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Chapter 11: Problem 69

The molecule formamide, \(\mathrm{HCONH}_{2}\), has the approximate bond angles \(\mathrm{H}-\mathrm{C}-\mathrm{O}, 123^{\circ} ; \mathrm{H}-\mathrm{C}-\mathrm{N}, 113^{\circ}\) \(\mathrm{N}-\mathrm{C}-\mathrm{O}, 124^{\circ} ; \mathrm{C}-\mathrm{N}-\mathrm{H}, 119^{\circ} ; \mathrm{H}-\mathrm{N}-\mathrm{H}, 119^{\circ}\) The \(\mathrm{C}-\mathrm{N}\) bond length is \(138 \mathrm{pm}\). Two Lewis structures can be written for this molecule, with the true structure being a resonance hybrid of the two. Propose a hybridization and bonding scheme for each structure.

Short Answer

Expert verified
The bonding scheme for the first structure involves an \(sp^2\) hybridization at the Carbon (C) forming a sigma bond with Nitrogen (N) which is \(sp^3\) hybridized in this case and a sigma bond and pi bond with oxygen which is \(sp^2\) hybridized. In case of the second structure, the hybridization at Carbon (C) is still \(sp^2\), but it forms a sigma bond with the \(sp^2\) hybridized Nitrogen (N) and sigma bond with the \(sp^3\) hybridized oxygen.

Step by step solution

01

Draw the Lewis Structures

Lewis structure for Formamide, \(HCONH_{2}\) can function in two ways: with oxygen double bonded to carbon and nitrogen single bonded to carbon, or with nitrogen double bonded to carbon and the oxygen single bonded to carbon. In both cases, each hydrogen atom forms a single bond with either carbon or nitrogen. In Lewis notation: 1. Structure 1: \(H-C=O / \backslash NH_{2}\) 2. Structure 2: \(H-C-N=O / \backslash H\)
02

Proposed hybridization scheme for Structure 1

In this case, the carbon atom is bonded to oxygen through a double bond and to nitrogen via a single bond. This suggests that the hybridization at Carbon (C) is \(sp^2\), with one of the \(sp^2\) orbitals forming a sigma bond with the nitrogen's \(sp^3\) hybrid orbital (N has an \(sp^3\) because it forms three sigma bonds and one lone pair), and two \(sp^2\) orbitals and the remaining p orbital overlaping with oxygen's orbitals to form a sigma and two pi bonds respectively. Thus, for Oxygen (O), the hybridization is \(sp^2\) (two sigma bonds and one lone pair), while for Nitrogen (N), it is \(sp^3\).
03

Proposed hybridization scheme for Structure 2

In the second structure, the carbon atom is bonded to nitrogen through a double bond and to oxygen via a single bond. This suggests that the hybridization at Carbon (C) is \(sp^2\), much like in Structure 1. For Nitrogen (N), one of the \(sp^2\) orbitals forms a sigma bond with the carbon's \(sp^2\) hybrid orbital, and two \(sp^2\) orbitals along with the remaining p orbital overlap with the hydrogen's orbitals to form sigma bonds, hence the Nitrogen (N) is also \(sp^2\) hybridized. In case of oxygen, it forms two sigma bonds (with carbon and hydrogen) and therefore is \(sp^3\) hybridized.

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Most popular questions from this chapter

A conjugated hydrocarbon has an alternation of double and single bonds. Draw the molecular orbitals of the \(\pi\) system of 1,3,5 -hexatriene. If the energy required to excite an electron from the HOMO to the LUMO corresponds to a wavelength of \(256 \mathrm{nm},\) do you expect the wavelength for the corresponding excitation in 1,3,5,7 -octatetraene to be a longer or shorter wavelength?

For each of the species \(\mathrm{C}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{F}_{2}^{+},\) and \(\mathrm{NO}^{+}\) (a) Write the molecular orbital diagram (as in Example \(11-6)\) (b) Determine the bond order, and state whether you expect the species to be stable or unstable. (c) Determine if the species is diamagnetic or paramagnetic; and if paramagnetic, indicate the number of unpaired electrons.

Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.

Magnesium is an excellent electrical conductor even though it has a full \(3 s\) subshell with the electron configuration: [Ne]3s^. Use band theory to explain why magnesium conducts electricity.

Why does the hybridization \(s p^{3} d\) not account for bonding in the molecule BrF \(_{5} ?\) What hybridization scheme does work? Explain.
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