Question

Write a Lewis structure for each of the following polyatomic ions. Show all bonding valence electron pairs as lines and all nonbonding valence electron pairs as dots. For those ions that exhibit resonance, draw the various possible resonance forms. a. chlorite ion, \(\mathrm{ClO}_{2}^{-}\) b. perbromate ion, \(\mathrm{BrO}_{4}^{-}\) c. cyanide ion, CN \(^{-}\)

Short answer

The Lewis structures for the given polyatomic ions are as follows: a. Chlorite ion, \(\mathrm{ClO}_{2}^{-}\): \[ \mathrm{O-Cl-O} \] \[ ..\:^{..-}\] b. Perbromate ion, \(\mathrm{BrO}_{4}^{-}\): \[ \mathrm{BrO}_{4}^{-} \] \[ .. \: ^{-} \] c. Cyanide ion, \(\mathrm{CN}^{-}\): \[ \mathrm{C \equiv N^{-}} \: \] \[ .. \: .. \]

Step-by-step solution

a. Chlorite ion, \(\mathrm{ClO}_{2}^{-}\)#

1. Count the total number of valence electrons for \(\mathrm{ClO}_{2}^{-}\). Chlorine (Cl) has 7 valence electrons, each oxygen (O) atom has 6 valence electrons, and the negative charge contributes 1 additional electron. Therefore, there are a total of 7 + 6*2 + 1 = 20 valence electrons in \(\mathrm{ClO}_{2}^{-}\). 2. Connect the central atom (Cl) to the oxygen atoms. Make a single bond between the Cl and each O atom. This uses 4 valence electrons, so we have 16 left to distribute. 3. Fulfill the octet of each atom. Start with the terminal oxygen atoms: they need 6 more electrons to fulfill their octets. This consumes 12 electrons, leaving 4 electrons. Place those 4 electrons on the central Cl atom. 4. Check the octet rule. Each oxygen atom has 8 electrons (2 from the bond with the Cl atom and 6 nonbonding electrons), and the chlorine atom has 10 electrons (2 from each O bond, and 4 nonbonding electrons). 5. Show the Lewis structure. The final structure shows a molecule with a Cl atom in the center, single bonds connecting to two O atoms, and sets of lone pairs on each oxygen atom and the central chlorine atom. The charge is indicated by the negative sign on the chlorine atom. \[ \mathrm{O-Cl-O} \] \[ ..\:^{..-}\]

b. Perbromate ion, \(\mathrm{BrO}_{4}^{-}\)#

1. Count the total number of valence electrons for \(\mathrm{BrO}_{4}^{-}\). Bromine (Br) has 7 valence electrons, each oxygen (O) atom has 6 valence electrons, and the negative charge contributes 1 additional electron. Therefore, there are a total of 7 + 6*4 + 1 = 32 valence electrons in \(\mathrm{BrO}_{4}^{-}\). 2. Connect the central atom (Br) to the oxygen atoms. Make a single bond between the Br and each O atom. This uses 8 valence electrons, so we have 24 left to distribute. 3. Fulfill the octet of each atom. Place 6 more electrons on all four O atoms to complete their octets. This consumes 24 electrons, leaving no electrons available. 4. Check the octet rule. Each oxygen atom has 8 electrons (2 from the bond with the Br atom and 6 nonbonding electrons), and the central Br atom has 8 electrons (2 from each O bond). 5. Show the Lewis structure. The final structure shows a molecule with a Br atom in the center, single bonds connecting to four O atoms, and sets of lone pairs on each oxygen atom. The charge is indicated by the negative sign on the bromine atom. \[ \mathrm{BrO}_{4}^{-} \] \[ .. \: ^{-} \]

c. Cyanide ion, \(\mathrm{CN}^{-}\)#

1. Count the total number of valence electrons for \(\mathrm{CN}^{-}\). Carbon (C) has 4 valence electrons, nitrogen (N) has 5 valence electrons, and the negative charge contributes 1 additional electron. Therefore, there are a total of 4 + 5 + 1 = 10 valence electrons in \(\mathrm{CN}^{-}\). 2. Connect the carbon (C) and nitrogen (N) atoms with a triple bond. This uses 6 of the 10 valence electrons, leaving 4 to distribute. 3. Fulfill the octet of each atom. Add 3 more electrons around the N atom and 1 electron around the C atom. No more electrons are left. 4. Check the octet rule. Carbon has 8 electrons (6 from the triple bond with N and 2 nonbonding electrons), and nitrogen has 8 electrons (6 from the triple bond with C and 2 nonbonding electrons). 5. Show the Lewis structure. The final structure shows a molecule with a C atom connected to an N atom by a triple bond and the charge on the nitrogen atom. \[ \mathrm{C \equiv N^{-}} \: \] \[ .. \: .. \]

Key concepts

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. They are crucial for understanding chemical bonding because they are the electrons that participate in forming chemical bonds. For polyatomic ions such as the chlorite ion \(\mathrm{ClO}_{2}^{-}\), perbromate ion \(\mathrm{BrO}_{4}^{-}\), and cyanide ion \(\mathrm{CN}^{-}\), counting the total number of valence electrons is the first step in drawing a Lewis structure.

In the case of the chlorite ion, we consider the 7 valence electrons from chlorine, 12 from the two oxygen atoms (6 each), and 1 additional electron due to the negative charge, resulting in 20 valence electrons. Similarly, perbromate ion has 32 valence electrons and cyanide has 10. These valence electrons are then arranged to form bonds and lone pairs, ensuring that each atom (with some exceptions) reaches a stable electronic configuration, often resembling that of the noble gases.

Resonance Forms

Resonance forms refer to the different possible arrangements of electrons (particularly the valence electrons) that depict the same molecular structure. Different resonance structures are used to represent molecules where the actual distribution of electrons is a hybrid or an average of the resonance structures. In molecules like the chlorite ion, there can be multiple valid ways to place the double bonds and lone pairs.

Electrons are not static; they move around and can be shared in different ways between atoms, which can't be fully captured by a single Lewis structure. Therefore, resonance forms are supposed to illustrate the flexibility in electron sharing. Each resonance form individually obeys the rules of Lewis structures but none of them alone accurately depicts the true nature of the ion. Students should consider all possible resonance forms, especially for molecules that do not fit exactly into simple structural guidelines.

Octet Rule

The octet rule is a principle that states that atoms tend to bond in such a way that they each have eight electrons in their valence shell, giving them the same electronic configuration as a noble gas. The rule is a useful guideline for predicting and explaining the majority of chemical bonding situations.

In our examples, every attempt is made to give each atom in \(\mathrm{ClO}_{2}^{-}\), \(\mathrm{BrO}_{4}^{-}\), and \(\mathrm{CN}^{-}\) an octet. This is achieved through the arrangement of electrons into bonding and non-bonding pairs. Chlorine in \(\mathrm{ClO}_{2}^{-}\) is shown to exceed the octet, which is acceptable for elements in the third period and beyond as they have d-orbitals that can accommodate extra electrons. The octet rule is not absolute, but rather a helpful starting point for understanding the distribution of valence electrons in a Lewis structure.