Question

You are given four compounds in which an element is kept in square brackets.

$[Mn{]}_2{O}_3$, $H\left[N\right]O_3$, $K\left[Mn\right]O_4$ and $H_2\left[C\right]O_3$

Select the option that represents the correct order of the compounds on the basis of increasing oxidation number value of the element in the square bracket that is higher the value of the oxidation number of the element in the square bracket higher its placement.

• A. $[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$ < $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$
• B. $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$ < $[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$
• C. $H\left[N\right]O_3$ < $[Mn{]}_2{O}_3$ < $K\left[Mn\right]O_4$ < $H_2\left[C\right]O_3$
• D. $H_2\left[C\right]O_3$ < $[Mn{]}_2{O}_3$ < $K\left[Mn\right]O_4$ < $H\left[N\right]O_3$

Answer 1

The correct option is A

$[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$ < $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$

Let's calculate the oxidation number of all the elements given in the square brackets.

(1) $[Mn{]}_2{O}_3$, let the oxidation number for Mn be $x$.

$2\times x+3\times (-2)=0$ oxidation number of Mn is +3.

(2) $H\left[N\right]O_3$, let the oxidation number for N be $x$.

$+1+x+3\times (-2)=0$ oxidation number of N is +5.

(3) $K\left[Mn\right]O_4$, let the oxidation number for Mn be $x$.

$+1+x+4\times (-2)=0$ oxidation number of Mn is +7.

(4) $H_2\left[C\right]O_3$, let the oxidation number for C be $x$.

$2\times (+1)+x+3\times (-2)=0$ oxidation number of C is +4.

So if the elements were to be arranged in increasing order of oxidation number value the correct order is

$[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$ < $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$