Arrange ICl,HI,I_2,and HIO_4 in the decreasin | vedclass

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Question

(C) To find the oxidation number $(x)$ of iodine in each compound:$ICl$: $x + (-1) = 0 \Rightarrow x = +1$$HI$: $(+1) + x = 0 \Rightarrow x = -1$$I_2$

Vedclass · Accepted Answer

$HIO_4 > ICl > I_2 > HI$

Vedclass · Answer

(C) To find the oxidation number $(x)$ of iodine in each compound:$ICl$: $x + (-1) = 0 \Rightarrow x = +1$$HI$: $(+1) + x = 0 \Rightarrow x = -1$$I_2$: $2x = 0 \Rightarrow x = 0$$HIO_4$: $(+1) + x + 4(-2) = 0$ $\Rightarrow x - 7 = 0$ $\Rightarrow x = +7$Comparing the oxidation numbers: $+7 (HIO_4) > +1 (ICl) > 0 (I_2) > -1 (HI)$.Thus,the decreasing order is $HIO_4 > ICl > I_2 > HI$.

Arrange $ICl$,$HI$,$I_2$,and $HIO_4$ in the decreasing order of the oxidation number of iodine.

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