In the case of alkali metals, the covalent character decreases in the order.

Options

A. MCl gt MI gt MBr gt MF
B. MF gt MCl gt MBr gt MI
C. MF gt MCl gt MI gt MBr
D. MI gt MBr gt MCl gt MF

Explanation

To determine the order of covalent character in the compounds formed by alkali metals (M) with halogens, we can analyze the size of the halogens and the resulting covalent character based on their atomic sizes. Here’s a step-by-step solution:

### Step 1: Understanding Covalent Character
Covalent character in ionic compounds arises due to the polarization of the electron cloud of the anion by the cation. The smaller the size of the anion, the greater the polarization and thus the greater the covalent character.

### Step 2: Identifying the Halogens
The halogens in question are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). Their sizes increase in the order:
- Fluorine (smallest)
- Chlorine
- Bromine
- Iodine (largest)

### Step 3: Analyzing Alkali Metal Halides
Now, let's consider the alkali metal halides formed with sodium (Na) as an example:
- NaF
- NaCl
- NaBr
- NaI

### Step 4: Comparing Covalent Character
1. **NaI**: Iodine is the largest halogen, leading to the least effective nuclear attraction on its electrons, resulting in the highest covalent character.
2. **NaBr**: Bromine is smaller than iodine but larger than chlorine, giving it a lower covalent character than NaI but higher than NaCl.
3. **NaCl**: Chlorine is smaller than bromine, leading to an even lower covalent character than NaBr.
4. **NaF**: Fluorine is the smallest halogen, resulting in the highest ionic character and the lowest covalent character.

### Step 5: Conclusion
Thus, the order of covalent character in alkali metal halides decreases as follows:
**NaI > NaBr > NaCl > NaF**

### Final Answer
The decreasing order of covalent character for alkali metal halides is:
**NaI > NaBr > NaCl > NaF**