Order of Covalent bond,
A. KF > KI, LiF > KF
B. KF < KI, LiF > KF
C. `SnCl_4` > `SnCl_2` , CuCl >NaCl
D. LiF > KF, CuCl < NaCl
E. KF < KI, CuCl > NaCl
Choose the correct answer from the options given below.

Options

A. A, B only
B. B, C only
C. B, C, E only
D. C, E only

Answer

To determine the order of covalent bonds in the given compounds, we will analyze each pair based on the factors affecting covalent character, particularly focusing on ionic compounds and the size of ions involved.

### Step-by-Step Solution:

1. **Analyzing KF and KI:**
- Both KF and KI are ionic compounds. According to Fajan's rule, the covalent character increases with the size of the anion.
- Fluoride (F) is smaller than Iodide (I). Therefore, KI will have a higher covalent character than KF.
- **Conclusion:** KI > KF.

2. **Analyzing LiF and KF:**
- Again, both are ionic compounds. Here, the anion (F) is the same, but we compare the cations: Li+ and K+.
- Lithium ion (Li+) is smaller than Potassium ion (K+). The smaller the cation, the higher the covalent character.
- **Conclusion:** LiF > KF.

3. **Analyzing SnCl4 and SnCl2:**
- For SnCl4, Sn has a +4 oxidation state, while in SnCl2, Sn has a +2 oxidation state.
- Higher oxidation state leads to a greater covalent character due to increased polarizing power.
- **Conclusion:** SnCl4 > SnCl2.

4. **Analyzing CuCl and NaCl:**
- Here, we compare Cu+ and Na+. Sodium (Na) is an alkali metal, while Copper (Cu) is a transition metal.
- The size of Cu is smaller than Na, leading to a higher covalent character for CuCl compared to NaCl.
- **Conclusion:** CuCl > NaCl.

5. **Summarizing the Findings:**
- From the analysis:
- KI > KF
- LiF > KF
- SnCl4 > SnCl2
- CuCl > NaCl

6. **Evaluating the Options:**
- A. KF > KI, LiF > KF (Incorrect)
- B. KF < KI, LiF > KF (Correct)
- C. SnCl4 > SnCl2, CuCl > NaCl (Correct)
- D. LiF > KF, CuCl < NaCl (Incorrect)
- E. KF < KI, CuCl > NaCl (Correct)

### Final Answer:
The correct options are B, C, and E.

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