Question

Carbon is in lowest oxidation state in:
A.$C{H_4}$
B.$CC{l_4}$
C.$C{F_4}$
D.$C{O_2}$

Answer 1

Hint: Oxidation is a number assigned to an element in a chemical combination which represents the number of electrons lost or gained, by an atom of that element in the compound. If the oxidation number is positive, then this means that the atom loses electrons, and if it is negative, it means the atom gains electrons. If it is zero, then the atom neither gains nor loses electrons.

Complete answer:
Oxidation number, also called oxidation state, the total number of electrons that an atom either gains or losses in order to form a chemical bond with another atom.
The oxidation state of a free element (uncombined element) is zero. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. Note that the sum of the oxidation levels of all of the atoms in the molecule equals zero.
 Let’s calculate the oxidation number of each.
Let the oxidation number of carbon be $X$
Oxidation state of $H = + 1,Cl = F = - 1$ and $O = 12$
A.$C{H_4} = X + 4 \times 1 = 0$
$ \Rightarrow X = - 4$
B.$CC{l_4} = X + 4 \times ( - 1) = 0$
$ \Rightarrow X = + 4$
C.$C{F_4} = X + 4 \times ( - 1) = 0$
$ \Rightarrow X = + 4$
D.$C{O_2} = X + 2 \times ( - 2) = 0$
$ \Rightarrow X = + 4$
Lowest oxidation of carbon is $ - 4$ is in $C{H_4}$
So, the correct answer is (A) $C{H_4}$

Note:
The oxidation state of an atom does not represent the real formal charge on that atom, or any other actual atomic property. This is particularly true of high oxidation states, where the ionization energy required to produce a multiply positive ion is far greater than the energies available in chemical reactions. Additionally, the oxidation states of atoms in a given compound may vary depending on the choice of electronegativity scale used in their calculation.