4
$\begingroup$

According to HSAB theory, the smaller radius and the less polarizable an acid is, the harder it is. According to Wikipedia, the ionic radii of $\ce{ K+}$ is much larger than $\ce{Cu+}$, which makes sense because d-electrons are poor shielders. Also, $\ce{Cu+}$ has larger electron density but it has a higher effective nuclear charge in comparison to potassium ion because of the poor shielding ability of d-electrons. Yet why is $\ce{K+}$ ion considered hard while $\ce{Cu+}$ ions are considered soft?

$\endgroup$
  • $\begingroup$ I have misspelt harder in the title can any privileged user/OP take a look. $\endgroup$ – Avnish Kabaj Mar 20 '18 at 15:18
2
$\begingroup$

Cu+ has a kernel of K+, with an extra 10 protons in the nucleus and a 3d^10 electron shell around it. Cu+ is smaller because it has more positive charge pulling those electrons in. But size isn't everything!

The first ionization potential of potassium (4.34 v) is much lower than the first ionization potential of copper (7.73 v). But from there on, the situation changes: second K 31.63/Cu 20.29; third 45.72/36.83; fourth 60.91/55.2; fifth 82.66/79.9. It is not till the sixth ionization potential that potassium is again easier to ionize than copper: K 100.0/Cu 103.

So although the smaller size of the Cu+ ion suggests that it would be harder (after all, it's stuffed with more electrons!), it turns out that the Cu+ ion has a softer (i.e., more easily ionized, therefore more easily polarized) outer skin.

Perhaps this is not predictable unless you take into account more than just ionic radii, but at least it is explainable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.