I've noticed that chalcogens never form binary acids, and instead end up like water, with very little acidity. Why is this? Or am I wrong and there's a counterexample, if so please give it.
Answer
To start off, all chalcogens are known to form binary hydrides of the form $\ce{H2X}$ which can then behave as binary acids, however weak or strong, by the equation $\ce{H2X <=> HX^- + H+}$.
Now to compare the strength of the acids. The most straightforward way of doing this is comparing gas phase acidities, which removes the complication of solvation effects. For this, we look at the proton affinities of the conjugate bases $\ce{HX^-}$. The definition of a substance's proton affinity is the $\Delta G$ when it is protonated in the gas phase (always an exergonic reaction for neutral species and anions), so gas phase acidity is simply the inverse reaction, with the opposite sign for the $\Delta G$. According to these two sources, the gas phase acidities of the hydrogen chalcogenides are:
$$\ce{H2O_{(g)} <=> HO^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1635\ kJ\ mol^{-1},\ K^{25ºC} = 2.5\times 10^{-287}$$ $$\ce{H2S_{(g)} <=> HS^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1477\ kJ\ mol^{-1},\ K^{25ºC} = 1.2\times 10^{-259} $$ $$\ce{H2Se_{(g)} <=> HSe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G = +1417\ kJ\ mol^{-1},\ K^{25ºC} = 4.1\times 10^{-249} $$ $$\ce{H2Te_{(g)} <=> HTe^{-}_{(g)} + H^{+}_{(g)}}\ \ \ \ \ \ \ \ \ \ \ \ \Delta G \simeq +1386\ kJ\ mol^{-1},\ K^{25ºC} \simeq 1.1\times 10^{-243} $$
As you can see, when going down the group, deprotonation becomes less endergonic, so the binary hydrides definitely become stronger acids, even if against what one might expect from electronegativity considerations. Calculating the equilibrium constants allows a more direct comparison. Don't worry that the constants are all extremely small, what matters to us is the ratio of the equilibrium constants. Thus, in the gas phase, $\ce{H2S}$ is a whopping 28 orders of magnitude more acidic than $\ce{H2O}$, while $\ce{H2Se}$ and $\ce{H2Te}$ are 38 and 44 orders of magnitude more acidic than $\ce{H2O}$, respectively. It is certainly noteworthy that $\ce{H2O}$ is a much, much weaker acid than its heavier homologues.
With the data presented, we ask, why does acidity increase down the group? There are two main factors presented in different sources. One of them is that the chalcogen-hydrogen bond becomes weaker for heavier chalcogens, and this is ascribed to the poor overlap between the tiny $\ce{H}\ 1s$ orbital and the increasingly larger chalcogen orbitals which weakens the covalent bond and presumably makes ionization easier. This can be seen from the dissociation energies of the $\ce{X-H}$ bonds ($\ce{O-H} = 428\ kJ\ mol^{-1}$ | $\ce{S-H} = 344 \ kJ\ mol^{-1}$ | $\ce{Se-H} = 305\ kJ\ mol^{-1}$ | $\ce{Te-H} = 268\ kJ\ mol^{-1}$). The second factor is the size of the chalcogen atoms. The heavier atoms are larger, meaning they spread the same charge in the conjugate base $\ce{HX^-}$ over a larger volume, which makes reprotonation less favourable. This latter explanation seems to be the most common.
Up to now we've been talking about acidity in the gas phase, but now let's add the solvent, specifically water. The presence of a solvent stabilizes ions immensely relative to gaseous ions (especially protons), and water is a particularly good solvent for ions, so all the dissociation equilibria will be shifted massively to the right compared to the gas phase. Let us compare the acid dissociation constants in water:
$$\ce{H2O_{(l)} <=> HO^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ k_w^{25ºC} = 1.0\times 10^{-14}$$ $$\ce{H2S_{(aq)} <=> HS^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.0\times 10^{-7} $$ $$\ce{H2Se_{(aq)} <=> HSe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 1.3\times 10^{-4} $$ $$\ce{H2Te_{(aq)} <=> HTe^{-}_{(aq)} + H^{+}_{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ K_a^{25ºC} = 2.3\times 10^{-3} $$
Some things to note:
All the binary chalcogen hydrides still increase in acidity after solvation when going down the group. This is not necessarily always the case. A classic example of relative acidities changing on solvation is the dimethylammonium ($\ce{Me_2NH2^+}$) and trimethylammonium ($\ce{Me_3NH^+}$) cations; in the gas phase the former is more acidic, but in aqueous solution the latter is more acidic as the $\ce{Me_3NH^+}$ ion is not as well solvated.
As I mention in the comments, aqueous hydrogen telluride ($pK_a=2.6$) is about as acidic as aqueous phosphoric acid ($pK_{a_1}=2.15$). Compare also with a common weak acid such as acetic acid ($pK_a=4.76$), a borderline strong acid such as nitric acid ($pK_a \simeq -1$), and a recognized strong acid such as hydrochloric acid ($pK_a \simeq -6$).
The $K_a$ of water is a little tricky, since it is both the reactant and the solvent. Its value depends on the definition used. I quoted the $k_w$, but many people use a $pK_a$ of 15.74 which implies $K_a = 1.8\times 10^{-16}$. The difference is not too important here.
$\ce{H2O}$ is a much stronger acid than expected from gas phase data, relative to its heavier homologues. This is ascribed to the exceptionally strong aqueous solvation possible only for $\ce{OH^{-}}$ among the listed conjugate bases, stemming from strong hydrogen bonding interactions, which helps weigh the equilibrium to the right side.
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