Question

What is the correct name for $S{n_3}{\left( {P{O_4}} \right)_2}$?

Answer 1

Hint: Ionic compounds are neutral compounds that contain both positively charged cations and negatively charged anions. Transition metals are elements with a partially filled d sub-shell on their atoms or those that can produce cations with an incomplete d sub-shell.

Complete answer:
The most important thing to remember is that we are working with an ionic compound containing the transition metal tin, $Sn$. As we might be aware, transition metals may have several oxidation states, so we will need to use a Roman numeral to indicate the oxidation state of the transition metal in this compound.
The Criss cross rule is used to write ionic formulas that states that the cation's charge becomes the anion's subscript, and vice versa.
In this case, we have $S{n_3}{\left( {P{O_4}} \right)_2}$, since the charge of the anion will be the $3$ subscript of the cation, and the charge of the cation will be the $2$ subscript of the anion, we may claim that
$S{n_3}{\left( {P{O_4}} \right)_2}\,\, \Leftrightarrow \,\,S{n^{2 + }}PO_4^{3 - }$
Remember that since the charges must be balanced, a formula unit of this compound will contain
a total of the three $S{n^{2 + }}$ cations
a total of the two $PO_4^{3 - }$ anions
As a result, the tin cation is in its $ + 2$ oxidation state, which necessitates the use of the Roman numeral (II) in the compound's name.
Finally, the phosphate polyatomic ion is $PO_4^{3 - }$. Substituting it all together we get the compound's name Tin (II) phosphate.
The correct name for $S{n_3}{\left( {P{O_4}} \right)_2}$is Tin (II) phosphate.

Note:
Tin alloys such as soft solder, pewter, bronze, and phosphor bronze are significant. Superconducting magnets are made of a niobium-tin alloy.