Question

The oxidation number of carbon is zero in:
(This question has multiple answers)
A. ${\text{HCHO}}$
B. ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$
C. ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$
D. ${{\text{C}}_{12}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{11}}$

Answer 1

Hint:Oxidation number is the number of electrons gained or lost by the atoms. The charge of an atom represents the oxidation number of that atom. Some atoms show fixed oxidation states so, by adding or subtracting the oxidation number of known and unknown, we can calculate the oxidation number of the desired atom.

Complete step-by-step answer:
An atom loss or gain electrons to form a bond with other atoms. So, the atoms get charged known as ions. The superscript of the ions represents the oxidation number of that atom.
In neutral molecules, the sum of the oxidation number is equal to zero.
The oxidation number of carbon in ${\text{HCHO}}$ is as follows:
The oxidation number of oxygen is $ - 2$and hydrogen is $ + 1$.
$\,\left( {{\text{x}} \times 1} \right)\, + \,\left( { - 2 \times 1} \right)\,\, + \left( { + 1 \times 2} \right) = \,0$
\[\,{\text{x}} = \, + 2 - 2\]
${\text{x = }}\,{\text{0}}$
The oxidation number of carbon in ${\text{HCHO}}$ is zero.
The oxidation number of carbon in ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is as follows:
The oxidation number of chlorine is $ - 1$and hydrogen is $ + 1$.
$\,\left( {{\text{x}} \times 1} \right)\, + \,\left( { - 1 \times 2} \right)\,\, + \left( { + 1 \times 2} \right) = \,0$
\[\,{\text{x}} = \, + 2 - 2\]
${\text{x = }}\,{\text{0}}$
The oxidation number of carbon in ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is zero.
The oxidation number of carbon in ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ is as follows:
The oxidation number of oxygen is $ - 2$and hydrogen is $ + 1$.
$\,\left( {{\text{x}} \times 6} \right)\, + \,\left( { - 2 \times 6} \right)\,\, + \left( { + 1 \times 12} \right) = \,0$
\[6\,{\text{x}} = \, + 12 - 12\]
${\text{6x = }}\,{\text{0}}$
The oxidation number of carbon in ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ is zero.
The oxidation number of carbon in ${{\text{C}}_{12}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{11}}$ is as follows:
The oxidation number of oxygen is $ - 2$and hydrogen is $ + 1$.
$\,\left( {{\text{x}} \times 12} \right)\, + \,\left( { - 2 \times 11} \right)\,\, + \left( { + 1 \times 22} \right) = \,0$
\[\,12\,{\text{x}} = \, + 22 - 22\]
${\text{x = }}\,{\text{0}}$
The oxidation number of carbon in ${{\text{C}}_{12}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{11}}$ is zero.
So, the oxidation number of carbon is zero in ${\text{HCHO}}$, ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$, ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$, ${{\text{C}}_{12}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{11}}$.
Therefore, option (A) ${\text{HCHO}}$, (B) ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$, (C) ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$and (D) ${{\text{C}}_{12}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{11}}$all are correct.

Note:In charged molecules, the sum of the oxidation number is equal to the charge of the molecule. The atoms in elemental numbers have zero oxidation number. The alkali metals have the oxidation number$ + 1$. The transition metals show a variable oxidation number. We know that the general oxidation state shown by oxygen is $ - 2$. Oxygen also forms peroxide and superoxide. The oxygen atom in peroxides always has $ - 1$oxidation number. In superoxide oxygen is found as diatomic and the oxidation state of peroxide is ${\text{O}}_2^ - $ In a molecule, the more electronegative atom has a negative oxidation number, and less electronegative has a positive oxidation number.