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Chapter 8: Q7 E (page 446)

A friend tells you \({{\rm{N}}_{\rm{2}}}\)has three \({\rm{\pi }}\)bonds due to overlap of the three p-orbitals on each N atom. Do you agree?

Short Answer

Expert verified

In the given structure; \({N_2},N \equiv N\), there is one sigma and two pi bonds because one sigma bond is formed by the axial overlapping of two p-orbital of each nitrogen atom. After the formation of one sigma bond; remaining p-orbital overlaps by sidewise and will from two pi bond also.

Step by step solution

01

Definition of Concept

The sigma bond is formed when atomic orbitals involved in the formation of a molecular orbital overlap along the internuclear axis.

The bond formed is known as a pi-bond when p-orbitals involved in bond formation overlap sideways.

02

Explain \({{\rm{N}}_{\rm{2}}}\) has three \({\rm{\pi }}\) bonds due to overlap of the three p-orbitals on each N atom

Considering the given information:

\({N_2},N \equiv N\)

Only a single bond is always a sigma bond, whereas a double bond has one sigma bond and one pi bond, and a triple bond has one sigma bond and two pi bonds.

Therefore, one sigma bond is formed by the axial overlapping of two p-orbitals of each nitrogen atom, the structure \({N_2},N \equiv N\) has one sigma and two pi bonds. Following the formation of one sigma bond, the remaining p-orbital overlaps sideways, resulting in the formation of two pi bonds.

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Most popular questions from this chapter

Use valence bond theory to explain the bonding in\({{\rm{O}}_{\rm{2}}}\). Sketch the overlap of the atomic orbitals involved in the bonds in \({{\rm{O}}_{\rm{2}}}\).

In addition to \({\rm{N}}{{\rm{F}}_{\rm{3}}}\), two other fluoro derivatives of nitrogen are known: \({{\rm{N}}_{\rm{2}}}{{\rm{F}}_{\rm{4}}}\) and \({{\rm{N}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\). What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?

Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) \({\rm{ClNO}}\) (\({\rm{N}}\) is the central atom) (b) \({\rm{C}}{{\rm{S}}_{\rm{2}}}\) (c) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{CO}}\) (\({\rm{C}}\) is the central atom) (d) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{SO}}\) (\({\rm{S}}\) is the central atom) (e) \({\rm{S}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{S}}\) is the central atom) (f) \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{Xe}}\) is the central atom) (g) \({\rm{ClO}}{{\rm{F}}_{\rm{2}}}^{\rm{ + }}\) (\({\rm{Cl}}\) is the central atom).

Strike-anywhere matches contain a layer of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) and a layer of \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\). The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) contains the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion. \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) is an unusual molecule with the skeletal structure.

  1. Write Lewis structures for \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion.
  2. Describe the geometry about the \({\rm{P}}\) atoms, the \({\rm{S}}\) atom, and the \({\rm{Cl}}\) atom in these species.
  3. Assign a hybridization to the \({\rm{P}}\) atoms, the \({\rm{S}}\)atom, and the \({\rm{Cl}}\) atom in these species.
  4. Determine the oxidation states and formal charge of the atoms in \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion.

The main component of air is N2. From the molecular orbital diagram of N2, predict its bond order and whether it is diamagnetic or paramagnetic.
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