Lewis theory is satisfactory to explain bonding in the ionic compound \(\mathrm{K}_{2} \mathrm{O},\) but it does not readily explain formation of the ionic compounds potassium superoxide, \(\mathrm{KO}_{2}\), and potassium peroxide, \(\mathrm{K}_{2} \mathrm{O}_{2}\) (a) Show that molecular orbital theory can provide this explanation. (b) Write Lewis structures consistent with the molecular orbital explanation.

Short Answer

Expert verified
Molecular Orbital theory explains the formation of KO2 and K2O2 by considering that Potassium atoms donate their 4s electrons to the antibonding orbitals of the Oxygen molecules, forming ionic compounds. The Lewis structures confirm these ionic bonds, showing electrons' transfer from Potassium to Oxygen molecules.

Step by step solution

01

Understanding molecular orbital theory for KO2

For KO2, Potassium (K) has one valence electron in its 4s orbital. O2 (dioxygen) is a diradical and exhibits paramagnetism. It has corresponding π* (antibonding) orbitals populated by one electron each. When Potassium reacts with O2, the electron from 4s orbital of K goes to one of these antibonding orbitals, making O2 a full bonding species while giving K a +1 charge, forming an ionic compound, KO2.
02

Understanding molecular orbital theory for K2O2

For K2O2, each Potassium atom donates one electron to an Oxygen molecule. As a result, each K atom gets a +1 charge, and O2 gets a -1 charge. Since two K atoms donate one electron each, the overall charge forms as neutral, creating KO2.
03

Writing Lewis structures for KO2

For KO2, the Lewis structure will show that Oxygen (O2) molecule forms an ionic bond with Potassium (K+), represented by placing K+(s) Naear [O2]-. It indicates that one electron is transferred from Potassium to Oxygen molecule.
04

Writing Lewis structures for K2O2

For K2O2, the Lewis structure will show two Potassium ions (K+) forming an ionic bond with Peroxide (O22-), represented by placing two K+ near [O - O]2-. It indicates that an electron from each Potassium atom is transferred to the Oxygen molecule.

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Most popular questions from this chapter

Explain the important distinctions between the terms in each of the following pairs: (a) \(\sigma\) and \(\pi\) bonds; (b) localized and delocalized electrons; (c) bonding and antibonding molecular orbitals; (d) metal and semiconductor.

The energy gap, \(\Delta E\), for silicon is \(110 \mathrm{kJ} / \mathrm{mol}\). What is the minimum wavelength of light that can promote an electron from the valence band to the conduction band in silicon? In what region of the electromagnetic spectrum is this light?

Methyl nitrate, \(\mathrm{CH}_{3} \mathrm{NO}_{3}\), is used as a rocket propellant. The skeletal structure of the molecule is \(\mathrm{CH}_{3} \mathrm{ONO}_{2}\). The N and three O atoms all lie in the same plane, but the \(\mathrm{CH}_{3}\) group is not in the same plane as the \(\mathrm{NO}_{3}\) group. The bond angle \(\mathrm{C}-\mathrm{O}-\mathrm{N}\) is \(105^{\circ},\) and the bond angle \(\mathrm{O}-\mathrm{N}-\mathrm{O}\) is \(125^{\circ} .\) One nitrogen-to-oxygen bond length is \(136 \mathrm{pm},\) and the other two are \(126 \mathrm{pm}\) (a) Draw a sketch of the molecule showing its geometric shape. (b) Label all the bonds in the molecule as \(\sigma\) or \(\pi\), and indicate the probable orbital overlaps involved. (c) Explain why all three nitrogen-to-oxygen bond lengths are not the same.

What is the total number of (a) \(\sigma\) bonds and (b) \(\pi\) bonds in the molecule \(\mathrm{CH}_{3} \mathrm{NCO}\) ?

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