Question #c29fa

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Question #c29fa

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Answer 1 · 2017-03-23T23:23:33

The statement is false.

Explanation:

The trend of ionization energies for the 3rd period is shown below.

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The ionization energy of $Mg$ $"Mg"$ is 737.7 kJ/mol, and the ionization energy of $S$ $"S"$
is 999.6 kJ/mol.

The explanation

The electron configuration of $Mg$ $"Mg"$ is ${[Ne] 3s}^{2}$ $"[Ne] 3s"^2$ .

The electron configuration of $S$ $"S"$ is ${[Ne] 3s}^{2} {3p}^{4}$ $"[Ne] 3s"^2 "3p"^4$ .

Thus, in going from $Mg$ $"Mg"$ to $S$ $"S"$ , you are adding four valence electrons, but you are also adding four protons to the nucleus.

The other electrons screen a valence electron from the full attraction of the nucleus, but they don’t screen it completely.

Hence, as you add more protons, the valence electrons are more attracted to the nucleus.

The result: it takes more energy to remove an electron from an $S$ $"S"$ atom than from an $Mg$ $"Mg"$ atom, so the ionization energy of $S$ $"S"$ is higher than that of $Mg$ $"Mg"$ .

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Question #c29fa

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