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Chapter 5: Problem 40

If potassium and oxygen were to bind, describe with words and dot diagrams what would happen to the valence electrons.

Short Answer

Expert verified
When potassium (K) and oxygen (O) atoms bind, the valence electron of potassium transfers to the oxygen atom, resulting in the formation of potassium ion (K⁺) and oxide ion (O²⁻). Their dot diagrams are K• for potassium and :O:: for oxygen. After bonding, the dot diagrams for the ions formed are K⁺[ ] for the potassium ion and [O]: for the oxide ion. The final compound formed is potassium oxide (K₂O), consisting of two potassium ions and one oxide ion held together by ionic bonds.

Step by step solution

01

1. Identify the electronic configuration

First, we'll find the electronic configuration for both potassium (K) and oxygen (O). Potassium has 19 electrons in total, and its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹, meaning that it has one valence electron in its outermost shell (the 4s orbital). Oxygen has 8 electrons in total, with the electronic configuration 1s² 2s² 2p⁴, meaning that it has six valence electrons, four of which are in the 2p orbital and two in 2s.
02

2. Draw dot diagrams for the atoms

Next, we draw the dot diagrams for the potassium and oxygen atoms, which involve placing the valence electrons around the symbol for each atom. For potassium, there is one valence electron in the 4s orbital, so the dot diagram will show a single dot near the "K" symbol. For oxygen, there are six valence electrons, so we'll place four dots around the "O" symbol to represent the 2p electrons, and two dots to represent the 2s electrons. Potassium: K• Oxygen: :O::
03

3. Describe the bonding process

The bonding process between potassium and oxygen results in a transfer of the valence electron from the potassium atom to the oxygen atom. This occurs because potassium has a low electronegativity and would rather lose one electron to achieve a stable noble gas configuration, while oxygen has a higher electronegativity and is looking to gain two electrons to achieve a stable configuration as well.
04

4. Draw the dot diagrams for the ions formed

After bonding occurs, the potassium atom loses its valence electron and becomes a potassium ion (K⁺) with a positive charge. The oxygen atom gains two electrons, one from bonding with potassium and one from bonding with another potassium atom (since it needs to gain two electrons for stability), forming an oxide ion (O²⁻) with a negative charge. Here are the dot diagrams for the ions: Potassium ion: K⁺[ ] Oxide ion: [O]:
05

5. Describe the final compound formed

The potassium ions and the oxide ions combine through ionic bonding, which occurs due to the electrostatic attraction between oppositely charged ions. The final compound formed is potassium oxide (K₂O), which consists of two potassium ions (K⁺) and one oxide ion (O²⁻) held together by ionic bonds. Thus, when potassium and oxygen bind, the valence electrons are transferred from the potassium atoms to the oxygen atom leading to the formation of potassium oxide (K₂O), which can be visualized using the given dot diagrams.

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Most popular questions from this chapter

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Arrange the following sets of bonds in order of increasing ionic character. Use the symbols \(\delta+\) and \(\delta-\) to indicate partial charges, if any, in the bonds: \(\mathrm{O}-\mathrm{Cl}, \mathrm{C}-\mathrm{F}, \mathrm{N}-\mathrm{Cl}, \mathrm{O}-\mathrm{H}, \mathrm{S}-\mathrm{O}\)

What is a shortcut rule for determining the number of covalent bonds a representative element from group IVA, VA, VIA, VIIA, or VIIIA can form?

A compound consisting of one oxygen atom and two nitrogen atoms is written \(\mathrm{N}_{2} \mathrm{O}\) and is called dinitrogen monoxide. It is not written as \(\mathrm{ON}_{2}\), and it is not called oxygen dinitride. Explain why.
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