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(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

Expert verified
One electron is transferred; potassium loses an electron to bromine.

Step by step solution

01

Identify the valence electrons

Potassium (K) is in Group 1 of the periodic table, so it has one valence electron. Bromine (Br) is in Group 17 and thus has seven valence electrons.
02

Represent Lewis symbols

Write the Lewis symbol for potassium as 'K.' with a single dot representing its one valence electron. For bromine, write 'Br' surrounded by seven dots.
03

Formation of ions

In order to achieve a stable electron configuration, potassium will lose its one valence electron, forming a cation (K鈦). Bromine will gain this electron to complete its octet, forming an anion (Br鈦).
04

Draw a Lewis symbol for KBr

After transferring the electron, the Lewis symbols represent K鈦 and Br鈦. Write K with no dots (since it has lost its electron) and Br with a complete octet and a negative charge. Enclose each ion in brackets and indicate their charges: [K]鈦 and [Br]鈦.
05

Determine number of electrons transferred

Potassium loses one electron, which bromine gains, resulting in a transfer of one electron.
06

Identify the atom that loses electrons

Potassium is the atom that loses its one valence electron during the reaction, forming a K鈦 ion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis symbols
Lewis symbols are a great way to visualize the valence electrons of an atom. They use dots positioned around the symbol of an element to represent these outer shell electrons. This is especially handy for predicting how atoms will interact in chemical reactions, like in the formation of ionic bonds.
For example, in the case of potassium (K), which is in Group 1 of the periodic table, the Lewis symbol is represented by the letter 'K' followed by a single dot. This dot signifies its one valence electron. Bromine (Br), on the other hand, is in Group 17, and its Lewis symbol features seven dots around 'Br', corresponding to its seven valence electrons. By clearly displaying valence electrons, Lewis symbols help us to foresee how an atom might gain, lose, or share electrons during bond formation.
Electron transfer
Electron transfer is the core process of forming an ionic bond. This process involves the movement of electrons from one atom to another, typically from a metal to a non-metal.
When potassium reacts with bromine, electron transfer is the key step. Potassium, with its single valence electron, tends to lose this electron due to its desire to achieve a stable configuration similar to noble gases. Bromine, being one electron short of a full octet, readily accepts this electron.
  • The result is the formation of a positively charged potassium ion (K鈦), due to the loss of an electron.
  • Bromine becomes a negatively charged ion (Br鈦) because it gains an electron.
This gain and loss of electrons lead to the creation of an ionic bond, forming the compound potassium bromide (KBr). The ionic bond is strong due to the electrostatic attraction between the positively charged K鈦 and negatively charged Br鈦 ions.
Valence electrons
Valence electrons are the outermost electrons in an atom and are crucial for chemical bonding. They are the electrons involved in forming bonds with other atoms. The number of valence electrons determines an element's chemical behavior during bond formation.
Taking potassium and bromine as examples:
  • Potassium has one valence electron, making it highly reactive as it seeks to lose this electron to attain a stable electron configuration.
  • Bromine has seven valence electrons, which drives its tendency to gain one electron to complete its valence shell.
Understanding the concept of valence electrons helps predict how atoms will interact in reactions, such as which atom will lose or gain electrons. In this case, potassium loses its one valence electron to bromine, facilitating the creation of the ionic compound KBr. This demonstrates that the behavior of atoms in the periodic table is largely influenced by their valence electrons, laying the groundwork for predicting chemical reactions.

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Most popular questions from this chapter

(a) Draw the dominant Lewis structure for the phosphorus trifluoride molecule, \(\mathrm{PF}_{3}\). (b) Determine the oxidation numbers of the \(\mathrm{P}\) and \(\mathrm{F}\) atoms. \((\mathbf{c})\) Determine the formal charges of the \(\mathrm{P}\) and \(\mathrm{F}\) atoms.

Write the electron configuration for each of the following ions. and determine which ones possess noble-gas configurations: (a) \(\mathrm{Be}^{2+}\) (b) \(\mathrm{Mn}^{2+},(\mathbf{c}) \mathrm{Cd}^{2+}\) (d) \(\mathrm{Fe}^{3+}\), (e) \(\mathrm{Tl}^{+}\), (f) \(\mathrm{At}^{-}\).

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound \(1: 40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \%\) M, \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of M? (b) What is the empirical formula for Compound 2 and the identity of M? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. \((\mathbf{e})\) Are compounds 1 and 2 ionic or molecular?

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) Al and Cl, (d) Li and \(O\). (b) \(\mathrm{Mg}\) and \(\mathrm{O},(\mathbf{c}) \mathrm{Zn}\) and \(\mathrm{Cl},\)

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Iodine has valence electrons and makes ___________ bond(s) in compounds. (b) Silicon has ____________ valence electrons and makes _________ bond(s) in compounds. (c) Phosphorus has ___________ valence electrons and makes ________ bond(s) in compounds. (d) Sulphur has __________ valence electrons and makes __________ bond(s) in compounds.

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