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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?

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  • $\begingroup$ I have misspelt harder in the title can any privileged user/OP take a look. $\endgroup$ – Avyansh Katiyar Mar 20 '18 at 15:18
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$\ce{Cu+}$ has a kernel of $\ce{K+}$, with an extra 10 protons in the nucleus and a $\mathrm{3d^{10}}$ electron shell around it. $\ce{Cu+}$ is smaller because it has more positive charge pulling those electrons in. But size isn't everything!

The first ionization potential of potassium ($\pu{4.34 V}$) is much lower than the first ionization potential of copper ($\pu{7.73 V}$). But from there on, the situation changes: Second ionization potential: $\pu{31.63 V}$ for $\ce{K}$ vs $\pu{20.29 V}$ for $\ce{Cu}$; third ionization potential: $\pu{45.72 V}$ for $\ce{K}$ vs $\pu{36.83 V}$ for $\ce{Cu}$; fourth ionization potential: $\pu{60.91 V}$ for $\ce{K}$ vs $\pu{55.2 V}$ for $\ce{Cu}$; and fifth ionization potential: $\pu{82.66 V}$ for $\ce{K}$ vs $\pu{79.9 V}$ for $\ce{Cu}$. It is not till the sixth ionization potential that potassium is again easier to ionize than copper: $\pu{100.0 V}$ for $\ce{K}$ vs $\pu{103 V}$ for $\ce{Cu}$.

So, although the smaller size of the $\ce{Cu+}$ ion suggests that it would be harder (after all, it's stuffed with more electrons!), it turns out that the $\ce{Cu+}$ 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.

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