# How to assess the strength of the acid?

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.

## 2 Answers

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.

• in the last part you mean the charge in Se- is more stable right? Btw, your explanation made so much sense!!! thank you so much Aditya my professor could not explain it at all! – Jx1 Oct 22 '15 at 7:15
• Yes. I will edit it. – Aditya Dev Oct 22 '15 at 7:17
• $-1$ The resonance concept much like the concept of double bonds for these structures (which would imply d-orbital contribution) is challenged and likely wrong. – Jan Oct 22 '15 at 11:26
• I had the same doubt. But this is how my teacher explained it to me. It does give the correct order of acidic strength. – Aditya Dev Oct 22 '15 at 11:45
• Wikipedia mentions resonance. – Aditya Dev Oct 22 '15 at 11:49

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 $-10$.

• Is this correct that less is the difference between size of atoms better is the resonance? – user161158 Mar 7 '18 at 4:50
• Like in case of ClO3- and BrO3- , resonance is better in former one due to less difference in size of O and Cl, so that is more stabilised. – user161158 Mar 7 '18 at 4:51
• @user161158 It is correct that orbital overlap and thus resonance is improved if the atom size is similar — which is usually read as ‘all elements are 2nd period’. So the resonance participation of sulphur in thioacetate is less than that of oxygen in acetate. However, you have managed to choose two examples (bromate and chlorate) in which there is no resonance to be observed. All three — as drawn in my answer — display only single bonds. – Jan Mar 9 '18 at 0:55
• is this because of the same reason that resonance is not observed in these molecules? – user161158 Mar 9 '18 at 1:49
• But the resonance should be there to stabilise these anions. – user161158 Mar 9 '18 at 1:51