Question

Find oxidation number of carbon in:
a) $C{{H}_{3}}C{{H}_{2}}OH$
b) $C{{H}_{3}}COOH$

Answer 1

Hint: The oxidation number also known as oxidation state which describes the degree of oxidation i.e. loss of electrons of an atom in a chemical compound. Conceptually the oxidation state may be positive, negative or zero.

Complete answer:
Oxidation states are typically represented by integers which may be positive, zero, or negative. In some cases, the average oxidation state of an element is a fraction.
The substance which donates electrons during the oxidation process is known as oxidized substance. To determine in which compound nitrogen have greatest oxidation state we have to calculate the oxidation state of nitrogen in the given atom which can be calculated as follows:
a)$C{{H}_{3}}C{{H}_{2}}OH$is known by the name ethanol which represents the presence of alcohol in this. Now we know oxidation state of oxygen is always two and hydrogen is one now only one oxygen atom is present and 6 hydrogen atoms are present in it so the oxidation state of carbon can be calculated by using the following method:
Let us suppose the oxidation state of carbon is x.
Now on carbon one three hydrogen atoms are there i.e. $C{{H}_{3}}=x+3=o;x=-3$ and on the 2nd carbon atom two hydrogen and one oxygen atom along with hydrogen is there, therefore oxygen will carry charge 1 and carbon will carry -3 charge.
Hence in $C{{H}_{3}}C{{H}_{2}}OH$ both carbon carries -3 charge
b) $C{{H}_{3}}COOH$ this can also be calculated in similar manner i.e.
Now the value of x can be calculated by on first carbon as discussed above i.e. it would be -3 and on second carbon atom 2 oxygen atom along with hydrogen so the overall charge will be 3 and carbon carries charge -3.
Hence in $C{{H}_{3}}COOH$ also both carbon carry -3 charge.

Note:
The increase in oxidation state of an atom, through a chemical reaction, is known as an oxidation; a decrease in oxidation state is known as a reduction. Such reactions involve the formal transfer of electrons: a net gain in electrons being a reduction, and a net loss of electrons being an oxidation. For pure elements, the oxidation state is zero.