Lewis Structures, Resonance, Formal Charge, Bond Properties, and VSEPR Theory

Lewis Structures and Related Concepts

Review: Drawing Lewis Structures for Ions

• Ammonium Ion ($NH_4^+$)

  • Step 1: Calculate total valence electrons.
    • Each Hydrogen (H) brings $1$ valence electron, and there are $4$ Hydrogens: $4 imes 1 = 4$ electrons.
    • Nitrogen (N) brings $5$ valence electrons.
    • Initial sum = $4 + 5 = 9$ electrons.
  • Step 2: Account for ionic charge.
    • The ion has a $+1$ charge, meaning it has lost $1$ electron.
    • Total electrons = $9 - 1 = 8$ electrons.
  • Step 3: Draw the Lewis structure.
    • Place Nitrogen in the center, bonded to four Hydrogens.
    • Each single bond represents $2$ electrons. Four single bonds make $4 imes 2 = 8$ electrons.
    • The structure correctly uses $8$ electrons.
  • Rule for Ions: Always draw brackets around the entire structure with the charge indicated on the outside (e.g., $[NH_4]^ +$) to denote electron loss or gain.

• Bond Representation:

  • A single line (single bond) represents $2$ electrons.
  • A double bond (two lines) represents $4$ electrons.
  • A triple bond (three lines) represents $6$ electrons.

• Nitrate Ion ($NO_3^-$)

  • Step 1: Determine the central atom.
    • Nitrogen (N) is less electronegative than Oxygen (O). Generally, the least electronegative atom (excluding Hydrogen) goes in the center. Fluorine is the most electronegative element; Nitrogen is further from Fluorine on the periodic table than Oxygen, making Nitrogen less electronegative.
    • Place Nitrogen in the middle surrounded by three Oxygen atoms.
  • Step 2: Calculate total valence electrons.
    • Each Oxygen (O) brings $6$ valence electrons, and there are $3$ Oxygens: $3 imes 6 = 18$ electrons.
    • Nitrogen (N) brings $5$ valence electrons.
    • Initial sum = $18 + 5 = 23$ electrons.
  • Step 3: Account for ionic charge.
    • The ion has a $-1$ charge, meaning it has gained $1$ electron.
    • Total electrons = $23 + 1 = 24$ electrons.
  • Step 4: Draw initial Lewis structure with single bonds and lone pairs.
    • Draw single bonds between the central N and each O (using $3 imes 2 = 6$ electrons).
    • Distribute remaining electrons as lone pairs to satisfy octets for outer atoms first, then the central atom.
    • If the initial structure has more electrons than needed (e.g., $26$ when only $24$ are required, as in the example), form double bonds by moving lone pairs from connected atoms into the bonding region.
  • Step 5: Adjust for correct electron count.
    • If the initial structure has $26$ electrons but only $24$ are needed, remove $2$ electrons ($1$ lone pair) from an outer Oxygen and $2$ electrons ($1$ lone pair) from the central Nitrogen and form a double bond between them. This reduces the total count by $2$. (More specifically, for a structure with too many electrons, you identify atoms with lone pairs and available bonding slots, then convert a lone pair and a lone pair from the central atom into a new bond, which often reduces the overall electron count while satisfying octets and formal charges better).
    • In $NO_3^-$, creating one double bond and having two single bonds, with appropriate lone pairs on oxygen and no lone pairs on nitrogen, satisfies the $24$ electron count and octets.
  • Step 6: Final check and ionic notation.
    • Ensure all atoms have an octet (where applicable) and the total electron count is correct.
    • Place brackets around the structure with the $-$ charge on the outside: $[NO_3]^ -$ .

Resonance Structures and Formal Charge

• Resonance Structure Defined: Occurs when two or more valid Lewis structures can be drawn for the same compound that differ only in the position of electrons, not the arrangement of atoms. Both structures are considered