Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of 2-fluoroethanol is

For any monosubstituted or disubstituted ethane, three principal staggered/eclipsed arrangements about the $$C-C$$ bond are considered:

• Eclipsed - dihedral angle $$0^{\circ}$$ (maximum torsional strain)
• Gauche - dihedral angle $$\approx 60^{\circ}$$ (staggered)
• Anti - dihedral angle $$\approx 180^{\circ}$$ (staggered)

In simple alkanes (e.g. butane) the usual stability order is
$$\text{Anti} \gt \text{Gauche} \gt \text{Eclipsed}$$
because anti keeps bulky groups far apart. 2-Fluoroethanol, $$HO-CH_2-CH_2-F$$, deviates from this rule for two reasons.

Reason 1 : Intramolecular hydrogen bonding
In the gauche orientation the -OH hydrogen can point toward the fluorine atom. Fluorine is highly electronegative and can accept a hydrogen bond from the hydroxyl hydrogen. The resulting five-membered ring (O-H···F) gives extra stabilisation that is absent in the anti form.

Reason 2 : Gauche effect (σ→σ* hyperconjugation)
When two electronegative substituents (here, $$O$$ and $$F$$ bearing partial -ve charge) are gauche, overlap of a $$\sigma_{C-H}$$ bonding orbital with a $$\sigma^{*}_{C-F}$$ or $$\sigma^{*}_{C-O}$$ antibonding orbital is possible. This hyperconjugative donation also favours the gauche arrangement.

Combining both effects:

• Gauche is the most stable.
• Anti is less stable (no H-bond, weaker hyperconjugation).
• Eclipsed is least stable due to torsional strain, regardless of any electronic factors.

Therefore the increasing order of stability is
$$\text{Eclipsed} \lt \text{Anti} \lt \text{Gauche}$$

Option C which is: Eclipse, Anti, Gauche