A) \[BO_{2}^{-}\]
B) \[BE_{6}^{3-}\]
C) \[BH_{4}^{-}\]
D) \[B(OH)_{4}^{-}\]
Correct Answer: B
Solution :
Due to absence of low lying vacant d orbital in B. \[s{{p}^{3}}{{d}^{2}}\] hybridization is not possible hence \[B{{F}_{6}}^{3-}\] will not formed.
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