401. A Period 2 comparison lists the following approximate first ionization enthalpies:
| Element | Outermost configuration | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| N | \(2s^22p^3\) | 1402 |
| O | \(2s^22p^4\) | 1314 |
| F | \(2s^22p^5\) | 1681 |
| Ne | \(2s^22p^6\) | 2081 |
A student concludes that neon must have a greater ordinary Pauling electronegativity than fluorine because neon has the greater first ionization enthalpy. Which assessment is correct?
ⓐ. A higher first ionization enthalpy predicts greater conventional Pauling electronegativity even for a noble gas with limited bonding.
ⓑ. The conclusion is valid because neon's completed shell makes it the strongest ordinary electron-pair attractor in chemical bonds.
ⓒ. Ionization measures gaseous-atom electron removal, not bonding electronegativity; neon lacks a conventional Pauling value.
ⓓ. The conclusion is unsupported only because oxygen has a lower first ionization enthalpy than nitrogen.
Correct Answer: Ionization measures gaseous-atom electron removal, not bonding electronegativity; neon lacks a conventional Pauling value.
Explanation: \(\textbf{Interpret the ionization data:}\)
Neon has a completed \(2s^22p^6\) outermost shell.
Its electrons are strongly bound.
Consequently, its first ionization enthalpy is exceptionally high.
The table correctly demonstrates that electron removal from neon is more difficult than from fluorine.
\(\textbf{Define the relevant electronegativity concept:}\)
Electronegativity concerns the attraction exerted by an atom for shared electrons in a chemical bond.
The Pauling scale is associated with bond-energy comparisons.
Therefore, it does not represent the same physical process as isolated-atom electron removal.
\(\textbf{Examine fluorine:}\)
Fluorine readily forms ordinary chemical bonds.
Its very high attraction for shared bonding electrons gives it the highest conventional Pauling electronegativity.
\(\textbf{Examine neon:}\)
Neon has a completed valence shell and exhibits extremely low ordinary chemical reactivity.
It does not provide the same range of ordinary stable chemical bonds used in conventional Pauling electronegativity comparisons.
Consequently, a straightforward numerical ordering of neon and fluorine on that scale is not established by the supplied ionization enthalpies.
\(\textbf{Identify the conceptual error:}\)
A high first ionization enthalpy means electron removal is energetically difficult.
It does not independently establish how strongly an atom attracts shared electrons in a bond.
The distinction is especially important at the halogen–noble-gas boundary.
The correct assessment separates isolated-atom ionization from bonding-context electronegativity.
402. An idealized experiment irradiates isolated neon, argon and xenon atoms with photons having an energy of \(1500\,\mathrm{kJ}\) per mole of photons. Their first ionization enthalpies are approximately:
| Atom | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| Ne | 2081 |
| Ar | 1521 |
| Xe | 1170 |
Assume a simplified one-photon ionization-threshold model, with no additional energy transfer between particles. Which prediction correctly identifies the energetically permitted first-ionization process and the ground-state electronic character of its resulting monocation?
ⓐ. Neon alone can be ionized; \(\mathrm{Ne^+}\) has a completed \(2p^6\) subshell and is diamagnetic.
ⓑ. Argon and xenon can both be ionized because their first ionization enthalpies are lower than neon's; their monocations are diamagnetic.
ⓒ. None of the three can be ionized because noble-gas atoms possess completed valence shells.
ⓓ. Only xenon meets the photon-energy threshold; ground-state \(\mathrm{Xe^+}\) has one unpaired \(p\) electron.
Correct Answer: Only xenon meets the photon-energy threshold; ground-state \(\mathrm{Xe^+}\) has one unpaired \(p\) electron.
Explanation: \(\textbf{Identify the threshold principle:}\)
The model requires a single photon to provide at least the first ionization energy of an isolated atom.
The supplied photon energy and ionization enthalpies use the same molar energy units.
Therefore, they can be compared directly.
\(\textbf{Test neon:}\)
Neon's first ionization enthalpy is \(2081\,\mathrm{kJ\,mol^{-1}}\).
The photon energy is only \(1500\,\mathrm{kJ\,mol^{-1}}\).
It is therefore insufficient for neon's first ionization under the stated one-photon model.
\(\textbf{Test argon:}\)
Argon's threshold is \(1521\,\mathrm{kJ\,mol^{-1}}\).
The supplied photon energy is slightly below this value.
Therefore, the ionization threshold is not satisfied for argon.
\(\textbf{Test xenon:}\)
Xenon's threshold is \(1170\,\mathrm{kJ\,mol^{-1}}\).
The photon energy exceeds this value.
Therefore, xenon ionization is energetically permitted.
\(\textbf{Examine the product configuration:}\)
Neutral xenon has the outermost configuration \(5s^25p^6\).
Formation of \(\mathrm{Xe^+}\) removes one electron from the filled outer \(p\) subshell.
The resulting configuration is \(5s^25p^5\).
It contains one unpaired \(p\) electron.
Thus, the ordinary ground-state monocation is paramagnetic.
\(\textbf{Interpret the first-element anomaly:}\)
The high ionization enthalpy of the smaller noble gases contributes to their low ordinary chemical reactivity.
The lower ionization enthalpy of xenon makes electron removal comparatively more accessible.
However, exceeding an energy threshold establishes energetic possibility, not a guaranteed ionization event.
Actual probabilities also depend on the relevant absorption and ionization processes.
The correct result combines the threshold comparison with the electronic configuration of the product.
403. Water and hydrogen sulphide are bent molecules containing two identical E–H bonds. For a simplified dipole-vector model, the following data are supplied.
| Property | \(\mathrm{H_2O}\) | \(\mathrm{H_2S}\) |
|---|
| Individual E–H bond-dipole magnitude | \(1.50\,\mathrm{D}\) | \(0.60\,\mathrm{D}\) |
| H–E–H bond angle | \(104.5^\circ\) | \(92^\circ\) |
| \(\cos(\theta/2)\) | 0.612 | 0.695 |
Assume that the molecular dipole moment is obtained solely by vector addition of the two specified bond dipoles. Which conclusion is correct?
ⓐ. Water has a resultant dipole moment of \(3.00\,\mathrm{D}\), while hydrogen sulphide has \(1.20\,\mathrm{D}\), because bond angles do not affect dipole addition.
ⓑ. Water has a resultant dipole moment of \(0.918\,\mathrm{D}\), while hydrogen sulphide has \(0.417\,\mathrm{D}\), because each bond contributes only half its dipole.
ⓒ. Hydrogen sulphide has the greater resultant dipole moment because its bond angle is closer to \(90^\circ\).
ⓓ. Water: about \(1.84\,\mathrm D\); hydrogen sulphide: about \(0.834\,\mathrm D\), owing mainly to bond polarity.
Correct Answer: Water: about \(1.84\,\mathrm D\); hydrogen sulphide: about \(0.834\,\mathrm D\), owing mainly to bond polarity.
Explanation: \(\textbf{Identify the molecular geometry:}\)
Oxygen and sulphur belong to Group 16.
Each central atom forms two bonds and retains two lone pairs in the simple hydride.
Consequently, both molecules are bent.
Their bond-dipole vectors must therefore be added rather than treated as independent scalar quantities.
\(\textbf{Establish the resultant relation:}\)
For two equal bond dipoles separated by angle \(\theta\):
\[
\mu_{\mathrm{resultant}}=2\mu_{\mathrm{bond}}\cos\left(\frac{\theta}{2}\right)
\]
This relation accounts for the vector cancellation associated with the molecular angle.
\(\textbf{Calculate the water dipole:}\)
\[
\mu_{\mathrm{H_2O}}=2(1.50)(0.612)
\]
\[
\mu_{\mathrm{H_2O}}=1.836\,\mathrm{D}
\]
\(\textbf{Calculate the hydrogen sulphide dipole:}\)
\[
\mu_{\mathrm{H_2S}}=2(0.60)(0.695)
\]
\[
\mu_{\mathrm{H_2S}}=0.834\,\mathrm{D}
\]
\(\textbf{Compare the two results:}\)
\[
\frac{\mu_{\mathrm{H_2O}}}{\mu_{\mathrm{H_2S}}}
=\frac{1.836}{0.834}
\]
\[
\approx2.20
\]
The simplified model therefore predicts a molecular dipole for water approximately \(2.20\) times that of hydrogen sulphide.
\(\textbf{Interpret the first-member distinction:}\)
Oxygen is more electronegative than sulphur.
The O–H bond consequently exhibits greater polarity than the S–H bond.
Although the wider H–O–H angle produces somewhat greater cancellation between equal bond dipoles, the larger individual O–H dipoles dominate the comparison.
\(\textbf{Recognize the model limitation:}\)
Actual molecular dipole moments reflect the complete electronic charge distribution, including contributions not represented by a simple two-bond vector model.
The supplied calculation is therefore an idealized prediction.
Within that model, both bond polarity and geometry are essential.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{\mathrm{H_2O}}\approx1.84\,\mathrm{D},\quad
\mu_{\mathrm{H_2S}}=0.834\,\mathrm{D},\quad
\frac{\mu_{\mathrm{H_2O}}}{\mu_{\mathrm{H_2S}}}\approx2.20}
\]
404. Fluorine and chlorine exhibit an unusual electron-gain-enthalpy comparison. Consider the following simplified two-stage processes, beginning with isolated gaseous halogen atoms and ending with hydrated halide ions.
| Contribution | Fluorine \((\mathrm{kJ\,mol^{-1}})\) | Chlorine \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| \(\mathrm{X(g)+e^-\rightarrow X^-(g)}\) | -328 | -349 |
| \(\mathrm{X^-(g)\rightarrow X^-(aq)}\) | -505 | -360 |
Assume these are the only energetic contributions to the specified processes. Which conclusion correctly compares electron attachment before and after hydration?
ⓐ. Cl electron attachment releases \(21\,\mathrm{kJ\,mol^{-1}}\) more, but hydrated F formation releases \(124\,\mathrm{kJ\,mol^{-1}}\) more.
ⓑ. Fluorine's isolated-atom electron attachment is more exothermic by \(21\,\mathrm{kJ\,mol^{-1}}\), and hydration preserves this advantage.
ⓒ. Chlorine's complete hydrated-ion formation is more exothermic by \(145\,\mathrm{kJ\,mol^{-1}}\), because chloride is the larger ion.
ⓓ. Both complete processes have identical enthalpy changes because both halogens form mononegative ions.
Correct Answer: Cl electron attachment releases \(21\,\mathrm{kJ\,mol^{-1}}\) more, but hydrated F formation releases \(124\,\mathrm{kJ\,mol^{-1}}\) more.
Explanation: \(\textbf{Separate the two physical stages:}\)
First electron-gain enthalpy concerns electron attachment to an isolated gaseous atom.
Hydration concerns transfer of a gaseous ion into an aqueous environment.
The second process includes interactions between the ion and surrounding water molecules.
These two enthalpy contributions must not be confused.
\(\textbf{Compare the gas-phase electron attachments:}\)
\[
\Delta H_{\mathrm{eg}}(\mathrm F)=-328\,\mathrm{kJ\,mol^{-1}}
\]
\[
\Delta H_{\mathrm{eg}}(\mathrm{Cl})=-349\,\mathrm{kJ\,mol^{-1}}
\]
The gas-phase advantage of chlorine is:
\[
(-328)-(-349)=+21\,\mathrm{kJ\,mol^{-1}}
\]
Therefore, chlorine's electron-gain process is more exothermic by \(21\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Add fluoride hydration:}\)
\[
\Delta H_{\mathrm{total,F}}=-328+(-505)
\]
\[
=-833\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Add chloride hydration:}\)
\[
\Delta H_{\mathrm{total,Cl}}=-349+(-360)
\]
\[
=-709\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the complete processes:}\)
\[
\Delta H_{\mathrm{total,F}}-\Delta H_{\mathrm{total,Cl}}
=-833-(-709)
\]
\[
=-124\,\mathrm{kJ\,mol^{-1}}
\]
The complete process for fluorine is therefore more exothermic by \(124\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Explain the reversal:}\)
Fluorine's compact \(2p\) orbitals produce substantial electron–electron crowding during gas-phase electron attachment.
This contributes to chlorine's more negative first electron-gain enthalpy.
However, the smaller fluoride ion interacts strongly with polar water molecules.
The larger hydration contribution more than compensates for the gas-phase difference in the supplied data.
\(\textbf{Preserve the chemical distinction:}\)
The calculation begins with isolated gaseous atoms, not molecular \(\mathrm{F_2}\) or \(\mathrm{Cl_2}\).
Therefore, the result must not be identified directly with a standard aqueous molecular-halogen reduction potential.
The appropriate comparison is between the explicitly defined two-stage processes.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{F,total}}=-833,\quad
\Delta H_{\mathrm{Cl,total}}=-709\ \mathrm{kJ\,mol^{-1}};\quad
\text{fluorine advantage}=124\,\mathrm{kJ\,mol^{-1}}}
\]
405. Three unidentified elements X, Y and Z belong to consecutive p-block groups in Period 3. Their ordinary highest-oxidation-state fluorides have empirical formulas \(\mathrm{XF_3}\), \(\mathrm{YF_4}\) and \(\mathrm{ZF_5}\), respectively. A fourth Period 3 element W immediately follows Z and forms a characteristic acidic trioxide, \(\mathrm{WO_3}\). Which identification and trend interpretation satisfy all the information?
ⓐ. X = B, Y = C, Z = N and W = O; the progression demonstrates increasing metallic character across the period.
ⓑ. X = Al, Y = Si, Z = P, W = S; oxide acidity increases with non-metallic character.
ⓒ. X = Al, Y = P, Z = S and W = Cl; the increasing number of fluorine atoms proves that group numbers increase by two.
ⓓ. X = Ga, Y = Ge, Z = As and W = Se; these elements belong to Period 3 because their valence shells contain \(p\) electrons.
Correct Answer: X = Al, Y = Si, Z = P, W = S; oxide acidity increases with non-metallic character.
Explanation: \(\textbf{Interpret the fluoride stoichiometries:}\)
Fluorine has oxidation state \(-1\) in the specified compounds.
The central elements therefore have oxidation states +3, +4 and +5, respectively.
These are characteristic higher positive oxidation states for Groups 13, 14 and 15.
\(\textbf{Apply the period constraint:}\)
The Period 3 Group 13 element is aluminium.
The succeeding Group 14 element is silicon.
The Group 15 element is phosphorus.
Thus, X, Y and Z must be aluminium, silicon and phosphorus.
\(\textbf{Identify the fourth element:}\)
Sulphur immediately follows phosphorus in Period 3.
It belongs to Group 16.
Sulphur forms \(\mathrm{SO_3}\), in which sulphur has oxidation state +6.
Sulphur trioxide exhibits acidic oxide character.
\(\textbf{Interpret the electronic progression:}\)
The valence configurations change from \(3s^23p^1\) through \(3s^23p^4\).
The valence-electron count increases successively.
The characteristic higher positive oxidation states are consistent with the permitted formula patterns.
\(\textbf{Interpret the chemical trend:}\)
Metallic character generally decreases across this portion of Period 3.
Aluminium oxide is amphoteric.
The representative higher oxides of silicon, phosphorus and sulphur exhibit acidic character.
This trend must not be interpreted as proof that all oxides or all oxidation states behave identically.
\(\textbf{Check the identification:}\)
The formulas, periodic positions and oxide character independently support the same sequence.
The result illustrates how electronic configuration, oxidation-state accounting and chemical trends can be used together to reconstruct an unidentified part of the p-block.
406. Consider the following hypothetical change to the electronic structure of second-period atoms.
The energetic advantage associated with effective sideways overlap between compact \(2p\) orbitals is selectively reduced. Nuclear charges, valence-electron counts and the maximum occupancy of the \(2s\) and \(2p\) subshells remain unchanged. The modification does not directly alter intermolecular hydrogen bonding or the electron-pair repulsion within ordinary single bonds.
Which prediction most directly follows for the first members of Groups 14–17?
ⓐ. Weaker sideways overlap would reduce carbon multiple bonding and N≡N stability, but not fluorine's second-shell limit.
ⓑ. The N≡N triple bond would strengthen because reducing sideways overlap increases the contribution of \(\pi\) bonding.
ⓒ. Changing \(p\pi\) overlap would remove the second-shell orbital restrictions and allow higher coordination.
ⓓ. Fluorine would become less electronegative than chlorine because its nuclear charge would decrease.
Correct Answer: Weaker sideways overlap would reduce carbon multiple bonding and N≡N stability, but not fluorine's second-shell limit.
Explanation: \(\textbf{Identify the selectively changed mechanism:}\)
Sideways overlap between suitable \(p\) orbitals contributes to \(\pi\) bonding.
The hypothetical modification specifically reduces the stabilization from this interaction.
Therefore, its most direct effects should concern bonding arrangements with important \(\pi\)-bonding contributions.
\(\textbf{Examine carbon:}\)
Carbon's compact \(2p\) orbitals support effective multiple bonding with suitable atoms.
Reducing sideways overlap would weaken the additional bonding contribution associated with such multiple bonds.
However, the modification would not remove carbon's four valence electrons.
\(\textbf{Examine nitrogen:}\)
The N≡N triple bond contains one sigma-bonding contribution and two \(\pi\)-bonding contributions in the elementary bonding description.
Reducing the effectiveness of the \(\pi\) contributions would tend to weaken the triple bond.
This could alter the relative stability of molecular nitrogen compared with alternative bonding arrangements.
\(\textbf{Examine oxygen and fluorine:}\)
Multiple bonding involving oxygen can also depend on effective \(2p\) sideways overlap.
However, the question specifically excludes direct changes to ordinary single-bond electron-pair repulsion.
Therefore, one cannot assume that all oxygen or fluorine single-bond anomalies would disappear.
\(\textbf{Check what remains invariant:}\)
The capacity of the second principal shell has not changed.
No additional \(d\) orbitals have been created in that shell.
The nuclear charges and valence-electron counts also remain unchanged.
Consequently, changes in multiple-bonding strength do not automatically imply changes in ordinary coordination limits or electronegativity ordering.
The correct prediction distinguishes an orbital-overlap effect from unrelated electronic properties.
407. Three neutral atoms X, Y and Z are consecutive Period 2 p-block elements with outermost configurations \(2s^22p^2\), \(2s^22p^3\) and \(2s^22p^4\), respectively. Consider the following observations:
Statement I: X and Y can exhibit effective multiple bonding involving compact \(2p\) orbitals.
Statement II: The first ionization enthalpy of Y exceeds that of Z despite Z having the greater nuclear charge.
Statement III: The higher electron count of Z requires its neutral atom to contain fewer paired \(2p\) electrons than Y.
Statement IV: Removing one \(2p\) electron from Z produces the same \(2p\)-subshell occupancy as neutral Y.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, II and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I, II and IV only
Explanation: \(\textbf{Identify the elements:}\)
The configurations correspond to carbon, nitrogen and oxygen, respectively.
They occupy Groups 14, 15 and 16.
Their outermost electrons belong to the second principal shell.
\(\textbf{Examine Statement I:}\)
Carbon and nitrogen possess compact \(2p\) orbitals.
These orbitals can overlap effectively in suitable multiple-bonding arrangements.
Statement I is correct.
\(\textbf{Examine Statement II:}\)
Nitrogen has the half-filled \(2p^3\) configuration.
Oxygen has \(2p^4\), with one paired set of \(p\) electrons.
Electron-pair repulsion contributes to the lower first ionization enthalpy of oxygen.
Therefore, nitrogen's first ionization enthalpy exceeds oxygen's despite oxygen's greater nuclear charge.
Statement II is correct.
\(\textbf{Examine Statement III:}\)
The \(2p^3\) configuration has three singly occupied orbitals.
The \(2p^4\) configuration contains one doubly occupied orbital.
Thus, oxygen has more paired \(2p\) electrons, not fewer.
Statement III is incorrect.
\(\textbf{Examine Statement IV:}\)
Removing one \(2p\) electron from oxygen changes its occupancy from \(2p^4\) to \(2p^3\).
This matches the neutral nitrogen atom's \(2p\)-subshell occupancy.
The resulting species still differ in nuclear charge and total electron count, so they are not identical atoms.
Statement IV is correct.
\(\textbf{Integrate the result:}\)
The comparison requires electronic identification, interpretation of ionization-enthalpy exceptions and precise orbital accounting.
The correct combination is Statements I, II and IV.
408. An isolated neutral helium atom and a neutral neon atom are each promoted into a hypothetical excited electronic configuration by supplying suitable energy. Assume that no electron leaves either atom and neither nucleus changes. Which conclusion correctly distinguishes excitation, ionization and the chemical significance of the ground-state noble-gas configurations?
ⓐ. Excitation changes the atomic numbers of helium and neon, so their positions in Group 18 may change.
ⓑ. Excitation necessarily produces \(\mathrm{He^+}\) and \(\mathrm{Ne^+}\), because any increase in electronic energy requires electron removal.
ⓒ. Both atoms retain electron count and oxidation state zero; excitation changes electronic stability and pairing.
ⓓ. Both atoms must remain in completed-shell configurations because Group 18 membership prevents occupation of higher-energy orbitals.
Correct Answer: Both atoms retain electron count and oxidation state zero; excitation changes electronic stability and pairing.
Explanation: \(\textbf{Identify the ground-state configurations:}\)
Helium has the configuration \(1s^2\).
Its first principal shell is completely occupied.
Neon has the configuration \(1s^22s^22p^6\).
Its second principal shell is completely occupied.
These closed-shell ground states contribute to their exceptionally low ordinary chemical reactivity.
\(\textbf{Distinguish electronic excitation:}\)
Excitation promotes an electron to a higher-energy orbital or electronic state.
It need not remove the electron from the atom.
Therefore, the total electron count can remain unchanged.
The atomic number also remains fixed because the nucleus has not changed.
\(\textbf{Distinguish ionization:}\)
Ionization requires removal or addition of an electron.
A neutral excited atom is not automatically a positive ion.
Consequently, the stipulated process does not necessarily produce \(\mathrm{He^+}\) or \(\mathrm{Ne^+}\).
\(\textbf{Interpret oxidation state:}\)
Both species remain neutral elemental atoms.
Their formal oxidation states remain zero.
The energy change is an electronic excitation rather than a chemical redox process.
\(\textbf{Examine the excited configurations:}\)
An excited electron arrangement can differ from the ground state in orbital occupancy and electron pairing.
The exceptional stability of the noble-gas ground state does not mean that higher-energy states are forbidden.
Such states may return to lower energies through appropriate relaxation processes.
This distinction is especially important when interpreting the first noble gas's completed \(1s^2\) shell.
The correct conclusion preserves the elemental identities while recognizing that a stable ground-state configuration need not be maintained during excitation.
409. Two Group 13 elements, boron and aluminium, form representative trioxides. Under suitable conditions, boron trioxide is acidic, while aluminium oxide is amphoteric. An unknown oxide E reacts with both an acid and a strong base. A student concludes that E must contain aluminium rather than boron. Which assessment is best justified?
ⓐ. The conclusion is logically certain because no oxide of any other element can be amphoteric.
ⓑ. Aluminium oxide fits the observation, but amphoterism alone does not uniquely identify the element.
ⓒ. The observation proves that the central element has oxidation state +1 because amphoteric oxides cannot contain trivalent atoms.
ⓓ. The observation is incompatible with aluminium oxide because aluminium is a metal and must form exclusively basic oxides.
Correct Answer: Aluminium oxide fits the observation, but amphoterism alone does not uniquely identify the element.
Explanation: \(\textbf{Interpret the chemical tests:}\)
An amphoteric oxide can participate in characteristic reactions with suitable acids and bases.
Aluminium oxide is a familiar example.
Its reactions with strong acids and alkalis support its amphoteric classification under appropriate conditions.
\(\textbf{Compare boron trioxide:}\)
Boron is the first member of Group 13.
Its trioxide exhibits acidic character.
This differs from the amphoteric behaviour of aluminium oxide.
The difference is consistent with aluminium's greater metallic character.
\(\textbf{Identify the logical limitation:}\)
Several oxides outside this specific boron–aluminium comparison can exhibit amphoteric behaviour.
Therefore, observing amphoterism does not uniquely identify an unknown oxide as aluminium oxide.
The inference would be stronger if independent evidence established that the unknown contained either boron or aluminium.
\(\textbf{Check oxidation-state reasoning:}\)
Aluminium has oxidation state +3 in \(\mathrm{Al_2O_3}\).
Amphoterism does not require a central-element oxidation state of +1.
Nor does metallic character require an oxide to be exclusively basic.
\(\textbf{Interpret the broader lesson:}\)
A chemical test can support a classification without uniquely identifying a substance.
Periodic trends provide plausible predictions but do not replace composition evidence.
The correct assessment distinguishes compatibility with an observation from proof of chemical identity.
This is particularly important when using first-element anomalies to identify unknown p-block compounds.
410. A researcher compares the stability of representative +3 and +5 oxidation states in Group 15. Phosphorus(V) is stable in suitable oxygen-containing compounds, whereas bismuth(III) is relatively favoured over bismuth(V) in many ordinary environments. Consider the following hypothetical change: the outermost \(ns^2\) electrons of heavy Group 15 atoms become more available for bonding, while their valence-electron counts and nuclear charges remain unchanged. Which consequence is most directly expected?
ⓐ. Heavy Group 15 elements would favour +5 more because outer \(s^2\)-pair participation would be easier.
ⓑ. The +3 oxidation state would necessarily disappear from every Group 15 compound because the neutral atoms would lose two valence electrons.
ⓒ. Bismuth would become a Group 13 element because its outermost \(s^2\) electrons would now participate more readily.
ⓓ. All Group 15 elements would acquire identical oxidation-state stabilities, irrespective of their bonding partners.
Correct Answer: Heavy Group 15 elements would favour +5 more because outer \(s^2\)-pair participation would be easier.
Explanation: \(\textbf{Identify the Group 15 electronic framework:}\)
Group 15 atoms have the general valence configuration \(ns^2np^3\).
They possess five valence electrons.
The formal +3 state can be associated with participation of the \(np^3\) electrons while retaining the outermost \(ns^2\) pair.
The +5 state involves participation of all five outermost electrons.
\(\textbf{Interpret the inert-pair effect:}\)
Among heavier p-block elements, the outermost \(ns^2\) pair becomes comparatively less favourable for bonding participation.
This contributes to the preference for lower positive oxidation states.
Bismuth(III) illustrates this tendency within Group 15.
\(\textbf{Apply the hypothetical modification:}\)
The question makes the \(ns^2\) pair more available for bonding.
Therefore, the energetic disadvantage of involving that pair is reduced.
This would tend to improve the relative accessibility of suitable +5 bonding arrangements.
\(\textbf{Identify what remains unchanged:}\)
The number of valence electrons does not change.
The nuclear charge also remains fixed.
Therefore, the element does not move to another periodic group.
\(\textbf{Recognize the qualification:}\)
Greater availability of the pair does not prohibit the +3 state.
Nor does it make the +5 state equally stable in every compound.
The identity of the bonding partner and the total compound-formation energetics remain important.
The correct prediction is therefore a shift in relative oxidation-state stability rather than an absolute elimination of one oxidation state.
411. Oxygen ordinarily exhibits oxidation state -2 in water and +2 in oxygen difluoride. Now consider hypofluorous acid, HOF, whose connectivity is H–O–F. The Pauling electronegativities are H = 2.20, O = 3.44 and F = 3.98. Which conclusion correctly assigns the oxidation states and explains the unusual result?
ⓐ. Oxygen has oxidation state -2 in all three compounds because it belongs to Group 16.
ⓑ. Oxygen has oxidation state +1 in HOF because fluorine withdraws one electron from the oxygen atom.
ⓒ. In HOF, H is +1, F is \(-1\), and O is zero because the two O–H and O–F assignments offset.
ⓓ. Oxygen has oxidation state +2 in HOF because every oxygen–fluorine bond makes oxygen divalent.
Correct Answer: In HOF, H is +1, F is \(-1\), and O is zero because the two O–H and O–F assignments offset.
Explanation: \(\textbf{Examine the electronegativity order:}\)
The supplied order is F > O > H.
Formal oxidation states are assigned by allocating bonding electrons to the more electronegative atom.
Therefore, hydrogen is assigned oxidation state +1 in HOF.
Fluorine is assigned oxidation state -1.
\(\textbf{Apply charge conservation:}\)
The molecule is electrically neutral.
Let oxygen's oxidation state be \(x\).
The oxidation-state sum must satisfy:
\[
(+1)+x+(-1)=0
\]
Hence oxygen has oxidation state zero.
\(\textbf{Interpret the bonding:}\)
In the O–H bond, the formal electron assignment favours oxygen.
In the O–F bond, it favours fluorine.
The two contributions give oxygen a net oxidation-state assignment of zero.
This does not mean that oxygen is present as an isolated elemental atom or that the bonds are non-polar.
\(\textbf{Compare the related compounds:}\)
In water, both hydrogen atoms are less electronegative than oxygen, giving oxygen oxidation state -2.
In \(\mathrm{OF_2}\), both fluorine atoms are more electronegative than oxygen, giving oxygen oxidation state +2.
The HOF example therefore demonstrates that oxygen's oxidation state depends on its bonding partners.
A fixed group number does not require a fixed oxidation state in every compound.
412. Chlorine forms the interhalogen compounds ClF and ICl. The electronegativity order is F > Cl > I. Both compounds contain one chlorine atom, and neither contains oxygen. Which statement correctly compares the formal oxidation state of chlorine and the direction of bond polarization?
ⓐ. Chlorine has oxidation state +1 in both compounds because it is a Group 17 element.
ⓑ. Chlorine has oxidation state -1 in ClF but +1 in ICl; fluorine is the more electropositive bonding partner.
ⓒ. Chlorine has oxidation state -1 in both compounds because all halogens behave identically in binary compounds.
ⓓ. Chlorine is +1 and electron-deficient in ClF, but \(-1\) and electron-rich in ICl.
Correct Answer: Chlorine is +1 and electron-deficient in ClF, but \(-1\) and electron-rich in ICl.
Explanation: The formal oxidation state of an atom depends on the atoms to which it is bonded.
Fluorine is more electronegative than chlorine.
Therefore, fluorine receives the formal oxidation-state assignment -1 in ClF.
Chlorine consequently has oxidation state +1.
Iodine is less electronegative than chlorine.
Therefore, chlorine is assigned -1 in ICl, leaving iodine at +1.
The polarity of the Cl–F bond is directed towards fluorine.
The polarity of the I–Cl bond is directed towards chlorine.
Thus, chlorine occupies opposite ends of the bond-polarization comparison in the two compounds.
Both species are heteronuclear diatomic molecules, so their individual bond dipoles do not cancel through molecular symmetry.
The contrast does not require chlorine to change its number of valence electrons as a neutral atom.
Its characteristic neutral-atom configuration remains \(ns^2np^5\).
The different formal oxidation states arise from changes in bonding partner and relative electronegativity.
This also illustrates why the exceptional electronegativity of fluorine distinguishes it from heavier halogens.
413. Tin exists in different elemental allotropes. Grey tin has a diamond-like structure and semiconducting character, whereas ordinary white tin is metallic. Grey tin is thermodynamically favoured below approximately \(13.2^\circ\mathrm{C}\), while white tin is favoured above this temperature at ordinary pressure. Assume sufficient time is allowed for equilibrium and ignore kinetic limitations. Which inference is correct?
ⓐ. The conversion from white tin to grey tin below \(13.2^\circ\mathrm{C}\) must involve oxidation of Sn(0) to Sn(II).
ⓑ. Grey tin has four valence electrons per atom, but white tin has only two because its bonding is metallic.
ⓒ. Both allotropes must possess identical electrical conductivity because their atoms occupy Group 14.
ⓓ. Tin can change crystal structure and conductivity without changing atomic identity or oxidation state.
Correct Answer: Tin can change crystal structure and conductivity without changing atomic identity or oxidation state.
Explanation: Tin is a Group 14 element with four valence electrons.
Both grey tin and white tin contain only tin atoms.
Therefore, tin has formal oxidation state zero in both allotropes.
The structural transformation is not inherently an oxidation–reduction reaction.
Grey tin possesses an extended structure associated with semiconducting electronic behaviour.
White tin has a metallic structure that permits electrical conduction.
The two structures have different electronic-energy arrangements despite containing the same element.
The specified transition temperature separates their relative thermodynamic stability under the stated conditions.
Below that temperature, grey tin is favoured; above it, white tin is favoured.
An actual transformation may nevertheless be slow, which is why the question explicitly removes kinetic limitations.
The number of neutral-atom valence electrons does not change during allotropy.
The example shows that electrical conductivity is not determined by group number alone.
It also demonstrates why a general increase in metallic character down Group 14 must be distinguished from the properties of individual allotropes.
414. The following average bond dissociation enthalpies are supplied.
| Bond | Bond dissociation enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| H–H | 436 |
| Cl–Cl | 243 |
| I–I | 151 |
| H–Cl | 431 |
| H–I | 299 |
Consider the following gas-phase reactions, with all species in their stated gaseous forms:
\[
\mathrm{H_2(g)+Cl_2(g)\rightarrow2HCl(g)}
\]
\[
\mathrm{H_2(g)+I_2(g)\rightarrow2HI(g)}
\]
Using the supplied average bond enthalpies, which conclusion is correct?
ⓐ. The chlorine reaction is \(172\,\mathrm{kJ\,mol^{-1}}\) more exothermic owing to its stronger H–Cl bonds.
ⓑ. The iodine reaction is approximately \(172\,\mathrm{kJ\,mol^{-1}}\) more exothermic because I–I is weaker than Cl–Cl.
ⓒ. Both reactions have identical enthalpy changes because two H–X bonds form in each.
ⓓ. The chlorine reaction is endothermic by \(183\,\mathrm{kJ\,mol^{-1}}\), whereas the iodine reaction is exothermic by \(11\,\mathrm{kJ\,mol^{-1}}\).
Correct Answer: The chlorine reaction is \(172\,\mathrm{kJ\,mol^{-1}}\) more exothermic owing to its stronger H–Cl bonds.
Explanation: \(\textbf{Identify the bond-energy relation:}\)
The approximate reaction enthalpy is obtained from the energy needed to break reactant bonds and the energy released when product bonds form.
\[
\Delta H\approx\sum D(\text{bonds broken})-\sum D(\text{bonds formed})
\]
\(\textbf{Calculate the chlorine reaction:}\)
One H–H bond and one Cl–Cl bond are broken.
Two H–Cl bonds are formed.
\[
\Delta H_{\mathrm{Cl}}=(436+243)-2(431)
\]
\[
=679-862
\]
\[
=-183\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the iodine reaction:}\)
One H–H bond and one I–I bond are broken.
Two H–I bonds are formed.
\[
\Delta H_{\mathrm I}=(436+151)-2(299)
\]
\[
=587-598
\]
\[
=-11\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the enthalpy changes:}\)
\[
\Delta H_{\mathrm{Cl}}-\Delta H_{\mathrm I}
=-183-(-11)
\]
\[
=-172\,\mathrm{kJ\,mol^{-1}}
\]
Thus, the chlorine reaction is more exothermic by approximately \(172\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Interpret the chemical competition:}\)
Breaking the weaker I–I bond requires less energy than breaking Cl–Cl.
However, the H–I bonds formed are considerably weaker than the H–Cl bonds.
The net reaction enthalpy therefore depends on both bond-breaking and bond-forming contributions.
The weaker elemental halogen bond does not by itself guarantee the more exothermic overall reaction.
\(\textbf{Recognize the conditions:}\)
The iodine reactant is explicitly gaseous.
The calculated enthalpy must not be confused with a standard reaction enthalpy beginning with solid iodine.
The calculation also describes reaction energetics, not the activation barrier or reaction rate.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{Cl}}=-183,\quad\Delta H_{\mathrm I}=-11\ \mathrm{kJ\,mol^{-1}};\quad\text{difference}=172\ \mathrm{kJ\,mol^{-1}}}
\]
415. Phosphorus forms the oxides \(\mathrm{P_4O_6}\) and \(\mathrm{P_4O_{10}}\). Under suitable oxidizing conditions, the lower oxide can be converted into the higher oxide. An initial mixture contains \(0.200\,\mathrm{mol}\) of pure \(\mathrm{P_4O_6}\) and \(0.300\,\mathrm{mol}\) of oxygen. Assume complete conversion of the limiting reactant according to:
\[
\mathrm{P_4O_6+2O_2\rightarrow P_4O_{10}}
\]
What are the amounts of \(\mathrm{P_4O_{10}}\) formed and \(\mathrm{P_4O_6}\) remaining?
ⓐ. \(0.200\,\mathrm{mol}\) and zero, respectively
ⓑ. \(0.150\,\mathrm{mol}\) and \(0.050\,\mathrm{mol}\), respectively
ⓒ. \(0.100\,\mathrm{mol}\) and \(0.100\,\mathrm{mol}\), respectively
ⓓ. \(0.150\,\mathrm{mol}\) and zero, respectively
Correct Answer: \(0.150\,\mathrm{mol}\) and \(0.050\,\mathrm{mol}\), respectively
Explanation: \(\textbf{Identify the oxidation-state change:}\)
Oxygen has oxidation state -2 in both ordinary phosphorus oxides.
In \(\mathrm{P_4O_6}\), phosphorus has average oxidation state +3.
In \(\mathrm{P_4O_{10}}\), phosphorus has oxidation state +5.
Each phosphorus atom therefore undergoes oxidation by two units.
\(\textbf{Examine the balanced equation:}\)
\[
\mathrm{P_4O_6+2O_2\rightarrow P_4O_{10}}
\]
The equation requires two moles of oxygen per mole of phosphorus(III) oxide.
\(\textbf{Calculate the oxygen requirement:}\)
For complete conversion of the initial phosphorus oxide:
\[
n_{\mathrm{O_2,required}}=2(0.200)
\]
\[
=0.400\,\mathrm{mol}
\]
Only \(0.300\,\mathrm{mol}\) of oxygen is available.
Therefore, oxygen is the limiting reactant.
\(\textbf{Determine the phosphorus oxide consumed:}\)
\[
n_{\mathrm{P_4O_6,consumed}}=\frac{0.300}{2}
\]
\[
=0.150\,\mathrm{mol}
\]
\(\textbf{Calculate product and remainder:}\)
The balanced equation produces one mole of \(\mathrm{P_4O_{10}}\) per mole of \(\mathrm{P_4O_6}\) consumed.
\[
n_{\mathrm{P_4O_{10},formed}}=0.150\,\mathrm{mol}
\]
The unreacted amount is:
\[
n_{\mathrm{P_4O_6,remaining}}=0.200-0.150
\]
\[
=0.050\,\mathrm{mol}
\]
\(\textbf{Check the redox consistency:}\)
Four phosphorus atoms each increase their oxidation state by two units per converted formula unit.
The total formal electron loss is therefore eight electrons per formula unit.
The two oxygen molecules collectively accept the corresponding eight electrons.
The balanced reaction is consistent with both oxygen-atom and electron accounting.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{P_4O_{10}}}=0.150\,\mathrm{mol},\qquad n_{\mathrm{P_4O_6,remaining}}=0.050\,\mathrm{mol}}
\]
416. Consider two hypothetical gas-phase charge-transfer processes involving noble-gas atoms and fluorine atoms:
\[
\mathrm{Ar(g)+F(g)\rightarrow Ar^+(g)+F^-(g)}
\]
\[
\mathrm{Xe(g)+F(g)\rightarrow Xe^+(g)+F^-(g)}
\]
The first ionization enthalpies of Ar and Xe are \(1521\) and \(1170\,\mathrm{kJ\,mol^{-1}}\), respectively. The first electron-gain enthalpy of fluorine is \(-328\,\mathrm{kJ\,mol^{-1}}\). Assume that no additional stabilization is included in the specified products. Which conclusion is correct?
ⓐ. The argon process is exothermic by \(1193\,\mathrm{kJ\,mol^{-1}}\), while the xenon process is exothermic by \(842\,\mathrm{kJ\,mol^{-1}}\).
ⓑ. Both processes are exothermic because fluorine has the greatest electronegativity.
ⓒ. Xenon's process is endothermic by \(842\,\mathrm{kJ\,mol^{-1}}\); argon's is endothermic by \(1193\,\mathrm{kJ\,mol^{-1}}\).
ⓓ. Both processes have the same enthalpy change because argon and xenon have completed valence shells.
Correct Answer: Xenon's process is endothermic by \(842\,\mathrm{kJ\,mol^{-1}}\); argon's is endothermic by \(1193\,\mathrm{kJ\,mol^{-1}}\).
Explanation: \(\textbf{Identify the required energetic stages:}\)
A noble-gas atom must lose one electron to form its monocation.
The electron is then accepted by a fluorine atom.
The enthalpy change is the sum of the relevant first ionization enthalpy and fluorine's first electron-gain enthalpy.
\(\textbf{Calculate the argon process:}\)
\[
\Delta H_{\mathrm{Ar}}=IE_1(\mathrm{Ar})+\Delta_{\mathrm{eg}}H(\mathrm F)
\]
\[
=1521-328
\]
\[
=+1193\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the xenon process:}\)
\[
\Delta H_{\mathrm{Xe}}=1170-328
\]
\[
=+842\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the processes:}\)
\[
\Delta H_{\mathrm{Ar}}-\Delta H_{\mathrm{Xe}}
=1193-842
\]
\[
=351\,\mathrm{kJ\,mol^{-1}}
\]
The xenon process requires \(351\,\mathrm{kJ\,mol^{-1}}\) less energy in the stipulated model.
\(\textbf{Interpret the noble-gas trend:}\)
The outermost electrons of xenon are less strongly bound than those of argon.
This contributes to xenon's greater chemical accessibility in suitable bonding environments.
However, both isolated gas-phase charge-transfer processes remain endothermic.
Therefore, neither result by itself establishes spontaneous formation of a stable compound.
\(\textbf{Recognize the omitted interactions:}\)
Actual ionic or covalent compound formation may involve additional stabilization.
The model deliberately excludes those contributions.
The calculation consequently compares isolated-atom electron-transfer energetics rather than the full thermodynamics of noble-gas compound formation.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{Ar}}=+1193,\quad\Delta H_{\mathrm{Xe}}=+842\ \mathrm{kJ\,mol^{-1}};\quad\text{difference}=351\ \mathrm{kJ\,mol^{-1}}}
\]
417. Helium and beryllium have electronic configurations \(1s^2\) and \(1s^22s^2\), respectively. Both have two electrons in their outermost occupied subshell. However, helium belongs to Group 18, while beryllium belongs to Group 2. Which explanation best identifies the decisive electronic difference?
ⓐ. Helium's first shell is full; beryllium's second shell has unfilled \(2p\) orbitals despite both having an \(s^2\) arrangement.
ⓑ. Helium belongs to Group 18 because it possesses eight valence electrons, whereas beryllium possesses two.
ⓒ. Beryllium belongs to Group 2 because it is a p-block element, while helium belongs to Group 18 because it is a d-block element.
ⓓ. Group assignment should depend on the occupied \(s^2\) subshell alone, placing helium and beryllium together.
Correct Answer: Helium's first shell is full; beryllium's second shell has unfilled \(2p\) orbitals despite both having an \(s^2\) arrangement.
Explanation: Helium has atomic number two and the configuration \(1s^2\).
The first principal shell contains only the \(1s\) orbital.
It can accommodate a maximum of two electrons.
Therefore, helium's first shell is completely filled.
Beryllium has atomic number four and the configuration \(1s^22s^2\).
Although its outermost occupied subshell is filled, its second principal shell also contains unoccupied \(2p\) orbitals.
Thus, the second principal shell of beryllium is not completely occupied.
The completed-shell character of helium contributes to its exceptionally low ordinary chemical reactivity.
Beryllium instead exhibits the characteristic chemical behaviour of a Group 2 element.
Helium's placement in Group 18 reflects its noble-gas character rather than an eight-electron outer shell.
An outer \(s^2\) configuration alone is therefore insufficient to assign two elements to the same chemical group.
The comparison illustrates why shell completion and subshell completion must be distinguished.
418. Consider the following two idealized layered covalent materials.
Material P contains a planar network of carbon atoms arranged in hexagonal rings. Every carbon is bonded to three neighbouring carbon atoms. Material Q contains a similarly arranged network in which boron and nitrogen alternate around the hexagons, so every boron is bonded to three nitrogen atoms and every nitrogen to three boron atoms.
The relevant valence-electron counts are B = 3, C = 4 and N = 5. The materials are identified as graphite-like carbon and hexagonal boron nitride, respectively. Which statement correctly compares their valence-electron totals and bond polarity?
ⓐ. A B–N pair supplies six valence electrons, whereas two carbon atoms supply eight; the network arrangements therefore cannot be compared electronically.
ⓑ. B–N and C–C pairs each total eight valence electrons, but only B–N bonds are polar.
ⓒ. The two materials must have identical electrical conductivity because equal valence-electron totals require identical electronic band structures.
ⓓ. B–N bonds are necessarily non-polar because boron and nitrogen are both second-period elements.
Correct Answer: B–N and C–C pairs each total eight valence electrons, but only B–N bonds are polar.
Explanation: \(\textbf{Count the valence electrons:}\)
Boron belongs to Group 13 and has three valence electrons.
Nitrogen belongs to Group 15 and has five.
A B–N pair therefore contributes eight valence electrons.
Carbon belongs to Group 14 and has four valence electrons.
Two carbon atoms also contribute eight.
\(\textbf{Compare the bonding partners:}\)
In the carbon network, each C–C bond joins atoms of the same element.
There is no electronegativity difference between the two bonded carbon atoms.
Therefore, an individual C–C bond has no permanent dipole arising from an electronegativity difference.
In the boron nitride network, nitrogen is more electronegative than boron.
The B–N bonds consequently possess polar character.
\(\textbf{Interpret the structural similarity:}\)
The two materials have comparable hexagonal connectivity in the supplied idealized descriptions.
They also contain comparable numbers of valence electrons per two-atom unit.
Nevertheless, their nuclear charges and distributions of electron density differ.
\(\textbf{Distinguish structural and electronic properties:}\)
Equal valence-electron totals do not establish identical band structures.
They therefore do not guarantee identical electrical conductivity.
The comparison shows how neighbouring p-block groups can form electronically related structures without producing chemically identical bonding.
419. Ordinary oxygen exists predominantly as \(\mathrm{O_2}\), whereas a familiar molecular form of elemental sulphur contains \(\mathrm{S_8}\) rings. Two separate samples each contain \(0.400\,\mathrm{mol}\) of atoms of the corresponding element, with all oxygen present as \(\mathrm{O_2}\) molecules and all sulphur as \(\mathrm{S_8}\) molecules. Assume intact molecules and their ordinary ground-state electronic structures. Which conclusion is correct?
ⓐ. Both samples contain the same number of molecules because they contain equal amounts of atoms.
ⓑ. Four times as many \(\mathrm{O_2}\) molecules; \(\mathrm{O_2}\) is paramagnetic but \(\mathrm{S_8}\) diamagnetic.
ⓒ. The sulphur sample contains four times as many molecules because sulphur has the larger atomic mass.
ⓓ. Both samples contain equal numbers of molecules and have identical magnetic behaviour because both elements belong to Group 16.
Correct Answer: Four times as many \(\mathrm{O_2}\) molecules; \(\mathrm{O_2}\) is paramagnetic but \(\mathrm{S_8}\) diamagnetic.
Explanation: \(\textbf{Identify the molecular structures:}\)
Oxygen and sulphur belong to Group 16.
However, their familiar elemental molecular forms differ.
Molecular oxygen is diatomic.
The specified elemental sulphur form contains eight sulphur atoms per molecule.
\(\textbf{Count oxygen molecules:}\)
\[
n(\mathrm{O_2})=\frac{0.400}{2}
\]
\[
=0.200\,\mathrm{mol}
\]
\(\textbf{Count sulphur molecules:}\)
\[
n(\mathrm{S_8})=\frac{0.400}{8}
\]
\[
=0.0500\,\mathrm{mol}
\]
\(\textbf{Calculate the molecular-number ratio:}\)
\[
\frac{N(\mathrm{O_2})}{N(\mathrm{S_8})}
=\frac{0.200}{0.0500}
\]
\[
=4
\]
The oxygen sample contains four times as many molecules.
\(\textbf{Compare the electronic properties:}\)
Ground-state \(\mathrm{O_2}\) contains two unpaired electrons in antibonding molecular orbitals.
It is therefore paramagnetic.
Ordinary molecular \(\mathrm{S_8}\) has a closed-shell electronic arrangement with paired electrons.
It is diamagnetic.
\(\textbf{Interpret the first-element anomaly:}\)
Oxygen's effective multiple bonding contributes to the stability of the diatomic molecule.
Sulphur has a greater tendency to form extended arrangements of S–S single bonds.
Equal group valence-electron counts therefore do not require identical elemental molecularity or magnetism.
Both elemental forms nevertheless contain their respective atoms in oxidation state zero.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{N(\mathrm{O_2})}{N(\mathrm{S_8})}=4}
\]
420. The following data are supplied for bromine and iodine monochloride.
| Property | \(\mathrm{Br_2}\) | \(\mathrm{ICl}\) |
|---|
| Molar mass \((\mathrm{g\,mol^{-1}})\) | 159.8 | 162.4 |
| Approximate normal boiling point \((^\circ\mathrm{C})\) | 58.8 | 97 |
| Molecular character | Homonuclear diatomic | Heteronuclear diatomic |
Iodine monochloride has a polar I–Cl bond, whereas bromine has a non-polar Br–Br bond. Which interpretation is most defensible?
ⓐ. The higher boiling point of ICl is impossible because its molar mass is approximately equal to that of \(\mathrm{Br_2}\).
ⓑ. ICl must contain fully separated \(\mathrm{I^+}\) and \(\mathrm{Cl^-}\) ions because it has a higher boiling point than molecular bromine.
ⓒ. ICl's permanent dipole and intermolecular attractions allow its higher boiling point despite similar molar mass.
ⓓ. \(\mathrm{Br_2}\) has the larger permanent dipole moment because its electron distribution is more polarizable.
Correct Answer: ICl's permanent dipole and intermolecular attractions allow its higher boiling point despite similar molar mass.
Explanation: Bromine exists as a homonuclear diatomic molecule.
Its two bonded atoms have the same electronegativity.
Therefore, \(\mathrm{Br_2}\) has no permanent bond dipole.
Iodine monochloride contains two different halogens.
Chlorine is more electronegative than iodine.
Consequently, an I–Cl molecule possesses a permanent dipole moment.
Both molecular substances experience London dispersion attractions.
Iodine monochloride can additionally experience orientation-dependent dipole–dipole attractions.
The supplied normal boiling points are consistent with stronger overall intermolecular attractions in ICl.
However, boiling point cannot be assigned to one interaction without considering the complete condensed-phase environment.
Similar molar masses do not require equal molecular polarizability, molecular shape or intermolecular attraction.
A permanent dipole also does not imply that the isolated molecule consists of fully separated ions.
The appropriate conclusion preserves the distinction between covalent bond polarity and the total forces responsible for vaporization.
