Which of the following statements is correct

Options

A. `SF_4` is polar and non-reactive
B. `SF_6` is non-polar and very-reactive
C. ` SF_6` is a strong fluorinating agent
D. `SF_4` is prepared by fluorinating `SCl_2` with NaF

Explanation

To determine which of the statements regarding SF4 and SF6 is correct, we can analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement:** SF4 is polar and non-reactive.

- **Analysis:**
- SF4 (sulfur tetrafluoride) has a seesaw molecular geometry due to the presence of one lone pair on the sulfur atom. This geometry leads to an uneven distribution of charge, making SF4 a polar molecule.
- However, the statement claims that SF4 is non-reactive, which is incorrect. SF4 is indeed reactive due to the presence of the highly electronegative fluorine atoms and can participate in various chemical reactions.

**Conclusion:** This statement is incorrect.

### Step 2: Analyze the second statement
**Statement:** SF6 is non-polar and very reactive because of steric hindrance.

- **Analysis:**
- SF6 (sulfur hexafluoride) has an octahedral geometry, which results in a symmetrical distribution of charge. Therefore, SF6 is a non-polar molecule.
- The claim that SF6 is "very reactive because of steric hindrance" is misleading. SF6 is actually quite stable and is not very reactive; it is known to be a strong fluorinating agent due to the presence of fluorine atoms, but steric hindrance is not the reason for its reactivity.

**Conclusion:** This statement is incorrect.

### Step 3: Analyze the third statement
**Statement:** SF6 is a strong fluorinating agent.

- **Analysis:**
- SF6 is indeed a strong fluorinating agent. Its stability and the presence of highly electronegative fluorine atoms allow it to effectively transfer fluorine in reactions, making it useful in various chemical processes.

**Conclusion:** This statement is correct.

### Step 4: Analyze the fourth statement
**Statement:** SF4 is prepared by fluorinating SCl2 with NaF.

- **Analysis:**
- The preparation of SF4 involves the reaction of sulfur dichloride (SCl2) with sodium fluoride (NaF). The balanced reaction is:
\[ \text{SCl}_2 + 2 \text{NaF} \rightarrow \text{SF}_4 + 2 \text{NaCl} \]
- This indicates that SF4 can indeed be synthesized using SCl2 and NaF.

**Conclusion:** This statement is also correct.

### Final Conclusion
- The correct statements are:
- SF6 is a strong fluorinating agent.
- SF4 is prepared by fluorinating SCl2 with NaF.

### Summary of Correct Statements:
1. SF6 is a strong fluorinating agent.
2. SF4 is prepared by fluorinating SCl2 with NaF.