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I understand that to compare relative acidity one must consider the stability of the conjugate bases.

Across a period the electronegativity of an element increases. And that is for example why $\ce{HF}$ is more acidic than $\ce{H2O}$, as the negative charge of the fluoride anion is drawn closer to the positively charged nucleus and thus more stabilized than that of an oxide anion.

Down the group the atom radius increases. And that is for example why $\ce{HI}$ is more acidic than $\ce{HF}$, as the negative charge of the iodide anion is distributed over a greater volume and thus more stabilized than that of a fluoride anion.

To me, these two trends are contradictory. How can the stability of an anion increase as the negative charge gets closer to the nucleus(observed trend across periods) but at the same time decreases as the negative charge gets further away from the nucleus(observed trend down groups)?

Is there a concept I have misunderstood?

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There are two factors determining acidity of a compound, the electronegativity factor and the size factor. Down the period, the size factor dominates, and the reason $\ce{HI}$ is more acidic than $\ce{HF}$ is because of this factor, as in the former molecule, the orbitals of iodide participating in bonding are larger and more diffuse, hence bond is weaker.

Across the period, however, size does not vary as significantly as down the group. Hence, we compare using the electronegativity factor. If electronegativity difference is higher, there is larger ionic character, and hence it is easier to deprotonate. Also, basicity of an ion is proportional to its charge density. So you misunderstood the first concept. You should compare fluoride with hydroxide ion, and not oxide.

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  • $\begingroup$ There is still something that I still can´t get my head around. You argued that HI is more acidic as its bond is weaker than that of HF. But this method of argument wouldn't support the observation that alkynes (which have stronger C-H bonds) are more acidic that alkanes (see source). So I believed that the only way of explaining the trend down a group is by talking about the bigger atom radius contributing to the stabilization of the negative charge. $\endgroup$
    – 0ItisMe0
    May 13 at 10:13
  • $\begingroup$ mentioned source: masterorganicchemistry.com/2018/01/19/… $\endgroup$
    – 0ItisMe0
    May 13 at 10:14
  • $\begingroup$ In case of HI and HF(diatomic molecules) hybridization doesn't apply, so so concept of 'higher s character' cannot be applied here. $\endgroup$ May 13 at 10:23
  • $\begingroup$ There is one more thing I would like to ask you. You argued that across a period size is not a significant contributing factor anymore. So that is the reason why we have to rely on electronegativity. But I believed that the increasing electronegativity across a period is based on the decreasing atomic radius and on the valence shell being more closer to the nucleus and thus keener on accepting an extra electron. $\endgroup$
    – 0ItisMe0
    May 13 at 10:27
  • $\begingroup$ Sure. Hybridization cannot indeed be applied here. But why can´t the concept of stronger bonds, in general, attributing to higher acidity be applied in the example of HI and HF as HF being the one having a stronger bond? $\endgroup$
    – 0ItisMe0
    May 13 at 10:32

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