Which has the largest second ionization energy?
Mg
Ca
Sr
Rb
Cs
Answer is Rb but I can't make sense of it. I understand the trend of first ionization energy...it increases diagonally from bottom left to top right, but is the trend for second IE different? I know it has to do with how easily it gives up electrons but I'm not getting anywhere...
Memorizing rules will only get you so far. Its better to understand the concepts; on the DAT, memorization will maybe get you 20%-30% of the questions, the rest are understanding.
Train of thought: What is ionization energy?
It is the amount energy required to remove an electron from an atom.
What is an electron?
A negatively charged particle?
What is holding the electron in the atom?
The charge between the nucleus (+ charged due to protons) and the negatively charged electron.
What affects the attraction between negatively and positively charged particles?
2 Things:
Distance and
value of the charge (i.e. higher (+) values or lower (-) values
Okay. Now to Ionization Energy. The train of thought listed above is very easy; but thats the thing, recalling simple concepts makes answering the tougher questions easier (Whether this question is tough or not, is dependent on the tester).
For a, b, & c (Mg, Ca, & Sr, respectively), after we remove the first electron, we are still in the valence shell. So removing the second electron will BE MORE DIFFICULT, since the ion is now positively charged (and slightly closer to the nucleus), but not by much. These can therefore be
eliminated.
The last two are more difficult, but understanding the basic facts listed above guides us through this rather easily. For both Cs and Rb, we are entering the core electrons, so these are definitly higher than a, b, & c.
Rubidium's electrons, however, are closer to the nucleus than Cesium's. This creates a higher attraction between the electrons and nucleus, so Rubidium's ionization will be higher (i.e. it will be more difficult to remove its electrons), whether 1st, 2nd, 3rd, 4th, etc.
