Question

Find the oxidation number of Carbon in Glucose ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

Answer 1

Step-1 Calculating the charge on ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

• Sugar is a neutral organic compound in nature.
• Therefore the overall charge on Sugar = $0$

Step-2 Calculating the oxidation number of Cabon in Glucose$C_6{H}_{12}{O}_6$-:

1. As per standard, the oxidation number on Hydrogen$\left(\mathrm{H}\right)$= $+1$
2. The oxidation number on Oxygen$\left(\mathrm{O}\right)$= $-2$
3. The oxidation number on Carbon $\left(\mathrm{C}\right)$ = $x$
   $FormalChargeonSugar=6\times OxidationnumberofC+6\times OxidationnumberofO+12\times OxidationnumberofH$

$0=6\times x+12\times 1+6\times \left(-2\right)$

$0=6x+12-12$

$$\begin{gathered} 6x=0 \\ x=0 \end{gathered}$$

Hence, the oxidation number of Carbon in $C_6{H}_{12}{O}_6$ is 0.