Question

The oxidation state of C in ${C_6}{H_{12}}{O_6}$is equal to the oxidation state of C in :
A.$HCOOH$
B.$HCHO$
C.$C{H_4}$
D.$CO$

Answer 1

To answer this question, recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and, therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in ${\text{C}}{{\text{O}}_{\text{2}}}$ would be $+ 4$ since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal to $+ 4$ .

Complete step by step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. The values can be zero, positive, or negative.
First find out the oxidation state of C in ${C_6}{H_{12}}{O_6}$. Let it be $x$. Then
$6x + 12 - 12 = 0$
$\Rightarrow x = 0$
Analysing each of the options systematically:
$HCOOH$ $\Rightarrow x + 2 + 2\left( { - 2} \right) = 0$
$\Rightarrow x = 2$
As this oxidation state of C is unequal to the one present in ${C_6}{H_{12}}{O_6}$, this option is wrong and can be eliminated
$HCHO$ $\Rightarrow x + 2 - 2 = 0$
$\Rightarrow x = 0$
As this oxidation state of C is equal to the one present in ${C_6}{H_{12}}{O_6}$, this option is correct.
$C{H_4}$ $\Rightarrow x + 4 = 0$
$\Rightarrow x = - 4$
As this oxidation state of C is unequal to the one present in ${C_6}{H_{12}}{O_6}$, this option is wrong and can be eliminated
$CO$ $\Rightarrow x - 2 = 0$
$\Rightarrow x = 2$
As this oxidation state of C is unequal to the one present in ${C_6}{H_{12}}{O_6}$, this option is wrong and can be eliminated.

Hence, the correct option is B.

Note:
In most of the compounds, the oxidation number of oxygen is $- 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $- 1$ . Example, $N{a_2}{O_2}$
Superoxide- Every oxygen atom is allocated an oxidation number of $- \dfrac{{{\text{ }}1}}{2}$ . Example, $K{O_2}$
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $+ 1$ .