What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?
(a) \( K\underline{I}_{3}\)
(b) \( H_{2}\underline{S}_{4}O_{6}\)
(c) \( \underline{Fe}_{3}O_{4}\)
(d) \( \underline{C}H_{3}\underline{C}H_{2}OH\)
(e) \( \underline{C}H_{3}\underline{C}OOH\)
1 Answer
Solution :-
(a) \( K\underline{I}_{3}\)
Let x be the oxidation no. of I.
Oxidation no. of K = +1
Then,
1(+1) + 3(x) = 0
1 + 3x = 0
x = \( -\dfrac{1}{3}\)
Oxidation no. cannot be fractional. Hence, consider the structure of \( KI_{3}\).
In \( KI_{3}\) molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.
_1638334978.bmp)
Therefore, in \( KI_{3}\) molecule, the oxidation no. of I atoms forming the molecule \( I_{2}\) is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.
_1638335033.bmp)
Two of the four S atoms have an O.N. of +5, whereas the other two have an O.N. of 0
(b) \( H_{2}\underline{S}_{4}O_{6}\)
Let x be the oxidation no. of S.
Oxidation no. of H = +1
Oxidation no. of O = -2
Then,
2(+1) + 4(x) + 6(-2) = 0
2 + 4x -12 = 0
4x = 10
x = \( +2\dfrac{1}{2}\)
However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule. The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.
(c) \( \underline{Fe}_{3}O_{4}\)
Let x be the oxidation no. of Fe.
Oxidation no. of O = -2
Then,
3(x) + 4(-2) = 0
3x -8 = 0
x = \( \dfrac{8}{3}\)
Oxidation no. cannot be fractional.
One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.
\( FeO^{+2} Fe_2O_3\)
(d) \(\underline{C}H_{3}\underline{C}H_{2}OH\)
Let x be the oxidation no. of C.
Oxidation no. of O= -2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 1(-2) = 0
2x + 4 -2 = 0
x = -2
Therefore, oxidation no. of C is -2.
(e) \( \underline{C}H_{3}\underline{C}OOH\)
Let x be the oxidation no. of C.
Oxidation no. of O= -2
Oxidation no. of H = +1
Then,
2(x) + 4(+1) + 2(-2) = 0
2x + 4 -4 = 0
x = 0
Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in \( CH_{3}COOH\)
Answered by Abhisek | 1 year agoRelated Questions
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