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 element has 6 valence electrons. The possible identities of the element include oxygen (O), sulfur (S), selenium (Se), and tellurium (Te). The compound formula formed with potassium would be K2X, where X represents the unknown element. The unknown element would have a smaller radius than barium and a smaller ionization energy than fluorine.

Step-by-step solution

a. Number of Valence Electrons

The valence electron configuration is given as \(n s^{2} n p^{4}\). To find the total number of valence electrons, simply add the exponents: \(2 + 4 = 6\). Thus, the element has 6 valence electrons.

b. Possible Identities of the Element

Elements with 6 valence electrons are typically found in Group 16 of the periodic table. Since the problem states that the element is a nonmetal, we can consider oxygen (O), sulfur (S), selenium (Se), and tellurium (Te) as possible identities for this element. Further information would be needed to narrow down the specific element.

c. Compound Formula with Potassium

Potassium (K) is an alkali metal in Group 1 and has a valence electron configuration of \(n s^1\). In general, these elements lose 1 electron to achieve a stable octet configuration, forming a +1 cation (K+). Our unknown element, with 6 valence electrons, will typically gain 2 electrons to achieve a full octet, forming a -2 anion (X2-). Thus, the compound formed between potassium and the unknown element would be K2X, where X represents the unknown element.

d. Radius Comparison with Barium

Barium (Ba) is an alkaline earth metal in Group 2, which has a relatively large atomic radius compared to nonmetals. The unknown element in question is a nonmetal, and nonmetals generally have smaller atomic radii compared to metals due to their higher effective nuclear charges. Therefore, the unknown element would have a smaller radius than barium.

e. Ionization Energy Comparison with Fluorine

Ionization energy is the energy required to remove an electron from an atom. Fluorine (F) has the highest ionization energy among the elements in Group 17 due to its small size and high effective nuclear charge. The unknown element is a nonmetal with 6 valence electrons from Group 16. In general, the ionization energy decreases as we move from right to left and down in the periodic table. As our unknown element is in Group 16 and Fluorine is in Group 17, the unknown element would have a smaller ionization energy than fluorine.

Key concepts

Nonmetals

Nonmetals are fascinating elements characterized by their diverse behavior and properties in the periodic table. Generally, they are poor conductors of heat and electricity and possess higher ionization energies than metals, which means they do not readily lose electrons. Unlike metals, nonmetals can gain electrons easily, allowing them to form anions or to participate in covalent bonding.
A notable property of nonmetals is their ability to form various types of compounds, especially with metals. When forming ionic compounds, nonmetals tend to gain electrons and acquire negative charges, balancing the positive charges of metals.
In the exercise, the unknown nonmetal with a valence electron configuration of \(n s^{2} n p^{4}\) suggests it belongs to Group 16. These elements form crucial compounds with metals and nonmetals alike, emphasizing their importance in numerous chemical reactions across nature and industry.

Periodic Table

The periodic table is an organized diagram that lists all known chemical elements in an ordered fashion. It is arranged according to increasing atomic number (number of protons) and electron configurations. Its structure allows us to predict the properties of elements based on their position.
Periods are the rows running horizontally, each representing a principal energy level of an atom's electron cloud. Groups, the columns, contain elements with similar valence electron configurations. This similarity allows the prediction of chemical behavior for groups of elements.
Group 16, where the nonmetals in the exercise are focused, contains elements such as oxygen, sulfur, selenium, and tellurium. Valence electrons \(n s^{2} n p^{4}\) indicate that members of this group typically have six valence electrons, making them reactive and capable of forming diverse compounds.

Ionization Energy

Ionization energy measures the amount of energy required to remove the most loosely bound electron from an atom in its gaseous state. It is a critical property affecting how easily elements can participate in chemical reactions.
Higher ionization energies mean it takes more energy to remove an electron, which is common in nonmetals due to their smaller atomic radii and higher effective nuclear charge. In contrast, metals have lower ionization energies, losing electrons more readily.
In comparing the unknown nonmetal with fluorine, the exercise explains that Group 17 elements like fluorine have particularly high ionization energies, with fluorine being among the highest. The unknown nonmetal, being from Group 16, would inherently have a lower ionization energy than fluorine, making it more likely to lose electrons in reactions involving higher energy inputs.

Atomic Radius

The atomic radius is a measure of the size of an atom, typically defined as the distance from the nucleus to the boundary of the surrounding electron cloud. It is crucial in understanding the behavior of elements in bonding and reactions.
Generally, as we move down a group in the periodic table, the atomic radius increases because additional electron shells are added. Conversely, as we move across a period from left to right, atomic radii tend to decrease due to increasing nuclear charge pulling electrons closer.
Barium, an earth alkaline metal, has a large atomic radius, while nonmetals from Group 16, like those in the exercise, are smaller. This is why the atomic radius of the unknown nonmetal is smaller than that of barium, reflecting its nonmetal character and typical positioning on the periodic table.

Chemical Formulas

Chemical formulas represent the composition of molecules and compounds. They use element symbols from the periodic table, with subscripts indicating the number of each type of atom present in the molecule.
Predominantly, compounds involving nonmetals form either ionic or molecular compounds. Ionic compounds arise when metals transfer electrons to nonmetals, leading to electrostatic attraction between cations and anions. Meanwhile, molecular compounds involve the sharing of electrons between nonmetals through covalent bonds.
In the exercise, the formation of a compound between the unknown nonmetal and potassium is considered. Potassium, a Group 1 metal, typically forms a \( K^{+} \) ion. The nonmetal, having six valence electrons, gains two electrons to form a \( X^{2-} \) ion, resulting in the chemical formula \( K_2X \). This formula exemplifies how valence electrons and chemical bonding principles govern compound formation.