Why is lead's first ionisation enthalpy greater than that of tin?

Asked Aug 19 '19 at 18:05

Active Aug 20 '19 at 13:42

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Using Slater's rule, I found that both lead and tin's $Z_\mathrm{eff}$ for 6p and 5p electrons equal. However, the ionisation energy is different. Is comparing ionisation enthalpy on the basis of $Z_\mathrm{eff}$ not correct?

Also, I found Clementi-Raimondi Effective Nuclear Charge which probably explains the trend, but why is this different to Slater's rule?

edited Aug 20 '19 at 13:42

Apoorv Potnis

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asked Aug 19 '19 at 18:05

Klyen Dave

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Perhaps Slater's rule is more of a suggestion. – Jon Custer Aug 19 '19 at 18:27
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This is an example of a relativistic effect, so Clementi-Raimondi effective nuclear charges (non-relativistic in nature) likely also cannot describe the behaviour adequately. You may find this question tangentially useful. – Nicolau Saker Neto Aug 20 '19 at 13:56

Why is lead's first ionisation enthalpy greater than that of tin?

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