[lmnmss]

Arrange the following compounds in increasing order of their Bronsted acidity and briefly explain your answer. H3PO4, H4SiO4, HBrO4, HOBr

So far, I've worked out (based on Pauling's rule), that HBrO4 is more acidic than H3PO4. I am certain that these two are more acidic than HOBr and H4SiO4, but I am not too certain on how to order the later two compounds. From what I understand, Si is more electronegative than Br since it is in a higher period than the latter, and thus it can draw electron density towards itself, weakening the OH bond and making it easier for the molecule to 'donate' the proton.

Because of that, I have tentatively worked out that HOBr is the least acidic, followed by H4SiO4, H3PO4 and HBrO4. Is it appropriate reasoning to explain that this order is 1. Due to Br being less electronegative than Si and thus, having a 'stronger' OH bond, making it harder for it to donate a proton, and 2. (with respect to the oxoacids) having more free oxygens allow for a greater polarisation of the O-H bond due to the electron withdrawing property of the free oxygens?

On a side note, I was wondering if it is 'correct' in a sense to picture the 'electron withdrawing' effect of a central atom such as Cl in HOCl as drawing the electron density on O towards itself, which is drawing electron density from the O-H bond -- since O itself is electronegative and can 'draw' electrons between itself and H towards itself and ultimately resulting in H having a positive partial charge.

Thanks!

[Dan]

So far, I've worked out (based on Pauling's rule), that HBrO4 is more acidic than H3PO4.

Can you elaborate on you logic here. Acidity order is correct, I am just not sure about your explanation.

Si is more electronegative than Br since it is in a higher period than the latter

Not true. Remember: electronegativity is also dependent on group.

[lmnmss]

I drew the structures for both HBrO4 and H3PO4. What I've learnt during lecture is that the pKa of an acid (O)pE(OH)q is pKa = 8-5p (Pauling's Rule). Since for HBrO4 this value is -7 vs 3 for H3PO4, HBrO4 is more acidic than H3PO4.

Not true. Remember: electronegativity is also dependent on group.

I understand that generally electronegativity increases across a period and decreases down a group, and since Br is a Halogen, it will be more electronegative than Si? Why isn't it less electronegative than Si (seeing that it is in a higher period)?

[Dan]

What I've learnt during lecture is that the pKa of an acid (O)pE(OH)q is pKa = 8-5p (Pauling's Rule).

Ok, I was not aware of that being Pauling's rule, but anyway - do you understand the theoretical basis of the rule? i.e. can you actually explain it?

I understand that generally electronegativity increases across a period and decreases down a group, and since Br is a Halogen, it will be more electronegative than Si? Why isn't it less electronegative than Si (seeing that it is in a higher period)?

OK, so why does electronegativity decrease down a group and increase across a period? What factors are at play?

[Borek]

What I've learnt during lecture is that the pKa of an acid (O)pE(OH)q is pKa = 8-5p (Pauling's Rule).

Way too approximate to be called a rule if you ask me.

[lmnmss]

Ok, I was not aware of that being Pauling's rule, but anyway - do you understand the theoretical basis of the rule? i.e. can you actually explain it?

I would think that the free oxygens on HBrO4 will be able to delocalise the charge on the conjugate base --> making it more stable and thus a preferable state for the molecule to exist in.

OK, so why does electronegativity decrease down a group and increase across a period? What factors are at play?

Effective nuclear charge decreases down a group due to the increased principal quantum shell, which effectively 'offsets' the increased nuclear charge due to the increased no. of protons. --> effective nuclear charge decrease, ability of atom to attract the electron density to itself decreases --> electronegativity decreases.

Across a period, effective nuclear charge increases. --> ability of atom to attract the electron density to itself increases --> electronegativity increases

[Dan]

I would think that the free oxygens on HBrO4 will be able to delocalise the charge on the conjugate base --> making it more stable and thus a preferable state for the molecule to exist in.

Fine, keyword: resonance stabilisation

Effective nuclear charge decreases down a group due to the increased principal quantum shell, which effectively 'offsets' the increased nuclear charge due to the increased no. of protons. --> effective nuclear charge decrease, ability of atom to attract the electron density to itself decreases --> electronegativity decreases.

Z_eff actually increases down a group, you will see this from any data table (example). However, the attractive force between an electron and the nucleus is dependent on both Z_eff and distance - it rapidly falls away with distance. Comparing properties between different periods based on Z_eff only is not a good idea because you do not take into account radius.

It is (roughly) true to say that the increase in radius as you move down a group offsets the increase in Z_eff. As a result electronegativity tends to decrease down a group.

Across a period, effective nuclear charge increases. --> ability of atom to attract the electron density to itself increases --> electronegativity increases

OK.

Note that there is no commonly used definition of electronegativity that states that electronegativity is proportional to Z_eff only. See http://en.wikipedia.org/wiki/Electronegativity

When thinking about electronegativity, I like to use the Mulliken scale, where Mulliken electronegativity is directly proportional to the sum of ionisation energy and electron affinity. (The calculations for the more common Pauling scale are more complicated and less intuitive).

Take Cl as a starting point. How do you expect the ionisation energy and electron affinity of Cl compare to Br? How do those properties compare between Cl and Si?

I suppose another way you could look at it, that may appeal to the approach you are taking already is: consider the combination of Z_eff and atomic radius. We already know Z_eff for Br is greater than Si, how do you think the atomic radii compare?