Problem 5 - Chemistry and Chemical Reactivity

Question

Draw a Lewis structure for each of the following molecules or ions. (a) \(\mathrm{NF}_{3}\) (b) \(\mathrm{ClO}_{3}^{-}\) (c) HOBr (d) \(\mathrm{SO}_{3}^{2-}\)

Step-by-Step Solution

Answer

NF3: N center with 3 F atoms as single bonds; ClO3-: Cl center with 1 double & 2 single bonded O; HOBr: O connects to H & Br; SO3^2-: S center with one double-bonded O.

Step 1: Determine Total Valence Electrons (NF3)

Count the valence electrons contributed by each atom. Nitrogen (N) contributes 5 electrons, and each Fluorine (F) atom contributes 7 electrons. Since there are three F atoms, calculate: 5 + (3 × 7) = 26 electrons.

Step 2: Draw the Skeleton Structure (NF3)

Place the less electronegative Nitrogen (N) atom at the center and attach three Fluorine (F) atoms to it with single bonds. Begin filling the octets for the surrounding F atoms.

Step 3: Assign Remaining Electrons and Check Octets (NF3)

Each Fluorine atom is satisfied with 3 lone pairs and one shared pair. Place any remaining electrons on the central nitrogen atom to complete its octet. The structure balances with N having one lone pair and all atoms have full octets.

Step 4: Determine Total Valence Electrons (ClO3-)

Count valence electrons: Chlorine (Cl) contributes 7 electrons, each Oxygen (O) contributes 6 electrons, and there's an extra electron due to the negative charge. Total: 7 + (3 × 6) + 1 = 26 electrons.

Step 5: Draw the Skeleton Structure (ClO3-)

Place Chlorine (Cl) in the center, and connect it to three Oxygen atoms with single bonds. This will set up a basic skeleton structure.

Step 6: Distribute Electrons and Create Double Bonds (ClO3-)

Assign lone pairs to oxygen atoms until their octets are complete. To satisfy Chlorine's expanded octet and use all electrons effectively, create one double bond with one of the oxygen atoms.

Step 7: Assign Formal Charge (ClO3-)

Check the formal charges for optimal distribution. Aim to have a formal charge of 0 or match the ion's overall charge. Here, give the central Cl a partial positive by completing its octet and assigning the extra electron where needed.

Step 8: Determine Total Valence Electrons (HOBr)

Hydrogen (H) has 1 valence electron, Oxygen (O) 6 electrons, and Bromine (Br) 7 electrons. Total is 1 + 6 + 7 = 14 electrons.

Step 9: Draw the Skeleton Structure (HOBr)

Place the less electronegative Oxygen (O) in the center with H on one side and Br on the other. Bond H—O—Br is the skeletal arrangement.

Step 10: Assign Remaining Electrons and Check Octets (HOBr)

Oxygen gets 2 lone pairs after fulfilling the bond requirements for Hydrogen and Bromine. Bromine has 3 lone pairs and satisfies its octet. Hydrogen is stable with 2 electrons in the bond.

Step 11: Determine Total Valence Electrons (SO3^2-)

Sulfur (S) contributes 6 electrons, each Oxygen (O) contributes 6 electrons, and there are 2 extra electrons due to the 2- charge. Total: 6 + (3 × 6) + 2 = 26 electrons.

Step 12: Draw the Skeleton Structure (SO3^2-)

Place Sulfur in the center and attach three Oxygen atoms with single bonds. Establish the basic structure before distributing the electrons.

Step 13: Distribute Electrons and Create Double Bonds (SO3^2-)

First, fill the octets of the outside oxygen atoms. To efficiently use both the valence and extra electrons, create a double bond with one of the oxygen atoms. The rest complete their octet with lone pairs.

Step 14: Assign Formal Charge and Make Adjustments (SO3^2-)

Ensure the total valence electrons account for the structure's charge. Aim for the formal charges on the oxygens to tally correctly with the ion's charge by using a single lone pair adjustment if needed.

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding. They are the electrons involved in forming bonds with other atoms. To understand how to draw Lewis structures, it's essential to know the number of valence electrons involved.

• For example, in the \[\text{NF}_{3}\] molecule, nitrogen (N) has 5 valence electrons, and each fluorine (F) has 7 valence electrons.
• The total count for \[\text{NF}_{3}\] is 5 (from N) + 3 × 7 (from three F atoms) = 26 valence electrons.

Valence electrons are determined by the atom's group number on the periodic table. For instance:

• Group 1 elements (like H) have 1 valence electron.
• Group 17 elements (like F, Cl) have 7 valence electrons.

Counting the correct number of valence electrons ensures accurate Lewis structure creation. It's always beneficial to double check this calculation to avoid errors in understanding the bonding framework of a molecule.

Formal Charge

Formal charge is a concept used to determine the most likely distribution of electrons in a molecule. It helps in assessing how electrons are shared among atoms in a molecule.

• The formula to calculate formal charge is:
  \[ \text{Formal charge} = \text{Valence electrons} - \text{Non-bonding electrons} - \frac{1}{2}\times\text{Bonding electrons} \]
• For example, in the \[\text{ClO}_{3}^{-}\] ion, we can use the formal charge to ensure that the distribution of electrons makes each part of the molecule stable. Adjustments can be made, such as creating a double bond, to reduce formal charges on individual atoms.

The goal in calculating formal charge is to have values as close to zero as possible or to reflect the overall charge in ionic species. Reducing formal charges makes a molecule more stable, which is crucial for accurate representation in Lewis structures.

Octet Rule

The octet rule is a principle stating that atoms tend to bond in ways that give them a full outer shell of 8 electrons, similar to noble gases. This rule is a key concept in understanding Lewis structures.
In Lewis structure diagrams:

• Each atom typically attempts to complete its octet by sharing, losing, or gaining electrons.
• For instance, in the \[\text{NF}_{3}\] molecule, the nitrogen atom shares electrons with three fluorine atoms to achieve a full outer shell.

While many elements follow the octet rule, there are exceptions:

• Hydrogen only needs 2 electrons, and some elements like boron may be stable with fewer than 8 electrons.
• Elements in period 3 or higher, like sulfur in \[\text{SO}_{3}^{2-}\], can exceed the octet rule due to available d-orbitals.

Understanding the octet rule and its exceptions allows for the development of more accurate and stable Lewis structures for representing molecules effectively.