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# In the reaction of ${H_2}S{O_4}$ with ${H_2}S$, the change in oxidation states of sulfur are:A. $+ 6$ to $+ 4$ and $- 2$ to $0$B. $+ 4$ to $+ 2$ and $+ 2$ to $0$C. $+ 6$ to $- 2$ and $+ 6$ to $4$D. $+ 2$ to $+ 6$ and $0$ to $- 2$

Last updated date: 13th Sep 2024
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Hint: We can define oxidation state as loss of an electron(s) in a chemical compound. We can calculate the oxidation state of an element in a compound with the rules of oxidation numbers.

The reaction that takes place between sulfuric acid and hydrogen sulfide, leads to the formation of sulfur dioxide, sulfur and water.
The chemical equation is given as,
${H_2}S{O_4} + {H_2}S\xrightarrow{{}}S{O_2} + S + 2{H_2}O$
Let us now calculate the oxidation state of sulfur in ${H_2}S{O_4}$
Let x be the oxidation state of sulfur in all the compounds of sulfur.
The oxidation state of hydrogen is $+ 1$.
The oxidation state oxygen is $- 2$.
In ${H_2}S{O_4}$, we can calculate the oxidation state of sulfur as,
$2\left( 1 \right) + x + 4\left( { - 2} \right) = 0$
$\Rightarrow$ $2 - 8 + x = 0$
$\Rightarrow$ $- 6 + x = 0$
$\Rightarrow$ $x = + 6$
The oxidation state of sulfur in ${H_2}S{O_4}$ is $+ 6$.
In ${H_2}S$, we can calculate the oxidation state of sulfur as,
$2\left( 1 \right) + x = 0$
$\Rightarrow$ $2 + x = 0$
$\Rightarrow$ $x = - 2$
The oxidation state of sulfur in ${H_2}S$ is $- 2$.
In $S{O_2}$, we can calculate the oxidation state of sulfur as,
$x + 2\left( { - 2} \right) = 0$
$\Rightarrow$ $x - 4 = 0$
$\Rightarrow$ $x = + 4$
The oxidation state of sulfur in $S{O_2}$ is $+ 4$.
The oxidation state of sulfur in its elemental state is zero.
The oxidation state of sulfur changes from$+ 6$ to $+ 4$and from $- 2$ to $0$.

So, the correct answer is “Option A”.

Note: We know that oxidation state is loss of an electron in a chemical compound. We can now see a few rules of oxidation numbers.
-A free element will be zero as its oxidation number.
-Monatomic ions will have an oxidation number equal to charge of the ion.
In hydrogen, the oxidation number is $+ 1$ when combined with elements having less electronegativity, the oxidation number of hydrogen is $- 1$.
-In compounds of oxygen, the oxidation number of oxygen will be$- 2$ and in peroxides, it will be$- 1$.
-Group $1$ elements will have $+ 1$ oxidation number.
-Group $2$ elements will have $+ 2$ oxidation numbers.
-Group $17$elements will have $- 1$oxidation number.
-Sum of oxidation numbers of all atoms in neutral compounds is zero.
-In polyatomic ions, the sum of the oxidation number is equal to the ionic charge.