Which element has the highest second ionization energy: Li, K, or Be?

Short Answer

Expert verified
The element with the highest second ionization energy among Li, K, and Be is lithium (Li).

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. The first ionization energy is the energy required to remove the first electron, and the second ionization energy would be the energy required to remove the second electron after the first has already been removed. Generally, the ionization energy increases from left to right across a period and decreases from top to bottom within a group.
02

Analyzing the Elements

Let's analyze the placement of Li, K, and Be in the periodic table: Li (Lithium) - Group 1, Period 2 K (Potassium) - Group 1, Period 4 Be (Beryllium) - Group 2, Period 2 All three elements are in the first two groups of the periodic table, which means they have relatively low ionization energies compared to elements on the right side of the table. The general trend is that ionization energy decreases as we go down a group, so K should have the lowest first ionization energy due to it being in a lower period than Li. Now considering the second ionization energy, after losing one electron, elements in Group 1 will have a complete octet (a noble gas configuration), making them incredibly stable and requiring much more energy for the removal of the second electron. On the other hand, Be after losing one electron will have an incomplete octet (still requiring more electrons to be stable), so it would require less energy to remove its second electron when compared to Li or K.
03

Determine the Highest Second Ionization Energy

Based on the analysis, it becomes apparent that after the first ionization, Group 1 elements Li and K would require more energy to remove their second electron when compared to Group 2 element Be. Between Li and K, Li is higher in the periodic table and more to the right compared to K, so Li should have a higher second ionization energy compared to K. Therefore, the element with the highest second ionization energy among Li, K, and Be is lithium (Li).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy
Ionization energy refers to the amount of energy needed to remove an electron from an atom or ion in its gaseous state. As the distance between the nucleus and the electron increases, the electron is less attracted to the nucleus, and therefore, it's easier to remove. The first ionization energy deals with removing the first electron, while the second ionization energy concerns the removal of the second electron after the first one has been removed. This process requires more energy because the remaining electrons are more strongly attracted to the nucleus, now less shielded by the electron cloud.

It's crucial to note that the second ionization energy is always higher than the first because you're trying to remove an electron from a positively charged ion, which has a greater hold on its remaining electrons. The concept of second ionization energy is particularly interesting when we examine elements that are just one electron away from having a noble gas configuration. After this electron is removed, the resulting ion holds onto its remaining electrons much more tightly, resulting in a very high second ionization energy.
Periodic Table Trends
When examining trends in the periodic table, we find predictable patterns in the properties of elements, particularly those that relate to ionization energies. The patterns observed include an increase in ionization energy as we move from left to right across a period due to increased nuclear charge resulting in a stronger attraction for the electrons.

Conversely, as we move down a group (vertically on the periodic table), the ionization energy generally decreases. This is because the outermost electrons are farther from the nucleus, and are therefore less tightly held, as a consequence of both increased distance and increased electron shielding from inner electron shells.

These trends are pivotal when deducing which element between lithium (Li), potassium (K), and beryllium (Be) has the highest second ionization energy. Position matters: elements located to the right and higher up on the periodic table typically have higher ionization energies. This is due to their small size and the comparatively greater nuclear charge per electron, which means electrons are held more tightly, and it takes more energy to remove them.
Electron Configuration
Understanding electron configuration is fundamental in predicting ionization energy trends. Electron configuration describes the distribution of electrons of an atom or molecule in atomic or molecular orbitals. For example, the noble gas configuration is a highly stable arrangement where the outer electron shell is full.

Elements tend to achieve this noble gas configuration, and after the first electron is removed from the elements under discussion—Li and K in Group 1—it results in a noble gas configuration. This makes the second ionization energy dramatically higher, as removing any further electrons would mean disrupting a stable configuration.

In contrast, Group 2 elements like Be still need to lose two electrons to achieve this stable state. Hence, its second ionization energy is not as high as that of a Group 1 element, because it's not yet achieved the optimal stability. Knowing the electron configurations of Li, K, and Be helps to clarify why Li, after losing one electron, will hold onto its second electron significantly more strongly than Be, leading to a higher second ionization energy for Li.

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Most popular questions from this chapter

Using only the periodic table, arrange each set of atoms in order of increasing radius: (a) Ba, Ca, Na; (b) In, Sn, As; (c) Al, Be, Si.

Which quantity must be determined experimentally in order to determine the bonding atomic radius of an atom? (a) The distance from the nucleus where the probability of finding an electron goes to zero. (b) The distance between the nuclei of two atoms that are bonded together. (c) The effective nuclear charge of an atom.

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\) . (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S\) , calculate \(Z_{\text { eff}}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S.\) (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

The first ionization energy of the oxygen molecule is the energy required for the following process: $$\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-}$$ The energy needed for this process is 1175 \(\mathrm{kJ} / \mathrm{mol}\) , very similar to the first ionization energy of \(\mathrm{Xe} .\) Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2} ?\) If so, suggest a product or products of this reaction.

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree?

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