How does the number of protons in the nucleus affect ionization energy trends?

• orgohell
In summary, the 3rd ionization energy for Beryllium is greater than the 3rd ionization energy for Oxygen because while the Zeff value for the 2p electron in Oxygen is higher, the number of protons in the nucleus is also higher, making it more difficult to remove the electron. This is due to the fact that the nucleus in Oxygen has 8 protons while the nucleus in Beryllium only has 4 protons.
orgohell
Hey guys, I am having trouble understanding/explaining some of the trends in ionization energy with calculated Zeff by slaters rules.
For example I was asked why is the 3rd ionization energy for Be greater then the 3rd ionization energy for O

My train of thought was to say its pulling an electron from a complete shell in Be (1s2) vs. pulling an electron from the 2p orbital...but how do I explain this with Zeff?

-Zeff of the 1s electron of Be2+ is=4-(1 x 0.35)=3.65
-Zeff of the 2p electron of O2+ is=8-(2 x .85)-(3 x 0.35)=5.25

so the p electron in the O still has a higher Zeff and wouldn't that mean its held tighter to the nucleus and more difficult to remove?

What am I missing here?

The key is to remember that it is not just the Zeff value that determines the ionization energy but also the number of protons in the nucleus. A higher Zeff value means that the electron is held more tightly to the nucleus, but if the nucleus has more protons then the extra positive charge will make it more difficult to remove the electron. In this case, Oxygen has 8 protons while Beryllium only has 4 protons, making it easier to remove electrons from Beryllium than Oxygen.

1. What is Zeff and how does it affect ionization energies?

Zeff, or effective nuclear charge, is the net positive charge experienced by an electron in an atom. It is determined by the number of protons in the nucleus and the shielding effects of inner electrons. A higher Zeff leads to a stronger attraction between the nucleus and the valence electrons, making it more difficult to remove an electron and increasing the ionization energy.

2. How does atomic radius affect ionization energies?

The atomic radius refers to the size of an atom, and it is directly related to ionization energies. As the atomic radius increases, the distance between the nucleus and the valence electrons increases, resulting in a weaker attraction and lower ionization energy. Similarly, a smaller atomic radius leads to a stronger attraction and higher ionization energy.

3. What is the trend of ionization energies on the periodic table?

The trend of ionization energies on the periodic table is an increase from left to right and bottom to top. This is because as you move across a period, the number of protons in the nucleus increases, leading to a higher Zeff and stronger attraction for valence electrons. Moving down a group, the atomic radius increases, resulting in a weaker attraction and lower ionization energy.

4. How does the number of valence electrons affect ionization energies?

The number of valence electrons refers to the outermost electrons in an atom's electron configuration. The more valence electrons an atom has, the easier it is to remove one and the lower the ionization energy. This is because valence electrons are shielded by inner electrons, reducing the Zeff and making them easier to remove.

5. What is the relationship between ionization energy and chemical reactivity?

Ionization energy is directly related to an atom's chemical reactivity. A higher ionization energy means that it is more difficult to remove an electron, making the atom less likely to form bonds and react with other elements. On the other hand, a lower ionization energy makes an atom more reactive as it can easily lose electrons and form bonds with other elements.

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