Practice problems drawn from Housecroft, JD Lee, House, and Descriptive IC texts. Select an answer or click “Show Answer” to reveal the explanation.
Lithium shows a well-known “diagonal relationship” with magnesium rather than behaving like its Group 1 congeners. What is the primary reason for this similarity?
The diagonal relationship arises because Li+ (r ≈ 76 pm) and Mg2+ (r ≈ 72 pm) have very similar ionic radii and therefore similar charge densities. This means they polarize anions to a comparable extent, leading to similar chemical behavior: both form covalent organometallic compounds, their carbonates decompose on heating, and their hydroxides are only sparingly soluble.
BeCl2 is predominantly covalent while MgCl2 is predominantly ionic. Applying Fajans’ rules, which factor best explains this difference?
Fajans’ rules state that covalent character increases when the cation is small and highly charged, and the anion is large and polarizable. Be2+ has an exceptionally small ionic radius (~27 pm for CN 4), giving it a very high charge density. This strongly distorts the electron cloud of Cl−, pulling electron density back toward the cation and producing substantial covalent character. Mg2+ is significantly larger (~72 pm), so it polarizes Cl− much less.
Lithium behaves anomalously compared to the other alkali metals in several ways. Which of the following is not an anomalous property of lithium?
Lithium’s anomalous behavior includes its high ionization energy, small size, tendency to form covalent compounds (such as organolithium reagents), and formation of the normal oxide Li2O rather than peroxides or superoxides. However, LiOH is actually sparingly soluble — unlike the freely soluble NaOH and KOH. Saying LiOH is “highly soluble” is incorrect and therefore not a genuine anomalous property.
What is the structure of BeCl2 in the gas phase compared to the solid phase?
In the gas phase, BeCl2 exists as a linear monomer with sp hybridization on beryllium (two bonding pairs, no lone pairs → 180°). In the solid phase, the electron-deficient Be center satisfies its octet by accepting lone pairs from neighboring chlorine atoms, forming a chain polymer with edge-sharing tetrahedra. Each beryllium becomes four-coordinate through chloride bridges.
Diborane (B2H6) contains an unusual type of bonding not found in simple Lewis structures. What type of bond holds the bridging hydrogen atoms, and why does boron form such bonds?
Boron has only 3 valence electrons and 4 valence orbitals, making it inherently electron-deficient — it cannot form enough conventional 2c-2e bonds to use all its orbitals. In B2H6, two bridging hydrogens each participate in a three-center two-electron (3c-2e) bond: one pair of electrons is shared across a B–H–B unit. This “banana bond” allows boron to achieve a closer approach to an octet despite its electron deficiency.
Nitrogen exists as N2 with a very strong triple bond (945 kJ/mol), while phosphorus forms P4 tetrahedra with single bonds. What accounts for this fundamental difference?
The key factor is orbital size and overlap efficiency. Nitrogen’s compact 2p orbitals achieve excellent lateral (π) overlap, making the N≡N triple bond extremely strong. Phosphorus uses diffuse 3p orbitals, which overlap poorly in the sideways orientation needed for π-bonds. Consequently, phosphorus prefers to maximize the number of σ-bonds instead, forming the tetrahedral P4 structure with six P–P single bonds.
Carbon exhibits several allotropes: diamond, graphite, and fullerenes. Which statement correctly relates their structures to their properties?
Diamond consists of sp3-hybridized carbon atoms in a rigid 3D tetrahedral network — every electron is localized in σ-bonds, making it extremely hard and an electrical insulator. Graphite has sp2 carbons arranged in planar hexagonal layers with one delocalized π-electron per carbon; these layers slide easily (lubricant) and the delocalized electrons conduct electricity. Fullerenes (e.g., C60) have curved sp2 surfaces with mixed single/double bond character.
The “inert pair effect” is an important trend in p-block chemistry. Which statement best describes this effect and gives correct examples?
The inert pair effect describes the increasing stability of the oxidation state that is two less than the group maximum as one descends a p-block group. The ns2 electrons become progressively less willing to participate in bonding. This is attributed to poor shielding by filled (n−1)d and (n−2)f subshells, which increases Zeff and contracts the ns orbital. Classic examples: Tl+ more stable than Tl3+, Pb2+ more stable than Pb4+, and Bi3+ more stable than Bi5+.
Sulfur can form SF6 (expanding its octet to 12 electrons), while oxygen cannot form OF6. What is the primary reason oxygen is limited to 8 valence electrons?
Oxygen is a period 2 element with only 2s and 2p valence orbitals available — there are no low-energy 2d orbitals. This limits it to a maximum of 4 electron pairs (octet). Additionally, oxygen’s small atomic radius creates severe steric crowding if six fluorine atoms attempt to coordinate. Sulfur, in period 3, is larger and can utilize its 3d orbitals (or, in modern MO descriptions, employ hypervalent bonding involving multi-center interactions) to accommodate up to 12 valence electrons.
The bond dissociation energies for the homonuclear dihalogen molecules are: F–F = 159, Cl–Cl = 242, Br–Br = 193, I–I = 151 kJ/mol. The F–F bond is unexpectedly weaker than the Cl–Cl bond. What is the explanation?
Fluorine atoms are extremely small (van der Waals radius ~1.47 Å), and each F atom carries three lone pairs. When two F atoms form F2, these lone pairs are forced into very close proximity, creating strong inter-electron repulsion that destabilizes the bond. Chlorine atoms are larger, so lone pair repulsion is much less severe, allowing the Cl–Cl bond to be stronger. From Cl2 onward, bond strength decreases normally as orbital overlap diminishes with increasing atomic size.
Interhalogen compounds are formed between different halogen elements. Which of the following correctly identifies an interhalogen compound and its VSEPR-predicted shape?
In interhalogen compounds, the larger halogen is the central atom. ClF3 has 5 electron pairs around Cl (3 bonding + 2 lone pairs); the electron geometry is trigonal bipyramidal, but the molecular shape is T-shaped (lone pairs occupy equatorial positions). BrF5 has 6 electron pairs around Br (5 bonding + 1 lone pair); the electron geometry is octahedral, giving a square pyramidal molecular shape. IF7 has 7 bonding pairs and adopts a pentagonal bipyramidal geometry.
HF is a weak acid in water (pKa ≈ 3.2), while HCl, HBr, and HI are all strong acids. Why is HF anomalously weak?
Acid strength in water depends on the balance between bond dissociation energy and solvation enthalpy of the resulting anion. The H–F bond (570 kJ/mol) is far stronger than H–Cl (432 kJ/mol), H–Br (366 kJ/mol), or H–I (298 kJ/mol). Although F− has a high hydration enthalpy, it is not sufficient to compensate for the very high bond energy. Additionally, HF engages in extensive hydrogen bonding in aqueous solution, forming species like HF2− that further reduce the availability of free H+.
Xenon forms several stable compounds (XeF2, XeF4, XeF6), but helium, neon, and argon form no stable compounds under normal conditions. Why can xenon form compounds while the lighter noble gases cannot?
Xenon’s ionization energy (1170 kJ/mol) is much lower than that of helium (2372), neon (2081), or argon (1521 kJ/mol), making it easier to promote electrons into bonding configurations. Xenon is also large and polarizable, facilitating orbital interactions with strongly electronegative partners like fluorine and oxygen. The lighter noble gases have ionization energies too high and atomic radii too small to accommodate the expanded valence electron counts needed for stable compound formation.
In Group 15, the relative stability of the +3 versus +5 oxidation state does not change monotonically — instead, an “alternation effect” is observed (N, As, and Bi prefer +3; P and Sb can readily access +5). What causes this alternation?
The alternation effect is rooted in the periodic structure of electron shells. Arsenic follows the first row of d-block elements (3d10), which provide poor shielding of the nuclear charge. This causes a contraction that stabilizes the 4s2 electrons, making them harder to ionize and favoring the +3 state. Similarly, bismuth follows both the 4f and 5d blocks. Phosphorus and antimony, which do not immediately follow a poorly-shielding filled subshell, more readily access the +5 state. Nitrogen’s preference for +3 is separate — it cannot expand its octet due to the absence of accessible d orbitals.
Across a period from left to right, the character of the elements changes from metallic to non-metallic. What is the fundamental electronic reason for this transition?
As nuclear charge increases across a period while electrons are added to the same valence shell, the effective nuclear charge Zeff rises steadily. Metals (left side) have low ionization energies and readily lose electrons. Moving right, increasing Zeff raises ionization energies and electron affinities, meaning elements increasingly prefer to gain electrons rather than lose them — the hallmark of non-metallic behavior. The metalloid diagonal (B–Si–Ge–As–Te) marks the approximate boundary of this transition.
Descriptive Chemistry — Main Group