Consider the following `S_(N^2)`j reactions
I. `RX+Y^(-)toR-Y+X^(-)`
II. `RX+Y^(-)toR-Y^(+)+X^(-)`
III. `RX^(+)+Y^(-)toR-Y+X`
IV. `RX^(+)+YtoR-Y^(+)+X`
In which reactions there is large increase and large decrease in rate of reaction respectively with increase in polarity of the solvent.

Options

A. II and III
B. II and IV
C. I and IV
D. IV and I

Explanation

To solve the problem, we need to analyze the given reactions in terms of how the polarity of the solvent affects the rate of reaction.

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- We have four reactions involving a haloalkane (RX) and a nucleophile (Y^(-) or Y) that can either be negatively charged or neutral.
- The reactions can be classified based on the charge of the reactants and products.

2. **Identifying the Nature of the Reactions**:
- **Reaction I**: `RX + Y^(-) → R-Y + X^(-)`
- Here, both reactants and products are neutral and negatively charged. The transition state is likely stable.
- **Reaction II**: `RX + Y^(-) → R-Y^(+) + X^(-)`
- This reaction goes from a neutral state to a positively charged product, indicating a transition from unstable to stable.
- **Reaction III**: `RX^(+) + Y^(-) → R-Y + X`
- The reactant is positively charged and the product is neutral. This suggests a transition from stable to unstable.
- **Reaction IV**: `RX^(+) + Y → R-Y^(+) + X`
- Similar to Reaction III, it starts from a positively charged reactant and ends with a positively charged product, indicating a transition from stable to unstable.

3. **Effect of Solvent Polarity**:
- In polar solvents, reactions that stabilize charged transition states will see an increase in reaction rates.
- Conversely, reactions that destabilize charged transition states will see a decrease in reaction rates.

4. **Analyzing Each Reaction**:
- **Reaction I**: No significant change in charge, so negligible effect on rate.
- **Reaction II**: Increase in polarity stabilizes the positively charged product (R-Y^(+)), leading to a large increase in rate.
- **Reaction III**: The positive charge on RX^(+) is destabilized in a polar solvent, leading to a large decrease in rate.
- **Reaction IV**: Similar to III, the positive charge is destabilized, leading to a large decrease in rate.

5. **Conclusion**:
- The reaction with a large increase in rate due to increased solvent polarity is **Reaction II**.
- The reactions with a large decrease in rate due to increased solvent polarity are **Reactions III and IV**.

### Final Answer:
- Large increase in rate: **Reaction II**
- Large decrease in rate: **Reactions III and IV**