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  • Can someone explain - p- d- and f-block elements - BITSAT

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

 

Answers (1)

best_answer

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 sp3  in \left [ BF_{3} OH\right ]^{^{-}}


Option 1)

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

This answer is correct

Option 2)

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

This answer is incorrect

Option 3)

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

This answer is incorrect

Option 4)

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

This answer is incorrect

Posted by

Aadil

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