Why is lithium aluminium hydride a better reducing agent than sodium borohydride?

Date: 18.09.2007

When I started learning organic chemistry in college I used to think quite often, as to what makes lithium aluminium hydride (Li[AlH4]) a better redcuing agent than sodium borohydride (Na[BH4]). As a student of Chemistry, I feel that it should be because of the following reasons:

Both lithium aluminium hydride and sodium borohydride are hydride complexes, in which, the former contains aluminium as the central metal ion and the latter contains boron as the central ion. They both serve as reducing agents by donating the hydride ions. Since the hydride ions (that are of interest to us now) are present in the coordination sphere, let us solely consider the constituents in the coordination sphere.

In sodium borohydride the 1s orbital of the hydride ions has to overlap with the 2p orbitals of boron whereas in the case of lithium aluminium hydride the 1s orbital of the hydride ions has to overlap with the 3p orbitals of aluminium. Overlap of an 1s orbital with 3p orbitals is obviously less favourable than the overlap of an 1s orbital with 2p orbitals. Consequently, the Al-H bonds are cleaved easily as compared to the B-H bonds. This easy cleavage of Al-H bonds actually facilitates the reducing power of the reagent.

One more thing is, both boron and aluminium belong to the same group in the periodic table, which is group-13A. Boron, the predecessor of aluminium in the group, is more electronegative. Thus boron, being more electronegative than aluminium, will have a tendency to retain more negative charge than does aluminium, which in this case is acquired from hydride ions. Thus, the difference in the electronegativities of the central atoms of the two complexes can also serve as one of the factors for the difference in reducing power between the two.