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

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

Writing Formulas Using the Crossover Method

Examples

In example (a), we’re asked to write the formula for magnesium chloride.

a) Write the formula for magnesium chloride

Mg2+ Cl–

MgCl2

Looking up Magnesium on the ion table or periodic table, we see its Mg with charge of 2+.


Mg2+ Cl–

MgCl2

And chloride is Cl with a charge of minus, or minus 1.


Mg2+ Cl–

MgCl2

We start the formula by writing down the symbols, Mg and Cl


Mg2+ Cl–

Mg1Cl2

Now we take the 2 from the charge on magnesium, (click) cross over, and (click) write it to next to the chloride.


Mg2+ 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.


Mg2+ Cl–

Mg1Cl2

So the formula has one Mg and two Cl’s.


Mg2+ Cl–

Mg1Cl2

So we’ll compact it and write the final formula for magnesium chloride as MgCl2.


Mg2+ Cl–

MgCl2

Formula

We’ll check the charges to verify that this formula is correct. (click) We have one Mg2+ ion and 2 Cl– ions


Mg2+ Cl–

MgCl2

Formula

2

So the total charge is positive 2, plus 2 × –1, or positive 2 plus negative 2,


Mg2+ Cl–

MgCl2

Formula

2+2 + 2(–1)

Which is equal to zero, so the formula MgCl2 is correct.


Mg2+ Cl–

MgCl2

Formula

2+2 + 2(–1) = 0

Example b asks us to write the formula for calcium oxide.

b) Write the formula for calcium oxide

Ca2+ O2–

Ca2O2

We look up calcium and its Ca with a charge of positive 2.


Ca2+ O2–

Ca2O2

And oxide is “O” with a charge of negative 2.


Ca2+ O2–

Ca2O2

We start the formula by writing down the symbols Ca and O.


Ca2+ O2–

Ca2O2

The charge on calcium is positive 2, so we (click) cross over and (click) write a 2 next to Oxygen in the formula


Ca2+ O2–

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


Ca2+ O2–

Ca2O2

Now, we have the formula Ca2O2.


Ca2+ O2–

Ca2O2

But you can see that both of the subscripts on this formula are divisible by 2.


Ca2+ O2–

Ca2O2

Both subscripts can be divided by

2

So we’ll simplify and divide both of these by 2 and get 1’s


Ca2+ O2–

Ca2O2

Divide both subscripts by 2

1 1

We do not keep the subscript 1 in a formula, so we (click) remove the 1’s, leaving us with the formula CaO.


Remove the 1’s

Ca2+ O2–

Ca2O21 1

Which we’ll compact and write as the final formula for calcium oxide.


Ca2+ O2–

CaO2

Formula

Checking charges, we have one Ca2+ ion and one O2– ion.


Ca2+ O2–

CaO2

Formula

So the total charge is positive 2 plus negative 2,


Ca2+ O2–

CaO2

Formula

+2 + –2

Which is equal to zero. So this formula is correct.


Ca2+ O2–

CaO2

Formula

+2 + –2 = 0

Question c asks us to write the formula for chromium(III) sulphide.

c) Write the formula for chromium(III) sulphide

Cr3+ S2–

Cr2S32

The chromium (III) ion is Cr 3 plus


Cr3+ S2–

Cr2S32

And the sulphide ion is S 2 minus


Cr3+ S2–

Cr2S32

We start the formula with the symbols Cr and S


Cr3+ S2–

Cr2S32

We take the 3 from the charge on chromium, (click) cross over and write (click) 3 next to sulphur in the formula.


Cr3+ S2–

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.


Cr3+ S2–

Cr2S32

There is no number other than 1, that will divide into 2 and 3, so this formula cannot be simplified.


Cr3+ S2–

Cr2S32

So the final formula for chromium III sulphide is Cr2S3.


Cr3+ S2–

Cr2S32

Formula

Checking charges, we have 2 Cr3+ ions and 3 S2– ions,


Cr3+ S2–

Cr2S32

Formula

32

So the total charge is 2 × positive 3, plus 3 × (–2), or positive 6 plus –6,


Cr3+ S2–

Cr2S32

Formula

322(+3) + 3(–2)

Which is equal to zero, so this formula is correct.


Cr3+ S2–

Cr2S32

Formula

322(+3) + 3(–2) = 0

Question d asks us to write the formula for palladium(IV) oxide.

d) Write the formula for palladium(IV) oxide

Pd4+ O2–

Pd2O42

Palladium(IV) is Pd with a charge of 4 plus


Pd4+ O2–

Pd2O42

And oxide is “O” with a 2 minus charge


Pd4+ O2–

Pd2O42

We start the formula with the symbols Pd and O


Pd4+ O2–

Pd2O42

We take the 4 from the charge on palladium, (click) crossover and (click) write 4 next to oxygen in the formula


Pd4+ O2–

Pd2O42

We take the 2 from the charge on the oxide ion, (click) crossover and (click) write 2 next to palladium in the formula


Pd4+ O2–

Pd2O42

Now, we have the formula Pd2O4.


Pd4+ O2–

Pd2O42

But we see that both of the subscripts 2 and 4 are divisible by 2.


Pd4+ O2–

Pd2O42

Both subscripts can be divided by

2

So we simplify by dividing both the 2 and the 4 by 2, giving us 1 and 2.


Pd4+ O2–

Pd2O42

Divide both subscripts by 2

1 2

We remove the subscript 1


Pd4+ O2–

Pd2O421 2

Remove the 1

Leaving us with the formula PdO2.


Pd4+ O2–

Pd2O421 2

Which we’ll compact and state as the final formula for palladium(IV) oxide


Pd4+ O2–

PdO2

Formula

Checking charges, we have one Pd4+ ion and two O2– ions


Pd4+ O2–

PdO2

Formula

2

So the total charge is positive 4 plus 2 times –2, or positive 4 plus negative 4,


Pd4+ O2–

PdO2

Formula

2 +4 + 2(–2)

Which is equal to zero, so PdO2 is the correct formula.


Pd4+ O2–

PdO2

Formula

2 +4 + 2(–2) = 0

Documents

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