Consider simply aqueous solutions of $\ce{H2S}$ and $\ce{HCl}$ . It is pretty much known that $\ce{HCl}$ is stronger than $\ce{H2S}$ in water. The reason behind that first is the electronegativity of $\ce{Cl}$ is higher than $\ce{S}$, so, in a polar medium, it becomes easier for the $\ce{H-Cl}$ bond to break than the other one. More ever , $\ce{Cl^-}$ ion being smaller in size is extensively hydrated in water than $\ce{HS^-}$ ion, which also favours in the increased acidity of $\ce{HCl}$.

But now consider acidity of $\ce{H2S}$ and $\ce{H2O}$ in water. If we apply the similar reasons as before, we will ultimately conclude water is a stronger acid than $\ce{H2S}$. But actually it’s the reverse. The only fact supporting the actual phenomenon is the weaker bond energy of $\ce{H-S}$, but if we apply this to the earlier case of $\ce{HCl}$ and $\ce{H2S}$, it will again contradict the actual phenomenon.

Why is this the case ? If my reasoning is incorrect, what is the actual reasoning that will support the correct acidity in both the cases ?

  • $\begingroup$ You could make a similar argument for comparison between the 'HX's (X = halogens). $\endgroup$ – Eashaan Godbole Jan 8 '19 at 18:42

You are right that electronegativity is an important consideration but another factor that is also very important is size of the atom. Although there is a decrease in electronegativity from $\ce {O}$ to $\ce {S}$, the size of the atom which the charge is mainly centred on is now larger. This allows the charge to be distributed over a larger volume of space, decreasing inter-electronic repulsions, providing greater stabilisation of the negative charge.

However, the greater size decreases charge density, decreasing the strength of ion-dipole interactions between the anion and the surrounding water molecules. Thus, the extent of solvation decreases for the larger anion.

Electronegativity is defined as the ability of the atom to attract a bonding pair of electrons. It merely serves as a proxy for the ability of the atom to stabilise negative charge. It should not be viewed as the most important factor that influences the atom's ability to stabilise negative charge.

Also, mentioned in your answer, the strength of the bond formed with the larger atom with more diffuse orbitals decreases. A bond that is more easily cleaved would facilitate bond cleavage and thus, favour the acid with the larger atom bonded to $\ce {H}$ as the stronger acid.

Consolidating the above discussion, there are really two factors on each side of the table below. Perhaps, the distribution of charge increases very significantly and the bond energy also decreases significantly while the decrease in extent of solvation and the electronegativity exhibit a relatively less significant effect. It would be great if someone is able to quantitatively show this...

\begin{array}{|c|c|c|c|} \hline \text{Favouring acids with larger atom} & \text{Favouring acids with smaller atom} \\ \hline \ce{Distribution of charge} & \ce{Extent of solvation}\\ \hline \ce{Bond strength} & \ce{Electronegativity}\\ \hline \end{array}

  • $\begingroup$ But you just restated my arguments only. You didn't proceed much towards the actual logic (if any) supporting both the cases . I don't see much new things in your answer. $\endgroup$ – Soumik Das Jan 7 '19 at 3:21
  • $\begingroup$ I have presented an additional important factor to consider - the greater extent of charge dispersal in conjugate bases with larger atoms having more diffuse orbitals. I have also presented clearly a framework for analysis of the acidity phenomena. I agree that my answer does not provide the most convincing answer to your question but it has provided a comprehensive framework (i.e. the consolidation of the 4 factors) on which we can work on. $\endgroup$ – Tan Yong Boon Jan 7 '19 at 8:33

It is not just electronegativity. You also have to look at how strongly the atom bonds to hydrogen. If a strong bond forms, it might convince the atom to act as a base even if it would be grumpy based on its electronegativity. If only a weak bond forms, even an atom with relatively low electronegativity might balk at being protonated.

As a rule, hydrogen forms stronger bonds with itself and with nonmetals in the second row ($\ce{C}$ through $\ce{F}$) than with just about anything else. It's due to favorable covalent overlap; take account of orbital sizes and nodes. Respecting this difference, treat $\ce{H,C,N,O,F}$ as one group and heavier nonmetals as another, and just compare electronegativities within each "class".


Acidity is a bulk property - it depends on the medium where a substance acts as an acid. For example, ammonia, a weak base in aqueous medium, can be a strong base in another medium.

Similarly, pure 100% dry $\ce{H2S}$ is theoretically a stronger acid than perfectly dry gaseous $\ce{HCl}$ due to a smaller bond dissociation energy and weaker dipole interaction. But, when it comes to aqueous medium, many factors are taken into account to decide the strength of acidity. To behave as a stronger acid, constituent ions need to be hydrated very well to facilitate the dissociation of $\ce{H+}$ ion or proton.

In case of $\ce{H2S}$, it is not so polar, (electronegativity difference between $\ce{H}$ and $\ce{S}$ is only about $0.4$ just like the case of $\ce{CH4}$), it doesn't dissolve very well in water. Also the dissolved $\ce{H2S}$ feebly dissociates to give proton or $\ce{H+}$ ion since the $\ce{HS-}$ is not well hydrated due to its bigger size. On the other hand, $\ce{HCl}$ dissolves (due to a bigger electronegativity difference, about 1.4) and is solvated very well, since $\ce{Cl-}$ ion is smaller in size when compared to $\ce{HS-}$ or $\ce{S^2-}$.

Hence, $\ce{HCl}$ is a stronger acid than $\ce{H2S}$ due to the environment in which it is present and also note that the electronegativity difference plays a minor role here.

  • $\begingroup$ +1 I've seen your contributions and they look good to me. Just a few tips in writing an answer. 1, Use paragraphs, it makes the whole thing look a lot neater. Two new lines breaks the word wall into paragraphs. 2. Use $\ce{}$ when formatting chemical compounds. It makes them look a lot prettier. $\endgroup$ – Safdar Faisal Jun 19 at 12:13
  • $\begingroup$ Why is solubility a factor in pKas...? The equation for Ka only involves aqueous species, i.e. molecules that have already dissolved. The question of solubility is quite separate. The case of $\ce{H2CO3}$, or $\ce{CO2}$, is similar: the "pKa of $\ce{H2CO3}$" already assumes that the molecule is formed from $\ce{CO2 + H2O}$, and it's only the "pKa of $\ce{CO2(aq)}$" where solubility / reactivity comes into play. The common confusion in this case only arises because $\ce{H2CO3}$ basically doesn't exist; but such arguments cannot be applied to $\ce{H2S}$. $\endgroup$ – orthocresol Jun 19 at 14:48
  • $\begingroup$ Thanks for the suggestion $\endgroup$ – Sudhagar Jun 20 at 7:02

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