Atomic Structure and Periodic Trends
What is effective nuclear charge, and how does it change across a period and down a group on the periodic table?
Effective nuclear charge is the net positive charge experienced by a valence electron after accounting for shielding by inner electrons.
Across a period: Zeff increases because the number of protons increases, while the shielding remains relatively constant.
Down a group: Zeff increases slightly, but the effect is less significant because the added core electrons shield the valence electrons more effectively.
Compare the lattice energies of MgO and NaCl. Which compound has the higher lattice energy and why?
Mg²⁺ and O²⁻ ions have charges of ±2
Na⁺ and Cl⁻ ions have charges of ±1
Magnesium ions have a smaller ionic radius compared to sodium ions.
Using VSEPR theory, predict the molecular geometry and bond angles for XeF₄ (xenon tetrafluoride). Is this molecule polar or nonpolar?
Molecular Geometry: Square planar (lone pairs occupy axial positions to minimize repulsion).
Bond Angles: 90° (between equatorial F atoms).
Polarity: Nonpolar because the symmetrical arrangement cancels out bond dipoles.
Compare the strength and nature of London dispersion forces (LDF), dipole-dipole forces, and hydrogen bonding. Provide an example of a molecule for each type.
sample molecules: Cl2, HBr, and H2O
The photoelectron spectrum (PES) of neon (Ne) shows peaks at binding energies of 84 MJ/mol, 54 MJ/mol, and 8 MJ.mol. Assign these peaks to the respective electron orbitals.
84 MJ/mol: 1s orbital
54 MJ/mol: 2s orbital
8 MJ/mol: 2p orbital
Explain why atomic radius decreases across a period but increases down a group.
Atomic radius decreases across a period because Zeff increases, pulling electrons closer to the nucleus without adding new energy levels.
Atomic radius increases down a group because new electron shells (energy levels) are added, increasing the distance between the nucleus and outer electrons.
Explain how the lattice energy and melting point of LiF compare to CsI.
LiF has a higher lattice energy and melting point than CsI.
Charges are equal, so compare the ionic sizes:
Li⁺ is much smaller than Cs⁺.
F⁻ is much smaller than I⁻.
Explain why H₂O has a bent shape with a bond angle of ~104.5° instead of the ideal tetrahedral angle (109.5°)
lone pairs compress the H–O–H angle
Why does ethanol (C₂H₅OH) have a higher boiling point than dimethyl ether (CH₃OCH₃), despite both having the same molecular formula (C₂H₆O)?
Ethanol can form hydrogen bonds (O–H group), while dimethyl ether only has dipole-dipole forces (no H-bonding).
H-bonds are much stronger, requiring more energy to break, hence ethanol’s higher boiling point (78°C vs. -24°C for dimethyl ether).
Why does the 2s peak in the PES of carbon (C) appear at a slightly higher binding energy than the 2p peak, even though they are in the same energy level?
Shielding effect: The 2s electrons experience less shielding from the nucleus than 2p electrons because they penetrate closer to the nucleus.
Higher effective nuclear charge (Z_eff) on 2s → higher binding energy.
Example: In carbon (1s² 2s² 2p²), the 2s peak is ~10 eV, while the 2p peak is ~6 eV.
Describe the trend in first ionization energy across a period and down a group. What are two key exceptions to this trend and why do they occur?
Across a period: First ionization energy increases due to rising Zeff, which holds electrons more tightly.
Down a group: Ionization energy decreases because outer electrons are farther from the nucleus and more shielded.
Exceptions:
Group IIA to IIIA: Ionization energy decreases because the outermost electron on the p orbital experiences the repulsion of the s orbital below it, making it easier to remove the electron.
Group VA to VIA: Ionization energy decreases because Group VIA involves removing an electron from a doubly-occupied orbital which increases the repulsion, making removal easier.
Predict and explain the shape and bond angles of ClF₃ using VSEPR theory. Why does this molecule deviate from the ideal domain geometry?
T-Shaped
90° (axial-equatorial) and 120° (equatorial-equatorial)
Lone pair repulsion: The 2 lone pairs exert stronger repulsion than bonding pairs, pushing the axial F atoms closer together.
Compare the polarity of CO₂ and SO₂.
CO₂: Linear shape cancels bond dipoles (O=C=O).
SO₂: Bent shape results in a net dipole (lone pair on S).
Explain why I₂ (iodine) is a solid at room temperature, while Cl₂ (chlorine) is a gas.
Both are nonpolar and only have London dispersion forces (LDF).
I₂ has more electrons (larger electron cloud) → stronger LDF → higher melting/boiling point (113°C vs. -101°C for Cl₂).
The PES of nitrogen (N) shows three peaks: 1s (40 MJ/mol), 2s (2.0 MJ/mol), and 2p (1.3 MJ/mol). Explain why the 2p peak is 1.5x taller than the 2s peak.
Electron configuration of N: 1s² 2s² 2p³.
Peak heights:
2s peak: 2 electrons → shorter peak.
2p peak: 3 electrons → 1.5x taller than 2s.
Binding energy order: 1s > 2s > 2p (due to orbital penetration and shielding).
What is electronegativity, and how does its trend compare to electron affinity across periods and down groups?
Electronegativity is an atom’s ability to attract electrons in a chemical bond.
Trends:
Across a period: Electronegativity increases as atoms more strongly attract electrons to complete their octets.
Down a group: Electronegativity decreases because atoms have more shielding and larger radii, weakening their ability to attract electrons.
This trend is similar to electron affinity, which also generally becomes more negative (exothermic) across a period and less negative down a group.
Name the following compounds using IUPAC rules:
a) FeCl3
b) P2O5
c) HNO2(aq)
d) SF6
For each compound, identify whether it is ionic, molecular (covalent), or an acid.
Iron(III) chloride Ionic
Diphosphorus pentoxide Molecular (covalent)
Nitrous acid Oxyacid
Sulfur hexafluoride Molecular (covalent)
Predict the shape and polarity of SF₄ (sulfur tetrafluoride). How do lone pairs affect its geometry?
lone pair occupies equatorial position. Instead of forming an ideal trigonal bipyramidal structure, it forms a seesaw structure instead.
Why does NH₃ (ammonia) have an unusually high boiling point for its molar mass (-33°C), while PH₃ (phosphine, -88°C) does not?
NH₃ forms hydrogen bonds (N–H bonds), while PH₃ only has dipole-dipole forces and LDF (P–H bonds are too weak for H-bonding).
H-bonds require H bonded to F, O, or N (not P), making NH₃’s IMFs significantly stronger.
How would the PES of fluorine (F) differ from that of chlorine (Cl)? Focus on peak positions, intensities, and number of peaks.
Cl’s peaks are at higher binding energies due to increased nuclear charge and electron shielding.
Given the following isoelectronic ions: N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, and Al³⁺, arrange them in order of increasing atomic radius and explain your reasoning.
Al³⁺ has the most protons (Z = 13) pulling on 10 electrons, creating the strongest attraction and smallest radius.
N³⁻ has the fewest protons (Z = 7) for the same number of electrons, so electron repulsion dominates, making it the largest.
Using the VSEPR model, predict the molecular geometry and bond angles for ammonia (NH3). Is NH3NH3 a polar or nonpolar molecule?
VSEPR Shape: Trigonal pyramidal (4 domains: 3 bonding pairs, 1 lone pair).
Bond Angles: ~107° (less than 109.5° due to lone pair repulsion).
Polarity: Polar because:
N–H bonds are polar (N is more electronegative).
Asymmetrical shape prevents dipole cancellation.
Why does BF₃ have a trigonal planar shape with 120° bond angles, while NH₃ is trigonal pyramidal with ~107° angles?
BF₃: 3 bonding pairs + 0 lone pairs → Trigonal planar (ideal 120°). No lone pair repulsion.
NH₃: 3 bonding pairs + 1 lone pair → Trigonal pyramidal. Lone pair repulsion reduces bond angles to ~107°.
Predict the order of boiling points for He, Ne, Ar, Kr, Xe and justify your ranking.
Boiling Point Order: He (-269°C) < Ne (-246°C) < Ar (-186°C) < Kr (-153°C) < Xe (-108°C).
Reason:
All are noble gases with only LDF.
Boiling point increases with atomic size/number of electrons (stronger LDF).
what are the hybridizations of carbon in alkane?
sp3