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Chapter 10: Problem 111

The molecule benzyne \(\left(\mathrm{C}_{6} \mathrm{H}_{4}\right)\) is a very reactive species. It resembles benzene in that it has a sixmembered ring of carbon atoms. Draw a Lewis structure of the molecule and account for the molecule's high reactivity.

Short Answer

Expert verified
The Lewis structure of benzyne shows four Carbon atoms with unpaired electrons. These unstable configurations contribute to its high reactivity as these free electrons often interact with other atoms or molecules to achieve stability.

Step by step solution

01

Drawing the Lewis structure.

Starting with the carbon atoms, as it is a six-membered ring similar to benzene, draw six Carbon (C) atoms in a hexagonal pattern. The Hydrogen (H) atoms will attach to two of these Carbon (C) atoms. The other Carbon atoms are connected directly to each other without any hydrogen. Each Carbon (C) atom has 4 valence electrons and takes 4 more to complete its octet. But since there are only 4 Hydrogen (H) atoms, only two of the Carbon atoms complete their octet and the other 4 Carbon (C) atoms will have two unpaired electrons each.
02

Accounting for the high reactivity.

The remaining four Carbon atoms left with unpaired electrons will make the molecule highly reactive. These unpaired electrons are often referred to as free radicals and are significant contributors to its reactivity. These free electrons often seek to bond with other atoms or molecules to make their electron configuration stable, hence leading to reactions.

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Most popular questions from this chapter

Nitryl fluoride (FNO \(_{2}\) ) is very reactive chemically. The fluorine and oxygen atoms are bonded to the nitrogen atom. (a) Write a Lewis structure for FNO \(_{2}\). (b) Indicate the hybridization of the nitrogen atom. (c) Describe the bonding in terms of molecular orbital theory. Where would you expect delocalized molecular orbitals to form?

Discuss the basic features of the VSEPR model. Explain why the magnitude of repulsion decreases in the following order: lone pair-lone pair \(>\) lone pair- bonding pair \(>\) bonding pair-bonding pair.

Draw a molecular orbital energy level diagram for each of the following species: \(\mathrm{He}_{2}, \mathrm{HHe}, \mathrm{He}_{2}^{+}\) Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.)

Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to \(\mathrm{Fe}^{2+}\) in the hemoglobin molecule. The molecular orbitals of CO have the same energy order as those of the \(\mathrm{N}_{2} \mathrm{~mol}-\) ecule. (a) Draw a Lewis structure of \(\mathrm{CO}\) and assign formal charges. Explain why CO has a rather small dipole moment of 0.12 D. (b) Compare the bond order of CO with that from molecular orbital theory. (c) Which of the atoms (C or O) is more likely to form bonds with the \(\mathrm{Fe}^{2+}\) ion in hemoglobin?

The geometry of \(\mathrm{CH}_{4}\) could be square planar, with the four \(\mathrm{H}\) atoms at the corners of a square and the \(\mathrm{C}\) atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral \(\mathrm{CH}_{4}\) molecule.
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