Naming Compounds with Polyatomic Ions Example 3 We’ll go over another example of naming an ionic compound that has a polyatomic ion in its formula.

Name the compound with the formula Sn(Cr2O7)2 We’re asked to name the compound with the formula Sn, Cr2O7 in brackets, 2.

Sn(Cr2O7)2 3 Elements Name the compound with the formula Sn(Cr2O7)2 We see that Sn(Cr2O7)2 has 3 elements (click) tin, chromium, and oxygen. Any ionic compound with more than two elements must contain a polyatomic ion.

Sn(Cr2O7)2 Name the compound with the formula Sn(Cr2O7)2 So we look for polyatomic ions in the formula Sn(Cr2O7)2

Sn(Cr2O7)2 Name the compound with the formula Sn(Cr2O7)2 We see that Cr2O7 with a negative 2 charge is called dichromate

Sn(Cr2O7)2 Name the compound with the formula Sn(Cr2O7)2 And the metal Sn is found (click) on positive ion section of this table.

Sn(Cr2O7)2 Name the compound with the formula Sn(Cr2O7)2 We see that Sn is tin and it is a multivalent metal. It can have a charge of either positive 2 or positive 4.

Sn(Cr2O7)2 Name the compound with the formula Sn(Cr2O7)2 This is verified by finding tin on the periodic table, we see, it has these two different possible charges

Name the compound with the formula Sn(Cr2O7)2 +4 –4 The formula tell us we have one Sn ion

Name the compound with the formula Sn(Cr2O7)2 +4 –4 And 2 Cr2O7 ions.

Name the compound with the formula Sn(Cr2O7)2 +4 –4 The ion table tells us the Cr2O7 is called dichromate,

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 So we can start the name by writing (click) tin dichromate

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate ? Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– ? Cr2O72– +4 –4 Tin is a multivalent metal with a charge of either positive 4 or positive 2, so we need a roman numeral (click) after tin in the name

Sn4+ Cr2O72– Cr2O72– +4 –4 ? tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate ? Sn4+ Cr2O72– Cr2O72– +4 –4 We don’t know what roman numeral to use at this point. We determine that by finding the charge the tin ion needs to have to balance the total negative charge on the dichromate ions

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 The ion table tells us that (click) each dichromate ion has a charge of negative 2

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 Total Positive Charge Total Negative Charge So the total negative charge is negative 2 plus negative 2 (click) which is negative 4

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 Total Positive Charge Total Negative Charge The charges must be balanced, (click) so the total positive charge is positive 4

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 Total Positive Charge Total Negative Charge The total positive charge is positive 4 and there is only one tin ion, (click) so the charge on one tin ion must be positive 4

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Sn4+ Cr2O72– Cr2O72– +4 –4 Total Positive Charge Total Negative Charge Therefore, the roman numeral we use is (click) “I” “V” for 4

Sn4+ Cr2O72– Cr2O72– +4 –4 tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 tin(IV) dichromate Name Sn4+ Cr2O72– Cr2O72– +4 –4 Total Positive Charge Total Negative Charge So the final name is tin four dichromate.

tin(IV) dichromate Name the compound with the formula Sn(Cr2O7)2 So the final answer to the question is is, the name of this compound (click) is tin four dichromate.