Question

Find the oxidation number of elements in each case.

$\underset{–}{S}$ in $N{a}_2{S}_2{O}_3,S_4,S_8$ and $N{a}_2{S}_2{O}_7$

• A. $+2,0,0,+6$
• B. $+1,1,0,+5$
• C. $0,0,1,+4$
• D. None of these

Answer 1

Answer: (A)

$+2,0,0,+6$

$N{a}_2{S}_2{O}_3$:

Let $S$ have $x$ oxidation state. $Na$ has fixed $+1$ and $O$ has also fixed $-2$ oxidation state.

So, $2(+1)+2x+3(-2)=0$
$2+2x-6=0$

$x=2$, so $S$ has $+2$ O.S.

$S_4$:
As this is an element so it must have 0 oxidation state.

$S_8$:
As this is also element so this must have 0 oxidation state.

$N{a}_2{S}_2{O}_7$
Let S have x oxidation state.

So, $2(+1)+2x+7(-2)=0$

$2+2x-14=0$

$2x=12$

$x=6$

So. $S$ has $+6$ oxidation state.