
Find oxidation number of Sulphur in ${H_2}{S_4}{O_6}$
Answer
550.8k+ views
Hint:In this question, students have to find the oxidation number i.e., oxidation state of sulphur. Oxidation number is defined as the absolute number of electrons that a particle either gains or loses to form a chemical bond with another molecule.
Complete step by step solution:
There are some rules that students need to remember while figuring the oxidation states in a molecule.
-Any Free component has an oxidation number equivalent to zero.
-For monatomic particles the oxidation number consistently has a similar value as the net charge comparing to the particle
-The hydrogen particle has an oxidation condition of $ + 1$. When the bonding takes place with less electronegativity it shows an oxidation number of $ - 1$
-When oxygen is bonded with peroxides, its oxidation state is $ - 2$.
-All group 1 elements and alkali metals have oxidation state $ + 1$.
-Similarly, alkaline earth metals have oxidation state $ + 2$
Now, let us consider oxidation number of sulphur be X
With the help of above rules, we can calculate oxidation number of sulphur as follows:
$2 \times (1) + 4 \times X + 6 \times ( - 2) = 0$
Here 1 is oxidation number of Hydrogen
And -2 is oxidation number of Oxygen
Further solving the above equation we have:
$
2 + 4x - 12 = 0 \\
\Rightarrow 4x = 10 \\
\Rightarrow x = 2.5 \\
$
Hence, from the above explanation we can find that the oxidation number of Sulphur in is ${H_2}{S_4}{O_6}$ 2.5.
Additional Information:${H_2}{S_4}{O_6}$is known as Tetra thionic acid and has unsymmetrical Sulphur atoms. As electronegativity of Oxygen is higher than that of Sulphur, so when the bond will break, Oxygen will have oxidation number -1 and Sulphur will have oxidation number +1.
Note:Students are advised to do all the calculations properly. Also, students should take care of the fact that the electronegativities differ for all the elements in different conditions, so they should abide by the rules thoroughly mentioned above while answering these types of questions. These questions are important for both boards as well as competitive exams.
Complete step by step solution:
There are some rules that students need to remember while figuring the oxidation states in a molecule.
-Any Free component has an oxidation number equivalent to zero.
-For monatomic particles the oxidation number consistently has a similar value as the net charge comparing to the particle
-The hydrogen particle has an oxidation condition of $ + 1$. When the bonding takes place with less electronegativity it shows an oxidation number of $ - 1$
-When oxygen is bonded with peroxides, its oxidation state is $ - 2$.
-All group 1 elements and alkali metals have oxidation state $ + 1$.
-Similarly, alkaline earth metals have oxidation state $ + 2$
Now, let us consider oxidation number of sulphur be X
With the help of above rules, we can calculate oxidation number of sulphur as follows:
$2 \times (1) + 4 \times X + 6 \times ( - 2) = 0$
Here 1 is oxidation number of Hydrogen
And -2 is oxidation number of Oxygen
Further solving the above equation we have:
$
2 + 4x - 12 = 0 \\
\Rightarrow 4x = 10 \\
\Rightarrow x = 2.5 \\
$
Hence, from the above explanation we can find that the oxidation number of Sulphur in is ${H_2}{S_4}{O_6}$ 2.5.
Additional Information:${H_2}{S_4}{O_6}$is known as Tetra thionic acid and has unsymmetrical Sulphur atoms. As electronegativity of Oxygen is higher than that of Sulphur, so when the bond will break, Oxygen will have oxidation number -1 and Sulphur will have oxidation number +1.
Note:Students are advised to do all the calculations properly. Also, students should take care of the fact that the electronegativities differ for all the elements in different conditions, so they should abide by the rules thoroughly mentioned above while answering these types of questions. These questions are important for both boards as well as competitive exams.
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