Search any question & find its solution
Question:
Answered & Verified by Expert
In which of the following compounds, sulphur show maximum oxidation number?
Options:
Solution:
1194 Upvotes
Verified Answer
The correct answer is:
All have same oxidation number for sulphur
All have same oxidation number for sulphate, i.e +6 .
(i)Oxidation number of S is
$\begin{aligned}
& \mathrm{H}_2 \mathrm{SO}_4=+2+x-8=0 \\
& \therefore x=+6
\end{aligned}$
(ii) Oxidation number of S in $\mathrm{SO}_3=x-6=0$
$\therefore x=+6$
$\begin{aligned} & \text { (iii) Oxidation number of } \mathrm{S} \text { in } \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \\ & =2+2 x-14=0\end{aligned}$
$\begin{aligned} \therefore 2 x & =+12 \\ x & =+6\end{aligned}$
[where, $x=$ Oxidation number sulphur $(S)$ ]
(i)Oxidation number of S is
$\begin{aligned}
& \mathrm{H}_2 \mathrm{SO}_4=+2+x-8=0 \\
& \therefore x=+6
\end{aligned}$
(ii) Oxidation number of S in $\mathrm{SO}_3=x-6=0$
$\therefore x=+6$
$\begin{aligned} & \text { (iii) Oxidation number of } \mathrm{S} \text { in } \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \\ & =2+2 x-14=0\end{aligned}$
$\begin{aligned} \therefore 2 x & =+12 \\ x & =+6\end{aligned}$
[where, $x=$ Oxidation number sulphur $(S)$ ]
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.