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Answer: 2. B

\(\mathrm{MF_6^{3-}}\) needs six \(\mathrm{F^-}\) around M, so M must use d orbitals to exceed an octet.

Boron is in period 2 and has no d orbitals; its maximum covalency is 4 (e.g. \(\mathrm{BF_4^-}\)). Al, Ga and In have vacant d orbitals and form \(\mathrm{AlF_6^{3-}}\), \(\mathrm{GaF_6^{3-}}\), \(\mathrm{InF_6^{3-}}\). So B cannot.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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