In which fo the following has the oxidation number of oxygen been arragned in increasing order ?

Options

A. `OF_(2) lt KO_(2) lt BaO_(2) lt O_(3)`
B. `BaO_(2) lt KO_(2) lt O_(3) lt OF_(2)`
C. `BaO_(2) lt KO_(2) lt OF_(2) lt KO_(2)`
D. ` KO_(2) lt OF_(2) lt O_(3) lt BaO_(2)`

Answer

To determine the increasing order of oxidation numbers of oxygen in the given compounds, we will calculate the oxidation state of oxygen in each compound step by step.

### Step 1: Calculate the oxidation state of oxygen in OF2
- Let the oxidation state of oxygen be \( x \).
- The oxidation state of fluorine (F) is \(-1\).
- The compound OF2 has 2 fluorine atoms, so the equation becomes:
\[ x + 2(-1) = 0 \]
- Simplifying this gives:
\[ x - 2 = 0 \implies x = +2 \]

### Step 2: Calculate the oxidation state of oxygen in KO2
- In KO2, potassium (K) has an oxidation state of \( +1 \).
- Let the oxidation state of oxygen be \( x \).
- The compound KO2 contains two oxygen atoms, so the equation becomes:
\[ +1 + 2x = 0 \]
- Simplifying this gives:
\[ 2x = -1 \implies x = -\frac{1}{2} \]

### Step 3: Calculate the oxidation state of oxygen in BaO2
- In BaO2, barium (Ba) has an oxidation state of \( +2 \).
- Let the oxidation state of oxygen be \( x \).
- The compound BaO2 contains two oxygen atoms, so the equation becomes:
\[ +2 + 2x = 0 \]
- Simplifying this gives:
\[ 2x = -2 \implies x = -1 \]

### Step 4: Calculate the oxidation state of oxygen in O3
- In O3 (ozone), all oxygen atoms are the same, and the oxidation state is:
\[ x + x + x = 0 \implies 3x = 0 \implies x = 0 \]

### Step 5: Arrange the oxidation states in increasing order
Now we have the oxidation states of oxygen in each compound:
- OF2: \( +2 \)
- KO2: \( -\frac{1}{2} \)
- BaO2: \( -1 \)
- O3: \( 0 \)

Arranging these in increasing order:
1. BaO2: \( -1 \)
2. KO2: \( -\frac{1}{2} \)
3. O3: \( 0 \)
4. OF2: \( +2 \)

Thus, the increasing order of oxidation numbers of oxygen is:
**BaO2 < KO2 < O3 < OF2**

### Final Answer
The correct option is **B**.