Q47 / 180 · AIIMS 2013 · Chemistry

In $\mathrm{BF}_{3^{\prime}}$ the $\mathrm{B}-\mathrm{F}$ bond length is $1.30 Å$, when $\mathrm{BF}_3$ is allowed to be treated with $\mathrm{Me}_3 \mathrm{~N}$, it forms an adduct, $\mathrm{Me}_3 \mathrm{~N} \rightarrow \mathrm{BF}_{3^{\prime}}$ the bond length of $\mathrm{B}-\mathrm{F}$ in the adduct is

  • Correct answer
    A
    greater than $1.30 Å$
  • B
    smaller than $1.30 Å$
  • C
    equal to $1.30 Å$
  • D
    none of these.

Solution

In $\mathrm{BF}_3$, there is back bonding in between fluorine and boron due to presence of $p$-orbital in boron. < smiles> < /smiles> back bonding imparts double bond characteristics. As $\mathrm{BF}_3$ forms adduct the back bonding is no longer present and thus double bond characteristic disappears. Hence, bond becomes a bit longer than earlier $(1.30 Å)$.

Hence, the correct answer is option: 'A'.

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