Source: F.21.210 Problem of the Day 01, Question 1.3(e)
Difficulty: Introductory-Intermediate | Prerequisites: Basic atomic structure, valence electrons, octet rule fundamentals Tags: Lewis structures, dot structures, ammonium hydroxide, NH4OH, closed shell, octet rule, ionic compounds, molecular construction, bonding, lone pairs, formal charge, CHEM 210, Organic Chemistry II, University of Michigan
This problem sits right at the start of Organic Chemistry II and tests whether you can work fluently with Lewis structures, not just draw them on command. The underlying skill is combinatorial: given a fixed set of atoms (N, H, H, H, H, O, H from ammonium hydroxide), how many valid molecules can you assemble? That means you need a solid grasp of valence rules, the octet/duet rule, lone pairs, and what makes an ionic compound valid. If you are shaky on how to count valence electrons or assign formal charges, revisit Chapter 1.1 before tackling this.
Ammonium hydroxide (NH₄OH) contains one nitrogen, five hydrogens, and one oxygen. The challenge is to rearrange subsets of these atoms into as many other valid molecules as possible, where every atom satisfies closed-shell (octet or duet) rules and any ionic compound includes both its cation and anion.
Lewis structure (electron dot structure)
A diagram showing every valence electron in a molecule as dots or lines (bonding pairs). Each line represents a shared pair of electrons; dots represent lone pairs.
In simple terms, it is a map of where the electrons live in a molecule.
Closed shell configuration
An arrangement where every atom in the molecule has a full outer electron shell: 8 electrons for second-row elements (the octet rule) or 2 electrons for hydrogen and helium (the duet rule).
Think of it as: every atom is "happy" because its valence shell is complete.
Valence electrons
The electrons in the outermost shell of an atom, which participate in bonding. Carbon has 4, nitrogen has 5, oxygen has 6, hydrogen has 1.
In simple terms, these are the electrons that do the chemistry.
Formal charge
The charge assigned to an atom in a Lewis structure, calculated as: valence electrons (free atom) minus lone pair electrons minus half of bonding electrons.
Think of it as: a bookkeeping tool that tells you whether an atom "owns" more or fewer electrons than it started with.
Octet rule
Second-row elements (C, N, O, F) tend to form bonds until they are surrounded by 8 valence electrons (four pairs). Hydrogen is the exception, needing only 2 (a duet).
Lone pair (non-bonding pair)
A pair of valence electrons on an atom that is not shared in a bond. Oxygen typically carries two lone pairs; nitrogen carries one.
In simple terms, electrons that belong to one atom alone and are not involved in holding atoms together.
Ionic compound
A compound composed of a positively charged ion (cation) and a negatively charged ion (anion) held together by electrostatic attraction. Ammonium hydroxide is ionic: NH₄⁺ is the cation, OH⁻ is the anion.
Think of it as: a partnership between a positively charged species and a negatively charged species.
Ammonium hydroxide (NH₄OH)
An ionic compound consisting of the ammonium cation (NH₄⁺) and the hydroxide anion (OH⁻). Its atom inventory: 1 N, 5 H, 1 O.
Question 1.3(e) asks you to draw the Lewis structure for ammonium hydroxide. The follow-up (POD 1) then asks: using that same set of atoms, how many other valid molecules can you build?
The atom inventory from NH₄OH:
1 nitrogen (N) with 5 valence electrons
5 hydrogen (H) atoms, each with 1 valence electron
1 oxygen (O) with 6 valence electrons
You do not need to use all seven atoms. Diatomic or triatomic molecules are fine.
Closed shell configurations: every atom must satisfy the octet rule (8 electrons for N, O) or duet rule (2 electrons for H). No radicals, no incomplete octets.
Ionic compounds must include both ions: you cannot list NH₄⁺ alone. If you write an ionic compound, it needs its cation and anion together.
Use only atoms from the inventory: you have 1 N, 5 H, and 1 O to work with, no more.
Work through the atoms by category:
Diatomic molecules (two atoms)
H₂ (hydrogen gas): H-H, each H gets its duet. The simplest valid molecule.
OH radical would be tempting, but it has an unpaired electron (not closed shell). However, consider whether any valid diatomic exists with O and H that satisfies both.
Triatomic molecules (three atoms)
H₂O (water): O with two bonds to H and two lone pairs. Closed shell for all atoms.
NH is not valid as a standalone neutral diatomic with closed shells.
Larger neutral molecules
NH₃ (ammonia): N bonded to 3 H atoms, one lone pair on N. Uses 1 N and 3 H.
H₂O₂ would require 2 oxygens, and you only have 1, so this is out.
CH compounds are out because you have no carbon.
Ionic compounds
NH₄OH itself (ammonium hydroxide): the original molecule.
NH₄⁺ paired with OH⁻: this is essentially the same compound written in ionic form.
Are there other valid ionic pairings from this atom set? Consider: H⁺ paired with OH⁻ is essentially water (a proton transfer), not a distinct ionic compound in the way the problem means.
Thinking about it carefully
The key molecules you can build from subsets of {N, 5H, O} with all closed shells:
H₂ (uses 2 H)
H₂O (uses 1 O, 2 H)
NH₃ (uses 1 N, 3 H)
NH₄OH (the original, uses all atoms)
Some students also identify:
Combinations like NH₃ + H₂O coexisting (both valid molecules from the same atom pool)
The ionic form NH₄⁺ OH⁻ versus the covalent form
The exact count depends on how your professor defines "different molecules" (structural isomers vs. simply different combinations of atoms). Count carefully and justify each one.
Formal charge formula:
\text{Formal Charge} = V - L - \frac{B}{2}Where V = valence electrons of the free atom, L = lone pair electrons, B = bonding electrons.
Valence electron counts to memorise:
H: 1 valence electron, needs 2 (duet)
N: 5 valence electrons, needs 8 (octet), typically forms 3 bonds + 1 lone pair (or 4 bonds with a +1 formal charge, as in NH₄⁺)
O: 6 valence electrons, needs 8 (octet), typically forms 2 bonds + 2 lone pairs (or 1 bond + 3 lone pairs with a -1 formal charge, as in OH⁻)
Quick bond-counting rule: an atom's typical number of bonds = 8 minus its group number (for second-row elements). Nitrogen (group 15) forms 3 bonds; oxygen (group 16) forms 2 bonds. Hydrogen always forms 1 bond.
Lewis structures are the foundation for predicting molecular shape (VSEPR theory), which in turn determines physical properties like boiling point, solubility, and reactivity. Pharmaceutical chemists use these principles daily: the shape and electron distribution of a drug molecule determines whether it will bind to its target protein. The ability to mentally construct valid molecules from a set of atoms is also central to retrosynthetic analysis, where organic chemists work backwards from a target molecule to figure out how to build it.
Students often forget that hydrogen can only form one bond. You cannot draw H with two bonds or a lone pair; it needs exactly 2 electrons total (a duet), not 8.
Students frequently draw OH as a standalone neutral molecule. Neutral OH has 7 valence electrons, leaving one unpaired. That is a radical, not a closed-shell molecule. OH⁻ (with 8 electrons) is the closed-shell form, but it is an anion, so you need a cation to pair it with.
Students sometimes list NH₄⁺ by itself as a valid answer. The problem specifies that ionic compounds must include both ions. A bare cation is not a complete ionic compound.
Students occasionally confuse the number of valence electrons with the total number of electrons. For nitrogen, the total is 7 (2 core + 5 valence), but only the 5 valence electrons matter for Lewis structures.
⚠️ Lewis structure questions appear on virtually every Organic Chemistry exam. The skill tested here (building molecules from atoms while respecting valence rules) is foundational for reaction mechanisms later in the course.
⚠️ Formal charge is a perennial exam favourite. If a Lewis structure gives an atom a formal charge of +2 or -2, the structure is almost certainly wrong or not the best resonance contributor.
⚠️ The "closed shell" requirement is a way of saying: no radicals. On an exam, if your drawn molecule has an odd number of total valence electrons and no charge, something is wrong.
⚠️ Professors love asking "how many valid structures can you draw" because it tests systematic thinking, not just memorisation. Practice being exhaustive rather than guessing.
True or False: Hydrogen can satisfy its valence requirements with just 2 electrons.
True. Hydrogen follows the duet rule, not the octet rule.
Fill in the blank: Nitrogen has ____ valence electrons and typically forms ____ covalent bonds in a neutral molecule.
5; 3.
True or False: A molecule with an odd total number of valence electrons can have all atoms with closed shells.
False. An odd electron count means at least one unpaired electron (a radical), which violates the closed-shell requirement.
Fill in the blank: In NH₄⁺, nitrogen forms ____ bonds and carries a formal charge of ____.
4; +1.
True or False: You can list OH⁻ by itself as a valid molecule for this problem.
False. The problem states that ionic compounds must include both ions.
Q: Draw the Lewis structure for ammonium hydroxide. Identify the formal charge on each atom.
A: NH₄⁺ has nitrogen at the centre bonded to four hydrogens. Nitrogen's formal charge: 5 - 0 - (8/2) = +1. Each hydrogen: 1 - 0 - (2/2) = 0. In OH⁻, oxygen is bonded to one hydrogen with three lone pairs. Oxygen's formal charge: 6 - 6 - (2/2) = -1. Hydrogen: 0.
Q: Using only the atoms from NH₄OH (1 N, 5 H, 1 O), list all valid neutral molecules you can construct where every atom has a closed-shell configuration.
A: H₂ (two hydrogens sharing a bond), H₂O (oxygen bonded to two hydrogens with two lone pairs), NH₃ (nitrogen bonded to three hydrogens with one lone pair). Each of these uses a subset of the available atoms and satisfies the octet/duet rule for every atom.
Q: Why is a standalone OH molecule not a valid answer to this problem?
A: Neutral OH has 7 total valence electrons (6 from O + 1 from H). An odd electron count means one electron is unpaired, making it a radical. Radicals do not have closed-shell configurations. OH⁻ (with 8 electrons) is closed-shell, but it is an anion and would need a cation partner.
Q: A student writes NH₅ as one of their answers. Explain why this is invalid.
A: Nitrogen has 5 valence electrons. If it formed 5 bonds to 5 hydrogens, that would place 10 electrons around nitrogen (5 bonding pairs). Second-row elements cannot exceed an octet because they have no accessible d orbitals. NH₅ violates the octet rule for nitrogen.
Q: If you use NH₃ and H₂O from the atom set, how many hydrogen atoms remain unused?
A: NH₃ uses 1 N and 3 H. H₂O uses 1 O and 2 H. Total used: 1 N, 5 H, 1 O. Zero atoms remain; you have used the entire inventory.
This material connects directly to VSEPR theory (Section 1.2+), where the Lewis structure you draw determines the 3D geometry of the molecule. The lone pairs you identify here will dictate bond angles and molecular shape.
It also sets the stage for understanding acid-base chemistry in organic reactions. The fact that NH₃ has a lone pair means it can act as a base (or nucleophile), while NH₄⁺ has no lone pair on nitrogen and cannot. Recognising which atoms carry lone pairs from a Lewis structure is essential for predicting reactivity throughout the rest of the course.
Formal charge shows up again heavily in resonance structures and reaction mechanisms, where you need to track electron flow with curved arrows.
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