Several books that I have read say that in $\ce{H2O2}$ the degree or extent of H-bonding is 6 as each H atom would form 1 and O atom would form 2 H-bonds respectively. However if we look at HF the no.of H-bonds formed by it is only 2 (because in overall mixture of molecules the extra lone pairs present in F would out no. the no. of available H atoms with partial positive charge)

My question is that going with the same logic even $\ce{H2O2}$ should form only 4 hydrogen bonds as the no. of lone pairs in O would outnumber the H atoms available then why is the degree of H bonding in $\ce{H2O2}$ 6 but not 4?

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    $\begingroup$ Avoid do over usage of markup, $\LaTeX$, don't use $\LaTeX$ in titles. Also I suspect your reasoning for lesser number of hydrogen bonds for HF (compared to $\ce{H2O}$). $\endgroup$ – Zenix May 9 '20 at 14:51
  • $\begingroup$ @Zenix So what would be your reaaon for that? $\endgroup$ – Physics freak May 9 '20 at 14:57
  • $\begingroup$ I think not one but there are many co-existing factors that act. Availability of number of hydrogens is an important factor, but, F is also, ig, too small to accommodate 2/3 hydrogen bonds. Availability and directionality of lp must also play their roles. $\endgroup$ – Zenix May 9 '20 at 15:07

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