Question #44143

The Inert Pair Effect typically renders the +5 oxidation state of Bismuth highly unstable, favoring Bi(III). However, in the presence of strong, highly electronegative ligands like $\ce{F^-}$, the 'd/f-orbital contraction' increases the effective nuclear charge, contracting the empty 6s/6p orbitals of Bi(V). This maximizes covalent overlap and bond enthalpy, making the high-valent state unexpectedly stable and reversing the expected direction of disproportionation.

Consider the spontaneous disproportionation of the transient Bismuth(IV) fluoro-complex at $300\text{ K}$: $2 \ce{[BiF6]^{2-}} \rightleftharpoons \ce{[BiF6]^{-}} + \ce{[BiF6]^{3-}}$

Given the thermodynamic parameters for the reaction: $\Delta H^\circ = -120\text{ kJ mol}^{-1}$ $\Delta S^\circ = -150\text{ J K}^{-1}\text{ mol}^{-1}$

Let $X$ be the magnitude of the standard free energy change ($\Delta G^\circ$) in $\text{kJ mol}^{-1}$. Let $Y$ be the total number of electrons residing in the $e_g$ (d-orbital derived) set of the central metal ions across both product complexes. Let $Z$ be the number of unpaired electrons in the central metal of the reactant complex.

Calculate the numerical value of $(X \times Z) + Y$.