= Solution
\b[Fluoride favors geminal disubstitution:] after the first substitution, the second F replaces the other chloride at that same phosphorus, giving a $\mathrm{PF_2}$ unit and two unchanged $\mathrm{PCl_2}$ units. The first highly electronegative F withdraws <electron> density inductively, maintaining or increasing the <electrophilicity> of its phosphorus and activating the neighboring P–Cl bond toward another <nucleophile>. The small fluoride ion also has little <steric hindrance> to attack there. In this context <geminal substitution> means both incoming substituents on one phosphorus; it does not describe the relative positions of different ring <nitrogen> atoms. Electronic activation at the substituted phosphorus competes with effects elsewhere in the ring, so the argument predicts the preferred pathway rather than an unconditional exclusive product.
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