problem 1.13

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juliefilice 4A
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problem 1.13

Postby juliefilice 4A » Sun Oct 14, 2018 12:06 am

why is the ionization energy for O lower than that of N or F when ionization energy usually increases from left to right across the periodic table?

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Re: problem 1.13

Postby Chem_Mod » Sun Oct 14, 2018 1:15 pm

This is a good question. To answer this, you'd have to look at how you fill the orbitals in the atom.

Ionization energy and electron affinity periodic table trends have some exceptions, whereas electronegativity, which is calculated from both electron affinity and ionization energy, generally doesn't.

Please ask the TA's after class or during discussion for a more in-depth answer.

Dina Geotas 4A
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Re: problem 1.13

Postby Dina Geotas 4A » Sun Oct 14, 2018 2:15 pm

Since nitrogen has the electron configuration of 1s^2 2s^2 2p^3, its orbitals are half filled, resulting in a stable condition. Oxygen has the electron configuration of 1s^2 2s^2 2p^4, which means it has two half filled orbitals and one orbital filled with two paired electrons. Since the atoms want to stay in stable condition, such as nitrogen's natural state, it will take more energy to remove an electron from the nitrogen atom than the oxygen atom. The oxygen atom will become more stable with the loss of one electron because it will result in a each orbital being half filled.

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Re: problem 1.13

Postby armintaheri » Sun Oct 14, 2018 2:17 pm

There are essentially two factors that determine ionization energy. There is the attraction between electrons and the nucleus, and the repulsion between electrons and other electrons. As you go from left to right on the periodic table, the number of protons in the nucleus increases, so the attraction between electrons and nucleus increases. The oxygen nucleus exerts a greater attractive force on its electrons than the nitrogen nucleus, as you would expect. But the p subshell has three orbitals, which can fit two electrons each. Nitrogen is relatively stable because each of those three orbitals has one electron in it. But oxygen has four electrons in the p subshell, which means one orbital contains two electrons. These electrons are really close together so they repel. This added repulsion overcomes the added attraction from the nucleus, so ionization energy decreases from nitrogen to oxygen, causing an exception in the trend.

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