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Chapter 8: Problem 76

Draw the Lewis structure for \(\mathrm{NO}^{+}\). Is the nitrogen-oxygen bond in \(\mathrm{NO}^{+}\) longer, shorter, or the same length as the nitrogen-oxygen bond in NO? Explain.

Short Answer

Expert verified
The Lewis structure for \(\mathrm{NO}^{+}\) is N=O with lone pairs on each atom. The nitrogen-oxygen bond in \(\mathrm{NO}^{+}\) is longer than the nitrogen-oxygen bond in NO, as the \(\mathrm{NO}^{+}\) ion has a double bond compared to the partial triple bond in neutral NO, which results in a shorter bond length due to its higher bond order.

Step by step solution

01

Determine the total number of valence electrons

In order to draw the Lewis structure, first we need to find the total number of valence electrons for \(\mathrm{NO}^{+}\). Nitrogen has 5 valence electrons, and Oxygen has 6. Since the ion has a positive charge, we need to remove one electron from the total valence electrons: Total valence electrons = 5 (from N) + 6 (from O) - 1 (charge of the ion) = 10 electrons.
02

Draw a skeleton structure

Now, we'll create a skeleton structure with Nitrogen and Oxygen atoms connected by a single bond. This is just a starting-point; we'll modify it as needed. We are using a single bond initially because it corresponds to the lowest bond order. ``` N - O ```
03

Distribute the remaining electrons

We used 2 of the 10 valence electrons to form the single bond between Nitrogen and Oxygen. We have 8 more electrons to distribute. Distribute these electrons as lone pairs around the atoms to fulfill the octet rule, which states that main-group elements will typically want 8 electrons in their valence shell. First, we'll complete the Oxygen atom's octet: ``` N - O :: ``` Now Oxygen has 8 electrons (2 in the single bond and 6 as lone pairs), so its octet is complete. We are left with 2 valence electrons, which we will place on the Nitrogen atom as a lone pair: ``` N - O : :: ```
04

Check and optimize the structure

At this point, we need to check if the octet rule is satisfied for both atoms. The Oxygen atom has a complete octet (a single bond and 3 lone pairs), but the Nitrogen atom does not (a single bond, 1 lone pair, and 1 unpaired electron). To address this, we can form a double bond between the Nitrogen and Oxygen atoms by moving one lone pair from Oxygen and sharing it with Nitrogen. Now Nitrogen has a complete octet. The optimized Lewis structure for \(\mathrm{NO}^{+}\) is: ``` N = O : : ```
05

Compare bond lengths

Now that we have the optimized Lewis structure for \(\mathrm{NO}^{+}\), we can compare its bond length to that of NO. For \(\mathrm{NO}^{+}\), we have a double bond, whereas in a neutral NO molecule, there is a partial triple bond (in reality NO has a bond order of 2.5, but for simplicity, we can compare a double bond to a triple bond). In general, shorter bond lengths are associated with higher bond orders. Since a triple bond is stronger than a double bond, the nitrogen-oxygen bond in the neutral NO molecule is shorter than the nitrogen-oxygen bond in \(\mathrm{NO}^{+}\). In conclusion, the nitrogen-oxygen bond in \(\mathrm{NO}^{+}\) is longer than the nitrogen-oxygen bond in NO.

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Most popular questions from this chapter

The iodine monobromide molecule, IBr, has a bond length of \(249 \mathrm{pm}\) and a dipole moment of \(1.21 \mathrm{D} .(\mathbf{a})\) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, \(e\).

Compare the \(\mathrm{O}-\mathrm{O}\) bond length in \(\mathrm{O}_{2}\) versus \(\mathrm{H}_{2} \mathrm{O}_{2}\). Which bond is shorter? Why?

There are many Lewis structures you could draw for sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (each \(\mathrm{H}\) is bonded to an \(\mathrm{O}\) ). (a) What Lewis structure(s) would you draw to satisfy the octet rule? (b) What Lewis structure(s) would you draw to minimize formal charge?

For each of the following molecules or ions of sulfur and oxygen, write a single Lewis structure that obeys the octet rule, and calculate the oxidation numbers and formal charges on all the atoms: (a) \(\mathrm{SO}_{2}\), (b) \(\mathrm{SO}_{3}\) (c) \(\mathrm{SO}_{3}^{2-}\). (d) Arrange these molecules/ions in order of increasing \(\mathrm{S}-\mathrm{O}\) bond length.

Which of these elements are unlikely to form ionic bonds? $\mathrm{Mg}, \mathrm{Al}, \mathrm{Si}, \mathrm{Br}, \mathrm{I}$.
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