Consider the bonding in nitrate ion, NO3^-. First draw resonance formulas of this ion. Now descr

step 1 Here we are looking at the NO3 minus. So we can look at the resonance structures. So we've got t

Consider the bonding in nitrate ion, NO3^-. First draw...

Consider the bonding in nitrate ion, $\mathrm{NO}_{3}^{-}$. First draw resonance formulas of this ion. Now describe the bonding of this ion in terms of molecular orbitals. (Refer to the delocalized bonding of the ozone molecule described in the text.) Suppose each atom uses $s p^{2}$ hybrid orbitals. How many $\pi$ molecular orbitals can you form from the $2 p$ orbitals that remain on these atoms? How many of these $\pi$ orbitals will be occupied?

Key Concepts

sp2 Hybridization

sp2 hybridization involves the mixing of one s orbital with two p orbitals to form three equivalent hybrid orbitals arranged in a plane. These sp2 orbitals are used to form the strong sigma bonds between atoms, while the remaining, unhybridized p orbital (orbitals) that lie perpendicular to this plane are available for forming pi bonds. This concept is fundamental in understanding how planar molecules such as nitrate achieve their bonding framework.

Resonance Structures

Resonance is the concept that a molecule can be represented by two or more valid Lewis structures, none of which completely describes the true electron distribution on its own. In the nitrate ion, several equivalent resonance forms can be drawn, each showing a different oxygen atom as double?bonded to the nitrogen. The true structure is a hybrid of these forms, meaning that the double?bond character is delocalized evenly over all the N–O bonds. This delocalization explains the experimentally observed uniform bond lengths.

Molecular Orbital Theory

Molecular orbital (MO) theory explains bonding in molecules by combining atomic orbitals to form molecular orbitals that extend over the entire molecule. In a system like nitrate, the unhybridized p orbitals on the nitrogen and oxygen atoms combine to give rise to a set of delocalized pi molecular orbitals. These MOs can be bonding, nonbonding, or antibonding, and their occupation by electrons determines the overall stability and bond order of the molecule.

Delocalized Pi Bonding

Delocalized pi bonding arises from the overlap of p orbitals that are not involved in sigma bonding. In the nitrate ion, each atom contributes one such p orbital; thus, four atomic p orbitals combine to form four pi molecular orbitals. The delocalization of electrons among these orbitals means that the pi electrons are spread over several atoms rather than localized between a single pair of atoms, which leads to bond orders that are fractional and bonds of equal magnitude.

Electron Counting in Delocalized Systems

In a delocalized pi system, the total number of electrons occupying the pi molecular orbitals is determined by adding the electrons available in the unhybridized p orbitals on each atom, plus any extra electrons due to ionic charge. For the nitrate ion, after each atom uses its sp2 hybrids for sigma bonding, the remaining p orbitals (four in total) combine to form four pi MOs. Electron?counting shows that six electrons are available to fill these pi orbitals—completely occupying the low?energy, bonding, and nonbonding orbitals while leaving the highest?energy, antibonding orbital empty. This electron distribution explains the closed?shell nature and equal bond order in the nitrate ion.

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