I'm so confused because I'm trying to draw $\ce{BeF2}$ … and the way it's drawn (something like $\ce{F-Be-F}$) just doesn't make sense to me because Be can only hold 2 valence electrons. But this structure shows it having 4.

  • $\begingroup$ It doesn’t exactly have four valence electrons the way you drew it, unless you assume the oxygen of water to hold eight valence electrons. $\endgroup$
    – Jan
    Sep 11, 2016 at 16:45

1 Answer 1


You are correct. $\ce{BeF2}$ is octet deficient.

To solve this problem, they would form polymers by themselves, but that would be out of syllabus.

(I cannot find the polymer of $\ce{BeF2}$, so here is the polymer of $\ce{BeCl2}$:)


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