13.1 General Chemistry: Atomic Structure, Bonding & Stoichiometry

13.1 General Chemistry: Atomic Structure, Bonding & Stoichiometry

The Center for Educational Measurement (CEM) NMAT Chemistry subtest is a 30-item block with a recommended ~30 minutes. Official content areas are General Chemistry, Analytical Chemistry, Organic Chemistry, and Biochemistry. This section builds the general-chemistry foundation: structure of matter, bonding and shapes, and quantitative reaction math at introductory college premed depth.

Items test understanding, applying, analyzing, evaluating, and synthesizing — not rote element-list trivia. You must connect electron configuration to periodic position, choose the right bond type, sketch a reasonable Lewis/VSEPR shape, and convert grams to moles without a calculator culture of careless arithmetic.

Quick frame: Atoms → electron arrangement → bonds and shapes → balanced equations and moles. Thermochemistry and gases sit on the same mole and energy bookkeeping skills you already used in Physics.

Atomic structure and electron configuration intro

An atom has a nucleus (protons + neutrons) and surrounding electrons. Key counts:

Isotopes share (Z) but differ in neutrons (e.g., (^{12}\mathrm{C}) and (^{14}\mathrm{C})). Average atomic mass on the periodic table is a weighted average of natural isotopes.

Electron configuration intro. Electrons occupy shells/subshells ((s), (p), (d), (f)) following Aufbau order, Pauli exclusion (max 2 electrons per orbital with opposite spin), and Hund’s rule (fill degenerate orbitals singly before pairing). Useful patterns for NMAT:

• Period number tracks principal shell for main-group valence electrons.
• Group number for main-group elements correlates with valence electron count (with known exceptions in transition metals).
• Noble-gas configuration and octet stability motivate ion formation and covalent bonding.

Worked example — particles. How many protons, neutrons, and electrons in (^{23}\mathrm{Na}^+)?

• (Z = 11) → 11 protons. Mass number 23 → neutrons = (23 - 11 = 12). The +1 charge means 10 electrons (lost one valence electron).

Periodic trends (high-yield)

Across a period (left → right): atomic radius generally decreases (higher effective nuclear charge pulls electrons in); first ionization energy and electronegativity generally increase.

Down a group: atomic radius generally increases (new shells); ionization energy and electronegativity generally decrease.

Metallic character increases toward the lower left; nonmetallic character toward the upper right (F, O, N, Cl region). Cations are smaller than their parent atoms; anions are larger — useful for comparing isoelectronic species (same electron count: more protons → smaller radius).

Exam tactic: If two trends compete, effective nuclear charge and principal quantum number usually decide. Do not invent “always” rules for every transition-metal exception; NMAT-style items favor main-group clarity.

Bonding: ionic, covalent, metallic

1. Ionic bonding: large electronegativity difference; electrons transfer → cations + anions; lattice of ions. Typical: metal + nonmetal (e.g., NaCl). Solids often high melting, conduct when molten or aqueous (mobile ions), brittle crystals.
2. Covalent bonding: electrons shared between nonmetals. Polar covalent if EN differs moderately; nonpolar if shared equally (or symmetric molecule cancels dipoles). Molecular substances often lower melting points than ionic lattices; conductivity depends on whether ions form (acids, etc.).
3. Metallic bonding: delocalized “sea” of electrons among metal cations — explains conductivity, malleability, and metallic luster at an intro level.

Lewis structures intro. Count valence electrons, place atoms (often least EN central except H), connect with single bonds, complete octets (duet for H), form multiple bonds if needed. Formal charge (FC = V - N_{\mathrm{nonbonding}} - \tfrac{1}{2}N_{\mathrm{bonding}}) helps choose among resonance forms when advanced items appear.

VSEPR shapes (high-yield). Electron domains around the central atom repel; geometry minimizes repulsion.

Bond angles: tetrahedral ≈ 109.5°; trigonal planar 120°; linear 180°. Lone pairs compress angles (water ≈ 104.5°, ammonia ≈ 107°).

Worked example — shape. What is the molecular geometry of (\mathrm{H_2O})? Four electron domains (2 bonding pairs + 2 lone pairs) → tetrahedral electron geometry, bent molecular shape.

Mole concept, molar mass, and balancing

One mole contains Avogadro’s number of entities: (N_A \approx 6.022 \times 10^{23},\mathrm{mol^{-1}}). Molar mass (M) (g/mol) equals the formula mass in amu numerically.

[ n = \frac{m}{M} \quad (\mathrm{mol} = \mathrm{g},/,\mathrm{g,mol^{-1}}) ]

Balancing equations conserves atoms (and charge in ionic equations). Coefficients scale relative moles; never change subscripts inside formulas.

Worked example — moles from mass. How many moles in 18.0 g of water? (M(\mathrm{H_2O}) = 18.0,\mathrm{g/mol}) → (n = 18.0/18.0 = 1.00,\mathrm{mol}) (exactly (N_A) molecules).

Worked example — balancing. Balance: (\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}).

• C: 3 → 3 (\mathrm{CO_2}); H: 8 → 4 (\mathrm{H_2O}); O atoms on right: (3\times 2 + 4\times 1 = 10) → 5 (\mathrm{O_2}).
• Balanced: (\mathrm{C_3H_8 + 5,O_2 \rightarrow 3,CO_2 + 4,H_2O}).

Limiting reactant and percent yield

Limiting reactant is consumed first and caps product amount. Method:

1. Convert each given reactant mass to moles.
2. Use stoichiometric ratios to find theoretical product from each reactant.
3. The smaller product amount identifies the limiting reactant.
4. Percent yield = ((\mathrm{actual\ yield}/\mathrm{theoretical\ yield}) \times 100%).

Worked example — limiting reactant. (\mathrm{2,H_2 + O_2 \rightarrow 2,H_2O}). Mix 4.0 g (\mathrm{H_2}) with 32.0 g (\mathrm{O_2}). Which limits? How many grams of water can form?

• Moles: (n(\mathrm{H_2}) = 4.0/2.0 = 2.0,\mathrm{mol}); (n(\mathrm{O_2}) = 32.0/32.0 = 1.0,\mathrm{mol}).
• Need 2 mol (\mathrm{H_2}) per 1 mol (\mathrm{O_2}). Available ratio is exactly stoichiometric — both are limiting together; product moles of (\mathrm{H_2O}) = 2.0 mol from either path.
• Mass water: (2.0 \times 18.0 = 36,\mathrm{g}).

Worked example — non-stoichiometric mix. Same reaction with 3.0 mol (\mathrm{H_2}) and 1.0 mol (\mathrm{O_2}).

• From (\mathrm{O_2}): 1.0 mol (\mathrm{O_2}) can make 2.0 mol (\mathrm{H_2O}) and needs 2.0 mol (\mathrm{H_2}).
• From (\mathrm{H_2}): 3.0 mol (\mathrm{H_2}) could make 3.0 mol (\mathrm{H_2O}) but would need 1.5 mol (\mathrm{O_2}).
• (\mathrm{O_2}) is limiting; theoretical (\mathrm{H_2O}) = 2.0 mol = 36 g. Excess (\mathrm{H_2}) left: (3.0 - 2.0 = 1.0,\mathrm{mol}).

Worked example — percent yield. Theoretical yield 36 g; student isolates 27 g. Percent yield = ((27/36)\times 100% = 75%).

Gas laws revisit (if helpful)

Ideal gas: (PV = nRT). At STP (classic 1 atm, 273 K), 1 mol occupies about 22.4 L. Combined gas law for fixed (n): (P_1V_1/T_1 = P_2V_2/T_2) with (T) in kelvin. Mole fraction and partial pressures (Dalton): (P_i = x_i P_{\mathrm{total}}).

Worked example. How many moles of ideal gas occupy 11.2 L at STP? (n = 11.2/22.4 = 0.50,\mathrm{mol}).

Thermochemistry conceptual: endo/exo and enthalpy

Exothermic processes release heat to surroundings ((\Delta H < 0) for the system’s enthalpy change under constant pressure convention); endothermic absorb heat ((\Delta H > 0)). Bond breaking requires energy; bond formation releases energy — net (\Delta H) depends on which dominates.

Enthalpy (H) is a state function; reaction enthalpy depends on states of reactants/products (standard states often implied). Hess’s law idea: path-independent sums of steps. For NMAT, recognize heat flow direction, sign of (\Delta H), and that catalysts change rate/path, not overall (\Delta H) of the net reaction.

Worked conceptual: Combustion of methane is exothermic — products ((\mathrm{CO_2}), (\mathrm{H_2O})) are lower enthalpy than reactants under standard comparison; heat is released to the surroundings.

Common traps

1. Confusing mass number with atomic number or molar mass.
2. Changing subscripts when balancing (illegal — changes identity).
3. Using the excess reactant to compute product without checking limiting reactant.
4. Percent yield using mass without ensuring same compound and pure actual yield.
5. °C in gas laws instead of K.
6. Assuming every molecule with polar bonds is a polar molecule (geometry can cancel dipoles, e.g., (\mathrm{CO_2})).

Exam tactics

• Convert all masses to moles before stoichiometry ratios.
• Write a mini mole map: g → mol → mol product → g product.
• For shapes: count domains (bonds + lone pairs on central atom) first, then molecular shape.
• For trends: locate both elements on the table; apply radius/IE/EN directions deliberately.

General Chemistry is the quantitative backbone of NMAT Chemistry. Solutions, acids/bases, and equilibrium (next section) reuse molarity and mole ratios constantly; analytical chemistry adds measurement discipline on top of the same arithmetic.