Why is the 2nd Ionization energy of Rb higher than that of Cu?
The electron configuration for Rb is [Kr] 5s1 while the electron configuration for Cu is [Ar] 3d10 4s1. I understand that the 2nd ionization means that 2 electrons are being removed and that the 2nd ionization energy increases greatly if you have to remove electrons from a shell closer to the nucleus, so shouldn't the 2nd ionization for Cu be higher than Rb?
- pisgahchemistLv 73 weeks ago
Second ionization energy ....
What we know:
.........Z ...e- config ............ 1st IE ............ 2nd IE
Cu ...29.. [Ar] 3d10, 4s1 ... 746 kJ/mol .... 1958 kJ/mol
Rb ...37.. [Kr] 5s1 ............. 403 kJ/mol .... 2633 kJ/mol
The valence electron lost for the first ionization energy of Rb is farther from that of Cu, it's in the next energy level, and more weakly attracted and therefore, at a higher energy. Thus, it takes less additional energy (the IE) to remove it from Rb than from Cu.
For the second electron to be removed, more energy is required to remove the electron from the 4p of rubidium than from the 3d of copper. Using Slater's rules we can compute the effective nuclear charge experienced by the outermost electron in each +1 ion. (Zeff = 7.85 for copper and Zeff = 9.95 for rubidium.) Despite having electrons in the next energy (4s and 4p) the Rb+ ion (with greater Zeff) has a smaller radius and thus more energy will be required to remove the second electron from Rb+ than from Cu+.