Question

(a) Using Lewis symbols, make a sketch of the reaction between potassium and bromine atoms to give the ionic substance KBr. (b) How many electrons are transferred? (c) Which atom loses electrons in the reaction?

Short answer

a) The reaction between potassium and bromine atoms using Lewis symbols is represented as: \(K \cdot + :Br: \rightarrow K^+ + :Br^- \rightarrow K^+Br^-\), forming the ionic substance KBr. b) The number of electrons transferred in the reaction is 1. c) Potassium (K) loses electrons in the reaction.

Step-by-step solution

Determine the electron configurations of potassium and bromine atoms

We can find the electron configurations by referring to the periodic table. Potassium has an atomic number of 19 (19 protons and 19 electrons), while Bromine has an atomic number of 35 (35 protons and 35 electrons). Thus, the electron configurations for potassium and bromine are: Potassium (K): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ Bromine (Br): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

Draw the Lewis symbols for potassium and bromine atoms

Lewis symbols show the valence electrons of an atom as dots arranged around the element symbol. For potassium, there is 1 valence electron in the 4s orbital, and for bromine, there are 5 valence electrons in the 4p orbitals. Their respective Lewis symbols are: Potassium (K): K• Bromine (Br): : Br :

Sketch the reaction between potassium and bromine using Lewis symbols and identify the ionic substance KBr

Potassium needs to lose 1 electron to attain a stable noble gas configuration, while bromine needs to gain 1 electron to achieve a stable noble gas configuration. Therefore, in the reaction, potassium will transfer its valence electron to bromine, resulting in the formation of the ionic compound KBr. The Lewis symbols for this reaction can be represented as: K• + :Br: → K⁺ + :Br⁻ → K⁺Br⁻

Determine the number of electrons transferred in the reaction

During the formation of the ionic bond between potassium and bromine, 1 electron is transferred from potassium to bromine as represented by the Lewis symbols. Therefore, the number of electrons transferred is 1.

Identify which atom loses electrons in the reaction

In the reaction, potassium (K) loses 1 electron to attain a stable noble gas configuration. Thus, potassium is the atom that loses electrons in the reaction. To summarize: a) The reaction between potassium and bromine atoms can be represented using Lewis symbols as K• + :Br: → K⁺ + :Br⁻ → K⁺Br⁻, forming the ionic substance KBr. b) The number of electrons transferred in the reaction is 1. c) Potassium (K) is the atom that loses electrons in the reaction.

Key concepts

Lewis Symbols

Lewis symbols are a simple way of representing the valence electrons of an atom. Valence electrons are the outermost electrons of an atom, which play a crucial role in chemical bonding. A Lewis symbol consists of the chemical symbol of an element surrounded by dots that represent its valence electrons.
For example, in the case of potassium (K) and bromine (Br), their Lewis symbols help us visualize the transfer of electrons. Potassium, with one valence electron, is represented as K•. Bromine, possessing seven valence electrons, appears as :Br:.
This visual representation underscores the process of ion formation, where potassium loses one electron and bromine gains an electron, forming an ionic bond.

Electron Configuration

Electron configuration is the arrangement of electrons in an atom's electron shells and orbitals. This configuration helps explain an element's chemical behavior.
Each element in the periodic table has a unique electron configuration, which can be determined from its atomic number. Potassium has an atomic number of 19, giving it an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. Bromine, with an atomic number of 35, has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.

• The distribution of electrons across different energy levels indicates how atoms engage in bonds.
• Potassium's lone electron in the 4s orbital signifies its readiness to lose this electron easily.
• Bromine's need for one more electron to complete its 4p subshell marks its tendency to gain an electron.

Understanding the electron configuration provides clarity on how and why potassium and bromine interact to form an ionic compound.

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. They are primarily responsible for the bonding behavior of the atom.

• For potassium, it has one valence electron in its 4s orbital.
• Bromine has seven valence electrons distributed in its 4p and one in 4s orbitals.

Valence electrons are significant because they are involved in forming chemical bonds. When atoms bond, they tend to do so in such a way to fill or empty their outermost shell, achieving a stable electron configuration similar to that of noble gases.
In the case of the reaction between potassium and bromine, potassium transfers its one valence electron to bromine. This allows both elements to achieve a stable electron arrangement: potassium attains a noble gas configuration similar to argon, and bromine achieves a configuration akin to krypton.

Electron Transfer

Electron transfer involves the movement of electrons from one atom to another. This process is essential in forming ionic bonds, where atoms become charged ions.
In our example of potassium (K) and bromine (Br), electron transfer is the key step in forming potassium bromide (KBr).

• Potassium loses its single 4s electron, becoming a positively charged ion (K⁺).
• Bromine gains this electron, turning into a negatively charged ion (Br⁻).

This transfer of electrons creates an electrostatic attraction between the oppositely charged ions, resulting in an ionic bond.
This bond is the foundation of the ionic compound KBr.
Understanding electron transfer helps explain the formation and stability of ionic compounds by highlighting the shift of electrons from metals to nonmetals.