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My textbook says that despite the large electronegativity difference $\ce{BeF2}$ is covalent since the beryllium ion will have too much charge density and it will attract the fluorine electron cloud and therefore forms polar covalent bonds.

I looked it up on Wikipedia and found out that it has a crystal lattice, is soluble in water, and has a quite high melting point.

What else does one need to call a compound ionic? Am I or the textbook right?

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    $\begingroup$ I am not putting this as an answer but bond between 2 different atoms is usually not completely covalent nor completely ionic: it is a mixture of them. We prefer to talk about the "covalent" or the "ionic" character of a bond. $\endgroup$
    – SteffX
    Aug 9, 2016 at 19:13
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    $\begingroup$ And BeF2 probably has more ionic then covalent character, not that it matters much. $\endgroup$
    – Mithoron
    Aug 9, 2016 at 19:16
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    $\begingroup$ Nearly anything has a crystal lattice. Many covalent compounds are soluble in water (and quite a few ionic ones are insoluble). As for the melting point, well, it's moderately high, so consider $\ce{BeF2}$ moderately ionic and stop pursuing this false dichotomy. Also, I suggest to close this question as a duplicate of AlCl3: ionic and covalent?, because though the title compounds are different, just about every argument applies equally well to both. $\endgroup$
    – Ivan Neretin
    Aug 9, 2016 at 19:40
  • $\begingroup$ related chemistry.stackexchange.com/questions/17064/… $\endgroup$
    – Mithoron
    Aug 9, 2016 at 20:31
  • $\begingroup$ @IvanNeretin "All bonds to fluorine, with the exception N-F, O-F, and F-F, have considerable ionic character, so the vast majority of AFn molecules are more appropriately described in terms of an ionic model rather than a covalent model." pubs.acs.org/doi/abs/10.1021/ic961315b BeF2 is much more extremely ionic than AlCl3 , with Be having a +1.8 charge. The AlCl3 answer does not apply well to BeF2. $\endgroup$
    – DavePhD
    Aug 10, 2016 at 13:02

2 Answers 2

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There is a very relevant article by Gillespie Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases? J. Chem. Educ., 1998, 75 (7), p 923.

According to the article, the charge on Be is +1.81 and the charges on the Fs are -0.91. (citing to his earlier article Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model Inorg. Chem., 1997, 36 (14), pp 3022–3030)

So while nothing is completely covalent or ionic, BeF2 is extremely toward ionic.

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    $\begingroup$ Thing is they used specifically "Almost Ionic Model" and therefore got what wanted. This is a reinterpretation, so I'm pretty sure other calculations shown much less charge separation. $\endgroup$
    – Mithoron
    May 18, 2020 at 12:29
  • $\begingroup$ @Mithoron "We have calculated the wave function and the electron density distribution ... The calculations were done using the Gaussian 94 package at the HF/6- 311++G(2d,2p) with 6d functions level and at the B3LYP/6-311++G- (2d,2p) level. ... The analysis of the electron density distributions was carried out using the AIMPAC suite of programs." $\endgroup$
    – DavePhD
    May 18, 2020 at 14:09
  • $\begingroup$ Well, they sure didn't pull it out of a sleeve, but different methods give different results, and this case may be an outlier. For example Geoff's answer here has lower % even for LIF. So, I wonder if other calculations also give that much charge separation... $\endgroup$
    – Mithoron
    May 18, 2020 at 17:48
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According to my chemistry book (Pearson Baccalaureate IB) beryllium forms covalent bonds due to it being one of the expectations to the intramolecular bonding.

Furthermore, according to the book the molecule should be non-polar because all the lone pairs are bonded and thus not exerting a repulsion and altering bond angles.

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    $\begingroup$ I think you are confusing polar bonds with dipole moment. Also, I doubt that any element will always form a covalent bond. Maybe you meant that beryllium forms covalent bonds with itself? $\endgroup$
    – CoffeeIsLife
    Jan 15, 2017 at 18:15
  • $\begingroup$ I didn't say always , furthermore no, I'm not confusing those two together $\endgroup$
    – Melodishhh
    Jan 17, 2017 at 20:36

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