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Chapter 9: Problem 106

Place the following molecules and ions in order from smallest to largest bond order: \(\mathrm{N}_{2}{ }_{2}^{2+}, \mathrm{He}_{2}{ }^{+}, \mathrm{Cl}_{2} \mathrm{H}_{2}^{-}, \mathrm{O}_{2}{ }^{2-}\).

Short Answer

Expert verified
The bond orders for the given molecules and ions are: \(BO_{N_{2}^{2+}}=2\), \(BO_{He_{2}^{+}}=0.5\), \(BO_{Cl_{2}}=3\), \(BO_{H_{2}^{-}}=0.5\), and \(BO_{O_{2}^{2-}}=3\). Therefore, the order from smallest to largest bond order is: \[ \mathrm{He}_{2}{ }^{+}, \mathrm{H}_{2}^{-} < \mathrm{N}_{2}{ }_{2}^{2+} < \mathrm{O}_{2}{ }^{2-}, \mathrm{Cl}_{2} \]

Step by step solution

01

Molecular orbital theory

Molecular orbital (MO) theory is a suitable method for determining the bond order. We will use the molecular orbital diagrams for \(1\sigma, 1\sigma^{*}, 2\sigma, 2\sigma^{*}, 1\pi, 1\pi^{*}, 2\pi, \text{ and } 2\pi^{*}\) orbitals.
02

Determine bond order

The bond order equation is given by:\(BO = \frac{(number\ of\ bonding\ electrons - number\ of\ antibonding\ electrons)}{2}\). We should determine the bond order for each of the given species.
03

Bond orders of individual species

To determine the bond order for each species, we have to calculate the electron count according to their electronic configurations. 1. \(\mathrm{N}_{2}{ }_{2}^{2+}\): \(N_2\) has a total of \(10\) electrons, and the \(2+\) means that it loses \(2\) electrons; thus, there are \(8\) electrons. Bond order for this is: \(BO_{N_{2}^{2+}}=\frac{(6-2)}{2}=2\) 2. \(\mathrm{He}_{2}{ }^{+}\): He has \(2\) electrons, and \(He_2\) will have \(4\) electrons. However, the \(+\) means that one electron is lost, leaving \(3\) electrons. So, the bond order for this is: \(BO_{He_{2}^{+}}=\frac{(2-1)}{2}=0.5\) 3. \(\mathrm{Cl}_{2}\): Cl has \(17\) electrons, and \(Cl_2\) will have \(34\) electrons. Bond order for this is: \(BO_{Cl_{2}}=\frac{(20-14)}{2}=3\) 4. \(\mathrm{H}_{2}^{-}\): H has \(1\) electron, and \(H_2\) would have \(2\) electrons. The negative charge implies an extra electron, making it \(3\) electrons. Bond order for this is: \(BO_{H_{2}^{-}}=\frac{(2-1)}{2}=0.5\) 5. \(\mathrm{O}_{2}{ }^{2-}\): \(O\) has \(8\) electrons, and \(O_2\) will have \(16\) electrons. The \(2-\) implies addition of \(2\) electrons, leaving \(18\) electrons. So, the bond order for this is: \(BO_{O_{2}^{2-}}=\frac{(12-6)}{2}=3\)
04

Order the species by bond order

Now that we have the bond orders, we can arrange the species from lowest to the highest bond order: \[ \mathrm{He}_{2}{ }^{+}, \mathrm{H}_{2}^{-} < \mathrm{N}_{2}{ }_{2}^{2+} < \mathrm{O}_{2}{ }^{2-}, \mathrm{Cl}_{2} \]

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Key Concepts

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

Bond Order
Bond order provides an insight into the stability and strength of a bond between atoms in a molecule. It's indicative of the number of bonds between a pair of atoms.
The formula to calculate bond order is straightforward:
  • Use the equation: \( BO = \frac{\text{number of bonding electrons} - \text{number of antibonding electrons}}{2} \).
A higher bond order typically means a more stable and stronger bond. Conversely, a low bond order might imply a weaker bond or even instability, as seen in cases with fractional bond orders.
A bond order of zero suggests no bond exists. Understanding the bond order helps predict molecular structure and reactivity, making it a powerful tool in chemistry.
Electron Configuration
Electron configuration is essential in determining how electrons are distributed in an atom or molecule's molecular orbitals. It's the step prior to understanding bond order. With Molecular Orbital Theory, these configurations tell us the energy placement of electrons between bonding and antibonding orbitals.
Determining electron configurations:
  • First, calculate the total number of electrons.
  • For charged species, add or subtract electrons based on the charge indicated.
Assign electrons to the molecular orbitals according to increasing energy order:
  • Bonding orbitals: Usually filled first, stabilizing the molecule.
  • Antibonding orbitals: Generally filled later, they can destabilize the molecule if too many are filled.
Knowing electron configurations allows chemists to analyze how molecules will interact, react, and behave.
Molecular Bonding
Molecular bonding describes the interactions holding atoms together within a molecule. It is crucial for the formation of molecules from individual atoms. In the context of Molecular Orbital Theory, these bonds are explained through the interaction between molecular orbitals formed by the overlap of atomic orbitals.
Types of bonds:
  • Covalent bonds: Arise from sharing electron pairs between atoms, leading to molecular stability.
  • Ionic bonds: Formed through electrostatic attraction between oppositely charged ions, though less common in molecular orbital discussions.
In molecular orbital theory, we discuss bonding in terms of:
  • Bonding orbitals: Formed when atomic orbitals with constructive interference combine, allowing electrons to be shared in a way that stabilizes the molecule.
  • Antibonding orbitals: Occur due to destructive interference in orbital overlap. Occupying these can weaken and potentially destabilize the molecule.
Understanding molecular bonding is vital for predicting the physical and chemical properties of substances.

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Most popular questions from this chapter

(a) Draw Lewis structures for chloromethane \(\left(\mathrm{CH}_{3} \mathrm{Cl}\right),\) chloroethene \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\right)\), and chloroethyne \(\left(\mathrm{C}_{2} \mathrm{HCl}\right) .(\mathbf{b})\) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule?

Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that the \(z\) -axis is defined as the metal-metal bond axis. (a) Which of the 3 d orbitals (Figure 6.23 ) is most likely to make a \(\sigma\) bond between metal atoms? (b) Sketch the \(\sigma_{3 d}\) bonding and \(\sigma_{3 d}^{*}\) antibonding MOs. (c) With reference to the "Closer Look" box on the phases oforbitals, explain why a node is generated in the \(\sigma_{3 d}^{*}\) MO. (d) Sketch the energylevel diagram for the \(\mathrm{Sc}_{2}\) molecule, assuming that only the \(3 d\) orbital from part (a) is important. (e) What is the bond order in \(\mathrm{Sc}_{2} ?\)

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

Would you expect the nonbonding electron-pair domain in \(\mathrm{NCl}_{3}\) to be greater or smaller in size than the corresponding one in \(\mathrm{PCl}_{3} ?\)
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