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$BF _{3}$ में तथा $BF _{4}^{-}$ में बंध लंबाई क्रमश: $130 \,pm$ तथा $143\, pm$ होने के कारण बताइए।
Solution
The $B-F$ bond length in $BF $$_{3}$ is shorter than the $B-F $ bond length in $B F_{4}^{-} $. $BF _{3}$ is an electron- deficient species. With a vacant $p$ -orbital on boron, the fluorine and boron atoms undergo $p$$n$$-pn$ back-bonding to remove this deficiency. This imparts a double bond character to the $B-F$ bond.
This double-bond character causes the bond length to shorten in $BF_3$ $(130\, pm ) .$ However, when $BF$ $_{3}$ coordinates with the fluoride ion, a change in hybridisation from $s p^{2}$ (in $BF _{3}$ ) to $s p^{3}\left(\text { in } B F_{4}^{-}\right)$ occurs. Boron now forms $4 \sigma$ bonds and the double-bond character is lost. This accounts for a $B-F$ bond length of $143\, pm$ in $B F_{4}^{-}$ ion.
