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In the ethylene molecule the two carbon atoms have the oxidation number
(A) -1, -1
(B) -2, -2
(C) -1, -2
(D) +2, -2


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Answer
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Hint: Alkene group has a general formula ${{C}_{n}}{{H}_{2n}}$ . The oxidation number of both C atoms in those molecules are the same as they are exactly similar to one another.

Complete step by step solution:
Firstly, we will see about the oxidation number.
Oxidation number-
It is a number given to an element or atom which represents the loss or gain of electrons; separately or in any molecule.
Key facts-
Oxidation number of;
-Free element = 0
-Monatomic and Polyatomic ion = charge of the ion
-Hydrogen = +1; -1 (if bonded with less electronegative element)
-Oxygen = -2; -1 (in peroxides)
-Group I elements = +1
-Group II elements = +2
-Group XVII elements = -1 (in binary compounds)
Sum of the oxidation number of all the atoms in the neutral molecule is zero.
Steps to find oxidation number in a molecule-
1. Sum up the constant oxidation state of atoms or ions comprising the molecule.
2. Equate the total oxidation state of atoms or ions to the total charge of a molecule.
Illustration-
Ethylene (${{C}_{2}}{{H}_{4}}$) –
Let us consider the oxidation number of unknown (C-atom) as x.
As stated above, the oxidation number of Hydrogen is +1. Thus,
$\left( 2\times x \right)+\left( 4\times \left( +1 \right) \right)=0$
$x=-2$
Oxidation number of both C-atoms is -2.
As, the charge on ${{C}_{2}}{{H}_{4}}$ is zero.

Therefore, option (B) -2, -2 is correct.

Note: According to the equality of oxidation number of similar atoms, option (C) and (D) would never be an answer. Only concentrate on the other two options while solving such questions.