Question

The oxidation number of O is in ${\text{OF}}_2$ and in ${\text{H}}_2\text{O}$.

• A. +2
• B. -1
• C. +1
• D. 0
• E. -2

Answer 1

Answer: (A), (E)

-2

In a neutral molecule, the sum of the oxidation numbers of all the elements is equal to $0$.
The oxidation number of F (Flourine) is -1 and H (Hydrogen) is +1
Let us assume that the oxidation number of $\text{O}$in ${\text{OF}}_2$ is $n$.
So, if we take the sum of oxidation numbers of the elements in ${\text{OF}}_2$, we get the equation
1 $\times$(O) + 2 $\times$ (F) = $0$
$1\times n+2\times (-1)=0$
$n=+2$
Hence, the oxidation number of oxygen in ${\text{OF}}_2$ is +2.

Let the oxidation number of O in ${\text{H}}_2\text{O}$ is $m$.
Similarly, writing the equation for ${\text{H}}_2\text{O}$ we get,
2 $\times$ (H) + 1 $\times$ (O) = $0$
$2\times (+1)+1\times m=0$
$m=-2$
Hence, the oxidation number of oxygen in ${\text{H}}_2\text{O}$ is -2.