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Chapter 15: Problem 4

Predict the \(\mathrm{C}-\mathrm{C}\) bond length for a molecule with a \(\mathrm{C}-\mathrm{C}\) bond order of \(1.5\).

Short Answer

Expert verified
The predicted C-C bond length for a molecule with a C-C bond order of 1.5 is 144 pm.

Step by step solution

01

Understanding bond order and bond length

Let's first define the terms we know. Bond order is the number of shared electron pairs between two atoms, indicating the strength and length of the bond. The higher the bond order, the stronger and shorter the bond. Bond length, on the other hand, is the distance between the nuclei of two bonded atoms in a molecule. It is measured in picometers (pm). In the case of Carbon-Carbon bond, we know that the average bond length for a single bond (\(C-C\)) is 154 pm, and for a double bond (\(C=C\)) it's 134 pm.
02

Interpolating the bond length

Since a bond order of 1.5 is halfway between a single bond and a double bond, we estimate that the bond length will be halfway between the length of a single bond and the length of a double bond. Therefore, we need to find out the average of these two lengths, which means we should sum them up and divide by 2.
03

Calculating the bond length for C-C bond

Applying the calculation to the known lengths, the bond length of a C-C bond with a bond order of 1.5 will be: \(\frac{154 \, \text{pm} + 134 \, \text{pm}}{2} = 144 \, \text{pm}\).

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Most popular questions from this chapter

Experimentally, we find that all six \(\mathrm{C}-\mathrm{C}\) bonds in benzene have the same bond strength (or bond energy), \(509 \mathrm{~kJ} /\) mole. Is this fact more consistent with the bond orders predicted by the Lewis structure or with the calculated bond orders for benzene?

How do the calculated bond orders for ethane, ethene, and ethyne compare to bond orders predicted by the Lewis structures?

How do the calculated bond orders for benzene compare to the bond orders predicted by the Lewis structure?
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