Step 1: Understanding the Concept:
Halic (V) acids are oxoacids of halogens where the halogen atom is in the $+5$ oxidation state.
The general formula for these acids is \( HXO_3 \).
Step 2: Detailed Explanation:
We examine the common halogens: Fluorine, Chlorine, Bromine, and Iodine.
1. Fluorine (F): Being the most electronegative element, fluorine does not show positive oxidation states. It forms only one oxoacid, \( HOF \) (fluoric (I) acid), where its oxidation state is effectively treated as $+1$ or $-1$ depending on convention, but it never reaches $+5$.
2. Chlorine (Cl): It forms chloric acid (\( HClO_3 \)), where the oxidation state of Cl is $+5$.
3. Bromine (Br): It forms bromic acid (\( HBrO_3 \)), where the oxidation state of Br is $+5$.
4. Iodine (I): It forms iodic acid (\( HIO_3 \)), where the oxidation state of I is $+5$.
Thus, the three halogens that form halic (V) acids are Chlorine, Bromine, and Iodine.
Step 3: Final Answer:
The number of halogens forming halic (V) acid is 3.