We’re being asked to determine the element in NaAlH4 with the lowest electronegativity.
Recall that the most electronegative element is fluorine (F) while the least electronegative element is francium (Fr).
This means the trend for electronegativity is as follows: it increases from left to right and decreases down a period in the periodic table.
A major challenge in implementing the “hydrogen economy” is finding a safe, lightweight, and compact way of storing hydrogen for use as a fuel. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. For example NaAlH4, can release 5.6% of its mass as H2 upon decomposing to NaH(s), Al(s), and H2(g). NaAlH4 possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds.
Which element in NaAlH4 is the least electronegative?
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