Answer
See the explanation
Work Step by Step
a. This element has a total of 6 valence electrons. This can be determined by adding up the number of electrons in the \(n_s\) and \(n_p\) orbitals, which in this case is \(2 + 4 = 6\).
b. Based on the valence electron configuration \(ns^2np^4\), some possible identities for this element could be oxygen (O), sulfur (S), selenium (Se), or tellurium (Te). These elements all have 6 valence electrons in their outermost energy level.
c. The compound this element would form with potassium would have the formula \(K_2X\), where \(X\) represents the unknown element. Since potassium (K) has one valence electron, it tends to lose that electron to achieve a stable electron configuration. The unknown element, with 6 valence electrons, tends to gain two electrons to achieve a stable configuration. Therefore, two potassium atoms would transfer their electrons to one unknown element, resulting in the compound \(K_2X\).
d. This element would have a smaller radius than barium (Ba). As we move across a period in the periodic table, the atomic radius generally decreases. Since the unknown element is a nonmetal, it is likely located to the right of barium in the periodic table, and therefore would have a smaller atomic radius.
e. This element would have a greater ionization energy than fluorine (F). Ionization energy is the energy required to remove an electron from an atom. As we move across a period in the periodic table, the ionization energy generally increases. Fluorine (F) has the highest ionization energy of all elements.
