Question

What are the oxidation numbers of C in $ {\text{C}}{{\text{H}}_{\text{4}}} $ , $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{Cl}} $ , $ {\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} $ , $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and $ {\text{CC}}{{\text{l}}_4} $ respectively?
(A) $ {\text{ + 4, + 2,0, - 2, - 4}} $
(B) $ {\text{ + 2, + 4,0, - 4, - 2}} $
(C) $ {\text{ - 4, - 2,0, + 2, + 4}} $
(D) $ {\text{ - 2, - 4,0, + 4, + 2}} $

Answer 1

Hint: In the above question, we are asked to find out the oxidation number of carbon atoms in the compounds listed. The charge on H atom is $ {\text{ + 1}} $ and Cl atom is $ - 1 $ and we can take the oxidation state of the carbon as x and by substituting the values of charge of H atom and Cl atom and comparing it to charge of the compound, we can find out the value of x.

Formula Used
Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)

Complete Step by step solution
Oxidation number which is also called an oxidation state can be defined as the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.
Let the oxidation state of carbon be x.
For $ {\text{C}}{{\text{H}}_{\text{4}}} $ :
Charge on H atom = 1
No. of H atom = 4
Charge of the molecule is 0.
Since, Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)
 $ \Rightarrow x = 0 - (1 \times 4) = - 4 $
So, oxidation state of carbon is $ - 4 $
For $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{Cl}} $ :
Charge on H atom = 1
No. of H atom = 3
Charge on Cl atom = $ - 1 $
No. of Cl atom = 1
Charge of the molecule is 0.
Since, Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)
 $ \Rightarrow x = 0 - (1 \times 3 - 1 \times 1) = - 2 $
So, oxidation state of carbon is $ - 2 $
For $ {\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} $ :
Charge on H atom = 1
No. of H atom = 2
Charge on Cl atom = $ - 1 $
No. of Cl atom = 2
Charge of the molecule is 0.
Since, Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)
 $ \Rightarrow x = 0 - (1 \times 2 - 1 \times 2) = 0 $
So, oxidation state of carbon is $ 0 $
For $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ :
Charge on H atom = 1
No. of H atom = 1
Charge on Cl atom = $ - 1 $
No. of Cl atom = 3
Charge of the molecule is 0.
Since, Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)
 $ \Rightarrow x = 0 - (1 \times 1 - 1 \times 3) = + 2 $
So, oxidation state of carbon is $ + 2 $
For $ {\text{CC}}{{\text{l}}_{\text{4}}} $ :
Charge on Cl atom = $ - 1 $
No. of Cl atom = 4
Charge of the molecule is 0.
Since, Oxidation number of an atom = Charge on the molecule – ( Charge on $ {{\text{i}}^{{\text{th}}}} $ element $ {{ \times }} $ no. of $ {{\text{i}}^{{\text{th}}}} $ atom)
 $ \Rightarrow x = 0 - ( - 1 \times 4) = + 4 $
So, oxidation state of carbon is $ + 4 $
Since, the oxidation numbers of C in $ {\text{C}}{{\text{H}}_{\text{4}}} $ , $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{Cl}} $ , $ {\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} $ , $ {\text{CHC}}{{\text{l}}_{\text{3}}} $ and $ {\text{CC}}{{\text{l}}_4} $ are $ - 4, - 2,0, + 2, + 4 $ respectively.
So, the correct option is option C.

Note
The oxidation number of hydrogen is $ {\text{ + 1}} $ when it is combined with non-metals such as in $ {\text{C}}{{\text{H}}_{\text{3}}} $ , $ {\text{HCl}} $ and others.
The oxidation number of hydrogen is $ - 1 $ when it is combined with a metal such as in $ {\text{Ca}}{{\text{H}}_{\text{2}}} $ , and $ {\text{LiAl}}{{\text{H}}_{\text{4}}} $ .