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I am asked to predict the correct molecular geometry of $\ce{BaH2}$ according to VSEPR theory. I assumed that it is linear due to barium being an alkaline earth metal and therefore not having any lone pairs when bonded to two hydrogen atoms, but apparently the bond angle between the two hydrogen atoms is about 119°. Why is that the case and how can one explain this observation with VSEPR theory?

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    $\begingroup$ Your logic is correct. VSEPR fails to predict the correct structure of BaH2. See pubs.acs.org/doi/10.1021/acs.jpca.9b10248 $\endgroup$
    – Andrew
    Mar 5 '21 at 21:42
  • $\begingroup$ isnt BaH2 ionic? does VSEPR applies to ionic compounds? $\endgroup$
    – maverick
    Mar 6 '21 at 4:50
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There is a d-Orbital-Contributions to HOMO in d0 MX2-Complexes.

All higher‐level calculations show SrH2 and BaH2 to be bent with angles of ∼140° and 120°, respectively, while CaH2 is linear with a flat potential‐energy surface for the bending motion.

The use of a core‐polarization potential together with the 2‐valence‐electron pseudopotential approach allows an investigation of the relative importance of core‐polarization vs direct d‐orbital bonding participation as reasons for the bent structures. The calculations strongly suggest that both contribute to the bending in SrH2 and BaH2. Thus, the bent structure of [BaH2] may be considered to arise from a stabilization of the b2 HOMO upon bending by mixing in metal dyz character in second-order perturbation theory.

Reference: https://core.ac.uk/download/pdf/232777714.pdf http://www.chimdocet-inorganica.it/SITO_ESERCIZI/Complementi/COMP1/VSEPREccezioni.pdf

This observation has been rationalized successfully by an increased involvement of (n - 1)d-orbitals in the bonding as M gets larger,19-21 and by the influence of core-polarization interactions between the M and X sites.2 Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs.

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