Question

What is the oxidation number of $C$ in ${C_{12}}{H_{22}}{O_{11}}$?
(i) $+ 2$
(ii) $0$
(iii) $- 1$
(iv) $+ 1$

Answer 1

Consider the oxidation number for $C$ to be $x$. For $H$ the oxidation number is $\left( { + 1} \right)$ and for $O$ the oxidation number is $\left( { - 2} \right)$. Multiply the oxidation number of each atom with the number of the atoms present in the given molecule. Summation of the obtained values will be $0$ as the molecule as a whole is considered to have no net charge.

Complete step-by-step solution: As we need to find the oxidation number for $C$, let us consider the oxidation number of $C$ in the given molecule $\left( {{C_{12}}{H_{22}}{O_{11}}} \right)$ to be $x$ .Now we know that for $H$ atom the oxidation state is generally $\left( { + 1} \right)$ and for $O$ the oxidation number is $\left( { - 2} \right)$ .Since in the given molecule, ${C_{12}}{H_{22}}{O_{11}}$ the number of $C$ atoms present is $12$,
$\therefore$Total charge of $C$ atoms in the molecule is $12x$.
Also in the given molecule, ${C_{12}}{H_{22}}{O_{11}}$ the number of $H$ atoms present is $22$,
$\therefore$Total charge of $H$ atoms in the molecule is $\left[ {22 \times \left( { + 1} \right)} \right] = + 22$.
And in the given molecule, ${C_{12}}{H_{22}}{O_{11}}$ the number of $O$ atoms present is $11$,
$\therefore$Total charge of $O$ atoms in the molecule is $\left[ {11 \times \left( { - 2} \right)} \right] = - 22$.
For any given molecule, we consider that the net charge of the molecule as a whole is $0$.Therefore, we can say
Total charge of $C$ atoms in the molecule $+$ Total charge of $H$ atoms in the molecule $+$ Total charge of $O$ atoms in the molecule $=$ $0$
$\Rightarrow 12x + 22 + \left( { - 22} \right) = 0$
$\Rightarrow 12x + 22 - 22 = 0$
$\Rightarrow 12x = 0$
$\Rightarrow x = 0$
Therefore, the oxidation number of the $C$ atom in the given molecule is $0$.

Hence the correct answer is (ii) $0$.

Note: The oxidation number of the atoms should be considered properly as an atom can have different oxidation numbers in different molecules. Also take proper care of the sign associated with the oxidation number of an atom. Always remember if no net charge is mentioned for the whole molecule consider it to be zero.