Which reaction involves a change in the electron-pair geometry (i.e. hybridisation) for the element mentioned in bracket?

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

(Boron)

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

(Nitrogen)

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

(Sulphur)

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

(Oxygen)

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Solution

The correct option is A $B F_{3} + F^{-} \to {B F_{4}}^{-}$ (Boron)
$B F_{3} (s p^{2}) + F^{-} \to B F_{4}^{-} (s p^{3})$
There is change in hybridisation from $s p^{2}$ to $s p^{3}$ .
In the rest of the reactions, there is no change in hybridisation.

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Similar questions

Which of the following conversions involve change in both shape and hybridisation?

a) CH4 -> C2H6

b) NH3 -> NH4+

c) BF3 -> BF4-

d) H2O ->H3O+

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

H F + S b F_{5} \to H S b F_{6}

P C l_{3} + C l^{-} \to P C l_{4}^{-}

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

H_{3} O + H^{+} \to H_{4} O^{+}

N H_{2}^{-}, N H_{3}

N H_{4}^{+}

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