Respuesta :
Answer:
A) Oxidation state of carbon is +2
B) Oxidation state of carbon is 0
Explanation:
In the first question , the compound is propanone.
While assigning oxidation states ,
- We must see to that , the electrons in the bond moves towards the most electronegative atom.
- If they are equally electronegative , each of them gets 0 oxidation state respectively.
- And if there is one bond , the oxidation state obtained by the electronegative atom is -1 and the other is +1.
- The order of electronegativity is as follows :
[tex]O>C>H[/tex]
- Thus [tex]O[/tex] always gets -1 and [tex]H[/tex] always gets +1 in these two compounds.
- The carbon - carbon bond will give 0 oxidation state.
A) According to these rules the Oxidation state of carbon is +2
B) According to these rules the Oxidation state of carbon is 0
- Diagram attached below
- The direction of the arrows in the diagram represents the movement of electron
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The oxidation number of the central carbon in molecule A is +2 while the oxidation number of the central carbon carbon in molecule B is 0.
In organic compounds, the oxidation state of carbon in organic compounds is as follows;
- In the C - C, carbon has an oxidation number of zero
- In the C - H, carbon has an oxidation mumber of -1
- When carbon is bonded to an electronegative atom, the oxidation number of carbon is +1
The first compound is acetone shown in image (1), the oxidation number of carbon bonded to two C H 3 groups and an O atom through a double bond is; 0 + 0 +2 = +2
In the second compound which is propan - 2- ol(image 2), the oxidation number of the central carbon atom is 0 + 0 + 1 - 1 = 0
Learn more: https://brainly.com/question/25610095
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