Question

The decreasing order of the second ionization potential of \(\mathrm{Mg}, \mathrm{Ca}\) and \(\mathrm{Ba}\) is(a) \(\mathrm{Mg}>\mathrm{Ca}>\mathrm{Ba}\) (b) \(\mathrm{Ca}>\mathrm{Ba}>\mathrm{Mg}\) (c) \(\mathrm{Ba}>\mathrm{Mg}>\mathrm{Ca}\) (d) \(\mathrm{Mg}>\mathrm{Ba}>\mathrm{Ca}\)

Short answer

The decreasing order of the second ionization potential for Mg, Ca, and Ba is \textbf{(a) Mg > Ca > Ba}.

Step-by-step solution

Understanding Ionization Potential

The ionization potential, or ionization energy, is the energy required to remove an electron from an atom or ion in its gaseous state. The second ionization potential refers to the energy needed to remove a second electron after the first one has already been removed. Generally, ionization potentials increase across a period and decrease down a group on the periodic table.

Considering Group Trends

Mg, Ca, and Ba are all alkaline earth metals located in Group 2 of the periodic table. As we move down the group from Mg to Ca to Ba, the size of the atoms increases, and the outermost electrons become further removed from the nucleus. This increase in atomic radius generally leads to a decrease in ionization energy because the valence electrons are less tightly bound to the atom. Therefore, we expect Ba to have the lowest second ionization potential and Mg the highest.

Identifying the Decreasing Order

By following the trend mentioned in Step 2, we would arrange the second ionization potentials in the order of the atomic numbers of Mg, Ca, and Ba, which decrease down the group. Hence, the correct order from highest to lowest second ionization potential is Mg > Ca > Ba, since Mg has the smallest atomic radius and the most tightly held electrons, followed by Ca and then Ba with the largest atomic radius and the least tightly held electrons.

Key concepts

Periodic Table Trends

Understanding the trends within the periodic table is crucial for explaining why certain elements behave the way they do. As you move across a period (from left to right), atoms tend to have higher ionization energies, as the positively charged nucleus exerts a stronger pull on the negatively charged electrons. This makes the electrons more difficult to remove. Conversely, as you move down a group, the ionization energy generally decreases. This happens because the atomic radius increases due to the addition of more electron shells; thus, the outermost electrons are further from the nucleus and more easily removed. Additionally, the increased number of electrons results in more electron shielding, reducing the effective nuclear charge experienced by the outermost electrons. These trends help us predict the behavior of elements in chemical reactions, including their propensity to lose or gain electrons.

Alkaline Earth Metals

The alkaline earth metals, which include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra), are found in Group 2 of the periodic table. These metals are known for having two electrons in their outermost shell, which they can lose to form divalent cations with a 2+ charge. When exploring trends such as ionization potential within this group, it’s important to note that these metals have relatively low ionization energies compared to the elements in Group 1. This is due to their slightly higher nuclear charge, which holds the outer electrons more tightly, making them a bit less reactive than their adjacent alkali metals. However, within the group of alkaline earth metals, ionization energy still decreases as we move from Be down to Ra.

Atomic Radius

The atomic radius is a measure of the size of an atom from the center of the nucleus to the boundary of the surrounding cloud of electrons. As you travel down a group on the periodic table, the atomic radius increases due to the addition of electron shells. This increase in distance weakens the attraction between the nucleus and the outermost electrons, making it easier for the atom to lose an electron and therefore lowering the ionization energy. When moving across a period, the atomic radius decreases because the number of protons in the nucleus increases, which enhances the nuclear charge and pulls electrons closer to the nucleus, increasing ionization energy. The atomic radius concept is directly tied to various properties of elements, like reactivity, ionization energy, and electronegativity.

Second Ionization Energy

The second ionization energy is the amount of energy required to remove a second electron from an atom after one electron has already been removed. It is always higher than the first ionization energy because, after the first electron is gone, the remaining electrons are held by the nucleus more strongly due to the reduced electron-electron repulsion and increased effective nuclear charge. In the context of the alkaline earth metals, Mg, Ca, and Ba, we must recognize that the second ionization energy follows the same periodic trend as the first. However, considering the second electron being removed from an already positively charged ion makes the process require more energy. Mg, having the smallest atomic radius among the three, will consequently have the highest second ionization energy, followed by Ca, with Ba having the lowest due to its largest radius and the electrons being furthest from the nucleus.