Periodic Trends – Ionization Energy

Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach everything you need to know about Periodic Trends and Ionization Energy. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!

Ionization energy is defined as the amount of energy required to remove an electron from an atom or an ion. As we have done for previous trends, consider ionization energy in the context of the electrostatic force. Remember that a stronger electrostatic force equates to the nucleus holding on to its valence electrons more tightly, which in turn will necessitate the addition of more energy to remove electrons. From this we can derive that the electrostatic force is directly proportional to that ionization energy. Since electrostatic force increases from left to right across the periodic table, and decreases going down each column from top to bottom, we can then also say that the ionization energy increases from left to right across each row and decreases from top to bottom down each column. Remember that this is describing the amount of energy needed to cause each atom to release an electron.

This trend describes the first ionization energy of each atom, or the energy required for the removal of the first electron from a neutral atom. However, atoms can continue to lose additional electrons after the first one, with each subsequent electron requiring increasingly more energy to remove. These values are known as subsequent ionization energies. As each successive electron removal leaves a more positive atom behind, thus resulting in a greater electrostatic attraction between the nucleus and remaining electrons, each further electron will be harder to remove, requiring a greater input of energy.

Subsequent Ionization Energy

While initial and subsequent ionization energies always require increasingly more energy, the specific change in those values is affected by the resulting electron configuration of the product ion, in particular it’s stability. Remember from our previous lessons that the most stable configuration an atom can assume is a full valence shell, and that in that configuration atoms are highly unreactive. As a result, we typically see a large spike in the ionization energy for any ion or atom in a full valence shell configuration, mirroring that of noble gases. As an example, consider sodium. In its neutral state, sodium has a first ionization energy of 496 kJ/mole. After the loss of the first electron sodium will exist in its common cationic state, Na+, which has the same electron configuration of neon, a noble gas. We would then expect the second ionization energy to be significantly higher, as it will require an exponentially greater amount of energy to disrupt the stable, full valence shell configuration of the Na+ cation. This is indeed what we observe: sodium’s second ionization energy is a much higher 4,563 kJ/mole.

In contrast, consider the element magnesium, directly to the right of sodium. Moving to the right along a period, we would expect magnesium’s first ionization energy to be higher than sodium’s. This is confirmed by magnesium having an initial ionization energy of 737 kJ/mole. This will result in the Mg+ cation, which has one total valence electron (one less than the two valence electrons held by neutral group 2 elements, such as neutral magnesium). Since this ion is not yet in a full valence shell configuration, it’s second ionization of 1450 kJ/mole, while still increasing, does not show the same exponential increase observed in sodium’s second ionization energy. However, the resulting ion is now the Mg2+ cation, which does have a full valence shell, or noble gas, configuration. Therefore, when we examine magnesium’s third ionization energy, which is measured at 7731 kJ/mole, we can expect to observe the same exponential increase. For the MCAT, be mindful of this trend: while subsequent ionization energies always increase, the largest increases occur once an ion has assumed that stable, full valence shell electron configuration, which occurs on the second ionization energy for alkali metals and on the third ionization energy for alkaline earth metals.

MEDSCHOOLCOACH

To watch more MCAT video tutorials like this and have access to study scheduling, progress tracking, flashcard and question bank, download MCAT Prep by MedSchoolCoach

#medschoolcoach #MCATprep #MCATstudytools

[jung]

6:00 isn't the electron configuration of sodium [Ne] 3s1? It looks like the one you have is the electron configuration for lithium