Medium Lewis Structure Practice Questions

Concept Explanation

A Lewis structure is a graphical representation of a molecule that shows the arrangement of valence electrons among the atoms, including bonded pairs and lone pairs. These diagrams serve as the foundational tool for predicting molecular geometry, reactivity, and physical properties. To master Medium Lewis Structure Practice Questions, you must move beyond simple octets and understand concepts like formal charge, resonance, and expanded octets for elements in the third period or below on the periodic table.

The process of drawing these structures follows a systematic approach: first, calculate the total number of valence electrons by summing the contributions from each atom (and adjusting for ionic charges). Second, select a central atom, usually the least electronegative element, and connect terminal atoms with single bonds. Third, distribute the remaining electrons to satisfy the octet rule, starting with the outer atoms. Finally, if the central atom lacks an octet, form multiple bonds or, if the central atom is from period 3 or higher, consider an expanded valence shell to minimize formal charges. Understanding electron configuration practice questions can help you quickly identify the number of valence electrons available for bonding.

Formal charge is a critical metric used to determine the most plausible Lewis structure when multiple variations are possible. It is calculated by subtracting the number of non-bonding electrons and half the bonding electrons from the atom's neutral valence count. The most stable structures typically have formal charges closest to zero and place negative charges on the most electronegative atoms.

Solved Examples

These examples demonstrate how to handle molecules that require formal charge analysis or involve polyatomic ions.

1. Draw the Lewis structure for the Nitrite Ion (NO₂⁻).

   1. Total valence electrons: N (5) + 2 × O (6) + 1 (for the negative charge) = 18 electrons.

   2. Place Nitrogen in the center and connect both Oxygens with single bonds. (4 electrons used, 14 left).

   3. Complete the octets for Oxygen. (12 electrons used, 2 left).

   4. Place the remaining 2 electrons on Nitrogen.

   5. Nitrogen only has 6 electrons. Move a lone pair from one Oxygen to form a double bond.

   6. Result: One N=O bond and one N-O bond with resonance. Formal charges: One O is -1, N is 0, other O is 0.

2. Draw the Lewis structure for Sulfur Dioxide (SO₂).

   1. Total valence electrons: S (6) + 2 × O (6) = 18 electrons.

   2. Place Sulfur in the center. Connect Oxygens with single bonds. (4 used, 14 left).

   3. Complete Oxygen octets and place the last pair on Sulfur.

   4. To satisfy the octet rule, form one double bond. However, to minimize formal charge, Sulfur (a period 3 element) can expand its octet.

   5. By forming two S=O double bonds, the formal charge on Sulfur and both Oxygens becomes 0. This is the preferred structure.

3. Draw the Lewis structure for Xenon Tetrafluoride (XeF₄).

   1. Total valence electrons: Xe (8) + 4 × F (7) = 36 electrons.

   2. Place Xe in the center and connect 4 Fluorines with single bonds. (8 used, 28 left).

   3. Complete the octets for all four Fluorine atoms. (24 used, 4 left).

   4. Place the remaining 4 electrons (2 lone pairs) on the central Xenon atom.

   5. The Xenon atom has an expanded octet with 12 electrons, which is permissible for noble gases in higher periods.

Practice Questions

Test your skills with these medium-difficulty problems. Ensure you consider resonance and formal charges for every molecule.

1. Draw the Lewis structure for the Carbonate ion (CO₃²⁻) and identify the number of resonance structures.

2. Determine the best Lewis structure for Phosphorus Pentachloride (PCl₅).

3. Draw the Lewis structure for the Chlorate ion (ClO₃⁻) and calculate the formal charge on Chlorine.

4. Draw the Lewis structure for Ozone (O₃) and specify the formal charges for each atom.

5. Draw the structure for Sulfur Hexafluoride (SF₆). Does it obey the octet rule?

6. Construct the Lewis structure for Carbonyl Sulfide (OCS). Which atom is in the center?

7. Draw the Lewis structure for the Ammonium ion (NH₄⁺).

8. Draw the Lewis structure for Bromine Trifluoride (BrF₃), accounting for lone pairs on the central atom.

9. Draw the Lewis structure for the Nitrate ion (NO₃⁻) and show all resonance contributors.

10. Draw the structure for I₃⁻ (Triiodide ion) and identify the number of lone pairs on the central Iodine.

Answers & Explanations

Detailed breakdowns for the practice questions provided above.

1. CO₃²⁻: Total electrons = 4 + (3×6) + 2 = 24. Carbon is central. One C=O double bond and two C-O single bonds are required to satisfy Carbon's octet. There are 3 resonance structures as the double bond can shift between any of the three Oxygens.

2. PCl₅: Total electrons = 5 + (5×7) = 40. Phosphorus is central, bonded to 5 Chlorines via single bonds. This uses all 40 electrons (8 per Cl). Phosphorus has 10 electrons (expanded octet), which is allowed for third-row elements.

3. ClO₃⁻: Total electrons = 7 + (3×6) + 1 = 26. To minimize formal charge, Cl forms two double bonds with two Oxygens and one single bond with the third. Cl has one lone pair. Formal charge on Cl = 7 - 2(lone) - 5(bonds) = 0.

4. O₃: Total electrons = 18. Central Oxygen is bonded to one Oxygen via a double bond and another via a single bond. The central Oxygen has one lone pair. Formal charges: Terminal double-bonded O (0), Central O (+1), Terminal single-bonded O (-1).

5. SF₆: Total electrons = 48. Sulfur is central with six single bonds to Fluorines. It does not obey the octet rule; it has 12 valence electrons (expanded octet).

6. OCS: Total electrons = 4 + 6 + 6 = 16. Carbon is the central atom (least electronegative). Structure is O=C=S. All atoms have a formal charge of 0.

7. NH₄⁺: Total electrons = 5 + (4×1) - 1 = 8. Nitrogen is central with four single bonds to Hydrogens. Nitrogen has a formal charge of +1.

8. BrF₃: Total electrons = 7 + (3×7) = 28. Bromine is central with three single bonds to Fluorines. This uses 24 electrons for octets. The remaining 4 electrons form two lone pairs on Bromine.

9. NO₃⁻: Total electrons = 24. Nitrogen central, one double bond to an Oxygen, two single bonds. Nitrogen has a +1 charge, two Oxygens have -1, and one has 0. There are 3 resonance structures.

10. I₃⁻: Total electrons = (3×7) + 1 = 22. Central Iodine is bonded to two terminal Iodines. After completing terminal octets (16 electrons used in bonds and lone pairs), 6 electrons remain. These 3 lone pairs are placed on the central Iodine.

Quick Quiz

Frequently Asked Questions

What is the octet rule in Lewis structures?

The octet rule is a chemical rule of thumb that reflects the observation that atoms of main-group elements tend to combine in such a way that each atom has eight electrons in its valence shell. This configuration mimics the stability of noble gases.

When can an atom have an expanded octet?

Atoms can have an expanded octet if they are in the third period or lower of the periodic table, as they have accessible d-orbitals to accommodate more than eight electrons. Examples include Phosphorus, Sulfur, and Chlorine.

How do you calculate formal charge?

Formal charge is calculated using the formula: [Valence Electrons] - [Non-bonding Electrons] - [1/2 × Bonding Electrons]. It helps identify the most stable electron arrangement among competing Lewis structures.

What are resonance structures?

Resonance structures are two or more valid Lewis structures for a single molecule that differ only in the position of electrons. The actual molecule is a hybrid of these structures, resulting in equivalent bond lengths and shared electronic character.

Why is Fluorine never a central atom?

Fluorine is the most electronegative element and only requires one electron to complete its octet, meaning it almost exclusively forms a single bond and never acts as a bridge or central hub. Its high electronegativity prevents it from sharing more than one electron pair effectively.

How does VSEPR theory relate to Lewis structures?

Lewis structures provide the connectivity and electron pair count needed to apply VSEPR geometry practice questions. Once the Lewis structure is drawn, VSEPR theory predicts the 3D shape based on the repulsion between electron domains.

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