301. The following observations concern two Group 13 elements, aluminium and thallium.
| Property | Aluminium | Thallium |
|---|
| General valence configuration | \(3s^23p^1\) | \(6s^26p^1\) |
| Characteristic higher oxidation state | +3 | +3 |
| Relative importance of +1 oxidation state | Limited | Considerable |
| Representative oxide behaviour | \(\mathrm{Al_2O_3}\) is amphoteric | \(\mathrm{Tl_2O}\) is basic |
A student concludes that thallium's greater preference for +1 must arise because it possesses fewer valence electrons than aluminium. Which alternative explanation best reconciles all the observations?
ⓐ. Thallium loses its \(6s^2\) electrons more readily than its \(6p\) electron, making +1 its preferred oxidation state.
ⓑ. The increasing stability of +1 is caused by the complete disappearance of the outermost \(s\) subshell in heavier Group 13 elements.
ⓒ. Thallium retains the outermost \(6s^2\) pair more readily in suitable compounds, while its greater metallic character is consistent with basic oxide behaviour.
ⓓ. Aluminium and thallium possess different group numbers, so their oxidation-state and oxide trends cannot be compared.
Correct Answer: Thallium retains the outermost \(6s^2\) pair more readily in suitable compounds, while its greater metallic character is consistent with basic oxide behaviour.
Explanation: \(\textbf{Establish the common electronic framework:}\)
Aluminium and thallium both belong to Group 13.
Their outermost configurations follow the general pattern \(ns^2np^1\).
Therefore, both neutral atoms possess three valence electrons.
The greater importance of +1 in thallium cannot be attributed to a smaller valence-electron count.
\(\textbf{Examine the oxidation-state difference:}\)
The +3 oxidation state formally involves participation of all three outermost electrons.
The +1 state can be described through participation of the \(np^1\) electron while retaining the outermost \(ns^2\) pair.
Among heavier p-block elements, this pair becomes comparatively less favourable for bonding participation.
The effect is commonly described as the inert-pair effect.
It contributes to the relative stability of thallium(I) compared with thallium(III).
\(\textbf{Interpret the oxide evidence:}\)
Aluminium oxide exhibits amphoteric behaviour.
Thallium(I) oxide has predominantly basic character.
Thallium is considerably more metallic than aluminium.
This difference is consistent with increasing metallic character down Group 13.
\(\textbf{Integrate the two observations:}\)
Oxidation-state stability and oxide basicity are different chemical properties.
The former involves the energetic availability of the outer electrons and stabilization within compounds.
The latter is influenced by metallic character and bonding.
Both trends are compatible with thallium's position near the bottom of Group 13.
The lower oxidation state does not mean the other two valence electrons have ceased to exist.
It means their participation is less energetically favourable in many thallium compounds.
302. Four consecutive Period 3 elements have the following approximate first ionization enthalpies.
| Element | Outermost configuration | \(IE_1\), \(\mathrm{kJ\,mol^{-1}}\) |
|---|
| Mg | \(3s^2\) | 738 |
| Al | \(3s^23p^1\) | 578 |
| Si | \(3s^23p^2\) | 786 |
| P | \(3s^23p^3\) | 1012 |
A student proposes two rules: (I) the first ionization enthalpy must increase at every successive atomic number, and (II) the introduction of a \(p\) electron must always make electron removal more difficult than from a filled \(s\) subshell. Which conclusion is correct?
ⓐ. Both rules fail because Mg has a greater nuclear charge than Al.
ⓑ. Rule I fails at Mg–Al; Rule II also fails because Al loses a \(3p\) electron more easily than Mg loses \(3s\).
ⓒ. Rule I is valid, but Rule II fails because aluminium has fewer valence electrons than magnesium.
ⓓ. Both rules are valid because silicon and phosphorus have greater ionization enthalpies than aluminium.
Correct Answer: Rule I fails at Mg–Al; Rule II also fails because Al loses a \(3p\) electron more easily than Mg loses \(3s\).
Explanation: \(\textbf{Identify the observed reversal:}\)
Magnesium has first ionization enthalpy \(738\,\mathrm{kJ\,mol^{-1}}\).
Aluminium has the lower value \(578\,\mathrm{kJ\,mol^{-1}}\).
Aluminium nevertheless has the greater atomic number.
Therefore, the proposed strictly increasing ionization-enthalpy rule is incorrect.
\(\textbf{Compare the electron configurations:}\)
Magnesium has the filled outermost \(3s^2\) subshell.
Its first electron is removed from a \(3s\) orbital.
Aluminium has the configuration \(3s^23p^1\).
Its differentiating electron occupies a \(3p\) orbital.
The \(3p\) electron is generally higher in energy and less penetrating than the \(3s\) electrons.
Consequently, it is comparatively easier to remove.
\(\textbf{Interpret the continuing trend:}\)
From aluminium to silicon and phosphorus, the first ionization enthalpy increases appreciably.
Increasing effective nuclear attraction contributes to this general rise.
The aluminium exception therefore does not eliminate the broad periodic trend.
It demonstrates that orbital energy and subshell occupancy can modify the effect of increasing nuclear charge.
\(\textbf{Distinguish the two proposed rules:}\)
Rule I incorrectly treats a general tendency as exception-free.
Rule II incorrectly predicts the effect of introducing an electron into a higher-energy \(p\) subshell.
The supplied data invalidate both propositions.
A scientifically sound explanation must account for effective nuclear charge and the energy of the orbital from which the electron is removed.
303. The oxides \(\mathrm{As_2O_3}\), \(\mathrm{As_2O_5}\) and \(\mathrm{Bi_2O_3}\) contain Group 15 elements. Under suitable conditions, arsenic(III) oxide is amphoteric, arsenic(V) oxide is acidic, and bismuth(III) oxide is predominantly basic. Which conclusion correctly separates the influence of oxidation state from the influence of position down the group?
ⓐ. Within arsenic oxides, higher oxidation state raises acidity; across +3 oxides, metallic character changes it.
ⓑ. Comparing the arsenic oxides shows that increasing oxidation state necessarily increases basicity; comparing arsenic with bismuth shows decreasing metallic character.
ⓒ. All three oxides should be equally amphoteric because their central elements belong to Group 15.
ⓓ. The change from arsenic(III) oxide to bismuth(III) oxide is caused entirely by an increase in the formal oxidation state of the central element.
Correct Answer: Within arsenic oxides, higher oxidation state raises acidity; across +3 oxides, metallic character changes it.
Explanation: \(\textbf{Separate the two comparisons:}\)
The three oxides allow two different chemical variables to be examined.
The two arsenic oxides contain the same element in different oxidation states.
Arsenic(III) oxide and bismuth(III) oxide contain different elements in the same oxidation state.
These comparisons must not be treated as if they test an identical periodic change.
\(\textbf{Compare the arsenic oxides:}\)
Oxygen is assigned oxidation state \(-2\).
Arsenic has oxidation state +3 in \(\mathrm{As_2O_3}\).
It has oxidation state +5 in \(\mathrm{As_2O_5}\).
The higher oxidation state is associated with greater acidic oxide character in this pair.
This is consistent with a common tendency among p-block oxides.
\(\textbf{Compare arsenic and bismuth:}\)
Arsenic and bismuth both belong to Group 15.
Metallic character increases considerably from arsenic towards bismuth.
Although both \(\mathrm{As_2O_3}\) and \(\mathrm{Bi_2O_3}\) contain the central element in oxidation state +3, their acid–base behaviour differs.
Bismuth(III) oxide exhibits more strongly basic behaviour.
\(\textbf{Interpret the combined evidence:}\)
Oxide character depends on both the oxidation state and the chemical nature of the central element.
A comparison that changes these variables simultaneously may not reveal which one is responsible.
The supplied pairs isolate the effects more effectively.
Thus, the correct conclusion recognizes two distinct influences rather than forcing every oxide into a single group-based classification.
304. Three elements X, Y and Z belong to consecutive Groups 14, 15 and 16 of the same period. Their neutral ground-state configurations terminate in \(np^2\), \(np^3\) and \(np^4\), respectively. An electron is removed from each atom. Which statement correctly compares the resulting ions and the relative stability of the initial electron arrangements?
ⓐ. All three ions contain three unpaired \(p\) electrons because one electron has been removed from each neutral atom.
ⓑ. Only Y produces a half-filled \(p\) subshell, because a half-filled neutral atom remains half-filled after ionization.
ⓒ. X gives \(np^1\), Y \(np^2\), Z \(np^3\); electron removal from Z forms stable half-filled \(p\) orbitals.
ⓓ. Z produces \(np^5\), making its first ionization less favourable than that of Y in every period.
Correct Answer: X gives \(np^1\), Y \(np^2\), Z \(np^3\); electron removal from Z forms stable half-filled \(p\) orbitals.
Explanation: \(\textbf{Identify the orbital occupancies:}\)
X has two electrons in its outermost \(p\) subshell.
According to Hund's rule, they occupy separate orbitals.
Y has three \(p\) electrons distributed singly across all three orbitals.
Z has four \(p\) electrons, so one orbital contains a pair.
\(\textbf{Follow the ionization processes:}\)
The first ionization removes one electron from the occupied \(p\) subshell.
X consequently changes from \(np^2\) to \(np^1\).
Y changes from \(np^3\) to \(np^2\).
Z changes from \(np^4\) to \(np^3\).
\(\textbf{Examine the half-filled arrangement:}\)
The neutral Y atom possesses a half-filled \(p\) subshell.
Ionization destroys that arrangement.
The neutral Z atom contains a paired set of \(p\) electrons.
Removal of one electron reduces the pairing repulsion and produces an \(np^3\) subshell.
\(\textbf{Interpret the periodic consequence:}\)
This difference can help explain why the first ionization enthalpy decreases slightly from certain Group 15 elements to the corresponding Group 16 elements in the same period.
The increase in nuclear charge still contributes to stronger attraction across the period.
However, the final electronic configuration and electron-pair repulsion also influence the energy required.
Thus, a simple orbital-count comparison would be insufficient without considering the electronic arrangements before and after ionization.
305. Three solids contain the Group 14 elements carbon, silicon and lead. Their representative structures are described as follows.
Structure P is a three-dimensional network in which each carbon atom is tetrahedrally bonded to four neighbouring carbon atoms. Structure Q is a three-dimensional network in which each silicon atom is tetrahedrally bonded to four neighbouring silicon atoms. Structure R is a metallic lead crystal with delocalized electronic states. All three structures are elemental solids, not chemical compounds.
Which conclusion is consistent with their electronic configurations and the differences in bonding?
ⓐ. All three solids must be electrical insulators because every Group 14 atom has four valence electrons.
ⓑ. P and Q have similar network connectivity but differing electronic properties; R is metallic, with oxidation state zero throughout.
ⓒ. R has lead in oxidation state +2 because delocalized metallic electrons are formally transferred between atoms.
ⓓ. P and Q must form discrete tetrahedral molecules rather than extended networks because their atoms have coordination number four.
Correct Answer: P and Q have similar network connectivity but differing electronic properties; R is metallic, with oxidation state zero throughout.
Explanation: \(\textbf{Establish the Group 14 similarity:}\)
Carbon, silicon and lead possess four valence electrons.
Their general valence-shell configuration is \(ns^2np^2\).
However, their atomic sizes and bonding tendencies differ substantially.
\(\textbf{Examine Structures P and Q:}\)
P represents diamond.
Q represents crystalline silicon.
Both contain extended tetrahedral covalent networks.
The coordination number four describes the number of directly bonded neighbours.
It does not require the solid to consist of discrete four-atom molecules.
\(\textbf{Compare electronic behaviour:}\)
Diamond and crystalline silicon have different electronic energy structures.
Diamond is a wide-gap electrical insulator under ordinary conditions.
Pure crystalline silicon exhibits semiconducting behaviour.
Comparable geometric connectivity therefore does not establish identical electronic properties.
\(\textbf{Examine Structure R:}\)
Lead is metallic under ordinary conditions.
Its electronic structure permits metallic conduction.
A metallic solid should not automatically be represented as containing isolated Pb(II) ions merely because lead can form Pb(II) compounds.
\(\textbf{Apply oxidation-state accounting:}\)
All three substances are elemental forms.
Their constituent elements therefore have formal oxidation state zero.
The differences in conductivity and bonding arise from electronic structure and atomic arrangement.
The comparison illustrates the broad transition from covalent-network behaviour towards metallic character down Group 14.
306. The first ionization enthalpies of three noble gases are approximately:
| Element | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) | Normal boiling point (K) |
|---|
| Ne | 2081 | 27 |
| Ar | 1521 | 87 |
| Xe | 1170 | 165 |
Suppose a student interprets the decreasing ionization enthalpy as proof that attractive forces between neutral noble-gas atoms also decrease down the group. Which assessment is correct?
ⓐ. The conclusion is valid because ionization enthalpy directly measures the strength of attractions between neighbouring atoms.
ⓑ. The conclusion is valid for neon and argon but must reverse at xenon because xenon can form compounds.
ⓒ. Both ionization enthalpy and boiling point must decrease down the group because the outermost configuration remains closed-shell.
ⓓ. The conclusion confuses ionization with dispersion; larger noble-gas atoms are more polarizable despite easier ionization.
Correct Answer: The conclusion confuses ionization with dispersion; larger noble-gas atoms are more polarizable despite easier ionization.
Explanation: \(\textbf{Interpret ionization enthalpy:}\)
The first ionization enthalpy measures the energy needed to remove an electron from an isolated gaseous atom.
The outermost electrons occupy successively higher principal shells from neon to xenon.
Greater shielding and increased distance generally make these electrons easier to remove.
Therefore, first ionization enthalpy decreases.
\(\textbf{Interpret the boiling-point data:}\)
The normal boiling point rises substantially from neon to xenon.
All three gases occur as monatomic substances.
Their atoms have no permanent molecular dipole moments.
The principal attractive forces between their atoms are London dispersion forces.
\(\textbf{Connect polarizability with boiling:}\)
The electron cloud becomes larger and more diffuse down the group.
It consequently becomes more polarizable.
Temporary induced-dipole attractions therefore become stronger.
More energy is required for vaporization, contributing to the increase in boiling point.
\(\textbf{Distinguish the two energetic processes:}\)
Removing an electron from an isolated atom is not the same process as separating neutral atoms from a condensed phase.
A low ionization enthalpy does not imply weak dispersion attractions.
Indeed, the two properties can change in opposite directions down a group.
\(\textbf{Interpret noble-gas reactivity:}\)
The lower ionization enthalpy of xenon also helps explain why suitable xenon compounds can exist.
However, its higher boiling point does not by itself prove that it forms chemical bonds in the pure condensed element.
The correct explanation requires separating atomic electron-removal energy from intermolecular attraction.
307. The following general changes occur when moving from boron towards thallium in Group 13:
Statement I: The outermost valence-electron count remains three.
Statement II: Metallic character generally becomes more pronounced.
Statement III: The +1 oxidation state generally becomes more significant among the heavier members.
Statement IV: The tendency of representative oxides to exhibit basic character must decrease because the atoms become larger.
Which combination is consistent with the broad periodic trends?
ⓐ. Statements I and IV only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and III only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and III only
Explanation: \(\textbf{Evaluate Statement I:}\)
Group 13 elements possess the general valence-shell configuration \(ns^2np^1\).
They therefore have three valence electrons.
Moving down the group changes the principal quantum number but not the characteristic valence-electron count.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Boron exhibits strongly covalent and non-metallic characteristics.
Aluminium and the familiar heavier members are metallic.
The general increase in metallic character down the group is consistent with changes in atomic size, electron attraction and bonding.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
The outermost \(ns^2\) pair becomes comparatively less favourable for bonding participation in heavier p-block elements.
This contributes to the increasing relative stability of lower positive oxidation states.
Thallium(I) provides a characteristic example.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
Oxide basicity generally becomes more pronounced as the metallic character of the parent element increases.
Boron trioxide is acidic, whereas aluminium oxide is amphoteric.
Still heavier Group 13 oxides can exhibit more basic character.
Therefore, Statement IV gives the trend in the wrong direction.
\(\textbf{Integrate the chemical evidence:}\)
A common valence-electron configuration does not imply identical oxidation-state preference or bonding character.
The changes in metallic nature, oxide behaviour and inert-pair influence can occur simultaneously.
Only Statements I, II and III are consistent with the broad down-group trends.
308. Carbon and silicon can each form tetrahedral compounds containing four hydrogen atoms, but their corresponding E–H bond properties differ. Consider the following hypothetical experimental observations.
Under identical heating conditions, a sample of silane decomposes more readily than a comparable sample of methane. Independent bond-energy measurements show that the ordinary Si–H bond is weaker than the C–H bond. The two hydrides are assumed to have no significant intermolecular hydrogen bonding.
Which interpretation is most defensible?
ⓐ. Weaker Si–H bonding can explain the difference, but decomposition rates also depend on pathways and activation barriers.
ⓑ. Si–H bonds are weaker, so silane must decompose more rapidly at any temperature, irrespective of the reaction pathway.
ⓒ. Methane must be more reactive because carbon is smaller and forms stronger covalent bonds.
ⓓ. The observations prove that silicon has oxidation state +2 in silane, while carbon has oxidation state +4 in methane.
Correct Answer: Weaker Si–H bonding can explain the difference, but decomposition rates also depend on pathways and activation barriers.
Explanation: \(\textbf{Establish the common structural framework:}\)
Carbon and silicon both belong to Group 14.
Their simple tetrahydrides are methane and silane.
Both have tetrahedral molecular geometries.
However, their central atoms differ considerably in size and orbital characteristics.
\(\textbf{Interpret the bond-energy evidence:}\)
The smaller carbon atom permits relatively effective overlap with hydrogen.
The resulting C–H bond is generally stronger than the corresponding Si–H bond.
The supplied observation is therefore consistent with different bonding strengths in the two hydrides.
\(\textbf{Distinguish bond strength from reaction rate:}\)
Bond dissociation enthalpy describes a specified bond-breaking process.
The rate of a decomposition reaction depends on its reaction mechanism and activation barrier.
These quantities are related to chemical bonding but are not identical.
Consequently, the weaker Si–H bond can help explain the observed reactivity difference without proving a universal rate ordering for every possible reaction.
\(\textbf{Examine formal oxidation states:}\)
Hydrogen is more electronegative than silicon but less electronegative than carbon.
Therefore, hydrogen has oxidation state \(-1\) in silane and +1 in methane.
Silicon consequently has oxidation state +4 in silane, while carbon has oxidation state \(-4\) in methane.
Neither central atom needs to possess oxidation state +2 in the stated compounds.
\(\textbf{Interpret the first-element distinction:}\)
Similar hydride formulas and molecular geometries do not guarantee identical bond strengths.
Carbon's compact valence orbitals contribute to its distinctive covalent bonding.
The correct conclusion preserves the distinction between bond energy, reaction kinetics and formal oxidation-state accounting.
309. An unidentified Period 4 p-block element X has the configuration \([\mathrm{Ar}]\,3d^{10}4s^24p^3\). It forms a representative trioxide in which X has oxidation state +3 and a higher oxide in which X has oxidation state +5. Compared with the first member of its group, X displays greater metallic character and different element–element bonding tendencies. Which identification and inference are correct?
ⓐ. X is selenium; its \(4p^3\) configuration explains its six valence electrons.
ⓑ. X is germanium; its trioxide establishes that Group 14 elements possess five valence electrons.
ⓒ. X is bromine; its common oxidation state must be +7 because it has three unpaired electrons.
ⓓ. X is Group 15 arsenic; greater metallic character than nitrogen does not alter its valence-electron count.
Correct Answer: X is Group 15 arsenic; greater metallic character than nitrogen does not alter its valence-electron count.
Explanation: \(\textbf{Determine the valence-electron count:}\)
The outermost configuration is \(4s^24p^3\).
It contains five valence electrons.
The \(p^3\) occupancy identifies Group 15.
The highest occupied principal shell is the fourth, so the element belongs to Period 4.
\(\textbf{Identify the element:}\)
The Period 4 Group 15 element is arsenic.
Its supplied configuration is consistent with atomic number 33.
The \(3d^{10}\) subshell forms part of the inner electronic structure.
It does not increase the ordinary valence-electron count from five to fifteen.
\(\textbf{Interpret the oxide formulas:}\)
Arsenic can exhibit oxidation states +3 and +5.
Its representative trioxide and pentoxide reflect those positive oxidation-state possibilities.
However, the stability and acid–base behaviour of these oxides cannot be inferred solely from the group number.
\(\textbf{Compare arsenic with nitrogen:}\)
Nitrogen is the first member of Group 15.
It is small and strongly non-metallic.
Arsenic is larger and exhibits more pronounced metalloid character.
Nevertheless, both possess five valence electrons.
\(\textbf{Interpret the bonding contrast:}\)
Nitrogen's compact \(2p\) orbitals support exceptionally strong multiple bonding in molecular nitrogen.
Arsenic has larger valence orbitals and a different balance of single- and multiple-bonding tendencies.
Thus, shared group membership preserves the general valence configuration without enforcing identical elemental structures or chemical behaviour.
The correct identification is arsenic.
310. The following diagram description concerns the valence orbitals of two p-block atoms.
Atom X has three \(p\) orbitals, each occupied by one electron with parallel spin. Atom Y also has three \(p\) orbitals, but one contains an electron pair while the remaining two each contain one electron. Both atoms possess a filled outermost \(s\) subshell and belong to the same period.
Suppose a reaction requires removal of one \(p\) electron from each atom. Which prediction correctly compares the resulting subshells and the qualitative electronic contribution to ionization?
ⓐ. X produces a half-filled \(p\) subshell, while Y produces a \(p^2\) subshell; X therefore must ionize more readily.
ⓑ. X produces \(p^2\), Y produces \(p^3\); removing Y's electron reduces pairing repulsion and yields a half-filled subshell.
ⓒ. X and Y both produce \(p^3\) because ionization always removes an electron from the outermost \(s\) subshell first.
ⓓ. Both atoms produce completely filled \(p\) subshells because the electrons rearrange to maximize the number of pairs.
Correct Answer: X produces \(p^2\), Y produces \(p^3\); removing Y's electron reduces pairing repulsion and yields a half-filled subshell.
Explanation: \(\textbf{Translate the orbital diagrams:}\)
Atom X has three singly occupied \(p\) orbitals.
Its subshell configuration is therefore \(p^3\).
Atom Y has one paired orbital and two singly occupied orbitals.
Its subshell configuration is \(p^4\).
\(\textbf{Identify the corresponding groups:}\)
Both atoms possess a filled outermost \(s^2\) subshell.
X therefore has five valence electrons and belongs to Group 15.
Y has six valence electrons and belongs to Group 16.
Since they lie in the same period, Y follows X in atomic number.
\(\textbf{Follow electron removal:}\)
Removing one occupied \(p\) electron from X gives \(p^2\).
Removing one from Y gives \(p^3\).
Therefore, Y acquires the half-filled configuration that X possessed initially.
\(\textbf{Explain the energetic consequence:}\)
The neutral \(p^4\) arrangement of Y contains one paired set of electrons.
The paired electrons experience additional repulsion.
Removing one electron can reduce this repulsion and produce a comparatively favourable half-filled arrangement.
\(\textbf{Reconcile the periodic trend:}\)
Y has the greater nuclear charge, which generally favours a higher first ionization enthalpy.
However, the difference in subshell occupancy may oppose that effect.
This explains a characteristic local ionization-enthalpy anomaly between Groups 15 and 16.
The reasoning requires interpreting the orbital diagram, predicting both ionic configurations and distinguishing the competing energetic contributions.
311. The acidic character of simple Group 16 hydrides generally increases from \(\mathrm{H_2S}\) to \(\mathrm{H_2Te}\), while the normal boiling points also increase down that part of the group. A student argues that both trends must be caused by progressively stronger E–H bonds. Which assessment is correct?
ⓐ. The argument is valid because stronger E–H bonds favour both proton loss and vaporization.
ⓑ. Down the group, H–E bonds weaken while higher polarizability strengthens intermolecular dispersion attractions.
ⓒ. The argument is valid because tellurium is more electronegative than sulphur.
ⓓ. The argument is invalid because the hydrides have different numbers of hydrogen atoms and cannot be compared.
Correct Answer: Down the group, H–E bonds weaken while higher polarizability strengthens intermolecular dispersion attractions.
Explanation: \(\textbf{Examine intramolecular bonding:}\)
Sulphur, selenium and tellurium belong to Group 16.
Their simple hydrides share the formula \(\mathrm{H_2E}\).
Atomic size increases down the group.
The E–H bond length generally increases, while the bond strength decreases.
Therefore, the bonds do not become progressively stronger.
\(\textbf{Interpret aqueous acidity:}\)
Proton release from a hydride depends on the complete acid–base equilibrium.
Weakening E–H bonds contributes to the increasing acidity of heavier Group 16 hydrides.
Consequently, \(\mathrm{H_2Te}\) is more acidic than \(\mathrm{H_2S}\) under suitable comparable aqueous conditions.
\(\textbf{Interpret the boiling-point change:}\)
Boiling a molecular hydride ordinarily separates intact molecules.
It does not require cleavage of every E–H covalent bond.
Moving down the group, the molecules become larger and more polarizable.
The resulting stronger London dispersion attractions contribute to increasing boiling points.
\(\textbf{Distinguish the two effects:}\)
Acid strength depends on proton-transfer energetics.
Boiling point depends on intermolecular interactions.
The same structural change down a group can therefore produce apparently opposing bond-strength and boiling-point trends.
The student's proposed common explanation is incorrect because it confuses intramolecular bond strength with intermolecular attractions.
The correct interpretation preserves the distinct mechanisms controlling the two properties.
312. Consider the following Assertion and Reason.
Assertion (A): An element with a closed outermost electronic shell need not have the highest normal boiling point among all elements in the same period.
Reason (R): Normal boiling point depends on the structure and intermolecular or interatomic interactions of the elemental substance, whereas valence-shell completion primarily concerns its electronic configuration.
Choose the correct option.
ⓐ. Both A and R are true, and R correctly explains A.
ⓑ. Both A and R are true, but R does not explain A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: \(\textbf{Examine the assertion:}\)
A noble gas has a completed outermost electronic shell.
This contributes to its low ordinary chemical reactivity.
However, chemical inertness does not require a high normal boiling point.
Noble gases commonly have very low boiling points.
Therefore, the assertion is true.
\(\textbf{Compare different elemental structures:}\)
Silicon can form an extended covalent network.
Phosphorus and sulphur can form molecular solids.
Chlorine exists as diatomic molecules.
Argon exists as monatomic atoms.
These differences strongly influence melting and boiling behaviour.
\(\textbf{Examine noble-gas attraction:}\)
A noble-gas atom has no permanent molecular dipole moment.
Condensation is supported mainly by weak London dispersion attractions.
The completed valence shell does not automatically produce the strong extended bonding present in a network covalent solid.
\(\textbf{Evaluate the reason:}\)
The reason correctly distinguishes an electronic configuration from a bulk physical property.
Boiling point depends on the energetic and structural conditions governing vaporization.
Valence-shell completion influences chemical behaviour but does not provide a universal boiling-point ranking.
Therefore, the reason is true.
\(\textbf{Establish the explanation:}\)
The dependence of boiling point on bonding and condensed-phase structure directly explains why a noble gas need not have the highest boiling point in its period.
The assertion and reason are both true, and the reason correctly explains the assertion.
313. A hypothetical p-block element X belongs to Period 5. Its two most common positive oxidation states differ by two units, and the lower state becomes relatively more stable in the corresponding Period 6 group member. The lower state can formally retain the outermost \(ns^2\) electron pair. The Period 5 element has four valence electrons. Which identification and prediction are correct?
ⓐ. X is indium, and its lower positive oxidation state is +1 because it belongs to Group 13.
ⓑ. X is antimony, and its lower positive oxidation state is +3 because it belongs to Group 15.
ⓒ. X is tin and its lower positive state is +2, which becomes more important in lead through the inert-pair effect.
ⓓ. X is tellurium, and its lower positive oxidation state is +4 because it belongs to Group 16.
Correct Answer: X is tin and its lower positive state is +2, which becomes more important in lead through the inert-pair effect.
Explanation: \(\textbf{Determine the group:}\)
The atom possesses four valence electrons.
For an ordinary p-block element, this corresponds to the valence configuration \(ns^2np^2\).
Therefore, X belongs to Group 14.
\(\textbf{Determine the period identity:}\)
The element belongs to Period 5.
The Period 5 Group 14 element is tin.
Its outermost configuration is \(5s^25p^2\).
\(\textbf{Identify the oxidation states:}\)
Tin commonly exhibits oxidation states +2 and +4.
These differ by two units.
The lower oxidation state can be described formally by participation of the outermost \(p\) electrons while retaining the \(s^2\) pair.
\(\textbf{Compare the heavier member:}\)
Lead lies below tin in Group 14.
Its outermost configuration is \(6s^26p^2\).
The outermost \(6s^2\) pair becomes comparatively less favourable for bonding participation.
This contributes to the enhanced stability of Pb(II) relative to Pb(IV).
\(\textbf{Interpret the distinction:}\)
The inert-pair effect does not change the neutral atom's valence-electron count.
It influences the relative energetics of participation in different oxidation states.
Both tin and lead can still exhibit more than one positive oxidation state under suitable chemical conditions.
The correct identification and prediction therefore require group recognition, period recognition and oxidation-state stability reasoning.
314. The following data describe the first ionization enthalpies and electron-gain enthalpies of three consecutive Period 3 elements.
| Element | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) | First electron-gain enthalpy |
|---|
| P | 1012 | Moderately negative |
| S | 1000 | Strongly negative |
| Cl | 1251 | More negative than that of S |
Which conclusion correctly interprets why the first ionization enthalpy decreases from P to S but rises from S to Cl, without assuming that electron-gain enthalpy must follow the same pattern?
ⓐ. The P–S ionization dip reflects \(3p\)-electron pairing, while S–Cl reflects stronger nuclear attraction; electron gain differs.
ⓑ. The P–S reversal proves that sulphur has fewer protons than phosphorus, while chlorine has more protons than sulphur.
ⓒ. Sulphur has a lower first ionization enthalpy because its valence electrons occupy a higher principal shell than phosphorus's.
ⓓ. The electron-gain and ionization processes are reverse reactions for the same pair of ionic species, so their enthalpy trends must be numerically identical.
Correct Answer: The P–S ionization dip reflects \(3p\)-electron pairing, while S–Cl reflects stronger nuclear attraction; electron gain differs.
Explanation: \(\textbf{Identify the valence configurations:}\)
Phosphorus has \(3s^23p^3\).
Sulphur has \(3s^23p^4\).
Chlorine has \(3s^23p^5\).
All three elements belong to Period 3.
Their outermost electrons occupy the same principal shell.
\(\textbf{Explain the phosphorus–sulphur reversal:}\)
Phosphorus possesses a relatively stable half-filled \(3p^3\) subshell.
Sulphur has an additional electron paired within one \(3p\) orbital.
This introduces electron–electron repulsion.
Removal of one electron from sulphur reduces that repulsion.
Consequently, its first ionization enthalpy is slightly lower than phosphorus's.
\(\textbf{Explain the sulphur–chlorine increase:}\)
Nuclear charge increases from sulphur to chlorine.
The additional valence electron still occupies the third principal shell.
Increasing effective nuclear attraction generally makes electron removal more difficult.
The supplied chlorine ionization enthalpy is therefore higher.
\(\textbf{Distinguish electron-gain enthalpy:}\)
First ionization enthalpy concerns electron removal from a neutral gaseous atom.
First electron-gain enthalpy concerns electron addition to a neutral gaseous atom.
These are not the forward and reverse processes for an identical pair of initial and final species.
Their numerical trends need not match.
\(\textbf{Interpret the complete data:}\)
Sulphur's ionization exception and chlorine's strongly favourable electron gain are compatible with their different subshell occupancies.
The observations cannot be explained by a single universal rule involving only atomic number.
The correct conclusion separates electron-removal mechanisms from electron-attachment energetics.
315. Three broad comparisons are made between the first members of Groups 14, 16 and 18 and their heavier congeners.
| Group | First member | Characteristic observation |
|---|
| 14 | C | Extensive catenation and effective multiple bonding |
| 16 | O | Strong multiple bonding but comparatively weak O–O single bonding |
| 18 | He | Complete first shell and exceptionally low ordinary chemical reactivity |
Which statement provides the most complete scientific interpretation without incorrectly assigning a single cause to all three anomalies?
ⓐ. All three anomalies can be attributed mainly to weak shielding of the outer \(2p\) orbitals.
ⓑ. Carbon's and oxygen's behaviour can be explained entirely by identical electron-pair repulsions, while helium's behaviour arises from its large polarizability.
ⓒ. Carbon favours orbital overlap for multiple bonds; oxygen has compact lone-pair repulsion; helium has a filled \(1s^2\) shell.
ⓓ. All three first members behave anomalously because they possess fewer valence electrons than the heavier members of their respective groups.
Correct Answer: Carbon favours orbital overlap for multiple bonds; oxygen has compact lone-pair repulsion; helium has a filled \(1s^2\) shell.
Explanation: \(\textbf{Examine carbon:}\)
Carbon is the first member of Group 14.
Its compact valence orbitals permit strong directional covalent bonding.
Effective overlap supports strong C–C single bonds and multiple bonds involving suitable atoms.
This contributes to carbon's pronounced catenation and extensive structural variety.
\(\textbf{Examine oxygen:}\)
Oxygen belongs to Group 16 and has six valence electrons.
Its compact \(2p\) orbitals can support effective multiple bonding.
However, ordinary O–O single bonding brings non-bonding electron pairs on neighbouring oxygen atoms close together.
The resulting repulsion weakens the single bond.
Therefore, oxygen's multiple-bonding tendency and single-bonding weakness must be explained through competing effects.
\(\textbf{Examine helium:}\)
Helium has the electronic configuration \(1s^2\).
The first principal shell can accommodate only two electrons.
Its shell is therefore completely filled.
Its compact electronic structure and very high ionization enthalpy contribute to exceptionally low ordinary chemical reactivity.
Helium does not require an octet to display noble-gas behaviour.
\(\textbf{Compare the anomaly mechanisms:}\)
Carbon's behaviour highlights effective covalent orbital overlap.
Oxygen's single-bond anomaly highlights electron-pair repulsion.
Helium's behaviour highlights first-shell completion and strong electron binding.
These mechanisms are different despite the small sizes of all three atoms.
The absence of valence \(d\) orbitals is not an adequate universal explanation.
Likewise, members of a group retain the same general valence-electron count as their heavier congeners, apart from helium's special configuration within Group 18.
Thus, the most complete interpretation assigns the appropriate electronic cause to each specific first-member anomaly.
316. The following approximate first ionization enthalpies are given for Group 13 elements.
| Element | B | Al | Ga | In | Tl |
|---|
| \(IE_1\) in \(\mathrm{kJ\,mol^{-1}}\) | 801 | 578 | 579 | 558 | 589 |
A student argues that the first ionization enthalpy must decrease at every successive step down Group 13 because the principal quantum number increases. Which assessment best reconciles the data with periodic theory?
ⓐ. The increases from Al to Ga and In to Tl show that the number of valence electrons increases down Group 13.
ⓑ. The increases from Al to Ga and In to Tl show that the first electron of the heavier atoms is removed from an inner \(d\) or \(f\) subshell.
ⓒ. Poor \(d,f\)-electron shielding and orbital-energy shifts produce irregular ionization trends.
ⓓ. The data show that increasing nuclear charge always reduces atomic radius, regardless of the number of occupied principal shells.
Correct Answer: Poor \(d,f\)-electron shielding and orbital-energy shifts produce irregular ionization trends.
Explanation: \(\textbf{Establish the broad group tendency:}\)
Group 13 elements possess the general valence-shell configuration \(ns^2np^1\).
Their first ionization removes an outermost \(p\) electron.
Moving down the group generally increases the principal quantum number of this electron.
Greater distance from the nucleus and additional shielding usually favour a reduction in first ionization enthalpy.
However, the table does not show a perfectly smooth decrease.
\(\textbf{Locate the departures:}\)
The value increases slightly from aluminium to gallium.
It also increases from indium to thallium.
These departures must be explained without changing the common Group 13 valence-electron count.
\(\textbf{Examine gallium:}\)
Gallium contains a filled inner \(3d\) subshell that aluminium lacks.
Inner \(d\) electrons do not shield the nuclear charge as effectively as an equally placed collection of \(s\) electrons.
Consequently, the increased nuclear charge is not completely offset by shielding.
This modifies the expected attraction experienced by gallium's outermost electron.
\(\textbf{Examine thallium:}\)
Thallium contains additional filled inner subshells, including \(4f\) electrons.
Incomplete shielding by these electrons contributes to contraction effects in heavier elements.
Changes in outer-orbital energies also influence its ionization behaviour.
Thus, thallium does not simply continue a perfectly decreasing sequence from indium.
\(\textbf{Interpret the complete trend:}\)
The general down-group principle remains useful, but it is not exception-free.
Neither the number of valence electrons nor the subshell from which the first electron is removed changes in the manner suggested by the incorrect options.
The appropriate explanation must account for both increasing shell extent and incomplete shielding.
The data therefore illustrate why ionization enthalpy depends on competing electronic influences.
317. The approximate H–E–H bond angles of three Group 15 hydrides are shown below.
| Hydride | Bond angle | Central-atom non-bonding pairs |
|---|
| \(\mathrm{NH_3}\) | \(107^\circ\) | 1 |
| \(\mathrm{PH_3}\) | \(93.5^\circ\) | 1 |
| \(\mathrm{AsH_3}\) | \(91.8^\circ\) | 1 |
Which explanation best accounts for the approach of the bond angles towards \(90^\circ\) down the series while preserving the common molecular geometry?
ⓐ. The central atoms gain an additional lone pair down the group, forcing the H–E–H angles towards \(90^\circ\).
ⓑ. Heavier atoms bond with more nearly perpendicular \(p\) orbitals; ammonia shows greater orbital mixing.
ⓒ. The number of E–H bonds decreases from ammonia to arsine, causing the molecular geometry to become linear.
ⓓ. The central atoms change from trigonal pyramidal to trigonal planar as their electronegativities decrease.
Correct Answer: Heavier atoms bond with more nearly perpendicular \(p\) orbitals; ammonia shows greater orbital mixing.
Explanation: \(\textbf{Identify the common electron-pair arrangement:}\)
Nitrogen, phosphorus and arsenic belong to Group 15.
Each neutral atom has five valence electrons.
In the simple trihydrides, the central atom forms three E–H bonds and retains one lone pair.
Therefore, all three molecules are trigonal pyramidal.
The observed angle differences do not indicate changes in the number of bonded hydrogen atoms.
\(\textbf{Examine ammonia:}\)
Nitrogen has compact \(2s\) and \(2p\) valence orbitals.
Its bonding arrangement produces an H–N–H angle considerably greater than \(90^\circ\).
Orbital mixing and the distribution of the bonding and non-bonding electrons contribute to this geometry.
\(\textbf{Examine the heavier hydrides:}\)
Phosphorus and arsenic have larger, more diffuse valence orbitals.
Their E–H bonding can be described as involving a greater contribution from comparatively unhybridized \(p\)-orbital character.
The three \(p\) orbitals are mutually perpendicular in the elementary orbital picture.
Consequently, the bond angles approach \(90^\circ\).
\(\textbf{Evaluate the lone-pair explanation:}\)
All three central atoms retain one lone pair.
Therefore, an increasing number of lone pairs cannot account for the trend.
The molecules also remain pyramidal throughout the series.
\(\textbf{Interpret the first-member distinction:}\)
Equal numbers of electron domains do not require identical bond angles.
The relative energies and spatial characteristics of the valence orbitals also matter.
Nitrogen's compact second-period orbitals produce a bonding arrangement distinct from those of heavier Group 15 elements.
The trend is therefore consistent with changing orbital contributions while retaining the same broad molecular geometry.
318. Four oxides of p-block elements are examined under suitable conditions.
| Oxide | Formal oxidation state of the non-oxygen element | Observed characteristic behaviour |
|---|
| \(\mathrm{CO}\) | +2 | Neutral oxide |
| \(\mathrm{NO}\) | +2 | Neutral oxide |
| \(\mathrm{CO_2}\) | +4 | Acidic oxide |
| \(\mathrm{SO_2}\) | +4 | Acidic oxide |
Which conclusion is supported by the complete set of observations?
ⓐ. Every oxide with a positive central-element oxidation state must be acidic.
ⓑ. Every oxide containing a Group 14 element must be neutral.
ⓒ. An oxide is amphoteric whenever its central element can exhibit more than one oxidation state.
ⓓ. Neither an element's identity nor its formal oxidation state alone determines oxide acid–base behaviour.
Correct Answer: Neither an element's identity nor its formal oxidation state alone determines oxide acid–base behaviour.
Explanation: \(\textbf{Distinguish oxide classifications:}\)
An acidic oxide exhibits characteristic acid-forming or base-reacting behaviour under suitable conditions.
A neutral oxide does not exhibit the ordinary acid–base reactions characteristic of acidic or basic oxides.
An amphoteric oxide reacts with suitable acids and bases.
These classifications describe chemical behaviour rather than merely the presence of oxygen.
\(\textbf{Compare carbon monoxide and carbon dioxide:}\)
Carbon has oxidation state +2 in CO.
Carbon monoxide is ordinarily classified as a neutral oxide.
Carbon has oxidation state +4 in \(\mathrm{CO_2}\).
Carbon dioxide is an acidic oxide.
Thus, two oxides of the same element can have different acid–base classifications.
\(\textbf{Examine nitric oxide:}\)
Nitrogen has oxidation state +2 in NO.
Nitric oxide is classified as a neutral oxide.
Its ability to undergo redox reactions does not automatically make it an acidic oxide.
Acid–base classification and redox behaviour must therefore be distinguished.
\(\textbf{Examine sulphur dioxide:}\)
Sulphur has oxidation state +4 in \(\mathrm{SO_2}\).
Sulphur dioxide exhibits acidic oxide character.
However, carbon dioxide has the same central-element oxidation state despite containing a different p-block element.
\(\textbf{Draw the comparative inference:}\)
The data do not support a universal classification based solely on a positive oxidation state.
They also do not support assigning one acid–base character to every oxide of a particular group.
Bonding, the identity of the central element and actual chemical reactions must be considered.
The correct conclusion preserves oxidation-state information without treating it as a substitute for chemical evidence.
319. Five consecutive Period 4 p-block elements are represented by their valence-shell configurations.
| Element | Configuration | Approximate Pauling electronegativity |
|---|
| Ga | \(4s^24p^1\) | 1.81 |
| Ge | \(4s^24p^2\) | 2.01 |
| As | \(4s^24p^3\) | 2.18 |
| Se | \(4s^24p^4\) | 2.55 |
| Br | \(4s^24p^5\) | 2.96 |
A student predicts that the metallic character decreases across this sequence and that a bond formed between each element and hydrogen must become progressively less polar. Which assessment is correct?
ⓐ. The metallic-character prediction is inconsistent with the configurations, but the bond-polarity prediction must hold because hydrogen has atomic number one.
ⓑ. Both predictions must hold because electronegativity and metallic character increase together across a period.
ⓒ. Metallic character decreases as expected, but E–H bond polarity depends on electronegativity differences, not that trend alone.
ⓓ. Neither prediction can be examined without first determining all five melting points.
Correct Answer: Metallic character decreases as expected, but E–H bond polarity depends on electronegativity differences, not that trend alone.
Explanation: \(\textbf{Interpret the across-period change:}\)
The five elements belong to consecutive p-block groups in Period 4.
Their nuclear charges increase while their outermost electrons occupy the same principal shell.
The general tendency is towards increasing effective nuclear attraction and increasing non-metallic character.
The supplied electronegativity values rise consistently across the sequence.
Therefore, decreasing metallic character is a reasonable broad prediction.
\(\textbf{Identify the fixed bonding reference:}\)
The student's second prediction concerns bonds to hydrogen.
Hydrogen has a Pauling electronegativity of approximately 2.20.
Bond polarity is related to the difference between the electronegativities of the two bonded atoms.
The relevant comparison is the magnitude of this difference, not merely the electronegativity of the p-block element.
\(\textbf{Evaluate the changing differences:}\)
Gallium is appreciably less electronegative than hydrogen.
Germanium is closer to hydrogen in electronegativity.
Arsenic has an electronegativity almost equal to hydrogen's.
Selenium and bromine are more electronegative than hydrogen.
Consequently, the magnitude of the electronegativity difference decreases towards arsenic and then increases.
\(\textbf{Interpret the direction change:}\)
The formal direction of bond polarization also reverses as the partner becomes more electronegative than hydrogen.
Thus, a monotonic increase in the electronegativity of one atom does not require a monotonic decrease in the polarity of its bond with a fixed reference atom.
\(\textbf{Connect the two trends:}\)
Metallic character is a broad elemental property.
Bond polarity is a comparison between two specified bonding partners.
The data support the first prediction but not the second.
The correct conclusion therefore separates a periodic elemental trend from a partner-dependent bonding property.
320. Phosphorus and bismuth both belong to Group 15 and can exhibit oxidation states +3 and +5. Under suitable chemical conditions, phosphorus(V) compounds are often comparatively stable, whereas bismuth(V) compounds tend to exhibit oxidizing character because bismuth(III) is relatively stable. Which interpretation correctly connects the contrasting redox tendencies with their electronic configurations?
ⓐ. Greater inert-pair character stabilizes Bi(III) relative to Bi(V), making suitable Bi(V) compounds oxidizing.
ⓑ. Bismuth(V) is oxidizing because the \(6p^3\) electrons are more reluctant to participate than its \(6s^2\) pair.
ⓒ. Phosphorus(V) is stable because phosphorus is metallic, while bismuth(V) is unstable because bismuth is non-metallic.
ⓓ. The difference proves that phosphorus and bismuth belong to different periodic groups.
Correct Answer: Greater inert-pair character stabilizes Bi(III) relative to Bi(V), making suitable Bi(V) compounds oxidizing.
Explanation: \(\textbf{Establish the common valence configuration:}\)
Phosphorus and bismuth possess the general Group 15 configuration \(ns^2np^3\).
Both therefore have five valence electrons.
The +3 oxidation state is associated formally with participation of three outermost electrons.
The +5 state involves participation of all five.
\(\textbf{Examine phosphorus:}\)
Phosphorus is a relatively small, non-metallic Group 15 element.
Suitable bonding environments can stabilize phosphorus in oxidation state +5.
Consequently, many phosphorus(V) compounds are chemically important.
The stability of a particular compound nevertheless depends on its bonding partners and conditions.
\(\textbf{Examine bismuth:}\)
Bismuth is considerably heavier and more metallic.
Its outermost \(6s^2\) pair becomes comparatively less favourable for bonding participation.
This is an example of the inert-pair effect.
Therefore, Bi(III) becomes relatively more stable than Bi(V).
\(\textbf{Connect stability with redox tendency:}\)
A suitable Bi(V) species may undergo reduction to Bi(III).
When it gains electrons in such a process, it acts as an oxidizing agent.
This does not imply that every Bi(V) compound reacts identically under every set of conditions.
\(\textbf{Interpret the broader trend:}\)
The difference is not caused by a decrease in the number of valence electrons down Group 15.
It reflects changing orbital energies and compound-formation energetics.
Thus, the same group can exhibit contrasting relative oxidation-state stabilities in its lighter and heavier members.
The correct interpretation connects the inert-pair effect with the redox behaviour of suitable higher-oxidation-state compounds.
