In oxygen difluoride `(OF_2)` and dioxygen difluoride `(O_2F_2` the oxygen is assigned an oxidation number of

Options

A. (+1)and (+2)respectively
B. (+2)and (+2) respectively
C. (+1) and (+1) respectively
D. (+2)and (+1)respectively

Answer

To determine the oxidation number of oxygen in oxygen difluoride `(OF_2)` and dioxygen difluoride `(O_2F_2)`, we can follow these steps:

### Step 1: Understand the Oxidation States
Fluorine is the most electronegative element and typically has an oxidation state of -1 in its compounds. Since we are dealing with compounds that contain fluorine, we will use this information to find the oxidation state of oxygen.

### Step 2: Calculate the Oxidation State in `OF_2`
Let the oxidation state of oxygen be \( x \).

In `OF_2`, there are two fluorine atoms, each with an oxidation state of -1. The overall charge of the molecule is neutral (0). Therefore, we can set up the equation:

\[ x + 2(-1) = 0 \]

This simplifies to:

\[ x - 2 = 0 \]

Solving for \( x \):

\[ x = +2 \]

Thus, the oxidation state of oxygen in `OF_2` is **+2**.

### Step 3: Calculate the Oxidation State in `O_2F_2`
Now, let’s determine the oxidation state of oxygen in `O_2F_2`.

Let the oxidation state of oxygen again be \( x \). In `O_2F_2`, there are two fluorine atoms, each with an oxidation state of -1. The overall charge of the molecule is neutral (0). Therefore, we can set up the equation:

\[ 2x + 2(-1) = 0 \]

This simplifies to:

\[ 2x - 2 = 0 \]

Solving for \( x \):

\[ 2x = 2 \]
\[ x = +1 \]

Thus, the oxidation state of oxygen in `O_2F_2` is **+1**.

### Summary of Results
- The oxidation state of oxygen in `OF_2` is **+2**.
- The oxidation state of oxygen in `O_2F_2` is **+1**.