Question

The chemical formula of magnesium sulphate is:
(A) $M{{g}_{2}}S{{O}_{4}}$
(B) $MgS{{O}_{4}}$
(C) $M{{g}_{2}}{{(S{{O}_{4}})}_{2}}$
(D) $Mg{{(S{{O}_{4}})}_{2}}$

Answer 1

Hint: Find out the valency of each element that is required for the chemical formula. They are magnesium, sulphur and oxygen (you can take the clue from the option itself). Cross the valencies so as to satisfy each atom.

Complete step by step solution:
Magnesium sulphate is a salt of an acid and base. The base part is magnesium hydroxide ($Mg{{(OH)}_{2}}$) and acid part is sulphuric acid (${{H}_{2}}S{{O}_{4}}$). From this we can make out that the sulphate group is actually “$S{{O}^{2-}}_{4}$”. Its valency is 2 with a negative charge, which implies this group is in a position to donate electrons as it has an excess of it.
On the other hand magnesium is a metal and therefore it is electropositive in nature. In its ionic form it will always have a positive charge which implies a deficiency of electrons and therefore it will always be ready to accept electrons. Magnesium belongs to group number 2 which is the group of alkali-earth metals and therefore it has a valency of 2.
It can be explained from its electronic configuration which is: \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}\]
The outermost “$s$” orbital contains two electrons which can be easily donated to achieve the state of neon (a noble element).
Now, as both of them have a valency of “2” it implies multiple atoms of any element or group is not required, only one atom of each would suffice.
The answer to this question is therefore option (B) $MgS{{O}_{4}}$
So, the correct answer is “Option B”.


Note: The valency of magnesium and sulphate group is 2. That does not mean the formula is$M{{g}_{2}}{{(S{{O}_{4}})}_{2}}$. This is a wrong way to interpret the use of valency. Valency is defined as the number of electrons that can be donated or accepted to reach a stable configuration.
As both have the same valency any one of the group cannot have multiple units such as $M{{g}_{2}}S{{O}_{4}}$or$Mg{{(S{{O}_{4}})}_{2}}$.