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Chapter 11: Problem 57

The Lewis structure of \(\mathrm{N}_{2}\) indicates that the nitrogento-nitrogen bond is a triple covalent bond. Other evidence suggests that the \(\sigma\) bond in this molecule involves the overlap of \(s p\) hybrid orbitals. (a) Draw orbital diagrams for the N atoms to describe bonding in \(\mathrm{N}_{2}\) (b) Can this bonding be described by either \(s p^{2}\) or \(s p^{3}\) hybridization of the \(\mathrm{N}\) atoms? Can bonding in \(\mathrm{N}_{2}\) be described in terms of unhybridized orbitals? Explain.

Short Answer

Expert verified
The triple bond in N2 is formed by one sigma bond and two pi bonds which are the result of sp hybridization of N atoms. The bonding in N2 cannot be accurately described using only unhybridized orbitals.

Step by step solution

01

Draw the Lewis Structure for N2

The nitrogen atom has 5 valence electrons. As there are two nitrogen atoms, we have a total of 10 electrons for N2. We'll put two electrons in the nitrogen-nitrogen bond forming a single covalent bond and then place pairs of electrons around each nitrogen atom. After fulfilling the octet for each nitrogen atom, we still have 2 more electrons, which we will place in the bond, forming a triple covalent bond. The resulting Lewis structure of N2 shows a triple bond between two nitrogen atoms.
02

Draw Orbital Diagrams for N atoms

Each Nitrogen atom (N) has two 1s, two 2s, and three 2p electrons. The 1s and 2s are core electrons and do not participate in bonding. The 2p electrons do. Nitrogen's 2s and three 2p orbitals hybridize (combine) to form four identical orbitals. These orbitals align themselves in space in such a way as to minimize repulsion between their electron clouds. Three of these orbitals form sigma bonds with a hydrogen atom's 1s orbital. The fourth hybrid orbital holds the remaining pair of nonbonding electrons.
03

Discuss the Type of Hybridization in N2

In N2, the bonding cannot be described by either sp2 or sp3 hybridization. Instead, it can be described by sp hybridization. This is because when a bond is formed, the s orbital and one p orbital of each nitrogen atom hybridize to form two sp hybrid orbitals. These sp orbitals overlap to form the sigma bond in N2. The remaining p orbitals on each nitrogen form two pi bonds. Therefore, the triple bond in N2 is composed of one sigma bond and two pi bonds, and is the result of sp hybridization. Also, it's important to note that the bonding in N2 cannot be accurately described using only unhybridized orbitals since the bonding primarily involves the sp hybrid orbitals.

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Most popular questions from this chapter

Think of the reaction shown here as involving the transfer of a fluoride ion from \(\mathrm{ClF}_{3}\) to \(\mathrm{AsF}_{5}\) to form the ions \(\mathrm{ClF}_{2}^{+}\) and \(\mathrm{AsF}_{6}^{-}\). As a result, the hybridization scheme of each central atom must change. For each reactant molecule and product ion, indicate (a) its geometric structure and (b) the hybridization scheme for its central atom. $$ \mathrm{ClF}_{3}+\mathrm{AsF}_{5} \longrightarrow\left(\mathrm{ClF}_{2}^{+}\right)\left(\mathrm{AsF}_{6}^{-}\right) $$

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.

Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\operatorname{COS} ;\) (c) \(\operatorname{SiCl}_{4} ;\) (d) \(\mathrm{NO}_{3}^{-}\);(e) AsF \(_{5}\)

Explain the essential difference in how the valencebond method and molecular orbital theory describe a covalent bond.

Explain why the concept of delocalized molecular orbitals is essential to an understanding of bonding in the benzene molecule, $\mathrm{C}_{6} \mathrm{H}_{6}$
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