Representing Ionic and Covalent Compounds
Ionic compounds are formed between metals and nonmetals. With this large electronegativity difference it is assumed the
metal will lose valence electrons to the nonmetal. Then the positive metal is electrostatically attracted to the negative
nonmetal, and an ionic bond is formed. One the other hand covalent compounds are formed between nonmetals. With their
small electronegativity difference it is assumed they share the electrons in bonds. Sharing electrons forms a covalent bond.
Because of the differences in making bonds, ionic and covalent compounds are represented on paper using different methods.
Ionic compounds use electron dot diagrams to show where the electrons have moved and the charges created by the moved
electrons. Lewis structures show bonds of shared electrons and dots of unshared electrons. It is important that the correct
representation be used on the correct type of compound.
In both ionic and covalent compounds the elements are seeking a position of lower chemical potential energy that comes with
having a filled valence energy level. Other than for hydrogen and helium, this valence level is filled when the s and p sublevels
are filled, thus it takes eight electrons. Combining in a way to fill the outer energy level with eight electrons is called the octet
rule. It is often said that Noble Gases do not bond because they already have a complete octet. This is a true statement, but is
only one of the many reasons that the Noble Gases do not easily bond.
How to draw Electron Dot Diagrams for ionic compounds:
1. Recognize it is an ionic compound - look for a metal with a nonmetal.
2. Using pencil (for erase-ability), write out the valence electron dot diagrams for each element, and in a symmetrical pattern
if necessary. Patterns do not have to be linear, but can also be T-shaped or + -shaped.
3. One at a time, erase valence electron dots from the metal(s) and add them to the nonmetal(s) until every metal is out of
valence electrons and every nonmetal has an octet.
4. Draw large square brackets around the nonmetal(s) to encompass the octet. Then place negative charges as a superscript
outside the brackets (equal to the number of gained electrons) and positive charges as superscripts on the metal (equal to
the number of lost electrons).
5. Be sure the compound is charge-balanced (zero charge) overall.
Examples:
NaCl
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
MgI2
Na is a metal and Cl is a nonmetal
●●
● Cl ●
Na ●
●
●●
●●
●
●
Na
● Cl ●
●●
+1
●●
●
●
Na
● Cl ●
●●
+1 + -1 = 0, it is zero charge overall
[
]
-1
Step 1:
Step 2:
Mg is a metal and I is a nonmetal
●●
●●
●
●
●
●
●
I
Mg
I ●
●
●
●●
●●
Step 3:
●●
●●
●
●
●
●
I
Mg
●
●
● I ●
●●
●●
Step 4:
●●
-1
+2
●●
-1
●
●
●
●
Mg
● I
●
● I ●
●●
●●
Step 5:
-1 + 2+ -1 = 0, it is zero charge overall
Certainly there is no need to show each step, but each step must be performed to arrive at the correct electron dot diagram. It is
highly recommended that pencil is used so dots can be erased in step 3, instead of trying to skip step 3 in your head.
At this time complete the “Electron Dot Diagrams Practice”.
[
]
[
]
How to draw Lewis Structures for covalent compounds:
1. Recognize it is a covalent compound - two or more nonmetals.
2. Total up the number of valence electrons for each element (as from their dot diagrams). Add for any negative
charges (more electrons) and subtract for any positive charges (less electrons).
3. Divide the total (from step 2) by 2. This pairs up electrons and electrons ALWAYS go in pairs.
4. Write the central atom with the outside atoms symmetrically around it. Shapes can be linear, triangular (better than
T-shaped) or +-shaped. If it is not easy to tell which one is in the center, it will be the least electronegative element,
except hydrogen can never be in the middle.
5. Draw a line between each outside atom and the central atom, and subtract the number of lines from the number of
pairs (from step 3).
6. Place three additional pairs of electrons (lone pairs) on each outside atom, except for hydrogen. Hydrogen gains a
noble gas electron configuration with only two electrons, so it never gets lone pairs. Subtract each lone pair from
the number of pairs (from step 3).
7. If there are any pairs still left after step 6, then place them on the central atom.
8. If the central atom is not carbon, nitrogen, oxygen, phosphorus, or sulfur, then check if there was a charge on the
compound. If so, place it in large brackets with the charge outside. If not, then the drawing is finished.
9. If the central atom is carbon, nitrogen, oxygen, phosphorus, or sulfur (BEWARE THE CNOPS) bonded to another
carbon, nitrogen, oxygen, phosphorus, or sulfur, check that these elements have at least 4 pairs of electrons. If
there is not at least 4 pairs, then remove a lone pair from the least electronegative outside CNOPS atom and form a
double bond instead. Repeat that process as needed until the central atom has at least 4 pairs. Hydrogen can never
form a double bond. Attempt to use double bonds before using a triple bond.
10. If a multiple bond is used, and there is more than one possible place for that multiple bond to have been placed, all
the possible drawing must be shown, (with large brackets around them and charges on the outside if needed) and
double sided arrows between. This shows each picture is an equal option for the compound and shows a
phenomenon in compounds called resonance, or resonance structures.
Examples:
CCl4 Step 1:
Step 2:
Step 3:
Steps 4 & 5:
16 – 4 = 12
C is a nonmetal and Cl is a nonmetal
4 + 7 + 7 + 7 + 7 = 32 electrons total
32  2 = 16 electron pairs
Cl

Cl  C

Cl

Cl
Step 6:
12 – 12 = 0


Step 7:
Step 8:
Step 9:
Step 10:




Cl 

Cl  C





Cl 

Cl 


Not needed
Not needed
C has at least 4 pairs, no further steps required
Not needed
Again there is no need to show each step individually, but do not attempt to skip steps in your head. Instead be sure to
follow the steps each time and write in pencil for the ability to erase lone pairs and turn them into multiple bonds.
H 2O
Step 1:
Step 2:
Step 3:
Steps 4 & 5:
4–2=2
H is a nonmetal and O is a nonmetal
1 + 1 + 6 = 8 electrons total
8  2 = 4 electron pairs
H

O

H
Step 6:
2–0=2
H

O

H
Step 7:
2 -2 = 0
H


H

O

CO2
Step 8:
Step 9:
Step 10:
Not needed
O has at least 4 pairs, no further steps required
Not needed
Step 1:
Step 2:
Step 3:
Steps 4 & 5:
8–2=6
C is a nonmetal and O is a nonmetal
4 + 6 + 6 = 16 electrons total
16  2 = 8 electron pairs
Step 6:
6–6=0
Step 7:
Step 8:
Step 9:
C needs 2
more pairs
Step 10:
HCN
O



C

C


O



Not needed
Not needed


O

Not needed
O
O



C


O





Step 1:
C is a nonmetal, N is a nonmetal, and H is a nonmetal
Step 2:
1 + 4 + 5 = 10 electrons total
Step 3:
10  2 = 5 electron pairs
Steps 4 & 5:
5–2=3
H  C  O
(C is least
electronegative)
Step 6:

3–3=0
H  C  O 
(H never has lone

pairs)
Step 7:
Not needed
Step 8:
Not needed
Step 9:

C needs 2

H  C 
 O 
more pairs
(H cannot make
multiple bonds)
Step 10:
Not needed
NO3-
Step 1:
Step 2:
Step 3:
Steps 4 & 5:
12 – 3 = 9
N is a nonmetal and O is a nonmetal
5 + 6 + 6 + 6 + 1 = 24 electrons total
(add for negative, subtract for positive)
24  2 = 12 electron pairs
Step 6:
9–9=0
Step 7:
Step 8:
Step 9:
Step 10:
Not needed
Not needed
N needs one more pair of electrons, but which of the three oxygens
should it come from?
Because all three oxygens are equal, the double bond could be in three
different places. In a situation like this resonance structures are used to
show all the possibilities, which in this case is three different structures.
Notice all three also have brackets and the charge, which is needed for
ANY structure with a negative or positive charge.
It is possible for the central atom to not have 4 pairs if it is not C, N, O, P, or S. Boron often only has three pairs of
electrons, and that’s correct for Boron. A few elements, typically S, P, and Xe, can also have more than 4 pairs of
electrons, and if that’s what it takes to complete the steps listed above then that is also correct for those elements. The
majority of the elements on the periodic table will only have 4 pairs of electrons because that completes the s and p
orbitals for that element, which contain the valence electrons, and can only hold up to 8 electrons. To go beyond 8
electrons the d orbitals get involved, but how that happens is a topic still under investigation and debate.
At this time complete the Lewis Structures part of the “Lewis Structures and VSEPR Practice”, that is everything except
below the “VSEPR information” heading.
Lewis structures are great for showing a quick drawing of a covalent compound with a lot of useful information, but it
fails to represent the actual shape the molecule would have in three dimensions. To determine the three dimensional
shape of a molecule, a simple rule is followed: bonds and lone pairs are comprised of electrons, and electrons repel each
other as far as possible. This concept is called Valence Shell Electron Pair Repulsion or VSEPR for short. To create a
correct three dimensional model of a molecule, placing the bonds and lone pairs as far apart as possible around the
central atom is all that is required.
For example:
 two pairs of electrons would arrange themselves on opposite sides of the central atom, that is 180° apart
 three pairs of electrons would arrange themselves into the corners of a triangle, that is 120° apart
 four pairs of electrons would arrange themselves into a pyramid, that is 109.5° apart
 five pairs and six pairs of electrons typically will be around 90° apart
To determine the correct VESPR shape from the Lewis structure, the total number of pairs around the central atom
(central pairs = CP) and the number of lone pairs on the central atom (lone pairs = LP) need to be counted. Be aware
that a double or triple bond still only counts as 1 CP. With those two numbers you can match up to the structures on the
next page.
EG = electron geometry. This is the name of the geometric figure made by the molecule if all the electron pairs (both
bonds and lone pairs) are considered.
MG = molecular geometry. This is the name of the geometric figure made by the molecule if only the bonds are
considered.
Hyb = Hybridization. When atoms combine, the central atom needs to use space in the s, p, and d and sublevels.
However, there is no distinction between these sublevels when scientists study the molecules in a lab. Thus the central
atom must be mixing the sublevels into a new type of sublevel, a process called hybridization. The amount of each
sublevel needed to form hybridized bonds depends on the number of electrons pairs on the central atom.
Number of Central Pairs (CP)
1
2
3
4
5
6
Sublevels needed
s
s+p
s+p+p
s+p+p+p
s+p+p+p+d
s+p+p+p+d+d
Hybridization
s
sp
sp2
sp3
sp3d
sp3d2
(Multiple bonds only count as 1 CP!)
Polarity
When different atoms form a covalent bond, there is rarely an equal sharing of electrons. This causes the covalent bond
to be polar. When multiple atoms bond this still occurs, but the 3 dimensional shape can also have an effect on the
overall polarity of the entire molecule.
CO2 is linear - Although the C = O bond is polar, the polarity to the
right balances the polarity to the left, and thus the entire molecule is
NONPOLAR
..
..
OH2 would seem just like CO2, but it is angular in shape, not linear This time the polarity to the right is balanced by the polarity to the
left, but the polarity down is not balance by any polarity up. Thus this
molecule is POLAR.
If a molecule has an odd number of lone pairs, or more than just two different elements, it is almost certain to be
POLAR. The only sure way to tell if a molecule is polar or nonpolar is to draw it out and look for balancing of
polarities.
At this time finish the “Lewis Structures and VSEPR Practice.”
CP
LP
EG
MG
2
0
Linear
Linear
3
0
Trigonal
Planar
Trigonal Planar
3
1
Trigonal
Planar
Angular (bent)
4
0
Tetrahedral
Tetrahedral
4
1
Tetrahedral
Trigonal Pyramidal
4
2
Tetrahedral
Angular (bent)
5
0
Trigonal
Bipyramidal
Trigonal Bipyramidal
5
1
Trigonal
Bipyramidal
Irregular Tetrahedral
5
2
Trigonal
Bipyramidal
T-Shaped
5
3
Trigonal
Bipyramidal
Linear
6
0
Octahedral
Octahedral
6
1
Octahedral
Square Pyramidal
6
2
Octahedral
Square Planar