periodic trends
Explain the trend in effective nuclear charge (ZeffZeff) across a period and down a group. Provide a reason for each trend.
Across a period (left to right): ZeffZeff increases because the number of protons increases while shielding by inner electrons remains relatively constant. This results in stronger attraction between the nucleus and valence electrons.
Down a group (top to bottom): ZeffZeff increases slightly because the additional electron shells are less effective at shielding the valence electrons from the nuclear charge, despite the increased distance.
Explain the difference between ionic and covalent bonding. Provide an example of each and describe one key property of the resulting compounds.
Ionic bonding: Occurs between metals and nonmetals via electron transfer (e.g., NaCl). Properties: High melting points, conduct electricity when molten/aqueous.
Covalent bonding: Occurs between nonmetals via electron sharing (e.g., H₂O). Properties: Lower melting points, poor electrical conductivity.
What are the key steps to drawing a Lewis structure? Illustrate using the example of carbon dioxide (CO₂).
Count valence electrons: CO₂ → C (4) + 2 × O (6) = 16 electrons.
Identify central atom: Carbon (least electronegative).
Form single bonds: C=O=O (uses 4 electrons).
Complete octets for terminal atoms: Oxygens each need 6 more electrons (12 total).
Check central atom octet: Carbon has only 4 electrons → form double bonds (O=C=O).
Final structure: O=C=O with no lone pairs on C and two lone pairs on each O.
Name the following ionic compounds:
a) Fe₂O₃
b) NaCl
c) CuCl₂
d) PbSO₄
a) Iron(III) oxide (Fe³⁺ and O²⁻)
b) Sodium chloride (Na⁺ and Cl⁻)
c) Copper(II) chloride (Cu²⁺ and Cl⁻)
d) Lead(II) sulfate (Pb²⁺ and SO₄²⁻)
Explain why the ionization energy of magnesium (Mg) shows a dramatic increase between the second (I₂ = 1451 kJ/mol) and third ionization energies (I₃ = 7733 kJ/mol). How does this relate to electron configurations and effective nuclear charge?
The large jump occurs because the first two electrons are removed from Mg's valence shell (3s²), while the third electron must be removed from a core shell (2p⁶). Core electrons experience much higher Zeff (less shielding) and are closer to the nucleus, requiring substantially more energy. This reflects the quantum mechanical principle that inner-shell electrons are bound more tightly.
Why does atomic size decrease across a period but increase down a group?
Across a period: Atomic size decreases because the increasing ZeffZeff pulls valence electrons closer to the nucleus, reducing atomic radius.
Down a group: Atomic size increases because additional electron shells are added, increasing the distance between the nucleus and valence electrons.
How does lattice energy relate to the strength of an ionic bond? Use Coulomb’s law to explain how ionic size and charge affect lattice energy.
Lattice energy measures ionic bond strength; higher values mean stronger bonds.
Coulomb’s law: Eel=kQ1Q2d2Eel=kd2Q1Q2.
Smaller ions (↓d): Increase attraction → higher lattice energy.
Larger charges (↑Q): Increase attraction → higher lattice energy (e.g., MgO > NaCl).
What is formal charge, and how is it calculated? Determine the formal charges for nitrogen (N) and oxygen (O) in the nitrate ion (NO₃⁻).
Formula: Formal Charge = Valence e⁻ − (Nonbonding e⁻ + ½ Bonding e⁻).
NO₃⁻ example:
Central N: 5 − (0 + ½ × 8) = +1.
Single-bonded O: 6 − (6 + ½ × 2) = −1.
Double-bonded O: 6 − (4 + ½ × 4) = 0.
Preferred structure: Minimizes formal charges (resonance spreads the −1 charge).
Write the formulas for the following compounds:
a) Aluminum sulfide
b) Calcium phosphate
c) Ammonium nitrate
d) Potassium permanganate
a) Al₂S₃ (Al³⁺ and S²⁻)
b) Ca₃(PO₄)₂ (Ca²⁺ and PO₄³⁻)
c) NH₄NO₃ (NH₄⁺ and NO₃⁻)
d) KMnO₄ (K⁺ and MnO₄⁻)
Predict which compound has higher lattice energy: MgO or KF. Justify your answer using Coulomb's Law and periodic trends in ionic size/charge.
MgO has higher lattice energy because:
Charge: Mg²⁺ and O²⁻ have double the charge of K⁺/F⁻ (Q₁Q₂ term in Coulomb's Law is 4× larger)
Size: Mg²⁺ is smaller than K⁺, and O²⁻ is smaller than F⁻ (smaller d → stronger attraction)
The combined effects dominate despite KF's ions both being in Period 3.
Compare the first ionization energies of sodium (Na), magnesium (Mg), and aluminum (Al). Explain any deviations from the general trend.
General trend: First ionization energy increases across a period (Na < Mg < Al).
Deviation: The ionization energy of Al (578 kJ/mol) is lower than Mg (738 kJ/mol) because Al’s valence electron is in a higher-energy pp orbital, which is farther from the nucleus and shielded by the ss electrons.
Compare sigma (σ) and pi (π) bonds. How many of each are present in a molecule of nitrogen (N₂)?
Sigma (σ): Head-on overlap, single bonds, free rotation.
Pi (π): Sideways overlap, double/triple bonds, no rotation.
N₂: Triple bond = 1σ + 2π bonds.
What is a resonance structure? Provide an example and explain why resonance stabilizes molecules.
Resonance: Multiple valid Lewis structures with the same atom arrangement but different electron distributions (e.g., ozone, O₃).
Stabilization: The actual structure is a hybrid of all resonance forms, averaging bond lengths/strengths (e.g., O₃ bonds are intermediate between single and double bonds, lowering energy).
Name the following covalent compounds:
a) N₂O₅
b) PCl₅
c) SF₆
d) CO
a) Dinitrogen pentoxide
b) Phosphorus pentachloride
c) Sulfur hexafluoride
d) Carbon monoxide
The carbonate ion (CO₃²⁻) has three resonance structures. Using formal charge analysis and molecular orbital theory, explain why the actual structure is more stable than any single resonance form.
Formal charges: Each resonance structure has one C=O (FC=0) and two C-O⁻ (FC=-1), but the real structure averages these bonds (all C-O bond orders = 1.33).
Delocalization: π electrons are distributed over three atoms in MO theory, lowering the overall energy.
Empirical evidence: All C-O bonds are identical (128 pm), intermediate between single (143 pm) and double (122 pm) bonds.
Why does fluorine (F) have the highest electronegativity value on the Pauling scale, while francium (Fr) has the lowest?
Fluorine: High electronegativity due to its small atomic size, high ZeffZeff, and strong attraction for electrons to complete its valence shell.
Francium: Low electronegativity because its large atomic size and low ZeffZeff make it difficult to attract electrons.
Why is CO₂ a nonpolar molecule despite having polar covalent bonds? How does this differ from H₂O?
CO₂: Linear shape cancels dipole moments (symmetrical).
H₂O: Bent shape results in a net dipole (asymmetric).
Key difference: Molecular geometry determines overall polarity.
How do you draw the Lewis structure for a molecule that violates the octet rule, such as sulfur hexafluoride (SF₆)?
Count valence electrons: S (6) + 6 × F (7) = 48 electrons.
Place S central, single-bonded to 6 F atoms (uses 12 electrons).
Complete octets for F atoms (36 more electrons).
Check central atom: Sulfur (Period 3) has 12 electrons (expanded octet using d orbitals).
Final structure: Six S−F single bonds, with 3 lone pairs on each F.
Classify and name the following acids:
a) HCl
b) H₂SO₄
c) HNO₂
d) H₃PO₄
a) Hydrochloric acid (binary acid)
b) Sulfuric acid (oxyacid, sulfate → -ic)
c) Nitrous acid (oxyacid, nitrite → -ous)
d) Phosphoric acid (oxyacid, phosphate → -ic)
Explain how sulfur in SF₆ violates the octet rule. Include in your answer: the type of hybridization used, the geometric arrangement of orbitals, and why this is impossible for oxygen in the same group.
Hybridization: sp³d² (uses 3s, 3p, and two 3d orbitals)
Geometry: Octahedral (90° bond angles)
Why oxygen can't do this:
Oxygen lacks accessible d-orbitals (2p valence shell can't expand)
Even if excited, the energy gap to 3d is prohibitively large (unlike Period 3+ elements)
Describe the discontinuities in electron affinity trends between Groups IIA and IIIA and between Groups VA and VIA.
Between IIA and IIIA: Group IIA elements (e.g., Be, Mg) have lower electron affinity because the added electron must occupy a higher-energy pp orbital, experiencing repulsion from ss electrons.
Between VA and VIA: Group VA elements (e.g., N, P) have lower electron affinity because the added electron must pair in a half-filled pp orbital, creating repulsion. Group VIA elements (e.g., O, S) readily accept electrons to fill their pp orbitals.
What is a resonance structure? Use ozone (O₃) as an example to explain how resonance affects bond length and stability.
Resonance: Multiple valid Lewis structures for the same molecule (e.g., O₃ has two forms with alternating double bonds).
Effect: Actual structure is a hybrid; bonds are intermediate in length/strength (shorter than single but longer than double bonds), increasing stability.
Compare the Lewis structures of BF₃ (boron trifluoride) and NF₃ (nitrogen trifluoride). Why does BF₃ have an incomplete octet while NF₃ follows the octet rule?
BF₃: Boron (Group 3A) has 3 valence e⁻ → forms 3 bonds (6 total e⁻, incomplete octet). Stable due to small size and trigonal planar geometry.
NF₃: Nitrogen (Group 5A) has 5 valence e⁻ → forms 3 bonds + 1 lone pair (8 e⁻, complete octet).
Key difference: Boron lacks enough valence electrons for an octet, while nitrogen naturally achieves it.
What are the names of the following polyatomic ions?
a) OH⁻
b) SO₄²⁻
c) NO₃⁻
d) ClO⁻
a) Hydroxide
b) Sulfate
c) Nitrate
d) Hypochlorite
Given these bond dissociation energies (kJ/mol): H-H (436), Cl-Cl (243), H-Cl (431). Calculate ΔH for the reaction H₂(g) + Cl₂(g) → 2HCl(g) and explain why the H-Cl bond is stronger than the average of H-H and Cl-Cl bonds despite chlorine's higher electronegativity.
Calculation:
ΔH = Σ(bonds broken) - Σ(bonds formed)
= (436 + 243) - 2(431) = -183 kJ/mol (exothermic)
Bond strength explanation:
Orbital overlap: H's 1s and Cl's 3p orbitals have better size-matching than Cl-Cl's 3p-3p overlap.
Polarity: The H-Cl dipole creates additional ionic-covalent resonance stabilization (though EN difference is only 0.9).
Electron density: Cl's lone pairs provide partial π-backbonding into H's empty 1s orbital.