1. Identify the molecule and the relevant conformations

   2-fluoroethanol is $\mathrm{HO{-}CH_2{-}CH_2{-}F}$.
   Rotation about the $\mathrm{C{-}C}$ bond gives three main conformations:

   • Eclipsed
   • Gauche: $\mathrm{OH}$ and $\mathrm{F}$ are staggered with dihedral angle $60^\circ$
   • Anti: $\mathrm{OH}$ and $\mathrm{F}$ are opposite, dihedral angle $180^\circ$

2. General stability of eclipsed vs staggered

   Eclipsed conformations are usually least stable because of:

   • torsional strain
   • unfavorable eclipsing interactions

   Therefore, $\text{Eclipsed is least stable}$

3. Compare Gauche and Anti in 2-fluoroethanol

   Normally, for simple substituted ethanes, anti is more stable than gauche due to less steric repulsion.
   However, in $\mathrm{HO{-}CH_2{-}CH_2{-}F}$, the gauche conformation is especially stabilized.

4. Reason for gauche stabilization

   In the gauche form, $\mathrm{OH}$ and $\mathrm{F}$ come close enough for favorable intramolecular interaction. This is commonly attributed to the gauche effect and can also be viewed as a weak intramolecular hydrogen-bonding/electrostatic stabilization between $\mathrm{O{-}H}$ and $\mathrm{F}$.

   Thus, $\text{Gauche is more stable than Anti}$

5. Overall increasing order of stability

   Combining the above: $\text{Eclipsed} < \text{Anti} < \text{Gauche}$

6. Match with options

   • A: Eclipse, Gauche, Anti $\to$ wrong
   • B: Gauche, Eclipse, Anti $\to$ wrong
   • C: Eclipse, Anti, Gauche $\to$ correct
   • D: Anti, Gauche, Eclipse $\to$ wrong

Therefore, the correct answer is: $\boxed{\text{C: Eclipse, Anti, Gauche}}$