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Square Planar Complexes

Which reactio...

Question

Which reaction involves a change in the electron-pair geometry for the underlined element?

B - - F_{3} + F^{-} \to {B - - F_{4}}^{-}

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N - - H_{3} + H^{+} \to N - - H_{4}^{+}

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2 S - - O_{2} + O_{2} \to 2 S - - O_{3}

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H_{2} O - - + H^{+} \to H_{3} {O - -}^{+}

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Solution

The correct option is A $B - - F_{3} + F^{-} \to {B - - F_{4}}^{-}$
$B F_{3}$ is sp $_{2}$ hybridised and trigonal planar in shape while $B F_{4}^{-}$ is sp $_{3}$ hybridised and tetrahedral in shape .

So both have different electron-pair geometry .

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