Question

Write Lewis structures for these ions. Show all valence electrons and all formal charges. (a) Amide ion, \(\mathrm{NH}_{2}{ }^{-}\) (b) Bicarbonate ion, \(\mathrm{HCO}_{3}{ }^{-}\) (c) Carbonate ion, \(\mathrm{CO}_{3}{ }^{2-}\) (d) Nitrate ion, \(\mathrm{NO}_{3}{ }^{-}\) (e) Formate ion, \(\mathrm{HCOO}^{-}\) (f) Acetate ion, \(\mathrm{CH}_{3} \mathrm{COO}^{-}\)

Short answer

Question: Draw the Lewis structures for the following ions: a) Amide ion: NH₂⁻ b) Bicarbonate ion: HCO₃⁻ c) Carbonate ion: CO₃²⁻ d) Nitrate ion: NO₃⁻ e) Formate ion: HCOO⁻ f) Acetate ion: CH₃COO⁻ Answer: a) The Lewis structure for the Amide ion (NH₂⁻) is [NH₂]⁻ with N having one single bond to each H and 2 lone pairs. b) The Lewis structure for the Bicarbonate ion (HCO₃⁻) is [HCO₃]⁻ with C having one single bond to H, one single bond to the 1st O, and one double bond to the 2nd O. c) The Lewis structure for the Carbonate ion (CO₃²⁻) is [CO₃]²⁻ with resonance between the Oxygen atoms. d) The Lewis structure for the Nitrate ion (NO₃⁻) is [NO₃]⁻ with resonance between the Oxygen atoms. e) The Lewis structure for the Formate ion (HCOO⁻) is [HCOO]⁻. f) The Lewis structure for the Acetate ion (CH₃COO⁻) is [CH₃COO]⁻, with Oxygen atoms having a negative formal charge due to resonance between the double bonds.

Step-by-step solution

Identify Valence Electrons

First, calculate the total number of valence electrons. Nitrogen has 5 valence electrons and each Hydrogen atom has 1 valence electron. Additionally, there is a 1- charge on the ion which means an extra electron is present. So, the total number of valence electrons is 5 + 1 + 1 + 1 = 8.

Draw Lewis Structure

Place the Nitrogen (N) atom in the center and connect the two Hydrogen (H) atoms to it with a single bond. Assign any remaining electrons to the central atom to satisfy the octet rule. In this case, N still needs 4 more electrons in order to have 8 (2 from the bonds to the H atoms). So, add those 4 remaining electrons as 2 lone pairs around the N atom. The resulting Lewis structure will be \(\mathrm{[NH}_{2}]^{-}\) with N having one single bond to each H and 2 lone pairs. #b)#Bicarbonate ion, \(\mathrm{HCO}_{3}{ }^{-}\)

Identify Valence Electrons

Count the total number of valence electrons. Carbon has 4, Oxygen has 6 each (total 18 for three Oxygen atoms), and Hydrogen has 1. Additionally, there is a 1- charge on the ion which means an extra electron is present. So, the total number of valence electrons is 4 + 18 + 1 + 1 = 24.

Draw Lewis Structure

Place the Carbon (C) atom in the center and connect one Oxygen (O) atom and one Hydrogen (H) atom to it using single bonds. Assign any remaining electrons to each atom to satisfy the octet rule. Carbon needs 4 more electrons, shared with 2 remaining Oxygen atoms. We end up with a single bond to the \(1^{st}\) Oxygen atom and a double bond to the \(2^{nd}\) Oxygen atom. The third Oxygen atom has another single bond to the \(1^{st}\) Oxygen atom and a single bond to the Hydrogen atom. The resulting Lewis structure will be \(\mathrm{[HCO}_{3}]^{-}\) with C having one single bond to H, one single bond to the \(1^{st}\) O, and one double bond to the \(2^{nd}\) O. The \(1^{st}\) O will also have two lone pairs, while the \(2^{nd}\) Oxygen has a negative formal charge. #c)#Carbonate ion, \(\mathrm{CO}_{3}{ }^{2-}\)

Identify Valence Electrons

Count the total number of valence electrons. Carbon has 4, Oxygen has 6 each (total 18 for three Oxygen atoms). Additionally, there is a 2- charge on the ion, which means two extra electrons are present. So, the total number of valence electrons is 4 + 18 + 2 = 24.

Draw Lewis Structure

Place the Carbon (C) atom in the center and connect each of the three Oxygen (O) atoms around it with a single bond. Assign any remaining electrons to each atom to satisfy the octet rule. The central Carbon atom needs 4 more electrons to complete its octet. We end up having two resonance structures: In the first one, two Oxygen atoms have a single bond with the Carbon atom, while one Oxygen atom has a double bond with Carbon. The second resonance structure has the double bond in a different Oxygen atom. Each single-bonded Oxygen atom has 3 lone pairs, and a negative formal charge. The resulting Lewis structure is \(\mathrm{[CO}_{3}]^{2-}\) with resonance between the O atoms. #d)#Nitrate ion, \(\mathrm{NO}_{3}{ }^{-}\)

Identify Valence Electrons

Count the total number of valence electrons. Nitrogen has 5, Oxygen has 6 each (total 18 for three Oxygen atoms). Additionally, there is a 1- charge on the ion which means an extra electron is present. So, the total number of valence electrons is 5 + 18 + 1 = 24.

Draw Lewis Structure

Place the Nitrogen (N) atom in the center and connect each of the three Oxygen (O) atoms around it with a single bond. Assign any remaining electrons to each atom to satisfy the octet rule. The central Nitrogen atom needs 2 more electrons to complete its octet. We end up having three resonance structures: In each structure, two Oxygen atoms have a single bond with Nitrogen, while one Oxygen atom has a double bond with Nitrogen. Each single-bonded Oxygen atom has 3 lone pairs, and a negative formal charge. The resulting Lewis structure is \(\mathrm{[NO}_{3}]^{-}\) with resonance between the O atoms. #e)#Formate ion, \(\mathrm{HCOO}^{-}\)

Identify Valence Electrons

Count the total number of valence electrons. Carbon has 4, Oxygen has 6 each (total 12 for two Oxygen atoms), and Hydrogen has 1. Additionally, there is a 1- charge on the ion which means an extra electron is present. So, the total number of valence electrons is 4 + 12 + 1 + 1 = 18.

Draw Lewis Structure

Place the Carbon (C) atom in the center and connect one Hydrogen (H) atom and both Oxygen (O) atoms to it using single bonds. Assign any remaining electrons to each atom to satisfy the octet rule. Carbon needs only 2 more electrons, shared with only one of the remaining Oxygen atoms. We end up with a double bond to one Oxygen atom and a single bond to the other, which has a negative formal charge. The remaining Oxygen has two lone pairs and Hydrogen has none. The resulting Lewis structure is \(\mathrm{[HCOO]}^{-}\). #f)#Acetate ion, \(\mathrm{CH}_{3} \mathrm{COO}^{-}\)

Identify Valence Electrons

Count the total number of valence electrons. Carbon has 4 each (total 8 for two Carbon atoms), Hydrogen has 1 each (total 3 for three Hydrogen atoms), Oxygen has 6 each (total 12 for two Oxygen atoms). Additionally, there is a 1- charge on the ion which means an extra electron is present. So, the total number of valence electrons is 8 + 3 + 12 + 1 = 24.

Draw Lewis Structure

Place one Carbon (C) atom in the center and connect it to the three Hydrogen (H) atoms and the other Carbon atom using single bonds. Assign any remaining electrons to each atom to satisfy the octet rule. The second Carbon atom still needs 3 more electrons, so it forms double bonds with the two Oxygen atoms. The resulting Lewis structure is \(\mathrm{[CH}_{3}\mathrm{COO]}^{-}\). The Oxygen atoms have a negative formal charge due to resonance between the double bonds.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial for predicting how elements will interact and bond with each other. They are the electrons available to be lost, gained, or shared in the formation of chemical compounds. For ions, like those mentioned in the exercise, the total count of valence electrons includes those that come from the constituent atoms plus or minus any electrons due to the overall charge on the ion. For example, with the amide ion ($NH\_2^-$), we count the 5 valence electrons from nitrogen, 1 from each hydrogen atom, and add an additional one since the ion carries a negative charge, making a total of 8 valence electrons.

Formal Charges

Formal charge is an essential concept used to estimate the way electric charge is distributed in a molecule or ion. It is calculated based on the number of valence electrons an atom has in its neutral state, the number of nonbonding electrons, and the electrons it 'owns' from covalent bonds (each bond counts for one electron). The formula is:

$Formal\; Charge\; =\; (Valence\; electrons)\; -\; (Nonbonding\; electrons)\; -\; (Bonding\; electrons/2)$.

In Lewis structures, atoms strive for formal charges as close to zero as possible, which often aligns with the most stable configuration of the ion.

Octet Rule

The octet rule is a chemical rule of thumb that reflects observation that atoms of main-group elements tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electron configuration as a noble gas. Exceptions to this rule include hydrogen (which prefers 2 electrons) and transition metals, which can have expanded octets. For ions, the goal is to draw a Lewis structure where each atom (except H) achieves or moves towards an octet. This may involve the existence of lone pairs or the formation of multiple bonds, as seen in the examples of carbonate and nitrate ions.

Resonance Structures

Resonance structures occur when there is more than one valid Lewis structure for a molecule or ion with the same arrangement of atoms. In such cases, the actual structure is a hybrid of the resonance forms, and electrons are delocalized. These different structures use curved arrows to denote the movement of electrons and indicate delocalization of charge within the compound. Carbonate and nitate ions provided in the exercise illustrate this principle, where the double bonds can be located between carbon and any of the three oxygens, resulting in different Lewis structures that are all valid resonance forms.