Problem 53 - Chemistry and Chemical Reactivity

Question

Answer the questions below about the elements A and \(\mathrm{B},\) which have the electron configurations shown. $$ \mathrm{A}=[\mathrm{Kr}] 5 s^{1} \quad \mathrm{B}=\left[\mathrm{Ar} | 3 d^{10} 4 \mathrm{s}^{2} 4 p^{4}\right. $$ (a) Is element \(A\) a metal, nonmetal, or metalloid? (b) Which element has the greater ionization energy? (c) Which element has the less negative electron attachment enthalpy? (d) Which element has the larger atomic radius? (e) What is the formula for a compound formed between A and B?

Step-by-Step Solution

Answer

(a) Metal; (b) B; (c) A; (d) A; (e) \(A_2B\).

Step 1: Identify Element A

Element A has an electron configuration of \([\text{Kr}] 5s^1\). This configuration indicates that element A belongs to Group 1 of the periodic table, which consists of alkali metals. Hence, element A is a metal.

Step 2: Identify Element B

Element B has an electron configuration of \([\text{Ar}] 3d^{10} 4s^2 4p^4\). This is the configuration for an element in Group 16 of the periodic table, known for chalcogens. Elements in this group, such as sulfur (S) and selenium (Se), are nonmetals. Hence, element B is a nonmetal.

Step 3: Compare Ionization Energy

Ionization energy increases across a period from left to right in the periodic table and decreases down a group. Since element A is an alkali metal while element B is a chalcogen, element B will have a higher ionization energy because it is farther to the right in the periodic table compared to alkali metals.

Step 4: Compare Electron Attachment Enthalpy

Electron attachment enthalpy becomes more negative across a period from left to right and is less negative going down a group. Nonmetals generally have more negative electron enthalpies than metals. Hence, element A has a less negative electron attachment enthalpy compared to element B.

Step 5: Compare Atomic Radius

Atomic radius decreases across a period from left to right and increases down a group. Since element A is in Group 1, it has a larger atomic radius than element B, which is in Group 16.

Step 6: Determine Compound Formula

Element A is a metal likely to form a cation \(A^+\), and element B is a nonmetal likely to form an anion \(B^{2-}\). Thus, the compound formed between elements A and B can be represented by the formula \(A_2B\) to balance the charges.

Key Concepts

Electron Configuration

Electron configuration refers to the arrangement of electrons in an atom's electron shells and subshells. This configuration is represented using notation that lists the occupied subshells and their respective electron counts. It gives valuable insights into an element's chemical behavior and properties. For example, element A has an electron configuration of \([ ext{Kr}] 5s^1\), indicating a single electron in its outermost shell, typical of alkali metals. On the other hand, element B has the configuration \([ ext{Ar}] 3d^{10} 4s^2 4p^4\), characteristic of chalcogens with six valence electrons.
Characteristics of electron configurations can help predict a variety of chemical behaviors, including:

• Valency and common oxidation states.
• Potential reactivity and chemical bonding patterns.
• The physical state of the element at room temperature.

By understanding electron configurations, we can better anticipate how elements will interact in compounds such as how elements A and B form \(A_2B\).

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. It is a crucial factor in determining an element's reactivity. Elements with high ionization energies tend not to lose electrons easily, making them less reactive and often more electronegative.
There are trends in ionization energy across the periodic table:

• It increases across a period from left to right due to increasing nuclear charge, making it harder to remove an electron.
• It decreases down a group due to additional electron shells reducing the nuclear attraction experienced by the outermost electrons.

In the case of our elements, B has a higher ionization energy compared to A, which is consistent with B being to the right of A in the periodic table. This makes B less likely to lose electrons and play a critical role in the formation of compounds with more electronegative elements.

Atomic Radius

The atomic radius is a measure of the size of an atom, typically the distance from the nucleus to the outermost stable electron shell. The atomic radius gives insight into the space the atom occupies and affects the element's physical and chemical properties.
The trend in atomic radii across the periodic table is:

• Decreases from left to right across a period because of increased nuclear charge pulling the electron cloud closer.
• Increases down a group as additional electron shells are added.

In the comparison of elements A and B, group differences become evident with element A having a larger atomic radius. Being in Group 1, A possesses fewer electrons in fewer shells compared to B in Group 16, whose tighter electron cloud results from higher nuclear charge. This subtle difference in electron cloud size changes how each atom interacts when forming bonds.

Chemical Bonding

Chemical bonding describes the forces that hold atoms together in compounds. These bonds can be covalent, ionic, metallic, or hydrogen bonds, depending on the elements involved and their electron configurations.
For elements A and B:

• Element A is an alkali metal, tending to lose one electron to achieve a stable noble gas configuration, forming a \(A^+\) cation.
• Element B, a chalcogen, tends to gain two electrons to fill its valence shell, forming a \(B^{2-}\) anion.

This electron loss and gain account for the ionic bond formed in the compound \(A_2B\). Ionic bonds generally result in compounds with high melting and boiling points due to the strong electrostatic forces between oppositely charged ions. Understanding such bonding principles helps explain the rationale behind compound formation and the resultant properties.