The number of halogen/(s) forming halic (V) acid is _________.

First, we recall that for any oxo-acid of a halogen the oxidation state of the halogen is obtained from the general expression

Oxidation state of X $$= \frac{\text{(Total charge of the acid)}-\sum \text{(known oxidation states of other atoms)}}{\text{Number of X atoms}}.$$

Now, a halic (V) acid is defined as an oxo-acid of the type $$\mathrm{HXO_3}$$ in which the halogen $$\mathrm{X}$$ possesses an oxidation state of $$+5 \, (\text{Roman numeral V}).$$

We examine each halogen one by one to check whether it can exhibit the $$+5$$ oxidation state and hence form $$\mathrm{HXO_3}.$$

1. Fluorine ($$\mathrm{F}$$):

Fluorine is the most electronegative element and never exhibits a positive oxidation state. Thus it cannot attain $$+5$$ oxidation state. Therefore no compound of the form $$\mathrm{HFO_3}$$ exists.

2. Chlorine ($$\mathrm{Cl}$$):

Using the formula for oxidation state in $$\mathrm{HClO_3}$$ we have

$$x + 3(-2) + 1(+1) = 0,$$

which simplifies step by step as

$$x - 6 + 1 = 0,$$

$$x - 5 = 0,$$

$$x = +5.$$

Since chlorine attains $$+5,$$ $$\mathrm{HClO_3}$$ is a genuine halic (V) acid.

3. Bromine ($$\mathrm{Br}$$):

For $$\mathrm{HBrO_3}$$ the same calculation gives

$$x + 3(-2) + 1(+1) = 0,$$

$$x - 6 + 1 = 0,$$

$$x - 5 = 0,$$

$$x = +5.$$

Hence bromine forms $$\mathrm{HBrO_3},$$ another halic (V) acid.

4. Iodine ($$\mathrm{I}$$):

For $$\mathrm{HIO_3}$$ we again obtain

$$x + 3(-2) + 1(+1) = 0,$$

$$x - 6 + 1 = 0,$$

$$x - 5 = 0,$$

$$x = +5.$$

Thus iodine also forms a stable halic (V) acid $$\mathrm{HIO_3}.$$

Summarising our findings, the halogens that can form halic (V) acids are $$\mathrm{Cl}, \mathrm{Br},$$ and $$\mathrm{I}.$$ Fluorine is excluded. Therefore the count of such halogens is

$$3.$$

Hence, the correct answer is Option C.