lewis structures

Introduction to Lewis Structures

  • Lewis structures are crucial in Chem 101 as they are foundational for understanding molecular shapes in three-dimensional space.

  • These structures relate closely to molecular formulas and structural formulas, exemplified by methanol.

  • Example Representations:

    • Methanol can be represented in both molecular and structural formulas.

    • The Lewis structure highlights both bonding and lone pairs of electrons.

Importance of Lewis Structures

  • Establishing a valid Lewis structure is the first step in determining a molecule's three-dimensional shape.

  • Subsequent lectures will build upon the Lewis structure to explore the 3D configurations of molecules, with ethanol serving as an example.

Pre-requisites for Understanding Lewis Structures

  • Familiarity with concepts such as:

    • Covalent bonding

    • Valence electrons

    • Electronegativity

  • Review of these concepts will occur through worked examples during the lecture.

Steps to Write a Lewis Structure

Step 1: Determine Valence Electrons

  • Method: Use the electron configuration or group number from the periodic table.

  • Example: Carbon (C) has an electron configuration of helium core 2s² 2p², resulting in 4 valence electrons.

  • Quick Method: Group number (for main group elements) determines valence electrons:

    • For Carbon: Group 14; Calculation: 14 - 10 = 4 or 4A indicates 4 valence electrons.

Overview of Strategy for Writing Lewis Structures

  • The following five-step strategy is recommended for successfully drafting Lewis structures, even for more complex examples than previously encountered.

Step 2: Identify Central Atom and Create Skeletal Structure

  • Guidelines:

    • Identify a central atom.

    • Use single bonds to connect to terminal atoms initially.

  • Example: For sulfur difluoride (SF₂):

    • Sulfur (S) in Group 6A has 6 valence electrons; Fluorine (F) in Group 7A has 7 valence electrons each.

    • Total valence electrons = 6 (from S) + 7*2 (from F) = 20 electrons.

    • Skeletal structure: S as central atom, terminal positions for the two F atoms.

Step 3: Calculate Remaining Electrons after Bonding

  • Bonds formed utilize electrons:

    • Each bond uses 2 electrons.

    • For SF₂: 2 bonds consume 4 electrons so 20 - 4 = 16 electrons remaining.

Step 4: Complete Octets on Terminal Atoms

  • Process: Use remaining electrons to satisfy the octet rule, first focusing on terminal atoms.

  • For SF₂:

    • Fluorine needs 8 electrons (octet):

      • Use 2, 4, 6, 8 electrons in bonds. Two fluorine atoms consume 12 of the remaining 16 electrons.

    • 4 electrons left for Sulfur; hence, S can complete its octet as it also ends with 8 electrons.

Step 5: Verify Octets and Resolve Any Remaining Electrons

  • Ensure all atoms have octets without leftover electrons, or determine the need for multiple bonding.

  • In the case of SF₂, all atoms are satisfied with their octet and no further adjustments are needed.

Example 1: Sulfur Difluoride (SF₂)

  • Valence Electrons Calculation: 20 total electrons.

  • Skeletal Structure: S central, F terminal.

  • Calculated Remaining Electrons: 16 after forming 2 S–F bonds.

  • Octets Completed: Each F gets 8 electrons, leaving 4 for S which also gets to 8.

  • Completeness: Final Lewis structure satisfied.

Example 2: Methanol (CH₃OH)

  • Valence Electrons Calculation: 2 (H) + 4 (C) + 6 (O) = 12 total electrons.

  • Skeletal and Central Atom Identification: C at the center, bonded to two H and one O.

  • Remaining Electrons: 6 after initial bonding consumes 6.

  • Populate Terminal Atom: O gains complete octet using leftover electrons and forming a double bond to satisfy C’s octet.

Example 3: Cyanide Ion (CN⁻)

  • Valence Electrons: C (4) + N (5) + 1 (extra from charge) = 10 total.

  • Skeletal structure: Only C and N, connect with a single bond.

  • Octets and Double Bonds: Transform lone pairs on N into triple bond with C to achieve octets for both.

Example 4: Nitrosyl Chloride (NClO)

  • Valence Electricity Calculation: N (5), Cl (7), O (6) = 18 total electrons.

  • Central Atom Selection: N as central atom due to equivalent electronegativity.

  • Creating Bonds: Connect N with Cl and O through single bonds with overall remaining calculation leading to octets filled by converting lone pair into double bonds.

Conclusion and Next Steps

  • Students are encouraged to practice Lewis structures as discussed in these examples before proceeding to the next lecture.

  • Next lecture will cover the concept of formal charges and how to evaluate the effectiveness of Lewis structures based on these charges.