The degree of dissociation $$(\alpha)$$ of a weak electrolyte, $$A_x B_y$$ is related to van't Hoff factor $$(i)$$ by the expression:

For an electrolyte $$A_xB_y$$ that dissociates in solution we can write

$$A_xB_y \;\rightleftharpoons\; x\,A^{y+} + y\,B^{x-}$$

If the initial amount of the electrolyte is taken as 1 mole, then after dissociation:

• Undissociated $$A_xB_y$$ remaining = $$1-\alpha$$ mole
• Ions $$A^{y+}$$ produced = $$x\alpha$$ mole
• Ions $$B^{x-}$$ produced = $$y\alpha$$ mole

Total number of moles present in the solution after equilibrium becomes

$$N_{\text{total}}=(1-\alpha)+x\alpha+y\alpha=1+(x+y-1)\alpha$$

By definition, the van’t Hoff factor $$i$$ is the ratio of the total moles present after dissociation to the moles taken initially:

$$i=\frac{N_{\text{total}}}{1}=1+(x+y-1)\alpha$$

Re-arranging for $$\alpha$$ gives

$$\alpha=\frac{i-1}{x+y-1}$$

Hence the correct expression is $$\boxed{\alpha=\dfrac{i-1}{(x+y-1)}}$$

Therefore, Option D is correct.