What are the oxidation number of the underlined elements in each of the following and how
do you rationalise your results ?
Answers
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
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H---O---------S⁺⁵----S⁰-----S⁰-------S⁺⁵--------O-----H
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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
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H------- C² ------------- C¹ ---- O----H [CH₃⁺ and CH₂OH⁻
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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
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H------C²----------C¹------O---H [CH₃⁺ and COOH⁻
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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.