The crossover method is a shortcut way of writing formulas for ionic compounds if
you know the charges on the ions in the compound. We’ll go over a few examples
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
MgCl2
In example (a), we’re asked to write the formula for magnesium chloride.
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
MgCl2
Looking up Magnesium on the ion table or periodic table, we see its Mg with charge
of 2+.
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
MgCl2
And chloride is Cl with a charge of minus, or minus 1.
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
Mg1Cl2
We start the formula by writing down the symbols, Mg and Cl
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
Mg1Cl2
Now we take the 2 from the charge on magnesium, (click) cross over, and (click)
write it to next to the chloride.
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
Mg1Cl2
Now we could take the 1 from the charge on chloride, (click) cross over, and write it next to the
magnesium, but we don’t use the subscript 1 in a formula, so we don’t need to write anything.
a) Write the formula for magnesium chloride
2+
Mg
Mg1Cl2
So the formula has one Mg and two Cl’s.
–
Cl
a) Write the formula for magnesium chloride
2+
Mg
–
Cl
MgCl2
Formula
So we’ll compact it and write the final formula for magnesium chloride as MgCl2.
a) Write the formula for magnesium chloride
2+
Mg
–
2Cl
MgCl2
Formula
We’ll check the charges to verify that this formula is correct. (click) We have one
Mg2+ ion and 2 Cl– ions
a) Write the formula for magnesium chloride
+2
2+
Mg
+
2(–1)
–
2Cl
MgCl2
Formula
So the total charge is positive 2, plus 2 × –1, or positive 2 plus negative 2,
a) Write the formula for magnesium chloride
+2
2+
Mg
+
2(–1)
–
2Cl
MgCl2
Formula
Which is equal to zero, so the formula MgCl2 is correct.
= 0
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
Example b asks us to write the formula for calcium oxide.
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
We look up calcium and its Ca with a charge of positive 2.
b) Write the formula for calcium oxide
2+
Ca
Ca2O2
And oxide is “O” with a charge of negative 2.
2–
O
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
We start the formula by writing down the symbols Ca and O.
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
The charge on calcium is positive 2, so we (click) cross over and (click) write a 2
next to Oxygen in the formula
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
The charge on oxide is negative 2, so we (click) cross over and (click) write a 2 next to
Calcium in the formula. Notice we always drop the negative when we write subscripts
b) Write the formula for calcium oxide
2+
Ca
Ca2O2
Now, we have the formula Ca2O2.
2–
O
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca2O2
Both subscripts can
be divided by 2
But you can see that both of the subscripts on this formula are divisible by 2.
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca12O21
Divide both
subscripts by 2
So we’ll simplify and divide both of these by 2 and get 1’s
b) Write the formula for calcium oxide
2+
Ca
2–
O
Ca12O21
Remove the 1’s
We do not keep the subscript 1 in a formula, so we (click) remove the 1’s, leaving
us with the formula CaO.
b) Write the formula for calcium oxide
2+
Ca
2–
O
CaO2
Formula
Which we’ll compact and write as the final formula for calcium oxide.
b) Write the formula for calcium oxide
2+
Ca
2–
O
CaO2
Formula
Checking charges, we have one Ca2+ ion and one O2– ion.
b) Write the formula for calcium oxide
+2
2+
Ca
+
CaO2
Formula
So the total charge is positive 2 plus negative 2,
–2
2–
O
b) Write the formula for calcium oxide
+2
2+
Ca
+
CaO2
Formula
Which is equal to zero. So this formula is correct.
–2 = 0
2–
O
c) Write the formula for chromium(III) sulphide
3+
Cr
2–
S
Cr2S32
Question c asks us to write the formula for chromium(III) sulphide.
c) Write the formula for chromium(III) sulphide
3+
Cr
Cr2S32
The chromium (III) ion is Cr 3 plus
2–
S
c) Write the formula for chromium(III) sulphide
3+
Cr
Cr2S32
And the sulphide ion is S 2 minus
2–
S
c) Write the formula for chromium(III) sulphide
3+
Cr
Cr2S32
We start the formula with the symbols Cr and S
2–
S
c) Write the formula for chromium(III) sulphide
3+
Cr
2–
S
Cr2S32
We take the 3 from the charge on chromium, (click) cross over and write (click) 3
next to sulphur in the formula.
c) Write the formula for chromium(III) sulphide
3+
Cr
2–
S
Cr2S32
And we take the 2 from the 2 minus charge on the sulphide ion and (click) cross
over and write (click) 2 next to chromium in the formula.
c) Write the formula for chromium(III) sulphide
3+
Cr
2–
S
Cr2S32
There is no number other than 1, that will divide into 2 and 3, so this formula
cannot be simplified.
c) Write the formula for chromium(III) sulphide
3+
Cr
2–
S
Cr2S32
Formula
So the final formula for chromium III sulphide is Cr2S3.
c) Write the formula for chromium(III) sulphide
3+
2Cr
2–
3S
Cr2S32
Formula
Checking charges, we have 2 Cr3+ ions and 3 S2– ions,
c) Write the formula for chromium(III) sulphide
2(+3)
3+
2Cr
+
3(–2)
2–
3S
Cr2S32
Formula
So the total charge is 2 × positive 3, plus 3 × (–2), or positive 6 plus –6,
c) Write the formula for chromium(III) sulphide
2(+3)
3+
2Cr
+
3(–2)
2–
3S
Cr2S32
Formula
Which is equal to zero, so this formula is correct.
= 0
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
Question d asks us to write the formula for palladium(IV) oxide.
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
Palladium(IV) is Pd with a charge of 4 plus
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
And oxide is “O” with a 2 minus charge
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
We start the formula with the symbols Pd and O
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
We take the 4 from the charge on palladium, (click) crossover and (click) write 4
next to oxygen in the formula
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
We take the 2 from the charge on the oxide ion, (click) crossover and (click) write 2
next to palladium in the formula
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
Now, we have the formula Pd2O4.
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd2O42
Both subscripts can
be divided by 2
But we see that both of the subscripts 2 and 4 are divisible by 2.
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd12O42
2
Divide both
subscripts by 2
So we simplify by dividing both the 2 and the 4 by 2, giving us 1 and 2.
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd12O42
2
Remove the 1
We remove the subscript 1
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
Pd12O42
2
Leaving us with the formula PdO2.
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
O
PdO2
Formula
Which we’ll compact and state as the final formula for palladium(IV) oxide
d) Write the formula for palladium(IV) oxide
4+
Pd
2–
2O
PdO2
Formula
Checking charges, we have one Pd4+ ion and two O2– ions
d) Write the formula for palladium(IV) oxide
+4
4+
Pd
+
2(–2)
2–
2O
PdO2
Formula
So the total charge is positive 4 plus 2 times –2, or positive 4 plus negative 4,
d) Write the formula for palladium(IV) oxide
+4
4+
Pd
+
2(–2)
= 0
2–
2O
PdO2
Formula
Which is equal to zero, so PdO2 is the correct formula.