The anion which cannot be formed by boron is:

B F_{6}^{3 -}

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B H_{4}^{-}

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B {(O H)}_{4}^{-}

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B O_{2}^{-}

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Solution

The correct option is B $B F_{6}^{3 -}$
Boron cannot form $B F_{6}^{3 -}$ anions due to absence of $2 d$ orbitals in boron. Since $2 d$ orbitals are not present in boron, it cannot expand its octet beyond 8. $B F_{6}^{3 -}$ anion will have 12 valence electrons around B.

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