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Get Answers to all your Questions
Q.57) Consider the following compounds :
$\mathrm{KO}_2, \mathrm{H}_2 \mathrm{O}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$.
The oxidation states of the underlined elements in them are, respectively:
A) $+4,-4$, and +6
B) $+1,-1$, and +6
C) $+2,-2$, and +6
D) $+1,-2$, and +4
Answers (1)
- KO? (O in KO? = –½ (superoxide ion, O??))
$x+2(-\frac{1}{2})=0 \Rightarrow x= 1 $
- H?O? (O in H?O? = –1 (peroxide))
$2(+1)+2 x=0 \Rightarrow 2+2 x=0 \Rightarrow x=-1$
- H?SO? (S in H?SO? = +6)
$2(+1)+x+4(-2)=0 \Rightarrow 2+x-8=0 \Rightarrow x=+6$
Hence the correct option is (2)
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