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.
Name the compound with the formula Sn(Cr2O7)2
Sn(Cr2O7)2
3 Elements
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.
Name the compound with the formula Sn(Cr2O7)2
Sn(Cr2O7)2
So we look for polyatomic ions in the formula Sn(Cr2O7)2
Name the compound with the formula Sn(Cr2O7)2
Sn(Cr2O7)2
We see that Cr2O7 with a negative 2 charge is called dichromate
Name the compound with the formula Sn(Cr2O7)2
Sn(Cr2O7)2
And the metal Sn is found (click) on positive ion section of this table.
Name the compound with the formula Sn(Cr2O7)2
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.
Name the compound with the formula Sn(Cr2O7)2
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
2–
4+
Sn
Cr2O7
2–
Cr2O7
+4
–4
The formula tell us we have one Sn ion
Name the compound with the formula Sn(Cr2O7)2
And 2 Cr2O7 ions.
2–
4+
Sn
Cr2O7
2–
Cr2O7
+4
–4
Name the compound with the formula Sn(Cr2O7)2
2–
4+
Sn
Cr2O7
2–
Cr2O7
+4
–4
The ion table tells us the Cr2O7 is called dichromate,
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
+4
–4
So we can start the name by writing (click) tin dichromate
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
+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
Name the compound with the formula Sn(Cr2O7)2
tin(IV)
? dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
+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
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
+4
–4
The ion table tells us that (click) each dichromate ion has a charge of
negative 2
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
Total
Positive
Charge
+4
–4
Total
Negative
Charge
So the total negative charge is negative 2 plus negative 2 (click) which is
negative 4
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
Total
Positive
Charge
+4
–4
Total
Negative
Charge
The charges must be balanced, (click) so the total positive charge is
positive 4
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
Total
Positive
Charge
+4
–4
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
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
2–
4+
Sn
Cr2O7
2–
Cr2O7
Total
Positive
Charge
+4
–4
Therefore, the roman numeral we use is (click) “I” “V” for 4
Total
Negative
Charge
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate Name
2–
4+
Sn
Cr2O7
2–
Cr2O7
Total
Positive
Charge
+4
So the final name is tin four dichromate.
–4
Total
Negative
Charge
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate
So the final answer to the question is is, the name of this compound
(click) is tin four dichromate.