Which of the following elements is least likely to form a cation that will then form an ionic bond with an ion of a Group 7A element?


S

K

Rb

Fr

electronegativity

K, Rb, Fr all readily form cations that will form ionic bonds with group 7A elements (Cl, F, Br, I). So S must be the answer.

To determine which element is least likely to form a cation that will then form an ionic bond with an ion of a Group 7A element, we need to consider the ionization energy and electronegativity trends of the elements.

Group 7A elements, also known as halogens, have a high electronegativity. This means they have a strong tendency to gain an electron to achieve a full valence shell. When halogens form an ionic bond, they gain an electron and become negatively charged ions (anions).

On the other hand, elements that are more likely to form cations are those that have a low ionization energy, meaning it is easier for them to lose electrons and become positively charged ions (cations).

Let's consider the given options:

S (sulfur): Sulfur is in Group 6A, and its ionization energy is relatively high. It is less likely to form a cation.

K (potassium): Potassium is an alkali metal in Group 1A. It has a low ionization energy and readily loses an electron to form a cation.

Rb (rubidium): Rubidium is also an alkali metal in Group 1A. Like potassium, it has a low ionization energy and readily forms a cation.

Fr (francium): Francium is another alkali metal in Group 1A. It has the lowest ionization energy of all elements and is the most likely to form a cation.

Based on these considerations, the least likely element to form a cation and ionic bond with a Group 7A element is S (sulfur).

To determine the element that is least likely to form a cation that will then form an ionic bond with an ion of a Group 7A element, we need to analyze the properties of each element.

Group 7A elements, also known as the halogens, are highly electronegative. They have a strong tendency to gain one electron to achieve a stable octet configuration, forming a negatively charged ion (anion). Therefore, when forming an ionic bond, they will attract a cation.

Looking at the given elements:
- S (sulfur): It is located in Group 6A, and it can gain two electrons to form a sulfide ion.
- K (potassium): It is located in Group 1A, and it readily loses one electron to form a potassium cation (K+).
- Rb (rubidium): It is also located in Group 1A, and it also readily loses one electron to form a rubidium cation (Rb+).
- Fr (francium): It is located in Group 1A and is the most reactive metal, readily losing one electron to form a francium cation (Fr+).

Based on this information, we can see that all the given elements can form cations that can potentially bond with ions of Group 7A elements, except for sulfur (S). Sulfur is not likely to form a cation in this scenario since it is an electronegative nonmetal and tends to gain electrons rather than lose them.

Therefore, the element least likely to form a cation that will then form an ionic bond with an ion of a Group 7A element is sulfur (S).