Question

Write Lewis structures for the following. Show all resonance structures where applicable. a. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}, \mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{~N}_{2} \mathrm{O}_{4}\right.\) exists as \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{NO}_{2} .\) ) b. \(\mathrm{OCN}^{-}, \mathrm{SCN}^{-}, \mathrm{N}_{3}^{-}\) (Carbon is the central atom in \(\mathrm{OCN}^{-}\) and \(\left.\mathrm{SCN}^{-} .\right)\)

Short answer

The Lewis structures and their resonance structures for the given molecules and ions are: a. For $\mathrm{NO}_{2}^{-}$: \[\begin{array}{cc} \mathrm{O}=\mathrm{N}-\mathrm{O}^{-} \longleftrightarrow \mathrm{O}^{-}-\mathrm{N}=\mathrm{O} \end{array}\] For $\mathrm{NO}_{3}^{-}$: \[\begin{array}{ccc} \mathrm{O}=\mathrm{N}-\mathrm{O}^{-} -\mathrm{O}^{-} & \longleftrightarrow & \mathrm{O}^{-}-\mathrm{N}=\mathrm{O} -\mathrm{O}^{-} \\ & \longleftrightarrow & \mathrm{O}^{-n}-\mathrm{O}^{-}-\mathrm{N}=\mathrm{O} \end{array}\] For $\mathrm{N}_{2}\mathrm{O}_{4}$: \[\begin{array}{c} (\mathrm{O}^{2-}\equiv \mathrm{N}^{\oplus}) - \mathrm{O} - \mathrm{N}^{\oplus}=\mathrm{O}^{-} \end{array}\] b. For $\mathrm{OCN}^{-}$: \[\mathrm{O} \equiv \mathrm{C} \equiv \mathrm{N}^{-}\] For $\mathrm{SCN}^{-}$: \[\mathrm{S}=\mathrm{C} \equiv \mathrm{N}^{-}\] For $\mathrm{N}_{3}^{-}$: \[\begin{array}{cc} \mathrm{N}^-=\mathrm{N}+\mathrm{N}^{2-}\equiv \longleftrightarrow \mathrm{N}^{2-}\equiv \mathrm{N}^{+}-\mathrm{N}^- \end{array}\]

Step-by-step solution

Draw the Lewis structures for NO2- and its resonance

To draw the Lewis structure for NO2-, we need to first determine the total number of valence electrons. Nitrogen contributes 5 electrons, and each Oxygen atom contributes 6 electrons. Since it is an anion with a -1 charge, we add one more electron. This gives us a total of 18 electrons. 1. Place Nitrogen atom at the center and surround it with two Oxygen atoms. 2. Connect each Oxygen atom to Nitrogen with a single bond, using 2 electrons each. This leaves 14 electrons. 3. Complete the octet for each Oxygen atom by adding 6 more electrons to them. This would consume 12 electrons, leaving 2 electrons. 4. Add the remaining 2 electrons to Nitrogen atom as a lone pair. Now, since the Nitrogen atom does not have a complete octet, the structure can benefit from resonance. We would move one of the double bonds in the Oxygen atoms to the central Nitrogen atom, creating a double bond. The two resonance structures of NO2- are: \[\begin{array}{cc} \mathrm{O}=\mathrm{N}-\mathrm{O}^{-} \longleftrightarrow \mathrm{O}^{-}-\mathrm{N}=\mathrm{O} \end{array}\]

Draw the Lewis structures for NO3- and its resonance

For NO3-, Nitrogen contributes 5 electrons, and each Oxygen atom contributes 6 electrons. As it is an anion with a -1 charge, we add one more electron. A total of 24 electrons are available. 1. Place Nitrogen atom at the center, surrounded by three Oxygen atoms. 2. Connect each Oxygen atom to Nitrogen with a single bond, using 2 electrons each. This leaves 18 electrons. 3. Complete the octet for each Oxygen atom by adding 6 more electrons to them, consuming 18 electrons. Now, since the Nitrogen atom has only 6 valence electrons, we can use resonance to form double bonds with the Oxygen atoms. The three resonance structures of NO3- are: \[\begin{array}{ccc} \mathrm{O}=\mathrm{N}-\mathrm{O}^{-} -\mathrm{O}^{-} & \longleftrightarrow & \mathrm{O}^{-}-\mathrm{N}=\mathrm{O} -\mathrm{O}^{-} \\ & \longleftrightarrow & \mathrm{O}^{-n}-\mathrm{O}^{-}-\mathrm{N}=\mathrm{O} \end{array}\]

Draw the Lewis structure for N2O4

For N2O4, follow these steps: 1. Place two Nitrogen atoms at the center connected to each other, surrounded by Oxygen atoms. Since N2O4 exists as O2N-NO2, we can break it down into two NO2 molecules and draw their structures. 2. Draw the structure of each NO2 molecule as shown in Step 1: O-N=O with a +1 charge on the central Nitrogen and -1 charge on the doubly bonded Oxygen. This creates O=N-O and O-N=O 3. Now, combine these two NO2 molecules to form N2O4 by connecting one of the singly bonded Oxygen atoms from each NO2 molecule. The Lewis structure for N2O4 is: \[\begin{array}{c} (\mathrm{O}^{2-}\equiv \mathrm{N}^{\oplus}) - \mathrm{O} - \mathrm{N}^{\oplus}=\mathrm{O}^{-} \end{array}\]

Draw the Lewis structures for OCN-, SCN-, and N3-

For OCN-, arrange the atoms in the order O-C-N with Carbon in the center. Connect each atom with triple bonds, satisfying the octet rule for all atoms. This gives a Lewis structure of O≡C≡N-. For SCN-, arrange the atoms in the order S-C-N with Carbon in the center. Connect a triple bond between C and N, and a double bond between C and S. The Lewis structure is S=C=N-. For N3-, follow these steps: 1. Place two Nitrogen atoms connected to each other, and have one of them connected to an extra Nitrogen atom. 2. Double bond one Nitrogen pair and triple bond the other pair. 3. Use a formal charge of -1 on the Nitrogen atom involved in the double bond, to account for the electron count as well as giving each nitrogen atom a stable octet. The Lewis structure, including its resonance, for N3- is: \[\begin{array}{cc} \mathrm{N}^-=\mathrm{N}+\mathrm{N}^{2-}\equiv \longleftrightarrow \mathrm{N}^{2-}\equiv \mathrm{N}^{+}-\mathrm{N}^- \end{array}\]

Key concepts

Resonance Structures

In chemistry, the concept of resonance structures provides a more accurate depiction of a molecule's electronic structure when a single Lewis structure is insufficient. Resonance structures are alternative Lewis structures for the same molecule, where the arrangement of atoms does not change, but the placement of electrons does. It is essential to remember that these structures are not isomers—they do not represent different compounds but different potential electron distributions within the same compound.

For example, in nitrite (O3-), we don't have just one static Lewis structure. Instead, the double bond can be located between the nitrogen atom and any one of the oxygen atoms. The actual molecule exists as a hybrid of these possibilities, with properties that are average of the resonance structures. It's crucial to use arrows (↔) to denote that the structures are resonating, indicating electron delocalization which stabilizes the molecule.

Importance in Understanding Molecules

Resonance is vital because it affects molecular properties such as shape, polarity, and reactivity. By understanding resonance structures, students can predict and explain a molecule's behavior in different chemical processes.

Valence Electrons

The valence electrons are the electrons in the outermost shell of an atom and are crucial in determining an element's chemical properties, including its reactivity and its ability to form bonds with other elements. The valence electrons are represented as dots in Lewis structures, symbolizing potential bond sites.

For example, in the step-by-step solution we provided for the nitrite (O3-) ion, we count the total number of valence electrons to decide how they can be distributed to give each atom a full octet, if possible. The counting of valence electrons must consider the molecule's overall charge, as seen when one extra electron is added for the negative charge of the ion. Knowledge of valence electrons is also crucial when predicting the formation of chemical bonds during the creation of compounds.

Key to Drawing Lewis Structures

Identifying valence electrons accurately is the first step in drawing a correct Lewis structure; it determines how atoms within a molecule are connected and reflects the molecule's actual electron distribution.

Octet Rule

The octet rule is a chemical guideline which posits that atoms tend to form bonds in such a way that each atom has eight electrons in its valence shell, hence gaining a configuration similar to noble gases. The octet rule is fundamental to understanding the formation of chemical bonds in main group elements and is especially relevant when drawing Lewis structures for molecules and polyatomic ions.

However, it's important to acknowledge exceptions to this rule, such as molecules with an odd number of electrons, molecules in which one or more atoms possess fewer or more than eight electrons, and molecules with extended octets involving elements from the third period onward on the periodic table.

Exception to the Rule

When dealing with ions like O2- or 3-, we see examples where the octet rule is not strictly followed: the central nitrogen in O2- obtains a full octet only through resonance structures, and 3- has a nitrogen atom with a non-typical electron configuration. Understanding these nuances helps in accurately representing molecules and predicting their chemical behavior.