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Question: What is the total number of ions present in the formula \(N{{a}_{2}}S{{O}_{4}}\)? A.2 B.3 C.4 ...

What is the total number of ions present in the formula Na2SO4N{{a}_{2}}S{{O}_{4}}?
A.2
B.3
C.4
D.6

Explanation

Solution

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 Na2SO4N{{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, Na+N{{a}^{+}} and SO42S{{O}_{4}}^{2-}. As the charge on sulfate ions is -2, therefore two ions of sodium will be needed to neutralize the charge. So, Na+N{{a}^{+}} are in a quantity of 2, while SO42S{{O}_{4}}^{2-} is in a quantity of 1 that makes a total of 3 ions.
Hence, the total number of ions present in Na2SO4N{{a}_{2}}S{{O}_{4}} are 3 ions.

So, option B is correct.

Note:
If one may think that Na2SO4N{{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 SO42S{{O}_{4}}^{2-} which is considered as a polyatomic ion.