Which reactions involves a change in the electron pair geometry (i.e. hybridisation) for the under lined geometry?

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 $B - - F_{3} + F^{-} \to B - - F_{4}^{-}$
$B F_{3} (s p^{2}) + F^{-} \to B F_{4}^{-} (s p^{3})$
$B F_{3}$ is $s p^{2}$ hybridized but $B F_{4}^{-}$ is $s p^{3}$ hybridized.
Both $N H_{3}$ and $N H_{4}^{+}$ are $s p^{3}$ hybridized.
Both $S O_{2}$ and $S O_{3}$ are $s p^{2}$ hybridized.
Both $H_{2} O$ and $H_{3} O^{+}$ are $s p^{3}$ hybridized.

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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

B

1 : 1

B F_{3}

N H_{3}

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