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Chapter 17: Problem 3

How was the \(\mathrm{N}-\mathrm{O}\) bond order of \(4 / 3\) calculated for \(\mathrm{NO}_{3}^{-}\) ?

Short Answer

Expert verified
The bond order of \(NO_{3}^{-}\) is calculated as 2.

Step by step solution

01

Formulation of Lewis structure

First, draw the Lewis structure for \(NO_{3}^{-}\). In this case, the nitrogen atom (N) is the central atom. It is linked to three oxygen atoms (O). The extra electron imparted by the negative charge is present on the nitrogen atom.
02

Counting of Electrons

Count the number of electrons involved in bonding. Bonding involves the sharing of electrons between atoms. A single bond involves 2 electrons, a double bond involves 4 electrons and a triple bond involves 6 electrons. Here, each Nitrogen-Oxygen bond is a double bond, so each has 4 electrons involved. Therefore, there are \(4*3 = 12\) bonding electrons in \(NO_{3}^{-}\).
03

Calculation of Bond Order

The formula for bond order is \[(Number\ of\ bonding\ electrons - Number\ of\ antibonding\ electrons) / 2\] In this case, there are no antibonding electrons. So, the bond order of \(NO_{3}^{-}\) is \[(12 - 0) / 2 = 6\] However, because there are 3 N-O bonds, the bond order per N-O bond in \(NO_{3}^{-}\) is \(6 / 3 = 2\).

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Most popular questions from this chapter

Why is a resonance hybrid representation of \(\mathrm{NO}_{3}^{-}\) better than just a single structure?

Why can \(\mathrm{C}, \mathrm{N}, \mathrm{O}\) and \(\mathrm{F}\) accommodate only eight electrons (sum of the bonding and lone-pair electrons)?

Why can atoms in the third (or greater) period accommodate more than eight electrons (sum of the bonding and nonbonding electrons)?
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