Question

Draw the best Lewis formula for each of the following. Include any necessary resonance forms. (a) \(\mathrm{ClO}_{4}{ }^{-}\) (c) \(\mathrm{NCO}^{-}\) (e) \(\mathrm{BF}_{3}\) (b) \(\mathrm{NO}_{2}^{-}\) (d) \(\mathrm{HCO}_{2}^{-}\)

Short answer

The Lewis structures depict the arrangement of atoms and distribution of electrons in given molecules or ions. In these structures, each dash '—' represents a bond and remaining valence electrons are distributed around atoms. The charge is written on the top right corner of the structure if applicable.

Step-by-step solution

Draw Lewis Structure for \(\mathrm{ClO}_{4}^{-}\)

Begin by calculating the total number of valence electrons present. Chlorine atoms have 7 valence electrons, each oxygen atom has 6 and one more for the -1 charge, giving us a total of \(7 + 4 * 6 + 1 = 32\) electrons. Chlorine will be the central atom with oxygens surrounding it. Draw single bonds between Cl and each O atom, which takes up 8 electrons. Distribute remaining electrons around the oxygens to complete their octets.

Draw Lewis Structure for \(\mathrm{NCO}^{-}\)

Calculate total number of valence electrons available; Nitrogen has 5, Carbon has 4, Oxygen has 6 and one more for the -1 charge - giving us 16 electrons to distribute. Carbon in the center and N and O on its sides. Draw a single bond between N and C and a double bond between C and O to use 10 electrons and distribute missing electrons to complete octets.

Draw Lewis Structure for \(\mathrm{BF}_{3}\)

Calculate valence electrons; Boron has 3 and each F atom has 7, which totals to 24 electrons. Boron in center and F atoms around. Draw single bonds between Boron and each Fluorine, consuming 6 electrons. Distribute remaining electrons around F atoms to complete their octets.

Draw Lewis Structure for \(\mathrm{NO}_{2}^{-}\)

Calculate valence electrons; Nitrogen has 5, each Oxygen has 6, and one from -1 charge, which sums to 18 electrons. N in center with O on sides. Draw a single bond between N and one O and a double bond between N and other O and distribute rest of the electrons.

Draw Lewis Structure for \(\mathrm{HCO}_{2}^{-}\)

Calculate valence electrons; Hydrogen has 1, Carbon has 4, each Oxygen has 6 and one more for the -1 charge, summing up to 18 electrons. Carbon in the center, O and H on one side and the other O on the other side. Draw single bonds between C and H, C and O and a double bond between C and other O and distribute rest of the electrons.

Key concepts

Resonance Forms

Resonance forms help to represent molecules where simple Lewis structures can't accurately depict all the features. Some molecules have electrons that are shared in different configurations, creating forms that are not independent. For example,

• **NO₂⁻ (Nitrite Ion):** It can have its double bond between nitrogen and either oxygen. Both forms are correct, showing electron distribution differently.

The resonance forms average these electron positions to describe a more accurate structure. If you ever feel stuck, try to find alternate placements for double or triple bonds to see other possible resonance forms. This can reveal a more comprehensive understanding of the molecule.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in forming bonds. They are responsible for the chemical behavior of the atom and are used when drawing Lewis structures.

• **Example:** For ClO₄⁻, you calculate the total valence electrons by counting 7 from Cl, 6 from each O, and 1 extra from the negative charge, resulting in 32.
• **Why it Matters:** The total number of valence electrons determines how they can be arranged to create a stable molecule. Using them properly ensures each atom achieves a stable electron configuration, facilitating the completion of Lewis structures.

Double-checking your count helps avoid errors while drawing the structure.

Octets

The octet rule states that atoms tend to combine in a way that they each have eight electrons in their valence shell, achieving a stable configuration similar to noble gases.

• **Example 1:** For HCO₂⁻, you'll want to ensure each atom (except hydrogen, which holds 2 electrons) has a full octet. Usually, distribute electrons first to peripheral atoms and adjust bonds, if necessary, to make sure the rule is fulfilled.
• **Exception:** Some elements, like Boron in BF₃, may not have a full octet. It's important to recognize these exceptions as it helps in understanding the unique nature of some molecules.

Applying the octet rule allows us to predict the stability and reactivity of molecules.