Question

What is the ionic compound name for $K_{2}S$ ?

Answer 1

Hint : Systematic name, most commonly known as IUPAC name are generally assigned to the compounds. IUPAC stands for International Union of Pure and Applied Chemistry. It is the world authority on the chemical nomenclature as well as terminology. In this question we have to tell the ionic compound name for $K_{2}S$ which is an inorganic compound and produced by heating $K_{2}S{O}_4$ in the presence of carbon (coke):
 $K_{2}S{O}_4+4C\to K_{2}S+4CO$

Complete Step By Step Answer:
IUPAC provides consistency to the names of organic compounds. It enables every compound to possess a unique name, which otherwise is not plausible with the common names. The chemical structure for $K_{2}S$ is shown below:

As clear from its chemical formula and structure, $K_{2}S$ comprises two metal ions and a non-metal ion. We know that any compound containing a metal and a non-metal is considered to be an ionic compound. Now let us identify the name for the given compound i.e. $K_{2}S$ . $K_{2}S$ consists of $S$ as non-metals after the metal i.e. $K$ so we'll use the Periodic Table to identify their names. Rules to write the names of Ionic Compounds are listed below:
1) Name the metal (i.e. cation) according to its appearance in the Periodic Table. For example: $N{a}^{+}=Sodium,M{g}^{2+}=Magnesium,A{l}^{3+}=Aluminum$
In the present case, $K$ stands for Potassium.
2) Find the name of the non-metal (i.e. anion) according to its appearance in the Periodic Table and replace its ending with ‘ide’ when you name a compound.
In the present case, $S$ stands for sulfur and it will be written as sulfide.
Hence after combining the name of metal and non-metal, the name of $K_{2}S$ is Potassium sulfide.

Note :
If you are given the systematic name of an ionic compound and you are asked to find the chemical formula of that compound, take into account the charges of each of the elements present in the compound. Make sure that the charges are balanced such that net charge for the compound must be zero.