Question

Match the compounds given in the I with the oxidation states of carbon given in column II and mark the appropriate choice.

Column I	Column II
(A) ${{C}_{6}}{{H}_{12}}{{O}_{6}}$	(i) +3
(B) $CHC{{l}_{3}}$	(ii) -3
(C) $C{{H}_{3}}C{{H}_{3}}$	(iii) +2
(D) ${{(COOH)}_{2}}$	(iv) 0

(A) (A)- (iv) , (B)- (iii) , (C)- (ii) , (D)- (i)
(B) (A)- (i) , (B)- (ii) , (C)- (iii) , (D)- (iv)
(C) (A)- (ii) , (B)- (iii) , (C)- (iv) , (D)- (i)

Answer 1

Hint: Oxidation tells us about the charge carried by the compound in the chemical reaction and if we know the oxidation numbers of other atoms in that compound, then by taking the oxidation number of that very atom as x, we can calculate its oxidation state. Now solve it.

Complete step by step answer:
First of let’s discuss the oxidation state. By the oxidation state ,we mean the charge present on the element or the compound when it undergoes a chemical reaction. The charge i.e. the oxidation state tells us whether that particular element or the compound in that chemical reaction has undergone oxidation i.e. loses electrons or reduction i.e. gain electrons.
Now considering the statement;
(A) ${{C}_{6}}{{H}_{12}}{{O}_{6}}$
Oxidation of carbon in this is as;
$\begin{align}
  & oxidation\text{ }state\text{ }of\text{ }carbon=6(x)+12\times 1+6\times (-2)=0 \\
 &\implies \text{ 6}x+12-12=0 \\
 &\implies \text{ 6}x+0=0 \\
 & \implies\text{ }x=-\frac{0}{6} \\
 &\implies \text{ x=0} \\
\end{align}$

So, the oxidation of carbon in ${{C}_{6}}{{H}_{12}}{{O}_{6}}$is $0$.
Similarly, we can find the oxidation states of carbon in the rest of the given compounds.
(B) The oxidation of carbon in $CHC{{l}_{3}}$is $+2$.
(C) The oxidation of carbon in $C{{H}_{3}}C{{H}_{3}}$i.e. ${{C}_{2}}{{H}_{6}}$ is $-3$.
(D) The oxidation state of carbon in ${{(COOH)}_{2}}$ i.e. ${{C}_{2}}{{O}_{4}}{{H}_{2}}$ is $+3$.
So, the Matching of the compounds given in the I with the oxidation states of carbon given in column II is as;

Column I	Column II
(A) ${{C}_{6}}{{H}_{12}}{{O}_{6}}$	(i) +3 (D)
(B) $CHC{{l}_{3}}$	(ii) -3 (C)
(C) $C{{H}_{3}}C{{H}_{3}}$	(iii) +2 (B)
(D) ${{(COOH)}_{2}}$	(iv) 0 (A)

Hence, option (A) is correct.

Note: If the oxidation state is positive , then that means that the compound has undergone oxidation and on the other hand, if the oxidation state is negative, then that means that the compound has undergone reduction.