Question

An unknown element is a nonmetal and has a valence electron configuration of \(n s^{2} n p^{4}\). a. How many valence electrons does this element have? b. What are some possible identities for this element? c. What is the formula of the compound this element would form with potassium? d. Would this element have a larger or smaller radius than barium? e. Would this element have a greater or smaller ionization energy than fluorine?

Short answer

The unknown nonmetal element has 6 valence electrons and can be identified as one of the following elements: oxygen (O), sulfur (S), selenium (Se), or tellurium (Te). When combined with potassium, the compound formed is K2X, where X is the unknown element. This element has a smaller atomic radius than barium and a smaller ionization energy than fluorine.

Step-by-step solution

1. Calculate the number of valence electrons

Since the given electron configuration is \(n s^{2} n p^{4}\), we can find the number of valence electrons by adding the superscripts. Valence electrons are electrons in the outermost shell that can be involved in forming chemical bonds or sharing with other atoms. Number of valence electrons = \(2 + 4 = 6\)

2. Determine the possible identities of the unknown element

We know the element is a nonmetal and has 6 valence electrons. The most common nonmetals sharing these characteristics are those in Group 16 of the periodic table, also known as the chalcogen group. They have an electron configuration ending in \(ns^2 np^4\). Possible identities of the unknown element include: oxygen (O), sulfur (S), selenium (Se), tellurium (Te).

3. Find the formula of the compound with potassium

Potassium (K) is an alkali metal in Group 1 of the periodic table and has one valence electron. It usually forms an ion with a +1 charge after losing its valence electron. The unknown nonmetal has 6 valence electrons, and it can achieve a full outer shell by gaining 2 more electrons. This will result in an ion with a -2 charge. Therefore, the formula of this compound is K2X, where X is the unknown element.

4. Compare the radius of the unknown element with barium

Barium (Ba) is an alkaline earth metal in Group 2 of the periodic table. In general, metallic elements have larger atomic radii compared to nonmetals. The atomic size usually decreases across a period (from left to right) and increases down a group (from top to bottom). Nonmetals are usually found on the upper right side of the periodic table and have smaller radii than metals. Since the unknown element is a nonmetal, its radius is expected to be smaller than barium's.

5. Compare the ionization energy of the unknown element with fluorine

Ionization energy is the energy required to remove an electron from an atom. In general, ionization energy increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table. Nonmetals usually have higher ionization energies than metals, as it is more challenging to remove an electron from them. Fluorine (F) is a nonmetal in Group 17 and has the highest ionization energy of all elements in its period. Our unknown element likely belongs to Group 16 and is found to the left of fluorine in the periodic table. Elements further to the left have lower ionization energy compared to those on the right. Therefore, the ionization energy of the unknown element is expected to be smaller than that of fluorine.