Question

What is the name for $S{{n}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}$ ?

Answer 1

Hint : We know that In order to solve this question we will first learn all the oxide state that tin can have then we will learn the phosphate of tin (II) phosphate and then we will calculate the molecular mass of the tin (II) phosphate by putting all the elements.

Complete Step By Step Answer:
The important thing to realize here is that you're dealing with an ionic compound that contains tin, $Sn,$ a transition metal. As you know, transition metals can exhibit multiple oxidation states, which implies that you're going to have to use a Roman numeral to indicate the oxidation state of the transition metal in this compound. So, ionic formulas are written based on the criss cross rule. Which states that the charge on the cation becomes the subscript of the anion, and vice versa.
In this case, you have $S{{n}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}$ Since the three subscript of the cation will be the charge of the anion, and the $2$ subscript of the anion will be the charge of the cation, you can say that $S{{n}_{3}}{{\left( P{{O}_{4}} \right)}_{2}}\Leftrightarrow S{{n}_{2}}+PO_{3}^{-4}$
Keep in mind that since the charges must be balanced, you can say that a formula unit of this compound will contain a total of three $S{{n}^{2+~}}$ cations, a total of two $PO_{3}^{-4~}$ anion. So, the tin cation is in its $+2~$ oxidation state, which means that you're going to have to use the Roman numeral (II) in the name of the compound. Finally, $PO_{3}^{-4~}$is the phosphate polyatomic ion.
Put all this together to get the name of the compound $Sn3\left( PO4 \right)2\to ~tin\left( II \right)\text{ }phosphate$

Note :
Remember that while solving these types of questions we should keep in mind that we have to put the value of the molecular mass of individual compounds correctly and first calculate the number of atoms present of the same type accurately.