At a certain temperature, only 50% HI is dissociated into H$$_2$$ and I$$_2$$ at equilibrium. The equilibrium constant is:

The dissociation reaction of HI is given as:

$$ 2HI \rightleftharpoons H_2 + I_2 $$

However, the equilibrium constant is often reported for the formation reaction in standard notation, which is:

$$ H_2 + I_2 \rightleftharpoons 2HI $$

We are told that 50% of HI is dissociated at equilibrium. Let the initial concentration of HI be $$ C $$ mol/L. Since 50% is dissociated, the concentration of HI dissociated is $$ 0.5C $$.

For the dissociation reaction $$ 2HI \rightleftharpoons H_2 + I_2 $$, the stoichiometry shows that 2 moles of HI produce 1 mole of $$ H_2 $$ and 1 mole of $$ I_2 $$. Therefore, the concentration of $$ H_2 $$ produced is half of the HI dissociated, and similarly for $$ I_2 $$.

Concentration of $$ H_2 $$ produced = $$ \frac{1}{2} \times 0.5C = 0.25C $$

Concentration of $$ I_2 $$ produced = $$ \frac{1}{2} \times 0.5C = 0.25C $$

Concentration of HI remaining = initial concentration - dissociated concentration = $$ C - 0.5C = 0.5C $$

At equilibrium:

$$ [HI] = 0.5C $$

$$ [H_2] = 0.25C $$

$$ [I_2] = 0.25C $$

For the formation reaction $$ H_2 + I_2 \rightleftharpoons 2HI $$, the equilibrium constant $$ K_c $$ is defined as:

$$ K_c = \frac{[HI]^2}{[H_2][I_2]} $$

Substituting the equilibrium concentrations:

$$ K_c = \frac{(0.5C)^2}{(0.25C)(0.25C)} $$

First, compute the numerator:

$$ (0.5C)^2 = 0.25C^2 $$

Next, compute the denominator:

$$ (0.25C) \times (0.25C) = 0.0625C^2 $$

Now, substitute these into the expression for $$ K_c $$:

$$ K_c = \frac{0.25C^2}{0.0625C^2} $$

The $$ C^2 $$ terms cancel out:

$$ K_c = \frac{0.25}{0.0625} $$

To simplify, divide 0.25 by 0.0625:

$$ \frac{0.25}{0.0625} = \frac{25/100}{625/10000} = \frac{25}{100} \times \frac{10000}{625} = \frac{25 \times 10000}{100 \times 625} = \frac{250000}{62500} $$

Dividing numerator and denominator by 2500:

$$ \frac{250000 \div 2500}{62500 \div 2500} = \frac{100}{25} = 4 $$

Alternatively, recognize that $$ 0.0625 = \frac{1}{16} $$ and $$ 0.25 = \frac{1}{4} $$, so:

$$ \frac{1/4}{1/16} = \frac{1}{4} \times 16 = 4 $$

Thus, $$ K_c = 4.0 $$.

Hence, the equilibrium constant is 4.0, which corresponds to option B.

Hence, the correct answer is Option B.