Question

The first five ionization energies of an element are as follows (in kJ/mol): 589.8, 1145, 4912, 6491, 8153. Which of the following elements is most likely to have these ionization energy values? A. K B. Ca C. Sc D. Ge E. As

Answer 1

The element with the lowest ionization energy values is most likely potassium (K).

Step-by-step explanation:

The ionization energy of an element is the energy required to remove an electron from a specific atom. As you move across a row on the periodic table, the ionization energies generally increase. This means that elements on the left side of the periodic table, such as alkali metals like potassium (K), have lower ionization energies compared to elements on the right side, such as the noble gases like neon (Ne).

Given the values provided, the element with the lowest ionization energy is most likely option A. K (potassium).

Answer 2

The provided ionization energies suggest Group 3 characteristics. The large increase after the third ionization energy points to Scandium (Sc) as the most likely element.

Step-by-step explanation:

To determine which element is most likely to have the provided ionization energy values, one must examine the significant jump in the ionization energies, which often indicates the removal of an electron from a new, closer energy level to the nucleus and generally a new shell. The large increase between the third and fourth ionization energies (from 4912 kJ/mol to 6491 kJ/mol) suggests that the first three electrons are removed from an outer shell, and the fourth electron removal comes from a more tightly bound inner shell. The given ionization energies significantly increase after the third ionization energy, indicative of electron removal from a new energy level. Considering this, the element is likely to be in Group 3 of the periodic table, where the first three electrons removed are from the outermost energy level, and the fourth electron comes from the next inner shell's d orbitals. Considering the options, only Scandium falls into group 3, where such a jump in ionization energies occurs after the removal of the third electron. This aligns with the known pattern of successive ionization energies for Group 3 elements. Therefore, the element is most likely Scandium.