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Chemical Bondingintermediate

Lewis Structure Practice: Molecules and Polyatomic Ions

Draw Lewis structures for molecules and polyatomic ions, applying formal charge minimization, resonance, and exceptions to the octet rule.

Problem Scenario

Draw the Lewis structure for each of the following: (a) NO₃⁻ (nitrate ion), (b) SF₄ (sulfur tetrafluoride), and (c) XeF₂ (xenon difluoride). For each, identify the molecular geometry using VSEPR theory.

Given Data

NO₃⁻5 + 3(6) + 1 = 24 valence e⁻

SF₄6 + 4(7) = 34 valence e⁻

XeF₂8 + 2(7) = 22 valence e⁻

Octet Rule8 electrons around each atom (2 for H)

Formal ChargeFC = valence e⁻ − lone pair e⁻ − ½ bonding e⁻

Requirements

Count total valence electrons for each species
Draw the Lewis structure showing all bonds and lone pairs
Calculate formal charges and identify the best structure
Identify resonance structures where applicable
Determine molecular geometry using VSEPR

Solution

Step 1:

NO₃⁻: 24 valence electrons. Place N at center bonded to three O atoms. Single bonds use 6 electrons, leaving 18. Place 6 electrons (3 lone pairs) on each O = 18 electrons. N only has 6 electrons — needs a double bond. Move one lone pair from an O to form a double bond: N has one double bond and two single bonds to O.

Step 2:

NO₃⁻ formal charges: For the double-bonded O: FC = 6 − 4 − ½(4) = 0. For single-bonded O: FC = 6 − 6 − ½(2) = −1. For N: FC = 5 − 0 − ½(8) = +1. Total: +1 + 0 + (−1) + (−1) = −1, matching the ion charge. There are 3 equivalent resonance structures (double bond rotates to each O). Geometry: trigonal planar (3 electron groups, no lone pairs on N).

Step 3:

SF₄: 34 valence electrons. S is the central atom bonded to 4 F atoms. Four single bonds use 8 electrons, leaving 26. Place 6 electrons on each F (24 total), leaving 2 electrons as a lone pair on S. S has 5 electron groups: 4 bonds + 1 lone pair. This is an expanded octet (10 electrons around S), which is allowed for Period 3+ elements.

Step 4:

SF₄ geometry: 5 electron groups = trigonal bipyramidal electron geometry. With 4 bonds and 1 lone pair, the molecular geometry is seesaw (or "sawhorse"). The lone pair occupies an equatorial position to minimize repulsion.

Step 5:

XeF₂: 22 valence electrons. Xe is the central atom bonded to 2 F atoms. Two single bonds use 4 electrons, leaving 18. Place 6 on each F (12 total), leaving 6 electrons = 3 lone pairs on Xe. Xe has 5 electron groups: 2 bonds + 3 lone pairs. This is another expanded octet (10 electrons around Xe).

Step 6:

XeF₂ geometry: 5 electron groups = trigonal bipyramidal electron geometry. With 2 bonds and 3 lone pairs, the molecular geometry is linear. The 3 lone pairs occupy all equatorial positions, and the 2 F atoms are at the axial positions (180° apart).

Final Answer

NO₃⁻: trigonal planar, 3 resonance structures, formal charges (+1 on N, −1 on single-bonded O). SF₄: seesaw geometry, expanded octet on S, 1 lone pair equatorial. XeF₂: linear geometry, expanded octet on Xe, 3 lone pairs equatorial in trigonal bipyramidal arrangement.

Key Takeaways

✓Always count valence electrons including ion charges before drawing
✓Formal charge = valence electrons − lone pair electrons − ½(bonding electrons)
✓Elements in Period 3+ can have expanded octets (more than 8 electrons)
✓Lone pairs on central atoms affect molecular geometry — VSEPR counts all electron groups

Common Errors to Avoid

✗Forgetting to add extra electrons for negative ion charges
✗Placing the most electronegative atom at the center (it should be the least electronegative)
✗Not recognizing expanded octets for Period 3+ elements
✗Confusing electron geometry with molecular geometry when lone pairs are present

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FAQs

Common questions about this problem type

Expanded octets are only possible for elements in Period 3 and below (S, P, Cl, Xe, etc.) because they have accessible d orbitals. Period 2 elements (C, N, O, F) never exceed 8 electrons. If distributing electrons normally leaves the central atom short, try forming double bonds before expanding the octet.

Equatorial positions have only two 90° interactions with neighboring groups, while axial positions have three. Lone pairs are larger and more repulsive than bonding pairs, so they occupy equatorial positions to minimize 90° repulsions.