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 \(\mathrm{Al}, \mathrm{Si}, \mathrm{P} ;(\mathrm{c})\) 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 Ba.

Short answer

(a) Oxygen (O) (b) Aluminum (Al) (c) Chlorine (Cl) (d) Fluorine (F)

Step-by-step solution

(a) Most electronegative element in Group 6A

Group 6A consists of O, S, Se, Te, and Po. As mentioned earlier, electronegativity increases as we move across a period and decreases as we move down a group. Therefore, the most electronegative element in Group 6A is Oxygen (O), which is at the top of the group.

(b) Least electronegative element in Al, Si, and P

As electronegativity increases from left to right in the periodic table, we can compare the electronegativity values of Al, Si, and P by their positions. Al is to the left of Si, which is to the left of P. Hence, the least electronegative element in this group is Aluminum (Al).

(c) Most electronegative element in Ga, P, Cl, and Na

In this group, we can again use the periodic table to determine electronegativity based on the elements' positions. Na is in the first period, Ga is in the third period, P is in the fifth period, and Cl is in the seventh period. Recall that electronegativity increases from left to right. Therefore, the most electronegative element in this group is Chlorine (Cl).

(d) Element most likely to form an ionic compound with Ba

To determine which element is most likely to form an ionic compound with Barium (Ba), we need to examine the electronegativity difference between the elements. For an ionic compound to form, there should be a significant electronegativity difference between the two elements. Considering the given elements, K is in the first period, C is in the second period, Zn is in the fourth period, and F is in the seventh period. Barium (Ba) is in the second group. Since electronegativity increases from left to right, Fluorine (F) has the highest electronegativity among the given elements. The electronegativity difference between F and Ba is greater than that for any of the other elements in the list. Therefore, Fluorine (F) is most likely to form an ionic compound with Barium (Ba).

Key concepts

Periodic Table Trends

Understanding the Periodic Table is crucial for predicting the chemical behavior of elements. One of the key trends is electronegativity, which is the measure of an atom's ability to attract and hold onto electrons. Electronegativity increases as you move from left to right across a period, due to the increasing positive charge in the nucleus that more effectively pulls electrons closer. Similarly, as you move down a group, electronegativity decreases because the extra shell of electrons for each row down diminishes the nucleus's pull on the valence electrons due to increased distance and electron shielding.

As seen in the exercise, these trends allow us to determine that Oxygen (O) in Group 6A is the most electronegative, as it is the furthest to the right and top of the group. In a trio of elements such as Al, Si, and P, Aluminum (Al) being to the far left makes it the least electronegative. These patterns make the Periodic Table a powerful tool for predicting how elements will interact in chemical reactions, even without knowing their exact electronegativity values.

Ionic Compounds

An ionic compound is formed when one atom donates an electron to another, creating ions that attract each other due to opposite charges. The formation of ionic compounds is directed by differences in electronegativity—the greater the difference, the more likely the atoms will form an ionic bond. In the exercise, we compare elements like K, C, Zn, and F to determine which would most likely form an ionic compound with Barium (Ba).

The 'winner' in this scenario is Fluorine (F), which is far to the right of the Periodic Table and thus highly electronegative. Barium, on the other hand, is a large, low electronegativity metal that readily loses electrons. Their significant electronegativity difference drives the formation of an ionic bond, leading to a compound such as barium fluoride (BaF\(_2\)). Understanding these interactions is key when investigating the chemistry of salts, minerals, and many other materials.

Chemical Properties

Chemical properties describe the characteristic ability of a substance to undergo a specific chemical change. They are deeply rooted in the attributes of atoms such as electronegativity, atomic size, and electron configuration. When an element’s atoms have high electronegativity, such as Chlorine (Cl), it means they have a tendency to attract electrons and participate in reactions leading to the acquisition of electrons, forming negative ions or covalent bonds where they share electrons with other atoms.

Elements with low electronegativity, such as Barium (Ba), are more likely to lose electrons and form positive ions. Their propensity to engage in such reactions is a fundamental aspect of their chemical properties and dictates the type of compounds they can form, be it ionic, covalent, or metallic. This shows the interconnection between periodic trends and the resulting chemical properties, which directly influence the substance's reactivity, corrosiveness, acidity, and many other relevant features for scientific and industrial applications.