# Acidity comparison of charged group 16 hydrides

March's Organic Chemistry (7th ed - page 337 under "Periodic Table Correlations" in chapter "Acids and Bases") states that:

$\ce{H3O+ > H3S+ > H3Se+}$ is the acidity order of these charged group 16 hydrides

I am trying to make out why this is true. I have seen the related questions but they are about the neutral hydrides and not charged ones.

My working was that the order of acidity should be the reverse order of the basicity of their conjugate bases. The conjugate bases are $\ce{H2O,H2S, H2Se}$. We know that basicity decreases down the group, as the lone pair becomes more dispersed over the large surface area (corroborated for group 15). Hence, according to me, order of basicity order is $\ce{H2O>H2S> H2Se}$. Hence, order of acidity should be $\ce{H3O+<H3S+<H3Se+}$. But, in March, the reverse order is given.

What is the reason for this order given by March? And what is wrong in my logic?

• The order is correct. – Mithoron Feb 21 '18 at 19:36
• @Mithoron What's the reason behind that? – Abcd Feb 22 '18 at 5:37

Your reasoning is somewhat flawed. It may not make sense to use conjugate base stability to analyse the acidity of these charged species since the conjugate bases are not charged and thus, reasons along the lines of "charge dispersal" would not be valid. The trend shown in March's Organic Chemistry can be easily rationalised using the the bond strength of the $\ce {H-X}$ bond ($\ce {X = O, S, Se}$).
The weaker the bond, the more easily the hydrogen is removed from the charged species. Going from oxygen to selenium, the $\ce {H-X}$ bond decreases in strength since the orbitals used in bonding from $\ce {X}$ becomes relatively more diffuse and also larger, thus there is decreasing effectiveness in orbital overlap between the chalcogen and the hydrogen atom. As a result, the acidity trend is as such.