Question

(a) Draw the best Lewis structure(s) for the nitrite ion, NO \(_{2}^{-}\) (b) With what allotrope of oxygen is it isoelectonic? (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

The nitrite ion (NO₂⁻) has two resonance structures: O-N=O and O=N-O. The isoelectronic allotrope of oxygen is ozone (O₃). The bond lengths in NO₂⁻ are shorter than in N-O single bonds but longer than in N=O double bonds due to the delocalized bonding having a bond order of approximately 1.5.

Step-by-step solution

(a) Drawing the Lewis Structure for NO₂⁻

To draw the Lewis structure for the nitrite ion, we need to consider the total number of valence electrons and arrange the atoms accordingly. 1. Count Valence Electrons: Nitrogen (N) has 5 valence electrons, Oxygen (O) has 6 valence electrons, and there is one additional electron due to the negative charge of the ion. So, the total number of valence electrons is (5 + 2*6 + 1) = 18. 2. Arrange the Atoms: Put the least electronegative atom in the center (N), and the remaining atoms (O) around it. Connect the central atom to surrounding atoms with single bonds. 3. Add Lone Pairs: Complete the octets for the surrounding atoms (O) by adding lone pairs. 4. Add Multiple Bonds: If the central atom (N) does not have a complete octet, create multiple bonds with one or more surrounding atoms. In the case of NO₂⁻, there are two possible resonance structures, which are equally important for the description of the bonding in the ion. The resulting two resonance structures can be represented as follows: O-N=O <-----> O=N-O

(b) Identifying the Isoelectronic Allotrope of Oxygen

An allotrope of oxygen that is isoelectronic with NO₂⁻ should have the same number of total electrons. Since NO₂⁻ has 5+6+6+1=18 electrons, the oxygen allotrope must also have 18 electrons. Ozone (O₃) is found to be isoelectronic with NO₂⁻, as each oxygen atom has 6 electrons, and hence O₃ has 3*6 = 18 electrons.

(c) Predicting the Bond Lengths in NO₂⁻ Relative to N-O Single and Double Bonds

Since NO₂⁻ has two resonance structures, O-N=O <-----> O=N-O, it shows that the N-O bonds are delocalized, which means they are somewhere between a single bond (N-O) and a double bond (N=O). A single bond (N-O) will typically be longer than a double bond (N=O), because there is only one shared pair of electrons in a single bond, causing less electrostatic attraction between the atoms. In NO₂⁻, the delocalized bonds mean that the electrons are shared over both N-O bonds, giving each bond a bond order of about 1.5. This partial double bond character will make the bond lengths in NO₂⁻ shorter than N-O single bonds, but longer than N=O double bonds.

Key concepts

Resonance Structures

Resonance structures are alternative ways of drawing a molecule or ion where the chemical connectivity remains the same, but the arrangement of electrons differs. For the nitrite ion ( O _{2}^{-} ), resonance structures help depict a more accurate picture of how electrons are distributed across the molecule.

Instead of choosing a single Lewis structure to represent O _{2}^{-} , we consider two resonance structures:

• The first: O-N=O
• The second: O=N-O

In both structures, the position of atoms doesn't change; only the location of the electron pairs changes.

This results in delocalization, where electrons are spread across different bonds or parts of the molecule. Delocalization is key in resonance because it often lends stability to the molecule beyond what any single structure could provide.

Valence Electrons

Valence electrons are those located in the outermost shell of an atom and are crucial in forming chemical bonds. To construct Lewis structures, it is essential to count all valence electrons properly.

For the nitrite ion ( O _{2}^{-} ), we calculate valence electrons by summing the contributions from each atom:

• Nitrogen (N): 5 valence electrons
• Two Oxygen atoms (O): each has 6 valence electrons, totaling 12
• Additional electron from the negative charge: 1

Thus, the total number of valence electrons available is 18.

These valence electrons are distributed in a manner that satisfies the octet rule, meaning most atoms will strive to have eight electrons in their outer shell. Resonance structures arise when there are multiple valid ways to distribute these electrons while still adhering to the octet rule.

Isoelectronic Species

Isoelectronic species are molecules or ions that have the same number of electrons or similar electronic structures. This concept is valuable for predicting properties and behaviors in chemical reactions.

For instance, NO _{2}^{−} is isoelectronic with ozone (O₃). Both have 18 electrons:

• NO _{2}^{−} has the combined electrons from nitrogen (5), two oxygens (12), and the extra electron due to the negative charge, making a total of 18
• Ozone (O₃) has three oxygen atoms, each contributing 6 electrons, thus also totaling 18

Being isoelectronic often implies similar bonding features or reactivity patterns, even when the atoms or molecules themselves differ.

Bond Lengths

Bond lengths refer to the average distance between the nuclei of two atoms bonded together in a molecule. These lengths provide insight into the strength and character of the bond.

In the nitrite ion ( O _{2}^{-} ), resonance indicates that both nitrogen-oxygen bonds are neither pure single bonds nor pure double bonds. Instead, they exhibit a mix, with characteristics between single and double bonds.

• N-O single bonds are longer due to lesser electron sharing
• N=O double bonds are shorter due to a greater share of electrons
• In O _{2}^{-} , due to resonance, bonds have a partial double character, making their length intermediate

This intermediate bond length is often quantified with a bond order, which is approximately 1.5 for each N-O bond in the ion. Consequently, the bond strength and length in the nitrite ion are uniquely defined by its resonance structures.