Simply: you are counting the electrons you have used and subtracting from
what you started with to find out how many you have left! This is a tally
you must always be aware of.
5 Starting with the atoms bonded TO the central atom (the ‘outside’ atoms),
distribute the electrons two at a time until all electrons from step 4 are used
OR until each atom has an octet (eight electrons around it); DO NOT give
hydrogen any of these ‘left-over’ electrons.
A good rule of thumb is to give EACH non-hydrogen two electrons at a
time until each outside atom has two extras, then give two more to each
non-hydrogen until each has four extra electrons and so on. Once all of
the ‘outside’ atoms have octets (except for the hydrogens), put remaining
electrons on the central atom(s) until all electrons are used or every atom
has an octet (or a duet, in the case of H).
.
AT THIS POINT: Any time you run out of electrons, look at
each atom and determine if it does or does not have an octet.
If all atoms DO have an octet go on to Step 8. If not, proceed
to Step 6
6 If there are too few electrons available to complete octets for all the atoms
that need them, make double and/or triple bonds between appropriate
atoms.
Remember that in doing this, no atom should lose electrons (i.e. double and
triple bonds SHARE electrons BETWEEN the two involved atoms and
as a result neither atom involved in the multiple bond loses electrons, so
do not move electrons FROM one atom to another).
a A major exception to this rule is one regarding electrically-neutral
Lewis acids containing Be, Al or B: it is possible to find Be with 4e–
around it (BeCl2), and B and Al with 6e– around them (AlCl3 and
BF3, for example)
7 If there are too many electrons available: first, RECOUNT the valence
electrons available (steps 2-4 above); then, after completing octets for all
the atoms that need them, place remaining electrons on the central atom IF
the central atom is a Period 3 or greater element.
8 After you have completed your Lewis structure, CHECK FOR OCTETS.
If all atoms have an octet, calculate formal charges for all atoms (see below
for method).
Remember: if you are drawing the Lewis structure of a cation or an anion,
it WILL have a charge (or charges) in it. NOTE: as you get better at
drawing Lewis structures, you will develop a ‘feel’ as to whether or not
an atom has a formal charge or not. (USUALLY, if an atom is not in
agreement with the guidelines given below in step 9, it may have a
charge and should ALWAYS be checked).
9.
Guidelines to follow in writing Lewis Structures (NOTE: there are
exceptions to these…)
i H follows the duet rule and will have a maximum of 2e- around it and
will therefore only form a single, single bond;
ii
C, N, O, F always obey the octet rule (But if they have less than an
octet they will carry a formal charge);
iii O Family: usually forms a TOTAL of two covalent bonds;
a) when this is the case, the atom should also have 2 unshared pairs
of electrons on it (called ‘lone pairs’) and should have no formal
charge associated with it;
iv N Family: usually forms a TOTAL of three covalent bonds;
a) when this is the case, the N should also have 1 unshared pair
(lone pair) of electrons on it and should have no formal charge
associated with it;
v
C Family: usually forms a TOTAL of four covalent bonds;
a) when this is the case, the C should also have NO unshared pairs
of electrons remaining and should have no formal charge
associated with it;
vi Halogen Family: Usually forms only one covalent bond to only one
atom at a time and is never the central atom;
a) when this is the case, the halogen should also have 3 unshared
pairs (lone pairs) of electrons on it and should have no formal
charge associated with it;
b) F will NEVER be a central atom;
c) Cl, Br, I can be a central atom under special conditions. If one of
these is the central atom this guideline is ignored.
vii B, Be and Al usually have less than an octet (and as a result they form
reactive compounds);
viii Second period elements never exceed the octet;
ix Third row elements and heavier usually satisfy the octet rule but CAN
expand their octet to 10 e or 12 e (this is usually to minimize formal
charge);
x
If there are an ODD number of valence electrons TOTAL (from step
1), one atom in the structure will have an unpaired electron, and that
atom will usually be the one that is the LEAST electronegative
EXCEPT H (of course…).
FORMAL CHARGE CALCULATION
Calculation of formal charge is the most common mistake made by
students at all levels of chemistry when drawing Lewis structures.
Formal charge is calculated for each atom independent of any others in
the molecule. The procedure is as follows:
1) Determine the number of valence electrons (VE) that the atom
of interest has; this is most easily done by using a periodic table;
2) Determine the number of unshared (or non-bonded) electrons
(NBE) that the atom of interest has;
3) Determine the number of bonds (B) that the atom of interest has
(remember that a double bond counts as two and a triple bond
counts as three);
4) Formal Charge is then calculated using the following formula:
FC = VE - B - NBE
Let’s calculate the formal charge for all the atoms in the Lewis
structure for SO2 shown below:
O
S
1
O
2
For Oxygen 1: FC = 6 - 1 - 6 = -1;
For Sulfur: FC = 6 - 3 - 2 = +1;
For Oxygen 2: FC = 6 - 2 - 4 = 0;
Thus the complete structure looks like the following:
O
1
S
O
2
Classical Bonding Motifs
Ionic: electrostatic (+ ⇔ −)
Covalent: shared electrons
Metallic: “sea” of electrons around metal “cation”
Coordinate Covalent: a shared arrangement not unlike pure covalent
“pure” ionic and “pure” covalent lie at ends of a spectrum
Polar Covalent
MORE
100% Ionic
LESS
100% Covalent
Electronegativity
Quantity (χ or Δ) defined to describe covalent bonding;
Tendency of an atom to polarize electrons in a bond when
attached to another atom
Two scales: Mulliken and Pauling
1
( IE1 + EA)
Mulliken:
!"
Pauling:
# $ E AB % #E AA #E BB
2
Although absolute values are different, same results are obtained by
both methods!
Increases up a group and left to right across a
period
FONClBrISCH
1
Estimating “Correctness” of
Lewis Structures
1. More covalent bonds, more stable and therefore
better the structure;
2. Charge separation will decrease stability;
3. Those structures where all atoms have an octet will
be the best ones;
4. Minimization of charge is best.
Rules of Resonance Structures
Ԫ Resonance is only a theory!
Ԫ Only electrons can be moved (π and l.p.);
Ԫ All structures must be proper Lewis
structures;
Ԫ All structures must have the same
number of unpaired electrons (if any);
Ԫ The actual energy of the molecule is
lower than any resonance contributor
2
Bonding Energetics-1
Na → Na+ + e−
Cl + e− → Cl−
Na + Cl → Na+ Cl−
ΔE = IE1 = +495.8 kJ/mol
ΔE = −EA = −349.0 kJ/mol
ΔEprocess = +146.8 kJ/mol
Q: Why does a bond form if it costs this much energy?
A: Stabilization that occurs on bonding must make up for it!
!ECoulomb =
( q+ )( q" )
4#$ o R
q = ion charge; charge × e (C)
εo = 8.854 × 10-12 C/J-m
R = atomic distance
Since one ion will always be an anion ΔE will be negative;
therefore Eoverall will be lowered
Covalent Bonding
Atoms of similar electronegativities share electrons and form
molecules
Characterize bonds according to LENGTH, ENERGY & ORDER
Bond Length: distance between two bonded nucleii (� );
•Increases with Z
Bond Energies: E req’d to break a bond;
•stable bonds = large values
•BDE or Ed: BE decreases with increasing Z
Bond Order: number of shared e- between two given atoms
3