Using principles of atomic structure, explain why the atomic radius of Te is less than the ionic radius of Te2− . Photoelectron spectroscopy data for the 1s sublevel of Te and the 1s sublevel of O are represented below. In terms of Coulomb's law and atomic structure, explain why the peak for O is positioned so far to the right of the peak for Te .

Using principles of atomic structure, the reason why the atomic radius of Te is less than the ionic radius of Te²− is because of the Electrostatic force of repulsion between the negatively charged electrons of Te²- thereby causing the ion to be bigger than the neutral atom.

By considering the second scenario, the element oxygen is positioned so far on the right because it has a very much smaller nuclear charge compared to Te.

As a result, in the PES spectrum, the 1s sublevel of oxygen lies far to the right of the 1s sublevel of Te.

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Explanation:

Let us recall that a negative ion is formed by addition of electrons to an atom. When electrons are added to the atom, greater interelectronic repulsion increases the size of the Te^2− hence it is greater in size than Te atom. Therefore, the ionic radius of Te^2− is greater than the atomic radius of Te.

In the second question, oxygen is positioned so far to the right because it has a far smaller nuclear charge compared to Te. Hence in the PES spectrum, the 1s sublevel of oxygen lies far to the right of that of Te.

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