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Test Bank — Unit 5

Descriptive Chemistry — Main Group

Practice problems drawn from Housecroft, JD Lee, House, and Descriptive IC texts. Select an answer or click “Show Answer” to reveal the explanation.

Group 1 & 2 — Alkali & Alkaline Earth

Questions 1–4
Housecroft & Sharpe, Chapter 11 Housecroft
Q1

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?

  • a Lithium and magnesium have identical electronegativities
  • b Lithium and magnesium have similar charge densities (charge-to-radius ratios)
  • c Both elements are found in the same mineral ores
  • d Both have the same number of valence electrons
Answer: (b)

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.

JD Lee, Chapter 9 JD Lee
Q2

BeCl2 is predominantly covalent while MgCl2 is predominantly ionic. Applying Fajans’ rules, which factor best explains this difference?

  • a Mg2+ is smaller than Be2+, so it polarizes Cl more
  • b Be has a noble-gas electron configuration that favors ionic bonding
  • c Be2+ is very small with high charge density, strongly polarizing the large Cl anion toward covalency
  • d MgCl2 has a higher lattice energy, forcing covalent character
Answer: (c)

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.

Housecroft & Sharpe, Chapter 11 Housecroft
Q3

Lithium behaves anomalously compared to the other alkali metals in several ways. Which of the following is not an anomalous property of lithium?

  • a Lithium forms a stable, highly soluble hydroxide LiOH
  • b Lithium forms a normal oxide Li2O on combustion, not a peroxide or superoxide
  • c Lithium has a much higher ionization energy than sodium
  • d Lithium forms covalent organolithium compounds (e.g., n-BuLi)
Answer: (a)

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.

House, Inorganic Chemistry, Chapter 12 House
Q4

What is the structure of BeCl2 in the gas phase compared to the solid phase?

  • a Tetrahedral monomer in both phases
  • b Chain polymer in both phases
  • c Bent monomer in gas; ionic lattice in solid
  • d Linear monomer (Cl–Be–Cl) in gas phase; chain polymer with bridging chlorides in solid phase
Answer: (d)

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.

Group 13–15 — Boron to Nitrogen Groups

Questions 5–8
Housecroft & Sharpe, Chapter 13 Housecroft
Q5

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?

  • a Normal 2-center 2-electron (2c-2e) B–H bonds; boron simply has enough electrons
  • b Three-center two-electron (3c-2e) bonds; boron is electron-deficient with only 3 valence electrons
  • c Hydrogen bonds between BH3 units
  • d Metallic bonding through delocalized electrons
Answer: (b)

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.

Housecroft & Sharpe, Chapter 15 Housecroft
Q6

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?

  • a Nitrogen’s small size allows effective 2p–2p π-overlap; phosphorus’s larger 3p orbitals overlap poorly for π-bonding
  • b Phosphorus has d orbitals available that prevent π-bonding
  • c Nitrogen is more electronegative, which forces triple bond formation
  • d P4 is actually held together by double bonds, not single bonds
Answer: (a)

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.

Housecroft & Sharpe, Chapter 14 Housecroft
Q7

Carbon exhibits several allotropes: diamond, graphite, and fullerenes. Which statement correctly relates their structures to their properties?

  • a Diamond has sp2 carbons in a layered structure, making it a good conductor
  • b Graphite has sp3 carbons in a 3D network, making it the hardest material
  • c Diamond has sp3 carbons in a 3D network (extremely hard); graphite has sp2 layers with delocalized π-electrons (electrical conductor, lubricant)
  • d Fullerenes are sp3 hybridized closed cages that are electrical insulators
Answer: (c)

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.

JD Lee, Chapter 12 JD Lee
Q8

The “inert pair effect” is an important trend in p-block chemistry. Which statement best describes this effect and gives correct examples?

  • a Light elements prefer lower oxidation states; e.g., N prefers +3 over +5
  • b The ns2 pair becomes increasingly reactive going down a group
  • c It only applies to Group 14 elements and involves d electrons
  • d Heavier p-block elements increasingly favor an oxidation state 2 less than the group maximum, as the ns2 pair becomes reluctant to participate in bonding (e.g., Tl+, Pb2+, Bi3+)
Answer: (d)

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+.

Group 16 & 17 — Chalcogens & Halogens

Questions 9–12
Housecroft & Sharpe, Chapter 16 Housecroft
Q9

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?

  • a Oxygen is too electronegative to share electrons with fluorine
  • b Oxygen has no energetically accessible d orbitals in its valence shell; its small size also prevents coordination of six fluorines
  • c Oxygen already has a full octet in O2 and cannot form more bonds
  • d The O–F bond is too weak to hold six fluorine atoms
Answer: (b)

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.

Housecroft & Sharpe, Chapter 17 Housecroft
Q10

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?

  • a Fluorine’s very small atomic radius causes severe lone pair–lone pair repulsion between the two fluorine atoms, weakening the bond
  • b Fluorine has the highest electronegativity, which destabilizes the F–F σ-bond
  • c The F–F bond has significant π-character that reduces its strength
  • d Chlorine benefits from d-orbital participation that strengthens Cl–Cl
Answer: (a)

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.

Descriptive Inorganic Chemistry, Rayner-Canham Descriptive IC
Q11

Interhalogen compounds are formed between different halogen elements. Which of the following correctly identifies an interhalogen compound and its VSEPR-predicted shape?

  • a ClF3 — trigonal planar
  • b BrF5 — trigonal bipyramidal
  • c ClF3 — T-shaped; BrF5 — square pyramidal
  • d IF7 — octahedral
Answer: (c)

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.

JD Lee, Chapter 16 JD Lee
Q12

HF is a weak acid in water (pKa ≈ 3.2), while HCl, HBr, and HI are all strong acids. Why is HF anomalously weak?

  • a Fluorine is not electronegative enough to stabilize F
  • b The H–F bond is exceptionally strong (570 kJ/mol), making it difficult to dissociate; additionally, F is poorly solvated compared to the energy needed to break the bond
  • c HF forms a covalent crystal in water that resists dissociation
  • d Water cannot solvate the fluoride ion due to its large size
Answer: (b)

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+.

Group 18 & Periodic Trends

Questions 13–15
Housecroft & Sharpe, Chapter 18 Housecroft
Q13

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?

  • a Xenon has unpaired electrons available for bonding
  • b The lighter noble gases have filled d orbitals that block bonding
  • c Xenon is a liquid at room temperature and reacts more readily
  • d Xenon has a sufficiently low ionization energy and large, polarizable electron cloud; its 5d orbitals are energetically accessible for bonding with highly electronegative atoms like fluorine
Answer: (d)

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.

House, Inorganic Chemistry, Chapter 14 House
Q14

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?

  • a Elements following a filled d-shell (As, after 3d) or f-shell have their ns electrons stabilized by poor shielding, making the +5 state less accessible at those positions
  • b The alternation is caused by differences in electronegativity alone
  • c Nitrogen prefers +3 because it can expand its octet easily
  • d The alternation is a random coincidence with no underlying electronic explanation
Answer: (a)

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.

Housecroft & Sharpe, Chapter 9 Housecroft
Q15

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?

  • a The number of electron shells increases across a period, pushing electrons outward
  • b Increasing Zeff across the period binds valence electrons more tightly, favoring electron gain over electron loss, shifting behavior from metallic to non-metallic
  • c Elements on the right have fewer valence electrons than those on the left
  • d Non-metals have completely filled d orbitals that prevent metallic bonding
Answer: (b)

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.

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Descriptive Chemistry — Main Group