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

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?

• Perhaps Slater's rule is more of a suggestion. – Jon Custer Aug 19 at 18:27
• 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 at 13:56