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junior college 2 | H2 Maths
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The ans key says the C2 oxidation no changes from +2 to 0. Shdnt it be +2 to -1? Since the C after reduction is bonded to an O
Date Posted: 6 years ago
Views: 393
But the electronegative oxygen will cause a change in the oxidation state isnt it?
Hello, I don't really agree with Tan Kiat Boon's answer.
To calculate the oxidation state of a Carbon in an organic compound, we need to consider the electronegativity of the atoms. Since the electronegativity of C is greater than H, for a compound like CH4, we assign Carbon to have an oxidation state of -4, and for CH3CH3, the oxidation state is -3 for both Carbons.
In this compound, to calculate the initial oxidation state of C2, we see that it is bonded to 2 C atoms and double bonded to an O atom. Hence, the initial oxidation state is +2 (due to the oxygen atom).
After reduction, C2 is now bonded to 2 Carbons, 1 O-H and 1 Hydrogen atom. The oxygen atom in this case would only give C2 a +1 oxidation state (single bond, +1 oxidation state to alcohol H). However, as C2 is bonded to a H, the oxidation state decreases by 1 (as electronegativity of C is greater than H). As such, the overall oxidation state of C2 after reduction is +1-1=0.
If you follow this pattern of logic, C1, C3 and C4 are reduced from +3 to -1.
Hope this explanation helps.
To calculate the oxidation state of a Carbon in an organic compound, we need to consider the electronegativity of the atoms. Since the electronegativity of C is greater than H, for a compound like CH4, we assign Carbon to have an oxidation state of -4, and for CH3CH3, the oxidation state is -3 for both Carbons.
In this compound, to calculate the initial oxidation state of C2, we see that it is bonded to 2 C atoms and double bonded to an O atom. Hence, the initial oxidation state is +2 (due to the oxygen atom).
After reduction, C2 is now bonded to 2 Carbons, 1 O-H and 1 Hydrogen atom. The oxygen atom in this case would only give C2 a +1 oxidation state (single bond, +1 oxidation state to alcohol H). However, as C2 is bonded to a H, the oxidation state decreases by 1 (as electronegativity of C is greater than H). As such, the overall oxidation state of C2 after reduction is +1-1=0.
If you follow this pattern of logic, C1, C3 and C4 are reduced from +3 to -1.
Hope this explanation helps.
I think Sao Chuan's answer make more sense. Do refer to his instead to avoid confusion. Thanks for the insight and explanation! I think my understanding for this part is wrong.
Ohh i see thank you!!
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