Girls Vs Boys Chemistry Challenge!

Don't worry, it's a sort of half right, half wrong answer.....

The question was answered, right? but I can't see a new one. Who should ask a new question? The game must go on! it's very interesting and instructive.

Question: Derive an expression for pH at the Half-neutralisation point of a titration.

Question: Derive an expression for pH at the Half-neutralisation point of a titration.

I was looking for a comparison between their bond lengths.

Since Deuterium bonds are shorter they require more energy to break. This leads to a decrease in the wave-number of the X-H stretch (which you stated).

EDIT: You get the point though.

I looked at the Henderson-Haselbalch equation for H-H

$pH = pKa +log\frac{base}{acid}$
Here the"base" and "acid" are the weak acid and its conjugate base.

At the half neutralisation point, the concentrations of the weak acid and its conjugate base are equal. So that ratio would be 1:1. The log of 1 is zero, so, the $pH = pKa$.



EDIT: The above method is probably better, this wasn't in the Scottish Higher chemistry course and it was answered from my recollection of the Henderson-Haselbalch equation. Is this still OK though?

A shorter bond length does not necessarily mean that a bond is harder to break! e.g. the silicon-silicon bond in analogues of ethane and ethene:

H₃Si-SiH₃; bond length 233 pm, dissociation enthalpy 314 kJ/mol.
H₂Si=SiH₂; 214 pm, 270 kJ/mol.

In terms of electron density, explain why the 4s orbital is filled before the 3d orbital in most cases.

In terms of electron density, explain why the 4s orbital is filled before the 3d orbital in most cases.

A larger proportion of the electron density in the 4s orbital penetrates the nucleus, so filling the 4s orbital before the 3d orbital is more energetically favourable?

ohhh, better than I was expecting! (seriously I not trolling, I'm just trying to get everyone to see the correct answer rather than the bad answers they give you in School chemistry) Now explain (for bonus marks) why there is this difference in orbital penetration.

In terms of electron density, explain why the 4s orbital is filled before the 3d orbital in most cases.

Electrons in the 3d orbitals are shielded by electrons in the 1s,2s and 2p orbitals so electrons in the 3d orbitals do not experience the full 'force' of the nuclear charge?

OK, well then why is the 4s orbital not as well shielded...... you may expect the 4s to be more shielded seeing as it is in a higher n level.

In terms of electron density, explain why the 4s orbital is filled before the 3d orbital in most cases.

Is it because the 4s orbital penetrates the nucleus more, so the charge from surrounding electrons is blocked out in a way?!.

I would say that 3d is more energetic-efficient than 4s, as 4s is a higher energy level than 3d. Electrons use the low-energy of the orbitals to fill them step by step. That is why the 3d orbital is 'refilled' completely before 4p orbital goes on.

OK, I'm looking for why it is more penetrating

Do you want me to talk about RDFs? I am not sure how much depth you want to the answer..

Yeah, thats as far as I was trying to get to. Obviously I could keep going and ask why but that is normally where we stop in Inorganic chem answers although we know the next level down.

The peak in an RDF plot represents the most probable distance for the electron from the nucleus. The peak on the plot for a 4s orbital is lower than that of the 3d, the small peak of the 4s orbital means the electrons are closer to the nucleus and hence the 4s orbital is more penetrative?!