Which of the following compounds are found to exist (General formula obtaining)?

Options

A. `MgF`
B. `Cu_(2)CI_(2)`
C. `NaF_(2)`
D. `K_(2)O`

Answer

To determine which of the given compounds exist in nature, we will analyze each compound based on the oxidation states of the elements involved and their typical behavior in forming compounds.

### Step-by-Step Solution:

1. **Analyze Compound A: MgF**
- Magnesium (Mg) is an alkaline earth metal and typically exhibits a +2 oxidation state.
- Fluorine (F) is a halogen and typically exhibits a -1 oxidation state.
- For the compound MgF to exist, the oxidation states must balance. The correct formula should be MgF₂ (since one Mg with +2 must balance with two F with -1 each).
- **Conclusion**: MgF does not exist; the correct compound is MgF₂.

2. **Analyze Compound B: Cu₂Cl₂**
- Copper (Cu) is a transition metal and can exhibit +1 and +2 oxidation states.
- In Cu₂Cl₂, if we assume both Cu are in +1 oxidation state, we have: 2x + 2(-1) = 0, leading to x = +1 for each Cu.
- This means the compound can exist as both Cu(I) and Cu(II) forms.
- **Conclusion**: Cu₂Cl₂ can exist.

3. **Analyze Compound C: NaF₂**
- Sodium (Na) is an alkali metal and typically exhibits a +1 oxidation state.
- In NaF₂, if Na were to have a +2 oxidation state, it would contradict its known behavior (as it can only lose one electron).
- **Conclusion**: NaF₂ does not exist.

4. **Analyze Compound D: K₂O**
- Potassium (K) is also an alkali metal and exhibits a +1 oxidation state.
- In K₂O, we have 2(+1) + (-2) = 0, which balances correctly.
- **Conclusion**: K₂O does exist.

### Final Answer:
- The compounds that exist are:
- **B: Cu₂Cl₂**
- **D: K₂O**