Arrange ICl, HI, I 2 and HIO 4 in decreasing order of oxidation state of iodine.

Question

Arrange ICl, HI, I 2 and HIO 4 in decreasing order of oxidation state of iodine. (A) HIO 4> ICl > HI > I 2 (B) ICl > HIO 4> I 2> HI (C) HIO 4> ICl > I 2> HI (D) ICl > HIO 4> HI > I 2

Tardigrade Admin · Accepted Answer

Oxidation number of iodine in oversetx-1ICl: x-1=0 ⇒ x=+1 overset+1 xHI :(+1)+x=0 ⇒ x=-1 stackrelxI2 : x=0 overset+1 x-2H I O 4:(+1)+x+4(-2)=0 ⇒ x=+7 Thus decreasing order of oxidation number of iodine is HIO 4> ICl > I 2> HI

Q. Arrange $I Cl, H I,$ $I_{2}$ and $H I O_{4}$ in decreasing order of oxidation state of iodine.

$H I O_{4} > I Cl > H I > I_{2}$

$I Cl > H I O_{4} > I_{2} > H I$

$H I O_{4} > I Cl > I_{2} > H I$

$I Cl > H I O_{4} > H I > I_{2}$

Q. Arrange ICl,HI, I2​ and HIO4​ in decreasing order of oxidation state of iodine.

HIO4​>ICl>HI>I2​

ICl>HIO4​>I2​>HI

HIO4​>ICl>I2​>HI

ICl>HIO4​>HI>I2​

Oxidation number of iodine in IClx−1:x−1=0⇒x=+1 HI+1x:(+1)+x=0⇒x=−1 Ix​2​:x=0 HIO4​+1x−2​:(+1)+x+4(−2)=0⇒x=+7 Thus decreasing order of oxidation number of iodine is HIO4​>ICl>I2​>HI

Q. Arrange $I Cl, H I,$ $I_{2}$ and $H I O_{4}$ in decreasing order of oxidation state of iodine.

$H I O_{4} > I Cl > H I > I_{2}$

$I Cl > H I O_{4} > I_{2} > H I$

$H I O_{4} > I Cl > I_{2} > H I$

$I Cl > H I O_{4} > H I > I_{2}$