Step 1: What we compare.
The metal $M$ is the same in every option, so only the halogen changes. We compare $F$, $Cl$, $Br$ and $I$.
Step 2: The guiding idea.
Fajan's rules say a bond becomes more covalent when the negative ion is big and easily distorted, and more ionic when the negative ion is small and holds its electrons tightly.
Step 3: Compare the sizes.
Going down the halogens the ions grow in size, $F^- < Cl^- < Br^- < I^-$. So iodide is the largest and fluoride is the smallest.
Step 4: Apply to covalent character.
The big iodide ion is easily pulled out of shape, so $MI$ is the most covalent.
Step 5: Apply to ionic character.
The tiny fluoride ion holds its electrons firmly and resists distortion, so $MF$ is the most ionic.
Step 6: Final choice.
The halide with the highest ionic character is $MF$, which is option 1.
\[ \boxed{MF} \]