Step 1: Frame the comparison.
We compare the acid strengths of the hydrogen halides $\mathrm{HF}$, $\mathrm{HCl}$, $\mathrm{HBr}$ and $\mathrm{HI}$ and find the weakest.
Step 2: Choose the right factor.
Down Group 17, acid strength of $\mathrm{HX}$ is governed mainly by the $\mathrm{H-X}$ bond dissociation enthalpy, not by electronegativity.
Step 3: Link bond strength to acidity.
A stronger $\mathrm{H-X}$ bond is harder to break, so less $\mathrm{H^+}$ is released and the acid is weaker.
Step 4: Spot the smallest halogen.
Fluorine is the smallest halogen, so the small hydrogen $1s$ overlaps tightly with the small fluorine $2p$, giving the strongest, shortest $\mathrm{H-F}$ bond.
Step 5: Deduce the trend.
Thus $\mathrm{HF}$ ionises least and is a weak acid, while $\mathrm{HCl}$, $\mathrm{HBr}$ and $\mathrm{HI}$ are strong acids: \[ \text{acid strength: } \mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}. \]
Step 6: Conclude.
Fluorine gives the hydride of lowest acidic strength, option (C).
\[ \boxed{\mathrm{F}} \]