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Chapter 7: Problem 159

The successive ionization energies for an unknown element are $$\begin{aligned} I_{1} &=896 \mathrm{kJ} / \mathrm{mol} \\ I_{2} &=1752 \mathrm{kJ} / \mathrm{mol} \\ I_{3} &=14,807 \mathrm{kJ} / \mathrm{mol} \\\ I_{4} &=17,948 \mathrm{kJ} / \mathrm{mol} \end{aligned}$$ To which family in the periodic table does the unknown element most likely belong?

Short Answer

Expert verified
The unknown element most likely belongs to Group 2 - the alkaline earth metals family, as it has 2 valence electrons indicated by the significant jump in ionization energy between \(I_2\) and \(I_3\).

Step by step solution

01

Identify the significant jump in ionization energy

The first step involves identifying where the significant jump in ionization energy occurs. Comparing the given ionization energy values, we can see that the jump between \(I_2\) and \(I_3\) is greater than the others.
02

Determine the number of valence electrons

As there is a considerable increase in ionization energy after the second ionization, it indicates that we have removed the valence electrons after the second ionization. Therefore, our unknown element must have 2 valence electrons.
03

Identify the family in the periodic table

Now that we know that the unknown element has 2 valence electrons, we can identify the family in the periodic table it most likely belongs to. Elements with 2 valence electrons belong to the family of alkaline earth metals, which is Group 2 in the periodic table. In conclusion, the unknown element most likely belongs to Group 2 - the alkaline earth metals family.

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Most popular questions from this chapter

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