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Place the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) in order of increasing bond length and increasing bond energy.

Short Answer

Expert verified
The order of increasing bond length for the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\). The order of increasing bond energy for these species is also: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).

Step by step solution

01

Calculate the number of valence electrons for each species

Count the valence electrons of each species to determine their electron configurations. - \(\mathrm{B}_{2}^{+}\): Each boron atom contributes 3 valence electrons, and there is a charge of +1, so we have a total of 3 + 3 - 1 = 5 valence electrons. - \(\mathrm{B}_{2}\): Each boron atom contributes 3 valence electrons, so we have a total of 3 + 3 = 6 valence electrons. - \(\mathrm{B}_{2}^{-}\): Each boron atom contributes 3 valence electrons, and there is a charge of -1, so we have a total of 3 + 3 + 1 = 7 valence electrons.
02

Calculate bond order for each species

Utilize the formula for bond order which is: \[Bond \, Order = \frac{1}{2} \times (number \, of \, bonding \, electrons - number \, of \, antibonding \, electrons)\] - For \(\mathrm{B}_2^+\), there are 5 valence electrons. According to molecular orbital theory, the electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^1\). Therefore, the bond order is \(\frac{1}{2}(3-2)=0.5\). - For \(\mathrm{B}_2\), there are 6 valence electrons. The electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\). Therefore, the bond order is \(\frac{1}{2}(4-2)=1\). - For \(\mathrm{B}_2^-\), there are 7 valence electrons. The electron configuration is \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma_{2p_x}^1\). Therefore, the bond order is \(\frac{1}{2}(5-2)=1.5\).
03

Arrange species in order of bond length

As mentioned, higher bond order corresponds to shorter bond length. Arrange the species based on their bond orders found in step 2: - \(\mathrm{B}_{2}^{+}\): Bond order = 0.5 (Longest bond length) - \(\mathrm{B}_{2}\): Bond order = 1 (Medium bond length) - \(\mathrm{B}_{2}^{-}\): Bond order = 1.5 (Shortest bond length) So the order of increasing bond length is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).
04

Arrange species in order of bond energy

Higher bond order also corresponds to greater bond energy. Using the bond orders calculated in step 2, arrange the species based on their bond energies: - \(\mathrm{B}_{2}^{+}\): Bond order = 0.5 (Lowest bond energy) - \(\mathrm{B}_{2}\): Bond order = 1 (Medium bond energy) - \(\mathrm{B}_{2}^{-}\): Bond order = 1.5 (Highest bond energy) So the order of increasing bond energy is: \(\mathrm{B}_{2}^{+} < \mathrm{B}_{2} < \mathrm{B}_{2}^{-}\).

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Most popular questions from this chapter

Show how a \(d_{x z}\) atomic orbital and a \(p_{z}\) atomic orbital combine to form a bonding molecular orbital. Assume the \(x\) -axis is the internuclear axis. Is a \(\sigma\) or a \(\pi\) molecular orbital formed? Explain.

Use the localized electron model to describe the bonding in \(\mathrm{H}_{2} \mathrm{O}\) .

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

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: $$ \mathrm{M}-\mathrm{C} \equiv \mathrm{O} $$ a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. 9.43 and 9.44.) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Consider the following molecular orbitals formed from the combination of two hydrogen 1s orbitals: a. Which is the bonding molecular orbital and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?

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