How to assess the strength of the acid?

Question

Which is the stronger acid in the given pairs?

1. $\ce{HClO3}$ and $\ce{HBrO3}$
2. $\ce{HClO2}$ and $\ce{HClO}$
3. $\ce{H2Se}$ and $\ce{H2S}$

1. given that the number of oxygen atoms is the same, the most electronegative atom should result in a stronger acid, thus $\ce{HClO3}$ is stronger.

2. $\ce{HClO2}$ is stronger since it has more oxygen atoms.

3. Here's where my logic fails, since the structure is the same (both have $\ce{2H+}$) then it boils down to 2 things to account for: electronegativity and the size of the atom. Going with electronegativity, sulfur is more electronegative thus stronger acid which was my original answer. But, the correct answer is $\ce{H2Se}$ is the stronger acid which implies in this case atomic size ($\ce{Se}$ is larger than $\ce{S}$) outweighs electronegativity. I do not understand why in this case $\ce{H2Se}$ is stronger than $\ce{H2S}$ given sulfur is more electronegative, and most importantly why the same logic used in 1. can't be applied here and how do I discern where to account for atomic size specifically or when to account for electronegativity?

Any detailed-dumbed down explanations on the logic are highly appreciated.

Answer 1

Let's start with $\ce{HXO_3}$.

In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,

Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl>Br>I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic. So acidic strength decreases as: $\ce{HClO_3>HBrO_3>HIO_3}$

Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order: $\ce{HClO_3>HClO_2>HClO}$. So its actually the number of resonance structures that affects the acidic strength.

Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O>S>SE>Te}$.

Hence, the acidic strength decreases as : $\ce{H_2Te>H_2Se>H_2S>H_2O}$

This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable.

The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.

Answer 2

The acidity of an acid is directly proportional to the stability of its conjugate base: The more stable a conjugate base is, the more stable the acid is. Thus for the first pairing, we need to consider the following more accurate Lewis structures (remember that main group elements observe the octet rule and the central halogen has a free electron pair resulting in an overall trigonal-pyramidal structure of the anions):

Comparing these two, we arrive at the conclusion that chlorine is more electronegative than bromine, thus allowing the oxygens to draw less charge towards them resulting in an overall lower charge density at each oxygen and therefore a greater tendency to displace a proton to gain more electron density.

For the second pairing, our picture is as follows:

In the chlorate anion, each oxygen effictively sees a net $+ \frac{2}{3}$ charge from the central chlorine atom. In chlorite’s case, they only see a $+\frac{1}{2}$ charge. Since $\frac{2}{3} > \frac{1}{2}$, the former is stabilised because of a greater partial charge at the central atom.

Finally, let’s consider hydrogen selenide versus hydrogen sulphide.

$$\ce{H-S-}~~~~~~\ce{H-Se-}$$

It is tougher here to decide which acid is stronger. Electronegativities do not help us, since sulphur and selenium only differ by $\Delta \mathrm{EN} \approx 0.03$ (values from Wikipedia) — almost nothing when compared even to the difference between chlorine and bromine which Wikipedia gives as $\Delta \mathrm{EN} \approx 0.2$. Both anions also have the same charge for the same number of atoms. However, selenium has an additional shell and is thus much larger. In a charge per volume description that means that the charge is more delocalised in selenium than it would be in sulphur and thus the hydrogen selenide anion is more stable and $\ce{H2Se}$ is the stronger acid.

Note how little electronegativity has to do with acidity: $\ce{HF}$ is the weakest hydrohalogenic acid and $\ce{H2O}$ is the weakest hydrochalcogenic acid with $\mathrm{p}K_\mathrm{a}$ values of $3.2$ and $15.7$, respectively. For comparison, the $\mathrm{p}K_\mathrm{a}$ value of $\ce{HI}$ is estimated to be $-10$.