Question #44325

The ability of oxygen to stabilize high oxidation states of transition metals often exceeds that of fluorine. This is because oxygen can form multiple bonds ($p\pi-d\pi$) with the metal, whereas fluorine stabilizes high oxidation states primarily through high lattice energy in the solid state.

Consider the highest stable binary oxide and the highest stable binary fluoride of Manganese at $298\text{ K}$. Let $O_{ox}$ = Oxidation state of Mn in its highest binary oxide. Let $O_{F}$ = Oxidation state of Mn in its highest binary fluoride. Let $N_{\pi}$ = Total number of $p\pi-d\pi$ bonds in one discrete molecule of the highest binary oxide. Let $Y$ = Number of unpaired electrons in the central metal ion of the highest binary fluoride. Let $C_{ox}$ = Coordination number of Mn in the highest binary oxide. Let $C_{F}$ = Coordination number of Mn in the solid-state structure of the highest binary fluoride.

Calculate the value of the expression: $(O_{ox} \times O_{F}) + (N_{\pi} \times Y) - C_{ox} + C_{F}$.