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Chapter 4: Problem 67

Using molecular orbital theory, explain why the removal of one electron in \(\mathrm{O}_{2}\) strengthens bonding, while the removal of one electron in \(\mathrm{N}_{2}\) weakens bonding.

Short Answer

Expert verified
In conclusion, using molecular orbital theory, removing one electron from O2 strengthens bonding because the bond order increases from 2 to 2.5. On the other hand, removing one electron from N2 weakens bonding because the bond order decreases from 3 to 3.5. The change in bond orders is due to the removal of electrons from the respective highest occupied molecular orbitals (HOMO) of O2 and N2.

Step by step solution

01

Know the molecular orbital diagrams and electronic configurations of O2 and N2

The molecular orbital diagrams for O2 and N2 are essential when explaining the effects of removing one electron from these molecules. O2's atomic number is 8, so it has 8 valence electrons, making a total of 16 electrons for O2. N2's atomic number is 7, so it has 7 valence electrons, making a total of 14 electrons for N2. The electronic configurations for O2 and N2 can be found as follows: O2: \(1\sigma_{g}^{2}2\sigma_{g}^{2}3\sigma_{g}^{2}1\pi_{u}^{4}1\pi_{g}^{2}\) N2: \(1\sigma_{g}^{2}2\sigma_{g}^{2}3\sigma_{g}^{2}1\pi_{u}^{4}2\sigma_{u}^{2}\)
02

Remove one electron from O2 and N2 and determine the new electron configurations

We now remove one electron from each O2 and N2 and determine their new electron configurations. For O2, when we remove one electron, we remove it from the HOMO, which is the \(1\pi_{g}\) orbital. Then, the new electron configuration becomes: O2+: \(1\sigma_{g}^{2}2\sigma_{g}^{2}3\sigma_{g}^{2}1\pi_{u}^{4}1\pi_{g}^{1}\) For N2, when we remove one electron, we remove it from the HOMO, which is the \(2\sigma_{u}\) orbital. The new electron configuration is: N2+: \(1\sigma_{g}^{2}2\sigma_{g}^{2}3\sigma_{g}^{2}1\pi_{u}^{4}2\sigma_{u}^{1}\)
03

Calculate the bond orders of O2, O2+, N2, and N2+

To relate the electron configurations to bonding strength, we need to calculate the bond orders. Bond order can be calculated using the formula: Bond Order = \(\frac{(\text{Number of bonding electrons - Number of antibonding electrons})}{2}\) Calculating bond orders for O2, O2+, N2, and N2+: O2: Bond Order = \(\frac{(10 - 6)}{2} = 2\) O2+: Bond Order = \(\frac{(10 - 5)}{2} = 2.5\) N2: Bond Order = \(\frac{(10 - 4)}{2} = 3\) N2+: Bond Order = \(\frac{(10 - 3)}{2} = 3.5\)
04

Analyze the change in bond orders and bonding strength

Now we can analyze the change in bond orders and bonding strength upon removing one electron from O2 and N2. For O2, the bond order of O2+ is greater than the bond order of O2 (2.5 > 2). This means that the removal of one electron from O2 strengthens the bonding between the oxygen atoms. For N2, the bond order of N2+ is less than the bond order of N2 (3.5 < 3). This means that the removal of one electron from N2 weakens the bonding between the nitrogen atoms. In conclusion, the removal of one electron from O2 strengthens the bonding due to an increase in bond order, while the removal of one electron from N2 weakens the bonding due to a decrease in bond order.

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Most popular questions from this chapter

A variety of chlorine oxide fluorides and related cations and anions are known. They tend to be powerful oxidizing and fluorinating agents. \(\mathrm{FClO}_{3}\) is the most stable of this group of compounds and has been studied as an oxidizing component in rocket propellants. Draw a Lewis structure for \(\mathrm{F}_{3} \mathrm{ClO}\) \(\mathrm{F}_{2} \mathrm{ClO}_{2}^{+},\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\). What is the molecular structure for each species, and what is the expected hybridization of the central chlorine atom in each compound or ion?

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 25 and \(26 .\) ) a. ICls b. \(\mathrm{XeCl}_{4}\) c. \(\mathrm{SeCl}_{6}\)

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) can be produced from the reaction of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) with water. Use both the localized electron and molecular orbital models to describe the bonding in the acetylide anion \(\left(\mathrm{C}_{2}^{2-}\right)\).

In the hybrid orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Many important compounds in the chemical industry are derivatives of ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right) .\) Two of them are acrylonitrile and methyl methacrylate. Complete the Lewis structures, showing all lone pairs. Give approximate values for bond angles \(a\) through \(f\). Give the hybridization of all carbon atoms. In acrylonitrile, how many of the atoms in the molecule must lie in the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in methyl methacrylate and acrylonitrile?
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