MCQ
$\text{CH}_4\ +\ 2\text{O}_2 \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\ \text{CO}_2\ +\ 2\text{H}_2\text{O}$The above reaction is
  • Oxidation.
  • B
    Decomposition reaction.
  • C
    Endothermic reaction.
  • D
    Double displacement reaction.

Answer

Correct option: A.
Oxidation.
Oxygen is reduced to water and Hydry in methane to $H+$ ion in water.
Because the $\ce{OXIDATION-NUMBER}$ of $\mathrm{O}_2$ in Molecule is $= 0$.
But in water it's $\ce{OXIDATION-NUMBER}$ is $= -2.$
So reduction. In methane the $\ce{OXIDATION-NUMBER}$ of $H- = -1$ and in water it's $\ce{OXIDATION-NUMBER}$ of Hydrogen is $= +1.$
So Oxidation.

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