I can't seem to remember. I know that they all want to fill their valence shells, but, aren't these different?
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Yes, lithium wants to lose electrons to be like helium because full valence shells are more stable states and all noble gases have full valence shells. So alkali metals lose one electron to achieve nearest noble gas configuration.
But beryllium is exceptional because of its small atomic and ionic radii. Due to this, its ionization potential is very high. So it mostly forms covalent bonds.
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Yes, they do. All alkali metals, in group 1 of the periodic table, display the tendency to lose electrons as well as group 2 alkaline earth metals. Instead of losing one electron like alkali metals, alkaline earth metals such as beryllium lose two!
The loss of an electron to some other element, say a halogen, is very favorable as it gives the potential for alkali metals or alkaline earth metals to form a full valence shell or stable octet.
Therefore it is very common to see ions such as $\ce{H+}$ as well as $\ce{Li^{+}}$ as well as $\ce{Be^{2+}}$. Thus, salts such as $\ce{NaCl}$ as well as other compounds such as hydroxides, $\ce{Li(OH)}$ are the product of these metals once they lose their electrons and react with another element.
First ionization energies:
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$\begingroup$ In general this is true. However, Beryllium is very special. Like no other (Main group) Metal it will form covalent bonds with non-metalls. This is due to the fact that the electron conformation is $\ce{1s^2 2s^2}$ and hence has already a full shell. The most stable oxidation state is +2 though. $\endgroup$– Martin - マーチン ♦Apr 1, 2014 at 6:25
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$\begingroup$ @Martin although the electron configuration for beryllium is stable and the atmo is quite small, the first and second ionization energies 9.32 eV, 18.21 eV respectively. $\endgroup$– user3735Apr 1, 2014 at 12:33