Why is the second IE for Mg less than the second IE for Al

Answer 1

The way I like to think about IE is how much energy is needed to remove an electron from an atom until it has a stable "happy" place. We know that the trend for IE is to increase as you move across the Periodic Table. Therefore, the first IE for Mg will be less than the Al. However, this will hold true for the second IE as well for the same reason. In both cases, the answer lies in how, as one of my professors called it, "happy" the element is. By removing a second electron, Mg²⁺ has reached its "happy" place where it has no electrons in its valence shell. To remove any more electrons will disrupt its "happiness", cause it to be unstable. Therefore, the third IE for IE is significantly high. When we look at Al²⁺, it has lost an electron but it still has not reached its "happy" stable place.

For a more technical explanation, please feel free to reach out to me. Good luck in your studies.