what happened when treating Boron trifluoride with LiAlH4 in diethyl ether?
Answers
The geometry of a molecule of BF3 is trigonal planar. Its D3h symmetry conforms with the prediction of VSEPR theory. The molecule has no dipole moment by virtue of its high symmetry. The molecule is isoelectronic with the carbonate anion, CO32−.
BF3 is commonly referred to as "electron deficient," a description that is reinforced by its exothermic reactivity toward Lewis bases.
In the boron trihalides, BX3, the length of the B-X bonds (1.30 Å) is shorter than would be expected for single bonds,[7] and this shortness may indicate stronger B-X π-bonding in the fluoride. A facile explanation invokes the symmetry-allowed overlap of a p orbital on the boron atom with the in-phase combination of the three similarly oriented p orbitals on fluorine atoms.[7] Others point to the ionic nature of the bonds in BF
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the reaction happen will be this
4BF 3 + 3 LiAlH 4 → 2B 2 H 6 + 3LiF + 3AlF 3