Help with formula for Tin IV Carbonate...thanks?

Question

I understand the crossover technique:

Fe2O3, where Fe's charge is +3 and O's is -2. But with Tin IV carbonate, Sn(Co3)2, it seems it fails me. What am I doing wrong? I know to multiply -2 and 2 to get -4 and simply offset that with 4 for Sn. Thanks for helping.

Answer 1

The crossover technique is kind of misleading in that it works for most ionic compounds but it's really not the proper way to think about this. You have to realize that the charges must balance and the overall change of the compound will be zero, you aren't just mindlessly switching numbers.

Use this formula and you'll always get it right:

a + x = 0

where "a" will be multiplied by the change on the metal (cation) and "x" will be multiplied by the change on the nonmetal (anion).

In this case the cation is Sn^4+ and the anion is CO3^2-:

(4a) + (-2x) = 0

we set it equal to zero because the charge on the compound will be 0

Then subtract the anion from both sides:

4a = 2x

and find the ratio of a/x

a/x = 2/4 which reduces to 1/2

so the formula is

Sn(CO3)2

Answer 2

Sn(iv) has a charge if +4

CO3 has a charge of -2

so for a balanced ionic structure you need two carbonates to every Tin atom. giving the formulae Sn(CO3)2.

Answer 3

The compound needs to be neutral.

You have Tin (IV) which has a charge of +4

You have CO3 which has a charge of -2

If you have 2 CO3 present then this has a charge of -4

As the charge of 2 CO3 balances out the +4 of Tin (IV) the compound is Sn(CO3)2