Question

Alkaline earth metals should exist as \(\mathrm{M}^{+}\) due to higher IE but they exist as \(\mathrm{M}^{2^{+}}\) ions. Explain.

Short answer

Alkaline earth metals form \( \mathrm{M}^{2^{+}} \) ions instead of \( \mathrm{M}^{+} \) ions, despite the higher ionization energy required, because by losing two electrons they achieve a stable electronic configuration similar to that of noble gases. This greater stability outweighs the additional energy required to remove the second electron.

Step-by-step solution

Understanding Ionization Energy and Ion Formation

Ionization energy is the energy required to remove an electron from an atom or ion. A metal atom becomes a positive ion by losing one or more of its electrons. The more the ionization energy, the more difficult it is to remove an electron. For an alkaline earth metal (M), forming a \( \mathrm{M}^{+} \) ion requires less energy than forming a \( \mathrm{M}^{2^{+}} \) ion, as removing more electrons requires more energy.

Considering the Electronic Structure of Alkaline Earth Metals

Alkaline earth metals have two electrons in their outermost shell. Metallic ions form to attain a stable electronic configuration akin to noble gases; that is, they tend to lose or gain electrons to have full outermost energy levels.

Explaining the Formation of \( \mathrm{M}^{2^{+}} \) Ions

Despite the higher ionization energy required, alkaline earth metals form \( \mathrm{M}^{2^{+}} \) ions instead of \( \mathrm{M}^{+} \) ions because by losing two electrons, they attain a stable electronic configuration similar to noble gases. The increased stability of the \( \mathrm{M}^{2^{+}} \) ion outweighs the greater energy required to remove the second electron.

Key concepts

Ionization Energy

Ionization energy plays a pivotal role in the formation of ions from atoms. It is the energy required to remove an electron from a neutral atom in the gas phase. When we talk about alkaline earth metals, which include elements like magnesium and calcium, understanding ionization energy becomes crucial for explaining why these metals form certain ions.

Firstly, these metals prefer to lose electrons to become more stable, as they tend to achieve an electronic configuration like noble gases. The first ionization energy is the energy needed to remove the first electron, and it's relatively lower compared to the second ionization energy, which is required to remove a second electron.' However, despite higher second ionization energies, alkaline earth metals tend to form \( \mathrm{M}^{2^{+}} \) ions instead of \( \mathrm{M}^{+} \) ions. This counterintuitive phenomenon is closely tied to the concept of electronic configuration and the overall stability of the resulting ion, topics that unfold further in the following sections.

Electronic Configuration

To grasp why alkaline earth metals typically form \( \mathrm{M}^{2^{+}} \) ions, we must delve into the electronic configuration of these elements. Alkaline earth metals occupy Group 2 of the periodic table and have two electrons in their outermost shell. This specific arrangement influences how these metals interact and form ions.

Attaining Noble Gas Configuration

Metals aim for stability by achieving a full outer electron shell, resembling the highly stable electronic configuration of noble gases. By removing two electrons, an alkaline earth metal transitions from having a reactive outer shell to a stable, full energy level in its new \( \mathrm{M}^{2^{+}} \) ionic form. This drive for electronic stability often outweighs the additional energy required for the second ionization process.

Metallic Ions Stability

The stability of metallic ions is central to understanding why alkaline earth metals form \( \mathrm{M}^{2^{+}} \) ions over \( \mathrm{M}^{+} \) ions. It's related not just to electron arrangements but also to other factors such as lattice energy and the energy released when ions form a crystal lattice.

Lattice Energy and Ionic Bonds

When metallic ions interact with oppositely charged ions, they can form ionic compounds with a crystal lattice structure. The formation of these structures releases lattice energy, which can compensate for the high ionization energy required to form \( \mathrm{M}^{2^{+}} \) ions. The resulting ionic compounds are usually more stable due to stronger electrostatic forces between the ions in the lattice.

In a nutshell, the formation of \( \mathrm{M}^{2^{+}} \) ions from alkaline earth metals is a balance of factors: the desire for a stable electronic configuration, the cost of ionization energy, and the stability imparted by the formation of ionic compounds. Even though it costs more energy up-front to create \( \mathrm{M}^{2^{+}} \) ions, the ultimate payoff is a more stable and lower energy state for the ion.