Question

What is the total number of ions present in the formula $N{{a}_{2}}S{{O}_{4}}$?
A.2
B.3
C.4
D.6

Answer 1

Hint: Ions exist in ionic compounds. Ions are formed when any neutral atom gains or loses electrons. A cation is the ion formed by losing electrons and has a positive charge. An anion is formed by gaining electrons that consist of a negative charge. Ions form bonds by transferring electrons.

Complete answer:
Ions are the constituent atoms that are present in an ionic compound. Ions may be positive or negative depending upon their ability to gain or lose electrons. An atom loses electrons, becoming a cation that has a positive charge. While, when an atom gains electrons, it becomes an anion and has a negative charge.
We have been given an ionic compound $N{{a}_{2}}S{{O}_{4}}$, we have to find the total number of ions in this compound. This compound is called sodium sulfate, and has sodium and sulfate as constituent ions. So, breaking the compound in constituent ions we have, $N{{a}^{+}}$ and $S{{O}_{4}}^{2-}$. As the charge on sulfate ions is -2, therefore two ions of sodium will be needed to neutralize the charge. So, $N{{a}^{+}}$ are in a quantity of 2, while $S{{O}_{4}}^{2-}$ is in a quantity of 1 that makes a total of 3 ions.
Hence, the total number of ions present in $N{{a}_{2}}S{{O}_{4}}$ are 3 ions.

So, option B is correct.

Note:
If one may think that $N{{a}_{2}}S{{O}_{4}}$ has sodium, sulfur and oxygen as ions, so 3 ions, then it may be wrong, as sodium (Na), sulfur (S), and oxygen (O) are elements and not ions. As ions are formed by gaining or losing electrons. Sulfate has $S{{O}_{4}}^{2-}$ which is considered as a polyatomic ion.