$BF_{3}$  reacts with water to give $HBF_{4}$ what is the change in hybridization Option 1) $Sp^{2}\, \, to \, \, Sp^{3}$ Option 2) $Sp^{3}\, \, to \, \, Sp^{2}$ Option 3) $Sp\, \, to \, \, Sp^{3}$ Option 4) $Sp^{2}\, \, to \, \, Sp^{3}d$

As we learnt in Concept

Reaction of Boron Trifluoride with water -

Does not undergo hydrolysis due to strong B-F bond, but reacts with water to form an adduct

- wherein

$BF_{3}+H_{2}O\rightleftharpoons H^{+}\left [ BF_{3}OH \right ]^{-}$

BF3 reacts with water to form an adduct

$BF_{3} +H_{2}O\leftrightharpoons H^{+}\left [ BF_{3} OH\right ]^{^{-}}$

The Hybridisation of B goes from sp2 in BF3  to sp in $\left [ BF_{3} OH\right ]^{^{-}}$

Option 1)

$Sp^{2}\, \, to \, \, Sp^{3}$

Option 2)

$Sp^{3}\, \, to \, \, Sp^{2}$

Option 3)

$Sp\, \, to \, \, Sp^{3}$

Option 4)

$Sp^{2}\, \, to \, \, Sp^{3}d$

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