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Chapter 10: Problem 57

Explain why the bond order of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+}\), but the bond order of \(\mathrm{O}_{2}\) is less than that of \(\mathrm{O}_{2}^{+}\)

Short Answer

Expert verified
The bond order of \(N_2\) is greater than that of \(N_2^+\) because \(N_2\) has a higher number of bonding electrons than \(N_2^+\), resulting in a higher bond order (3 compared to 2.5). However, the bond order of \(O_2\) is less than that of \(O_2^+\) because the removal of an electron in \(O_2^+\) occurs from an antibonding orbital, which decreases the number of antibonding electrons and increases the bond order from 2 to 2.5.

Step by step solution

01

Understanding the concept of bond order

Bond order is a measure of the number of shared electron pairs between two atoms in a molecule. It can be calculated by subtracting the number of electrons populating anti-bonding orbitals from the number populating bonding orbitals, and then dividing by two.
02

Calculating bond order of \(N_2\) and \(N_2^+\)

In \(N_2\) molecule, there are 10 electrons in bonding molecular orbitals and 4 in antibonding orbitals. Using the formula, the bond order is \((10-4)/2 = 3\). In \(N_2^+\) ion, there is one less electron, so there are 9 electrons in bonding molecular orbitals and 4 in antibonding orbitals, giving a bond order of \((9-4)/2 = 2.5\). Therefore, the bond order of \(N_2\) is greater than that of \(N_2^+\)
03

Calculating bond order of \(O_2\) and \(O_2^+\)

In \(O_2\) molecule, there are 8 electrons in bonding molecular orbitals and 4 in antibonding orbitals, giving a bond order of \((8-4)/2 = 2\). In \(O_2^+\) ion, there is one less electron, so there are 8 electrons in bonding molecular orbitals and 3 in antibonding orbitals. This gives a bond order of \((8-3)/2 = 2.5\). Therefore, the bond order of \(O_2\) is less than that of \(O_2^+\)

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Most popular questions from this chapter

Acetaminophen is the active ingredient in Tylenol. (a) Write the molecular formula of the compound. (b) What is the hybridization state of each \(\mathrm{C}, \mathrm{N},\) and \(\mathrm{O}\) atom? (c) Describe the geometry about each \(\mathrm{C}, \mathrm{N}\), and \(\mathrm{O}\) atom.

Both ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) contain the \(\mathrm{C}=\mathrm{C}\) bond. The reactivity of ethylene is greater than that of benzene. For example, ethylene readily reacts with molecular bromine, whereas benzene is normally quite inert toward molecular bromine and many other compounds. Explain this difference in reactivity.

Consider the reaction $$ \mathrm{BF}_{3}+\mathrm{NH}_{3} \longrightarrow \mathrm{F}_{3} \mathrm{~B}-\mathrm{NH}_{3} $$ Describe the changes in hybridization (if any) of the \(\mathrm{B}\) and \(\mathrm{N}\) atoms as a result of this reaction.

Predict the geometries of the following species using the VSEPR method: (a) \(\mathrm{PCl}_{3}\) (b) \(\mathrm{CHCl}_{3}\) (c) \(\mathrm{SiH}_{4},\) (d) \(\mathrm{TeCl}_{4}\).

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) has a tendency to lose two protons \(\left(\mathrm{H}^{+}\right)\) and form the carbide ion \(\left(\mathrm{C}_{2}^{2-}\right),\) which is present in a number of ionic compounds, such as \(\mathrm{CaC}_{2}\) and \(\mathrm{MgC}_{2}\). Describe the bonding scheme in the \(\mathrm{C}_{2}^{2-}\) ion in terms of molecular orbital theory. Compare the bond order in \(\mathrm{C}_{2}^{2-}\) with that in \(\mathrm{C}_{2}\).
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