Question

Write down the name of the compound represented by $A{l_2}{(S{O_4})_3}$.

Answer 1

Hint: For finding out the name of compound see the ions attached it means see what is metal and which anion is present. Here, anion is of sulphur and we know that sulphur can make three types of anions which are sulphide, sulfate and sulfite. Now if anyone knows how these three represent then can easily differentiate between them.

Complete step-by-step answer:
We have aluminum here which is an element of group $13$ of boron family, each element of boron family have a charge of $( + 3)$ so aluminum is also having charge of $( + 3)$ represented as $A{l^{ + 3}}$. Now comes to the anionic part it is a sulphur constituent, we know that sulphur forms three types of anions, one is sulphide in which it will exist alone having a charge of $ - 2$ on it, it can be represented as ${S^{2 - }}$.

Another two types of anions of sulphur are just the same by name, one is sulfate and another is sulfite. In sulfate having $ - ate$ as a suffix so in this type of anion it is represented as $SO_4^{2 - }$ . In another anion, in sulfite it can be represented as $SO_3^{2 - }$. In the above formula aluminum reacts with sulfate and forms aluminum sulfate. $A{l^{ + 3}}\,\,\,SO_4^{2 - }$

These two ions form compounds that give charge of each on another anion, thus we get $A{l_2}{(S{O_4})_3}$ .


Note: When there is compound formation, the cation and anion react and give up charge which is present on their own towards the other atom or group. Same happens here, aluminum gave up charge $( + 3)$ to sulfate while sulfate also gave up charge of $ - 2$ to aluminum and got a compound named aluminum sulfate.