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

Due to back -π bonding in ${\mathrm{BF}}_3$ the B-F bond gets double bond character, the B-F bond length in ${\mathrm{BF}}_3$ is $1.30 {\mathrm{A}}^0$ .
But when it form adduct with ${\mathrm{NMe}}_3$, it can't forms back π-bond with F-filled 2p-orbitals due to lack of empty p-orbital on  B-atom therefore in ${\mathrm{BF}}_3.{\mathrm{NMe}}_3$ adduct the B-F bond always single bond. That is in aduct the B-F bond length is$>1.30 {\mathrm{A}}^0$

B can't form back π-bond with "F" filled orbitals.