What is the correct order of decreasing oxida | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To determine the oxidation state of iodine $(I)$ in each compound:$1$. In $HIO_4$: Let the oxidation state of $I$ be $x$. $1 + x + 4(-2) = 0 \impl

Vedclass · Accepted Answer

$HIO_4, ICl_5, I_2, HI$

Vedclass · Answer

(B) To determine the oxidation state of iodine $(I)$ in each compound:$1$. In $HIO_4$: Let the oxidation state of $I$ be $x$. $1 + x + 4(-2) = 0 \implies x = +7$.$2$. In $ICl_5$: Let the oxidation state of $I$ be $x$. $x + 5(-1) = 0 \implies x = +5$.$3$. In $I_2$: The oxidation state of iodine in its elemental form is $0$.$4$. In $HI$: Let the oxidation state of $I$ be $x$. $1 + x = 0 \implies x = -1$.Thus,the decreasing order of oxidation states is $+7 (HIO_4) > +5 (ICl_5) > 0 (I_2) > -1 (HI)$.

What is the correct order of decreasing oxidation state of iodine in the following compounds?

Similar Questions

The oxidation state of nitrogen is highest in

$A$ metal ion $M^{3+}$ after the loss of three electrons in a reaction will have an oxidation number equal to

The oxidation number of carbon in $CH_2Cl_2$ is

In which of the following,the oxidation state of nitrogen is lowest?

Carbon is present in the highest oxidation number in

Vedclass Test Series

Exam Paper Generator

Online Exam Module