Chemistry, asked by hari66074, 6 months ago

What are the oxidation number of the underlined elements in each of the following and how

do you rationalise your results ?​

Answers

Answered by diyakhrz12109
3

Answer:

Explanation:

(a) KI₃

let the oxidation number of I is x

+1 + 3x = 0

   x = -1/3

hence, the average oxidation number of I is -1/3 .but we know, oxidation number can't be fractional. actually,

      the structure of KI₃,     K⁺ (I⁰------I⁰<----I⁻)⁻

in this structure, a co-ordination bond is formed between I₂ molecule and I⁻ ion. hence, oxidation of I₃ is -1

(b)H₂S₄O₆

let the oxidation number of S = x

+2+4x +6(-2)=0

 x = +5/2  

similarly, oxidation number of this compound depends upon structure of molecule.  

see structure,      O                       O

                             ||                        ||

         H---O---------S⁺⁵----S⁰-----S⁰-------S⁺⁵--------O-----H

                             ||                        ||  

                             O                      O

after seeing the structure, it is clear that oxidation number of Satoms  are +5,0,0, +5 respectively.

(c)Fe₃O₄

let the oxidation number of Fe is x  

then, 3x + 4(-2) = 0

   x = + 8/3  

hence, the average oxidation number of Fe is +8/3, but stoichiometrically, Fe₃O₄ is an equimolar mixture of FeO and Fe₂O₃.  

                The oxidation number of Fe in FeO is +2  

and the oxidation number of Fe in Fe₂O₃ is +3  

then, 2x + x = +6 + 2 = +8 ⇒ x = +8/3

(d) CH₃CH₂OH

let the o.n of C is x  

 x + 3 + x + 2 - 2 + 1 =0

     x = -2  

therefore, average o.n of C is -2  

now, see the structure of given molecule.

             H                H

              |                  |

 H------- C² ------------- C¹ ---- O----H   [CH₃⁺ and CH₂OH⁻

             |                   |

            H                  H

   now, oxidation number of C₁ atom : x+ 2 -2 +1 = -1 ⇒ x = -2

   oxidation number of C₂ atom : x + 3 = +1 ⇒ x = - 2

thats why average oxidation number of C - atoms is -2  

(e) CH₃COOH

    let the oxidation number of C is x  

x + 3 + x -2 -2 + 1 = 0

 2x = 0  ⇒x = 0

therefore, average oxidation number of C is zero.  

                 let see the structure for understanding  

         H           O

          |            ||

H------C²----------C¹------O---H    [CH₃⁺ and COOH⁻  

         |

        H

               now, oxidation number of C₁ atom : x - 2 -2 +1 =1 ⇒ x = +2

                       oxidation number of C₂ atom : x +3 = +1 ⇒ x = -2  

that's why average oxidation number of CH₃COOH is zero.

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