Question

A metal (M) when forms sulphate of formula ${{\text{M}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, what will be the formula of the compound when it forms bromine compound.

Answer 1

Hint:To answer this question, you must find the valency of the metal M. Majority of the physical and chemical properties of an element are determined by their valence electrons.

Complete answer:
The valency of an atom is equal to the number of electrons present in the valence shell of the atom. The valence electrons are the most loosely held electrons in the atom and thus they determine the properties of the element. Each atom tends to attain a stable noble gas- like electronic configuration. This stable electronic configuration can be attained by either losing, gaining or sharing electrons.
It is given to us that the metal forms sulphate of the chemical formula of the type ${{\text{M}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. We know that sulphate ion carries a negative charge of $ - 2$. Since the chemical formula is ${{\text{M}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, it means that the metal is existing as a cation of charge $ + 1$. Thus, the metal M has a valency of 1.
Now, considering the bromine compound of the metal, metals form bromides. Bromine is a halogen and needs one electron to attain stable noble- gas configuration. Thus, it has a valency of 1.
The bromine compound of the metal will be of the form ${\text{MB}}{{\text{r}}_{\text{2}}}$

Note:
Valency of an element represents the combining capacity of that element. Valency is not always necessarily equal to the number of valence electrons. For number of valence electrons less than 4, the valency is equal to the number of valence electrons and for number of valence electrons more than 4, the valency is given by subtracting the number of electrons from 8.