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In BF₃, the B—F bond length is 1.30 A. when BF₃ is allowed to be treated with Me3N it forms an adduct, Me₃N →BF₃, the bond length of B—F in the adduct is

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smaller than 1.30A

greater than 1.30A

equal to 1.30 A

none of these

answer is A.

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Detailed Solution

In BF₃, there is back bonding in between fluorine and boron due to presence of p-orbital in boron.

back bonding imparts double bond characteristics

As BF₃ forms adduct the back bonding is no longer present and thus double bond characteristic disappears. bond becomes a bit longer than earlier (1.30 A˚).

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