we’ll go over another example of naming an ionic compound that has a polyatomic ion in its formula
TRANSCRIPT
We’ll go over another example of naming an ionic compound that has a polyatomic ion in its formula.
Naming Compounds with Polyatomic Ions
Example 3
We’re asked to name the compound with the formula Sn, Cr2O7 in brackets, 2.
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
3 Elements
Name the compound with the formula 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
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
Sn(Cr2O7)
2
Name the compound with the formula Sn(Cr2O7)2
The formula tell us we have one Sn ion
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
And 2 Cr2O7 ions.
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
The ion table tells us the Cr2O7 is called dichromate,
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
So we can start the name by writing (click) tin dichromate
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
tin(IV) dichromate
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
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
tin(IV) dichromate
?
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
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
tin(IV) dichromate?
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
Sn4+ Cr2O72–
+4 –4Cr2O7
2–
tin(IV) dichromate
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
Sn4+ Cr2O72–
+4 –4Total Positive Charge
Total Negative Charge
Cr2O72–
tin(IV) dichromate
The charges must be balanced, (click) so the total positive charge is positive 4
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Total Positive Charge
Total Negative Charge
Cr2O72–
tin(IV) dichromate
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
Sn4+ Cr2O72–
+4 –4Total Positive Charge
Total Negative Charge
Cr2O72–
tin(IV) dichromate
Therefore, the roman numeral we use is (click) “I” “V” for 4
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Total Positive Charge
Total Negative Charge
Cr2O72–
tin(IV) dichromate
So the final name is tin four dichromate.
Name the compound with the formula Sn(Cr2O7)2
Sn4+ Cr2O72–
+4 –4Total Positive Charge
Total Negative Charge
Cr2O72–
tin(IV) dichromate Name
So the final answer to the question is is, the name of this compound (click) is tin four dichromate.
Name the compound with the formula Sn(Cr2O7)2
tin(IV) dichromate