The decreasing order of O - O bonds length in the following molecules is:

Options

A. `Na_(2)O_2 gt KO_2 gt O_2 gt O_2 (AsF_6)`
B. `KO_(2) = Na_(2)O_(2) gt O_(2) gt O_(2) (AsF_6)`
C. `KO_(2) gt Na_(2)O_(2) gt O_(2) gtO_(2) (AsF_6)`
D. `Na_2O_2 gt O_2 gt KO_2 gt O_2(AsF_6)`

Explanation

To determine the decreasing order of O-O bond lengths in the given molecules, we need to analyze the bond order of each molecule. The bond length is inversely related to the bond order; thus, a higher bond order corresponds to a shorter bond length.

### Step-by-Step Solution:

1. **Identify the Molecules**: The molecules we need to consider are O2, KO2, Na2O2, and O2ASF6.

2. **Determine the Total Electrons**:
- O2 has 16 electrons (8 from each oxygen).
- KO2 has 17 electrons (16 from O2 and 1 from the extra electron due to the negative charge).
- Na2O2 has 18 electrons (16 from O2 and 2 from the two sodium ions).
- O2ASF6 has 15 electrons (16 from O2 and 1 less due to the positive charge).

3. **Determine the Bond Order**:
- **For O2**:
- Bonding electrons = 10, Anti-bonding electrons = 6
- Bond order = (10 - 6) / 2 = 2
- **For KO2**:
- Bonding electrons = 10, Anti-bonding electrons = 7 (due to the extra electron in anti-bonding)
- Bond order = (10 - 7) / 2 = 1.5
- **For Na2O2**:
- Bonding electrons = 10, Anti-bonding electrons = 8 (due to two extra electrons in anti-bonding)
- Bond order = (10 - 8) / 2 = 1
- **For O2ASF6**:
- Bonding electrons = 10, Anti-bonding electrons = 5 (due to the loss of one electron)
- Bond order = (10 - 5) / 2 = 2.5

4. **Rank the Bond Orders**:
- O2ASF6: Bond order = 2.5
- O2: Bond order = 2
- KO2: Bond order = 1.5
- Na2O2: Bond order = 1

5. **Determine the Bond Length Order**:
- Since bond length is inversely proportional to bond order, the order of bond lengths from longest to shortest is:
- Na2O2 (longest)
- KO2
- O2
- O2ASF6 (shortest)

### Final Answer:
The decreasing order of O-O bond lengths is:
**Na2O2 > KO2 > O2 > O2ASF6**