The higher oxidation states of transition elements are found to be in the combination with A and B, which are:

Options

A. F, O
B. O, N
C. O, S
D. F, Cl

Answer

To solve the question regarding which combinations of elements A and B correspond to the higher oxidation states of transition elements, we can follow these steps:

### Step 1: Understand the Concept of Oxidation States
Oxidation state refers to the degree of oxidation of an atom in a chemical compound. Transition metals can exhibit multiple oxidation states, and their higher oxidation states are typically stabilized by highly electronegative ligands.

### Step 2: Identify the Given Options
The options provided are:
- A) F and O
- B) O and S
- C) F and Cl

### Step 3: Analyze the Electronegativity of the Elements
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The higher the electronegativity, the more likely the element can stabilize higher oxidation states of metals:
- Fluorine (F) is the most electronegative element.
- Oxygen (O) is also highly electronegative.
- Chlorine (Cl) is less electronegative than both F and O.
- Sulfur (S) is less electronegative than O.

### Step 4: Determine the Best Combination
Since transition metals can achieve higher oxidation states when combined with highly electronegative elements, we analyze the combinations:
- **F and O**: Both are highly electronegative, making this combination suitable for stabilizing higher oxidation states.
- **O and S**: While O is electronegative, S is not as effective in stabilizing high oxidation states.
- **F and Cl**: F is highly electronegative, but Cl is less effective than O.

### Step 5: Conclusion
Based on the analysis, the combination of elements that allows for the highest oxidation states of transition metals is **F and O**.

### Final Answer
The higher oxidation states of transition elements are found to be in combination with **F and O**.

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