Question

By referring only to the periodic table, select (a) the most electronegative element in group \(6 \mathrm{~A} ;\) (b) the least electronegative element in the group \(\mathrm{Al}, \mathrm{Si}, \mathrm{P} ;(\mathbf{c})\) the most electronegative element in the group \(\mathrm{Ga}, \mathrm{P}, \mathrm{Cl}, \mathrm{Na} ;\) (d) the element in the group \(\mathrm{K}, \mathrm{C}, \mathrm{Zn}, \mathrm{F}\) that is most likely to form an ionic compound with \(\mathrm{Ba}\).

Short answer

The short answer to the question is: (a) Oxygen (O) is the most electronegative element in group 6. (b) Aluminum (Al) is the least electronegative element in the group Al, Si, P. (c) Chlorine (Cl) is the most electronegative element in the group Ga, P, Cl, Na. (d) Fluorine (F) is the element in the group K, C, Zn, F most likely to form an ionic compound with Barium (Ba).

Step-by-step solution

Finding the most electronegative element in Group 6

In Group 6 on the periodic table, we have Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). As electronegativity decreases down a group, the most electronegative element is the one highest up in the group. In this case, it is Oxygen (O). So, the most electronegative element in group 6 is Oxygen (O).

Finding the least electronegative element in Al, Si, P

We have Aluminium (Al), Silicon (Si), and Phosphorus (P) in this group. They are present in the same period (Period 3) in the periodic table. Since electronegativity increases across a period, we can deduce that the least electronegative element in this group would be the one on the left. In this case, it's Aluminum (Al). So, the least electronegative element in Al, Si, P is Aluminum (Al).

Finding the most electronegative element in Ga, P, Cl, Na

We have Gallium (Ga), Phosphorus (P), Chlorine (Cl), and Sodium (Na) in this group. They are also present in the same period (period 3). Because electronegativity increases from left to right, the most electronegative element in this group is the one farthest to the right. In this case, it is Chlorine (Cl). So, the most electronegative element in Ga, P, Cl, Na is Chlorine (Cl).

Finding the element in K, C, Zn, F most likely to form an ionic compound with Ba

We have Potassium (K), Carbon (C), Zinc (Zn), and Fluorine (F) in this group. Barium (Ba) is an alkaline earth metal and has a tendency to lose 2 electrons to achieve a stable electron configuration. To form an ionic compound with Ba, the element should be able to accept 2 electrons. The most electronegative element among K, C, Zn, and F is Fluorine (F), which can accept 1 electron to achieve a stable electron configuration. However, two Fluorine atoms can each accept one electron from Ba, forming BaF₂, an ionic compound. So, the element in K, C, Zn, F most likely to form an ionic compound with Ba is Fluorine (F).

Key concepts

Periodic Table Trends

Understanding the periodic table trends is crucial for predicting the chemical behavior of elements. One such trend is electronegativity, which measures the tendency of an atom to attract a shared pair of electrons towards itself in a chemical bond. Electronegativity values generally increase across a period from left to right due to the increasing number of protons in the nucleus, which creates a stronger pull on the shared electrons. Conversely, electronegativity values tend to decrease down a group as the electrons are farther from the nucleus and more shielded by inner layers of electrons.

In the exercise, when students must identify the most or least electronegative element within a group or period, they utilize these trends to make accurate selections. For instance, in Group 6A, moving from Oxygen at the top to Polonium at the bottom, electronegativity decreases. Similarly, within a period, the element on the far left will have a lower electronegativity compared to elements on the right. So the least electronegative among Aluminum (Al), Silicon (Si), and Phosphorus (P) would be Aluminum (Al), which is farthest to the left.

Group 6A Elements

Group 6A of the periodic table, also known as the chalcogens, consists of the elements Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). These elements have six electrons in their outer shell, making them tend to gain two electrons to achieve a stable octet configuration.

When considering group 6A elements for their electronegativity, Oxygen (O) is the most electronegative as it's the first element of the group and has the highest attraction for electrons due to its small size and high nuclear charge. This is why in the given exercise, when tasked with finding the most electronegative element in Group 6A, the correct answer is Oxygen (O). Recognition of this fact helps students understand why Oxygen typically forms strong double bonds, as it does in molecules like O₂ and CO₂.

Ionic Compound Formation

Ionic compounds are formed when there is a complete transfer of electrons from one atom to another, leading to the formation of ions. This usually occurs between metals, which tend to lose electrons, and nonmetals, which tend to gain electrons. The resultant cations (positive ions) and anions (negative ions) are held together by the electrostatic forces of attraction, known as ionic bonds.

In the exercise, we analyze which element among Potassium (K), Carbon (C), Zinc (Zn), and Fluorine (F) is most likely to form an ionic compound with Barium (Ba), a metal that loses 2 electrons to form Ba²⁺ ions. Fluorine (F) is the most electronegative of the choices and seeks to gain 1 electron to achieve a stable electron configuration. Thus, two F atoms will each gain one electron from Ba to form BaF₂, in which the components are held together by strong ionic bonds. This logic is fundamental to predicting the formation of such compounds in real-world applications.