From last chapter, we saw how to use the periodic table to predict the charge on an element when
it forms an ion: +1, +2, +3, skip, -3, -2, -1, 0
Now let’s look at cations and anions in terms of electron configuration, valence electrons, and
electron dot structures.
Formation of Cations –
Let’s look at the electron configuration of sodium: 1s22s22p63s1
In order for sodium to achieve a complete octet, it needs to get to a complete shell:
1s22s22p63s1 -> 1s22s22p63s23p6 (same electron config as Ar, gains 7 electrons) or
1s22s22p63s1 -> 1s22s22p6 (Same electron config as Ne, loses 1 electron)
So, sodium loses 1 electron going to a complete 8 electron octet at the 2nd energy level.
What’s the charge on sodium? Well it still has the same number of protons, 11, and count the
electrons: 10. You have 11 +’s and 10 –‘s, so you have a net of one + for a charge of +1.
Draw electron dot structure for:
Na -> Na+ + eWhen Na loses one electron we write it as a balanced chemical equation:
Na -> Na+ + eNow let’s look at the electron dot for calcium:
Ca
For its balanced chemical equation, we write:
Ca -> Ca+2 + 2eIons with a charge of greater than three are extremely rare and unlikely.
Practice problems:
Write the balanced chemical equations using electron dot structures for the formation of cations
for the following elements. If the element does not form cations then write “no reaction” after
the reaction arrow.
1. Sr
2. Li
3. Ba
4. P
5. In
Transition Metal Cations –
Some ions formed by transition metals do not have a noble-gas configuration, and are thus
exceptions to the octet rule.
Let’s look at cadmium:
1s22s22p63s23p64s23d104p65s24d10
How does cadmium achieve a noble gas configuration? Lose 12 electrons? Gain 6 electrons?
Both options are very hard.
Cadmium loses its 2 5s2 electrons and has what we call a pseudo-noble gas configuration. Cd+2
has all of it’s energy sublevels complete and a 18 electrons in it’s 4th energy level.
1s22s22p63s23p64s23d104p65s24d10 -> 1s22s22p63s23p64s23d104p64d10
Practice Problem:
Silver violates the aufbau principal and has the following electron configuration:
1s22s22p63s23p64s23d104p65s14d10
1) How is this a violation of the aufbau principal?
2) How many valence electrons does silver have?
3) Write a balanced chemical equation using electron dot structures to show the formation
of a silver cation.
Formation of Anions –
Let’s look at the electron configuration of chlorine – Cl: 1s22s22p63s23p5
In order for chlorine to achieve a complete octet, it has to:
1s22s22p63s23p5 -> 1s22s22p6 (Same electron config as Ne, loses 7 electrons)
1s22s22p63s23p5 -> 1s22s22p63s23p6 (same electron config as Ar, gains 1 electron)
So, chlorine gains 1 electron going to a complete 8 electron octet at the 3rd energy level.
What’s the charge on chlorine? Well it still has the same number of protons, 17, and count the
electrons: 18. You have 17 +’s and 18 –‘s, so you have a net of one + for a charge of +1.
Draw electron dot structure for:
[draw on blackboard Cl with 7 dots]
When chlorine gains one electron we write it as a balanced chemical equation:
[Cl with 7 dots] + e- -> [Cl with 8 dots]Practice problems:
Write the balanced chemical equations for the formation of the following anions using electron
dot structures. If the element does not form anions, write “no reaction” after the arrow.
1)
2)
3)
4)
5)
I
S
B
N
O
Ionic Compounds
An ionic compound is a compound composed of cations and anions.


Ionic compounds are electrically neutral – the total of the positive charges on the
cation(s) has to balance the total of the negative charges on the anion(s)
typically a metal cation and a non-metal anion
Ionic Bond – electrostatic forces that hold ions together (the attraction of the + charge and the –
charge to each other).
Draw the electron dot for sodium and the electron dot for chlorine.
Circle the sodium electron and draw an arrow over to chlorine.
Draw reaction arrow.
Draw product ions.
NEXT draw the electron configurations for starting atoms, then electron configurations for the
product ions.
Step 1: Draw the atoms electron dot structures
Step 2: Determine the number of electrons that each atom will gain or lose to form it’s complete
valence shell.
Step 3: Find the least common multiple of the two charges
Step 4: Add enough electron dot structures of each atom to get each charge to the least common
multiple.
Step 5: Draw arrows from the electrons leaving the cations filling up the anions (every electron
from the cations should leave, and each anion should end up with a complete shell).
Step 6: Write the product ion electron dot structures
Step 7: Count up the cations, put the number as a subscript to the lower right hand side of the
cations. Count up the anions, put the number of anions as a subscript to the lower right hand side
of the anions.
Note: if a polyatomic ion, put the ion in parenthesis with the number after the parenthesis
Sample Problem
Use electron dot structures to predict the formula of the ionic compound formed from Ca and Br
Show example stepwise.
Practice Problems
Use electron dot structures to predict the formula of the ionic compound formed from:
1) K and O
2) B and F
3) Mg and N