The ionization energies for an element are listed in the table below. $\begin{array}{lllll}{\text { First }} & {\text { Second }} & {\text { Third }} & {\text { Fourth }} & {\text { Fifth }} \\ {8 \mathrm{eV}} & {15 \mathrm{eV}} & {80 \mathrm{eV}} & {109 \mathrm{eV}} & {141 \mathrm{eV}}\end{array}$ Based on the ionization energy table, the element is most likely to be (A) Sodium (B) Magnesium (C) Aluminum (D) Silicon

Short Answer

Expert verified
The element is most likely to be Magnesium (B).

Step by step solution

01

Identify the jump

In the table, we can see a jump from 15 eV to 80 eV between the second and third ionization energies. This means that the 3rd electron being removed is coming from an inner shell.
02

Match with the properties of the elements

If an element loses two electrons and the third electron comes from the next shell, it suggests that the outermost shell of that element has two electrons i.e., its valency is 2. Among the options given, Magnesium (Mg) is the element which has a valency of 2.
03

Verify

Magnesium's electron configuration is \(1s^2 2s^2 2p^6 3s^2\). If we remove two electrons from the \(3s^2\) shell, then the third electron will come from the \(2p^6\) shell, which is closer to the nucleus and thus require more energy to remove, consistent with the ionization energies given.

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Most popular questions from this chapter

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