Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Draw the best Lewis structure(s) for the nitrite ion, \(\mathrm{NO}_{2}^{-}\). (b) With what allotrope of oxygen is it isoelectronic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds and double bonds?

Short Answer

Expert verified
(a) The best Lewis structures for the nitrite ion, \(\mathrm{NO}_{2}^{-}\), are two resonance structures: (i) N atom connected to one O atom with a double bond, and to another O atom with a single bond, which has an extra lone pair of electrons, and (ii) N atom connected to one O atom with a single bond with an extra lone pair of electrons, and another O atom with a double bond. Both of these structures satisfy the octet rule. (b) The peroxide ion (O\(_2^{2-}\)) is isoelectronic with the nitrite ion, as they both have 18 electrons. (c) In the nitrite ion, the N-O bond lengths will be intermediate between the single bond length found in \(\mathrm{N}-\mathrm{O}\) single bond compounds and the double bond length found in \(\mathrm{N}=\mathrm{O}\) double bond compounds.

Step by step solution

01

Drawing the Lewis structure for nitrite ion

To determine the best Lewis structure(s) for \(\mathrm{NO}_{2}^{-}\), we will first count the total number of valence electrons. Nitrogen has 5 valence electrons, each oxygen atom has 6, and there is an additional electron due to the negative charge. Therefore, the total number of valence electrons is 5 + 6(2) + 1 = 18. Now, we will construct different resonance structures for \(\mathrm{NO}_{2}^{-}\) and verify if they fulfill the octet rule. Two possible structures are: (i) N atom connected to one O atom with a double bond, and to another O atom with a single bond, which has an extra lone pair of electrons. (ii) N atom connected to one O atom with a single bond with an extra lone pair of electrons, and another O atom with a double bond. Both of these structures satisfy the octet rule.
02

Determine the isoelectronic allotrope of oxygen

Isoelectronic species have the same number of electrons. In this case, we have \(5 + 6(2) + 1 = 18\) electrons in the nitrite ion. Now, we need to check which allotrope of oxygen has the same number of electrons. Oxygen exists in various allotropic forms such as O\(_2\) (16 electrons) and O\(_3\) (ozone, 24 electrons). None of these molecules have 18 electrons. However, if we consider a superoxide ion (O\(_2^{-}\)), it has 16 + 1 = 17 electrons, and a peroxide ion (O\(_2^{2-}\)) has 16 + 2 = 18 electrons. Therefore, the peroxide ion (O\(_2^{2-}\)) is isoelectronic with the nitrite ion.
03

Predicting the bond lengths of the nitrite ion

We have two possible Lewis resonance structures for \(\mathrm{NO}_{2}^{-}\), and the actual molecule is a hybrid of these two structures. In resonance hybrid structure, bond lengths are intermediate between single and double bonds. For the nitrite ion, the N-O bond lengths will be intermediate between the single bond length found in \(\mathrm{N}-\mathrm{O}\) single bond compounds and the double bond length found in \(\mathrm{N}=\mathrm{O}\) double bond compounds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial in determining how atoms interact with each other. For example, when drawing Lewis structures, valence electrons help us predict the bonding and geometry of molecules. The goal is to show how these electrons are used to bond atoms together and fulfill the octet rule, which suggests that atoms are most stable with eight electrons in their outer shell.
In the case of the nitrite ion, \(\mathrm{NO}_2^-\), the total number of valence electrons can be calculated. Nitrogen (N) normally has 5 valence electrons, each oxygen (O) has 6, and the additional negative charge on the ion provides one more electron. This totals to 18 valence electrons: 5 from nitrogen, 12 from two oxygen atoms, and 1 from the negative charge.
This allocation of electrons helps in constructing the Lewis structures for the molecule and allows us to distribute them across the molecule to achieve stability. Each structure, whether it uses single or double bonds with different oxygen atoms, will fulfill the octet rule for each atom in the ion, ensuring stability.
Resonance Structures
Resonance structures are different ways to draw the same molecule that show the possible arrangements of electrons. They help illustrate the delocalization of electrons within a molecule when a single, definitive structure doesn't fully describe the molecule's properties.
For the nitrite ion, \(\mathrm{NO}_2^-\), there are two main resonance structures. In one structure, nitrogen forms a double bond with one oxygen atom and a single bond with the other oxygen, which carries extra electrons as lone pairs. In the second structure, the roles of the oxygen atoms are reversed. This electron sharing creates two almost equivalent structures.
It’s important to understand that the actual representation of \(\mathrm{NO}_2^-\) is a resonance hybrid, where the true electron distribution is somewhere between these two structures. This means that, in reality, the bonds have characteristics that are intermediate between single and double bonds. These resonance structures make molecules more stable because they allow electrons to be more spread out over the molecule.
Isoelectronic Species
Isoelectronic species are atoms, ions, or molecules that have the same number of electrons and hence similar electronic structures but might contain different elements. This similarity results in comparable chemical and physical properties.
In our example, the nitrite ion \(\mathrm{NO}_2^-\) is compared to an allotrope of oxygen which has a matching electron count. By counting the electrons, we find that \(\mathrm{NO}_2^-\) has 18 electrons. If we look at different oxygen species, none of them naturally have 18 electrons. However, the peroxide ion, \(\mathrm{O}_2^{2-}\), also contains 18 electrons, matching that of \(\mathrm{NO}_2^-\).
Recognizing isoelectronic pairs is essential in chemistry because it helps predict molecule shapes, bond strengths, and reactivity patterns. Even if the elements in the species differ, their electronic similarity means they might behave similarly in chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fill in the blank with the appropriate numbers for both electrons and bonds (considering that single bonds are counted as one, double bonds as two, and triple bonds as three). (a) Iodine has ___________ valence electrons and makes ___________ bond(s) in compounds. (b) Silicon has ___________ valence electrons and makes ___________ bond(s) in compounds. (c) Phosphorus has ___________ valence electrons and makes ___________ bond(s) in compounds. (d) Sulphur has ___________ valence electrons and makes ___________ bond(s) in compounds.

The substance chlorine monoxide, \(\mathrm{ClO}(g)\), is important in atmospheric processes that lead to depletion of the ozone layer. The ClO molecule has an experimental dipole moment of \(1.24 \mathrm{D},\) and the \(\mathrm{Cl}-\mathrm{O}\) bond length is \(160 \mathrm{pm} .(\mathbf{a})\) Determine the magnitude of the charges on the \(\mathrm{Cl}\) and \(\mathrm{O}\) atoms in units of the electronic charge, \(e .(\mathbf{b})\) Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule? (c) Using formal charges as a guide, propose the dominant Lewis structure for the molecule. (d) The anion ClO \(^{-}\) exists. What is the formal charge on the Cl for the best Lewis structure for \(\mathrm{ClO}^{-} ?\)

Using Lewis symbols and Lewis structures, make a sketch of the formation of \(\mathrm{NCl}_{3}\) from \(\mathrm{N}\) and \(\mathrm{Cl}\) atoms, showing valence- shell electrons. (a) How many valence electrons does N have initially? (b) How many bonds Cl has to make in order to achieve an octet? (c) How many valence electrons surround the \(\mathrm{N}\) in the \(\mathrm{NCl}_{3}\) molecule? (d) How many valence electrons surround each Cl in the \(\mathrm{NCl}_{3}\) molecule? (e) How many lone pairs of electrons are in the \(\mathrm{NCl}_{3}\) molecule?

Draw the Lewis structures for each of the following ions or molecules. Identify those in which the octet rule is not obeyed; state which atom in each compound does not follow the octet rule; and state, for those atoms, how many electrons surround them: \((\mathbf{a}) \mathrm{HCl},(\mathbf{b}) \mathrm{ICl}_{5},\) (c) \(\mathrm{NO}\) (d) \(\mathrm{CF}_{2} \mathrm{Cl}_{2},(\mathbf{e}) \mathrm{I}_{3}^{-}\)

Consider the hypothetical molecule \(\mathrm{A}-\mathrm{A}=\mathrm{A}\) with a bent shape. Are the following statements true or false? (a) This molecule cannot exist. (b) If this molecule exists, it must possess an odd electron.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free