The Lewis Structures Of Four Compounds Are Given

Mastering Lewis Structures: A Step-by-Step Guide to Four Key Compounds

Understanding how atoms bond to form molecules is foundational to chemistry, and Lewis structures are the essential tool for visualizing this process. These diagrams, named after Gilbert N. Lewis, depict the arrangement of valence electrons around atoms, showing how they are shared or transferred to achieve stable electron configurations, typically following the octet rule. By mastering Lewis structures, you gain profound insight into molecular shape, reactivity, and properties. This guide will deconstruct the Lewis structures for four critical compounds—carbon dioxide (CO₂), sulfate ion (SO₄²⁻), ammonium ion (NH₄⁺), and phosphorus pentachloride (PCl₅)—illustrating core principles and important exceptions.

Compound 1: Carbon Dioxide (CO₂) – The Linear Molecule

Carbon dioxide is a simple yet perfect example of a molecule with multiple bonds and a linear geometry.

1. Count Total Valence Electrons:

   • Carbon (C) is in Group 14: 4 valence electrons.
   • Oxygen (O) is in Group 16: 6 valence electrons each. With two oxygens: 2 × 6 = 12.
   • Total = 4 + 12 = 16 valence electrons.

2. Skeleton Structure & Bonding:

   • Carbon is less electronegative than oxygen, so it becomes the central atom: O - C - O.
   • Place a single bond (2 electrons) between C and each O. This uses 4 electrons.
   • Remaining electrons = 16 - 4 = 12.

3. Complete Octets on Terminal Atoms First:

   • Each oxygen needs 6 more electrons to complete its octet (they already have 2 from the single bond). Place 6 electrons (3 lone pairs) on each oxygen.
   • Electrons used: 2 oxygens × 6 = 12. All 16 electrons are now placed.

4. Check the Central Atom & Form Multiple Bonds:

   • Carbon currently has only 4 electrons (two single bonds). It needs 4 more to complete its octet.
   • We can't add more lone pairs to the terminal oxygens (they already have octets). The solution is to convert lone pairs on oxygen into bonding pairs with carbon.
   • Take one lone pair from each oxygen and form a second bond (a double bond) with carbon.
   • Final Lewis Structure: O=C=O.
   • Each atom now has an octet: Carbon has two double bonds (4 bonding pairs = 8 electrons). Each oxygen has one double bond (4 electrons) and two lone pairs (4 electrons).

Key Takeaway: CO₂ demonstrates that when the central atom lacks an octet after all electrons are placed, forming double or triple bonds with terminal atoms is necessary.

Compound 2: Sulfate Ion (SO₄²⁻) – Resonance and Formal Charge

The sulfate ion is a polyatomic anion with resonance, meaning its true structure is an average of several contributing forms.

1. Count Total Valence Electrons:

   • Sulfur (S) is in Group 16: 6 valence electrons.
   • Oxygen (O): 6 × 4 = 24.
   • The ion has a 2- charge, meaning we add 2 extra electrons.
   • Total = 6 + 24 + 2 = 32 valence electrons.

2. Skeleton Structure:

   • Sulfur is the central atom. Connect the four oxygen atoms with single bonds: S with four single bonds to O.
   • Electrons used in bonds: 4 bonds × 2 electrons = 8.
   • Remaining electrons = 32 - 8 = 24.

3. Complete Octets on Terminal Atoms:

   • Place the remaining 24 electrons as lone pairs on the four oxygen atoms. Each oxygen gets 6 electrons (3 lone pairs).
   • Electrons used: 4 oxygens × 6 = 24. All electrons placed.

4. Check the Central Atom & Formal Charge:

   • Sulfur currently has only 8 electrons (four single bonds). It has an octet, but we must check formal charge.
   • Formal Charge Formula: (Valence electrons) - (Non-bonding electrons) - ½(Bonding electrons).
   • For Sulfur: 6 - 0 - ½(8) = 6 - 4 = +2.
   • For each Oxygen (with 3 lone pairs and 1 single bond): 6 - 6 - ½(2) = 6 - 6 - 1 = -1.
   • The sum of formal charges (+2) + 4(-1) = -2, matching the ion's charge. However, a +2 formal charge on sulfur is high.

5. Minimize Formal Charge with Resonance:

   • To reduce formal charge, we form double bonds. Convert one lone pair from two different oxygen atoms into a bonding pair with sulfur.
   • This creates two S=O double bonds and leaves two S-O single bonds (with the single-bonded oxygens each carrying a -1 formal charge).
   • Crucially, any two of the four oxygen atoms can form the double bonds. This gives us two major resonance structures. The true structure is a resonance hybrid with all S-O bonds being identical (order of 1.5).

Key Takeaway: For molecules/ions with multiple equivalent terminal atoms (like SO₄²⁻, NO₃⁻, CO₃²⁻), resonance is used to distribute charge and minimize formal charge, leading to bond lengths that are intermediate between single and double bonds.

**Compound 3: Ammonium Ion (NH₄⁺) – A Cation from Coordinate Covalency

Compound 3: Ammonium Ion (NH₄⁺) – A Cation from Coordinate Covalency

The ammonium ion illustrates a common scenario where a central atom with a lone pair bonds to a proton (H⁺), forming a coordinate covalent bond (also called a dative bond), where both bonding electrons originate from the same atom.

1. Count Total Valence Electrons:

   • Nitrogen (N) is in Group 15: 5 valence electrons.
   • Hydrogen (H): 1 × 4 = 4.
   • The ion has a 1+ charge, meaning we remove 1 electron.
   • Total = 5 + 4 - 1 = 8 valence electrons.

2. Skeleton Structure:

   • Nitrogen is the central atom. Connect the four hydrogen atoms with single bonds: N with four single bonds to H.
   • Electrons used in bonds: 4 bonds × 2 electrons = 8.
   • Remaining electrons = 8 - 8 = 0. All electrons are used in bonding.

3. Check Octets and Formal Charge:

   • Nitrogen is surrounded by 8 electrons (four bonds). It has a complete octet.
   • Each hydrogen has 2 electrons (a duet), satisfying the duet rule.
   • Formal Charge on Nitrogen: 5 - 0 - ½(8) = 5 - 4 = +1.
   • Formal Charge on each Hydrogen: 1 - 0 - ½(2) = 1 - 1 = 0.
   • The sum of formal