Question

${\text{MS}}{{\text{O}}_{\text{4}}}$is the formula of sulphate of metal ${\text{M}}$. Write down the formula of it:
a.Hydroxide
b.Chlorite
c.Chloride
d.Nitrate
e.Nitrite
f.Peroxide
g.Chromate
h.Phosphate

Answer 1

Hint: To solve this question, we need to have knowledge about how to form compounds with different valency or charge. When a cation and an anion with different valencies join to form a compound, the charge on those ions are usually cross-multiplied with each other so as to maintain electrical neutrality. So if we can find the charge on metal ion M, then we can cross multiply it with the charge on the anions given to form the formula of those compounds.

Complete step by step answer:
We shall first find out the charge present on the metal ion M. As we know that, sulphate ion is present in the form ${\text{SO}}_{\text{4}}^{{\text{2 - }}}$, i.e. with a charge of $ - 2$. Thus, to maintain electrical neutrality, we can conclude that the metal ion is present in the form ${{\text{M}}^{{\text{2 + }}}}$.
In option a, hydroxide ion is present at ${\text{O}}{{\text{H}}^{\text{ - }}}$, it has a charge of $ - 1$. So, with ${{\text{M}}^{{\text{2 + }}}}$ when we cross multiply the charges, we get the formula as ${\text{M}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ . In this way, the electrical neutrality of the compound is maintained.
In option b, chlorite ion is present in the form of ${\text{ClO}}_{\text{2}}^{\text{ - }}$ , a charge of $ - 1$. So, with ${{\text{M}}^{{\text{2 + }}}}$ ion, it will form a compound with the formula \[{\text{M}}{\left( {{\text{Cl}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}\] .
In option c, chloride is present in the form of ${\text{C}}{{\text{l}}^{\text{ - }}}$ , a charge of $ - 1$. So, with ${{\text{M}}^{{\text{2 + }}}}$ ion, it will form a compound with the formula ${\text{MC}}{{\text{l}}_{\text{2}}}$ .
In option d, nitrate is present as ${\text{NO}}_{\text{3}}^{\text{ - }}$, it has a charge of $ - 1$. So, with ${{\text{M}}^{{\text{2 + }}}}$ , the compound formed will have the formula ${\text{M}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ .
In option e, nitrite is present as ${\text{NO}}_{\text{2}}^{\text{ - }}$ , a charge of $ - 1$. So, it will form a compound with ${{\text{M}}^{{\text{2 + }}}}$ ion having the formula ${\text{M}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_{\text{2}}}$ .
In option f, peroxide is present in the form of ${\text{O}}_{\text{2}}^{{\text{2 - }}}$ , a charge of $ - 2$. So, when we cross multiply it with the charge of ${{\text{M}}^{{\text{2 + }}}}$, we get the formula as ${\text{M}}{{\text{O}}_{\text{2}}}$.
In option g, chromate is present as ${\text{CrO}}_{\text{4}}^{{\text{2 - }}}$, a charge of $ - 2$. So, with ${{\text{M}}^{{\text{2 + }}}}$ it will form a compound with the formula ${\text{MCr}}{{\text{O}}_{\text{4}}}$ .
In option h, phosphate ion is found as ${\text{PO}}_{\text{4}}^{{\text{3 - }}}$ , a charge of $ - 3$. So, when we cross multiply it with the charge on ${{\text{M}}^{{\text{2 + }}}}$, the formula of the compound will be ${{\text{M}}_{\text{3}}}{\left( {{\text{P}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}$ .

Note:
 In some cases, while forming compounds, if the charges on both the ions are same, they are said to cancel each other out and instead of writing it as ${{\text{M}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ we write is as ${\text{M}}{{\text{O}}_{\text{2}}}$ . Sometimes, a metal ion may also show variable oxidation state while bonding with the same anion. For example, iron forms compounds with ${\text{C}}{{\text{l}}^{\text{ - }}}$ ion as both ${\text{F}}{{\text{e}}^{{\text{2 + }}}}$ and ${\text{F}}{{\text{e}}^{{\text{3 + }}}}$, forming ${\text{FeC}}{{\text{l}}_{\text{2}}}$ and ${\text{FeC}}{{\text{l}}_{\text{3}}}$, respectively.