201. The first electron-gain enthalpies of two gaseous halogen atoms X and Y are \(-328\) and \(-349\,\mathrm{kJ\,mol^{-1}}\), respectively. A metal atom M has the same first ionization enthalpy when paired with either halogen. Let the lattice formation enthalpies of MX and MY be \(L_X\) and \(L_Y\), with both quantities defined as negative for exothermic lattice formation. Assume the reactions begin with isolated gaseous atoms:
\[
\mathrm{M(g)+X(g)\rightarrow MX(s)}
\]
\[
\mathrm{M(g)+Y(g)\rightarrow MY(s)}
\]
What condition must the lattice enthalpies satisfy for formation of MX to be more exothermic than formation of MY?
ⓐ. \(L_X-L_Y\gt +21\,\mathrm{kJ\,mol^{-1}}\)
ⓑ. \(L_X-L_Y=-21\,\mathrm{kJ\,mol^{-1}}\)
ⓒ. \(L_X-L_Y\lt -21\,\mathrm{kJ\,mol^{-1}}\)
ⓓ. \(L_X-L_Y\lt +21\,\mathrm{kJ\,mol^{-1}}\)
Correct Answer: \(L_X-L_Y\lt -21\,\mathrm{kJ\,mol^{-1}}\)
Explanation: \(\textbf{Identify the competing energetic contributions:}\)
Both reactions begin with one gaseous metal atom and one gaseous halogen atom.
The metal ionization step is identical in both pathways.
Consequently, the metal ionization enthalpy will cancel when the two reaction enthalpies are compared.
The remaining differences arise from the electron-gain enthalpies and lattice formation enthalpies.
\(\textbf{Write the first reaction enthalpy:}\)
Let the common metal ionization enthalpy be \(I\).
For MX:
\[
\Delta H_X=I-328+L_X
\]
\(\textbf{Write the second reaction enthalpy:}\)
For MY:
\[
\Delta H_Y=I-349+L_Y
\]
\(\textbf{Apply the required condition:}\)
For MX formation to be more exothermic, its reaction enthalpy must be more negative.
Therefore:
\[
\Delta H_X\lt \Delta H_Y
\]
Substituting:
\[
I-328+L_X\lt I-349+L_Y
\]
Cancelling the common ionization contribution:
\[
-328+L_X\lt -349+L_Y
\]
Rearranging:
\[
L_X-L_Y\lt -21\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the threshold:}\)
Y has the more negative first electron-gain enthalpy.
Therefore, electron attachment alone favours the MY pathway by \(21\,\mathrm{kJ\,mol^{-1}}\).
For MX to become more exothermic overall, its lattice formation must provide an advantage greater than that amount.
The lattice enthalpy difference must therefore be more negative than \(-21\,\mathrm{kJ\,mol^{-1}}\).
Equality would produce equal reaction enthalpies, not a strictly more exothermic MX reaction.
The comparison demonstrates why an isolated-atom electron-gain trend cannot independently determine the energetics of ionic compound formation.
\(\textbf{Final answer:}\)
\[
\boxed{L_X-L_Y\lt -21\,\mathrm{kJ\,mol^{-1}}}
\]
202. A periodic-property graph for five consecutive Period 3 p-block elements is described below.
The horizontal axis lists silicon, phosphorus, sulphur, chlorine and argon in increasing atomic-number order. The vertical axis represents first ionization enthalpy. The plotted values generally rise, except for a small downward step from phosphorus to sulphur. The final point, argon, is the highest. A second curve representing comparable covalent radii is shown only for silicon through chlorine and decreases across those four elements.
Which interpretation correctly explains both the small ionization-enthalpy reversal and the absence of an argon point on the covalent-radius curve?
ⓐ. The ionization reversal arises from a change in principal shell, while argon's covalent radius is zero.
ⓑ. The reversal is caused by a decrease in nuclear charge from phosphorus to sulphur, while argon has no electrons available for radius measurement.
ⓒ. The reversal proves that sulphur is more metallic than phosphorus, while argon must be assigned chlorine's covalent radius.
ⓓ. Paired-electron repulsion lowers sulphur's ionization enthalpy relative to phosphorus; argon's covalent radius is not defined comparably.
Correct Answer: Paired-electron repulsion lowers sulphur's ionization enthalpy relative to phosphorus; argon's covalent radius is not defined comparably.
Explanation: \(\textbf{Interpret the first-ionization-enthalpy curve:}\)
All five elements belong to Period 3.
Their valence electrons occupy the same principal shell.
The nuclear charge generally increases from silicon to argon.
This tends to increase effective attraction for the outermost electrons.
Consequently, first ionization enthalpy generally rises across the period.
\(\textbf{Explain the phosphorus–sulphur exception:}\)
Phosphorus has the outermost configuration \(3s^23p^3\).
Its three \(p\) electrons occupy different orbitals, giving a relatively favourable half-filled arrangement.
Sulphur has \(3s^23p^4\).
One of its \(p\) orbitals contains a pair of electrons.
The additional repulsion between these paired electrons makes removal of one electron relatively easier.
Therefore, sulphur can have a slightly lower first ionization enthalpy despite having a higher nuclear charge.
\(\textbf{Interpret the covalent-radius curve:}\)
A covalent radius is normally determined from internuclear distances in suitable covalently bonded arrangements.
Silicon, phosphorus, sulphur and chlorine permit meaningful covalent-radius comparisons under the stated convention.
Argon is a noble gas with a closed valence shell and does not ordinarily form the same type of covalently bonded elemental structure.
A corresponding ordinary covalent-radius value is therefore not available on the same basis.
Argon can instead be described using a van der Waals radius.
\(\textbf{Identify the measurement limitation:}\)
A van der Waals radius and a covalent radius refer to different physical situations.
Inserting argon's van der Waals radius into the covalent-radius curve would mix unlike measurements.
This could create a misleading apparent periodic trend.
The two graphs therefore require different kinds of interpretation.
The ionization-enthalpy anomaly is electronic, whereas the missing covalent-radius value reflects a measurement-convention limitation.
Both observations are consistent with periodic chemistry.
203. A mixture initially contains \(0.100\,\mathrm{mol}\) of \(\mathrm{Cl_2}\), \(0.080\,\mathrm{mol}\) of \(\mathrm{Br^-}\) and \(0.060\,\mathrm{mol}\) of \(\mathrm{I^-}\) in aqueous solution. Assume that chlorine oxidizes both halide ions completely, that the only halogen products are \(\mathrm{Br_2}\) and \(\mathrm{I_2}\), and that no further reactions occur. Under these stipulated conditions, which pair gives the amounts of unreacted \(\mathrm{Cl_2}\) and produced \(\mathrm{I_2}\), respectively?
ⓐ. \(0.030\,\mathrm{mol}\) and \(0.030\,\mathrm{mol}\)
ⓑ. \(0.070\,\mathrm{mol}\) and \(0.030\,\mathrm{mol}\)
ⓒ. \(0.030\,\mathrm{mol}\) and \(0.060\,\mathrm{mol}\)
ⓓ. \(0.020\,\mathrm{mol}\) and \(0.040\,\mathrm{mol}\)
Correct Answer: \(0.030\,\mathrm{mol}\) and \(0.030\,\mathrm{mol}\)
Explanation: \(\textbf{Identify the feasible displacement reactions:}\)
Chlorine is a stronger oxidizing agent than bromine and iodine under the stated aqueous comparison.
It can therefore oxidize bromide to bromine and iodide to iodine.
The relevant balanced equations are:
\[
\mathrm{Cl_2+2Br^-\rightarrow 2Cl^-+Br_2}
\]
\[
\mathrm{Cl_2+2I^-\rightarrow 2Cl^-+I_2}
\]
Each reaction consumes one mole of chlorine per two moles of halide ions.
\(\textbf{Calculate chlorine consumed by bromide:}\)
The initial bromide amount is \(0.080\,\mathrm{mol}\).
Using the first equation:
\[
n_{\mathrm{Cl_2,Br^-}}=\frac{0.080}{2}
\]
\[
=0.040\,\mathrm{mol}
\]
The bromine produced is:
\[
n_{\mathrm{Br_2}}=\frac{0.080}{2}
\]
\[
=0.040\,\mathrm{mol}
\]
\(\textbf{Calculate chlorine consumed by iodide:}\)
The initial iodide amount is \(0.060\,\mathrm{mol}\).
Using the second equation:
\[
n_{\mathrm{Cl_2,I^-}}=\frac{0.060}{2}
\]
\[
=0.030\,\mathrm{mol}
\]
The iodine produced is:
\[
n_{\mathrm{I_2}}=\frac{0.060}{2}
\]
\[
=0.030\,\mathrm{mol}
\]
\(\textbf{Determine the total chlorine demand:}\)
\[
n_{\mathrm{Cl_2,consumed}}=0.040+0.030
\]
\[
=0.070\,\mathrm{mol}
\]
The initial chlorine amount is \(0.100\,\mathrm{mol}\).
Therefore:
\[
n_{\mathrm{Cl_2,remaining}}=0.100-0.070
\]
\[
=0.030\,\mathrm{mol}
\]
\(\textbf{Verify the electron balance:}\)
Oxidation of \(0.080\,\mathrm{mol}\) of bromide releases \(0.080\,\mathrm{mol}\) of electrons.
Oxidation of \(0.060\,\mathrm{mol}\) of iodide releases \(0.060\,\mathrm{mol}\) of electrons.
The total electron amount released is:
\[
0.080+0.060=0.140\,\mathrm{mol}
\]
Reduction of \(0.070\,\mathrm{mol}\) of chlorine consumes:
\[
2(0.070)=0.140\,\mathrm{mol\ of\ electrons}
\]
Thus, the electron transfer is balanced.
The supplied assumption that both displacement processes proceed completely avoids ambiguity about competing reaction pathways.
\(\textbf{Final answer:}\)
\[
\boxed{n_{\mathrm{Cl_2,remaining}}=0.030\,\mathrm{mol},\qquad n_{\mathrm{I_2}}=0.030\,\mathrm{mol}}
\]
204. Four elements W, X, Y and Z belong to the same period. Their outermost configurations are \(ns^2np^1\), \(ns^2np^3\), \(ns^2np^5\) and \(ns^2np^6\), respectively. Consider the following statements.
Statement I: W and Y each possess one unpaired \(p\) electron in their isolated ground states.
Statement II: X possesses a half-filled \(p\) subshell and three unpaired \(p\) electrons.
Statement III: Addition of one electron to Y produces a closed valence shell, whereas addition of one electron to Z requires occupation beyond its initially closed shell.
Statement IV: The number of unpaired \(p\) electrons increases continuously from W to Z.
Which combination is correct?
ⓐ. Statements I and II 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: \(\textbf{Identify the group sequence:}\)
The configurations correspond to Groups 13, 15, 17 and 18.
The four elements belong to the same period but contain different numbers of outermost \(p\) electrons.
Their total valence-electron counts are three, five, seven and eight, respectively.
\(\textbf{Examine Statement I:}\)
W has an \(np^1\) subshell.
Its single \(p\) electron is unpaired.
Y has an \(np^5\) subshell.
In this arrangement, two \(p\) orbitals contain electron pairs and one contains a single electron.
Therefore, Y also has one unpaired \(p\) electron.
Statement I is correct.
\(\textbf{Examine Statement II:}\)
X has the \(np^3\) configuration.
According to Hund's rule, the three electrons occupy the three degenerate \(p\) orbitals singly.
This produces a half-filled subshell with three unpaired electrons.
Statement II is correct.
\(\textbf{Examine Statement III:}\)
Y has seven valence electrons.
Addition of one electron completes its \(np^6\) subshell.
Z already possesses the closed-shell configuration \(ns^2np^6\).
An additional electron cannot enter its filled valence \(s\) or \(p\) subshell.
It must occupy a higher-energy available orbital.
Statement III is correct.
\(\textbf{Examine Statement IV:}\)
The unpaired-electron count does not rise continuously with increasing \(p\)-electron occupancy.
It increases up to \(p^3\) and then falls as electrons pair.
The \(p^6\) configuration contains no unpaired electrons.
Therefore, Statement IV is incorrect.
\(\textbf{Interpret the broader consequence:}\)
Group identification requires the complete valence configuration, not merely the unpaired-electron count.
Likewise, electron-gain behaviour depends on the availability and energy of the receiving orbital.
The four statements must therefore be evaluated using both Hund's rule and shell-completion reasoning.
Only Statements I, II and III are correct.
205. A chemist compares the first ionization enthalpies and normal boiling points of argon and krypton. The approximate values are:
| Element | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) | Normal boiling point (K) |
|---|
| Ar | 1521 | 87 |
| Kr | 1351 | 120 |
Consider an initially gaseous equimolar mixture of argon and krypton that is cooled at \(1\,\mathrm{atm}\) to \(100\,\mathrm{K}\). Assume idealized equilibrium behaviour without significant gas-mixture complications. Which statement correctly predicts the relative tendency of the two components to condense and explains the ionization-enthalpy trend?
ⓐ. Argon condenses preferentially because it has the higher first ionization enthalpy.
ⓑ. Krypton has the greater tendency to condense, but its first ionization enthalpy must be higher because its atomic number is larger.
ⓒ. Both gases must remain uncondensed because their atoms have complete valence shells.
ⓓ. Krypton condenses more readily at \(100\,\mathrm K\) but has lower ionization enthalpy because of its larger, more shielded shell.
Correct Answer: Krypton condenses more readily at \(100\,\mathrm K\) but has lower ionization enthalpy because of its larger, more shielded shell.
Explanation: \(\textbf{Compare the boiling temperatures:}\)
Argon has a normal boiling point near \(87\,\mathrm{K}\).
Krypton has a normal boiling point near \(120\,\mathrm{K}\).
The specified temperature of \(100\,\mathrm{K}\) lies above argon's normal boiling point but below krypton's.
This indicates that krypton has the greater tendency to remain in a condensed phase under the simplified conditions.
Argon is comparatively more volatile.
\(\textbf{Explain the condensation difference:}\)
Both substances are monatomic noble gases.
They possess no permanent molecular dipole moments.
Their attractive interactions arise principally from London dispersion forces.
Krypton has a larger and more polarizable electron cloud.
Consequently, the attractions between krypton atoms are stronger than those between argon atoms.
This contributes to krypton's higher boiling point.
\(\textbf{Interpret the ionization-enthalpy comparison:}\)
Argon has the outermost configuration \(3s^23p^6\).
Krypton has \(4s^24p^6\).
Both possess completed valence shells, but krypton's outermost electrons occupy a higher principal shell.
The increased distance from the nucleus and additional shielding generally make removal of an outer electron easier.
Therefore, krypton has the lower first ionization enthalpy.
\(\textbf{Reconcile the properties:}\)
Stronger dispersion attractions do not imply stronger nuclear binding of the outermost electron.
Boiling point and ionization enthalpy describe different processes.
The former depends on attractions between neutral atoms in condensed matter.
The latter concerns removing an electron from an isolated gaseous atom.
The observed increase in boiling point and decrease in ionization enthalpy are therefore compatible.
Under the stated idealized cooling conditions, krypton shows the greater condensation tendency.
206. A neutral atom X possesses a \(4p^4\) subshell, while another neutral atom Y possesses a \(4p^5\) subshell. Both have the same filled inner subshells and the outermost \(4s^2\) pair. Which option correctly identifies the ions formed by adding one electron to each atom and compares their ground-state magnetic behaviour?
ⓐ. \(\mathrm{X^-}\) has \(4p^5\) with one unpaired electron, while \(\mathrm{Y^-}\) has \(4p^6\) with no unpaired electrons.
ⓑ. \(\mathrm{X^-}\) has \(4p^6\) with no unpaired electrons, while \(\mathrm{Y^-}\) has \(4p^5\) with one unpaired electron.
ⓒ. Both ions have \(4p^6\) because their neutral atoms belong to the same period.
ⓓ. \(\mathrm{X^-}\) has \(4p^3\) with three unpaired electrons, while \(\mathrm{Y^-}\) has \(4p^4\) with two unpaired electrons.
Correct Answer: \(\mathrm{X^-}\) has \(4p^5\) with one unpaired electron, while \(\mathrm{Y^-}\) has \(4p^6\) with no unpaired electrons.
Explanation: \(\textbf{Identify the initial orbital occupancies:}\)
X has the outermost configuration \(4s^24p^4\).
It therefore possesses six valence electrons and belongs to Group 16.
Y has \(4s^24p^5\).
It possesses seven valence electrons and belongs to Group 17.
Both atoms occupy Period 4.
\(\textbf{Follow electron gain by X:}\)
Adding one electron to the \(4p^4\) configuration produces \(4p^5\).
According to Hund's rule, a \(p^5\) subshell has two paired orbitals and one singly occupied orbital.
Therefore, the resulting mononegative ion has one unpaired \(p\) electron.
Its configuration remains one electron short of a completely filled \(p\) subshell.
\(\textbf{Follow electron gain by Y:}\)
Adding one electron to \(4p^5\) produces \(4p^6\).
All three \(p\) orbitals are then completely occupied.
The resulting mononegative ion has no unpaired electrons in this subshell.
Its outermost configuration is closed-shell.
\(\textbf{Connect electron configuration with magnetism:}\)
An isolated ground-state species containing an unpaired electron is paramagnetic.
A closed-shell species with all electrons paired is diamagnetic.
Thus, the formal mononegative ion of X has an unpaired electron, whereas the mononegative ion of Y is diamagnetic.
\(\textbf{Qualify the chemical interpretation:}\)
This is a comparison of stipulated isolated ionic electronic configurations.
It does not imply that the mononegative ion of every Group 16 element is the most stable form in ordinary compounds.
Group 16 elements frequently attain a closed shell by acquiring two electrons rather than one.
The result therefore distinguishes the electronic consequence of a single electron-gain event from the preferred ion or oxidation state in a chemical compound.
The correct option follows directly from the different initial \(p\)-electron occupancies.
207. Two gaseous p-block atoms P and Q are in the same period. P possesses the outermost configuration \(ns^2np^2\), while Q possesses \(ns^2np^5\). An electron is removed from each atom. A student claims that because both atoms initially have electrons in the same principal shell, removal must produce the same number of unpaired \(p\) electrons. Which assessment is correct?
ⓐ. The claim is valid because one-electron removal always creates exactly three unpaired \(p\) electrons.
ⓑ. The claim is valid because the principal quantum number alone determines orbital pairing.
ⓒ. The claim is invalid because P forms \(np^1\) with one unpaired electron, while Q forms \(np^4\) with two unpaired electrons.
ⓓ. The claim is invalid because P forms \(np^3\) and Q forms \(np^6\), producing different closed-shell arrangements.
Correct Answer: The claim is invalid because P forms \(np^1\) with one unpaired electron, while Q forms \(np^4\) with two unpaired electrons.
Explanation: \(\textbf{Identify the electron-removal process:}\)
P and Q belong to the same period, so their outermost electrons occupy the same principal shell.
However, the two atoms belong to different groups.
P has four valence electrons, whereas Q has seven.
Their initial \(p\)-electron occupancies are therefore different.
\(\textbf{Examine P:}\)
P begins with an \(np^2\) subshell.
The two \(p\) electrons occupy separate orbitals in the isolated ground state.
Removing one electron produces an \(np^1\) configuration.
The resulting ion contains one unpaired \(p\) electron.
\(\textbf{Examine Q:}\)
Q begins with an \(np^5\) subshell.
Removing one electron produces an \(np^4\) configuration.
The four electrons occupy three \(p\) orbitals according to Hund's rule.
One orbital contains an electron pair while two orbitals contain one electron each.
Therefore, the resulting ion has two unpaired \(p\) electrons.
\(\textbf{Identify the faulty assumption:}\)
The principal quantum number identifies the outermost shell but does not establish its exact electron occupancy.
Two atoms in the same period can have different numbers of \(p\) electrons.
Removing one electron from each consequently need not produce the same electron-pairing pattern.
The correct calculation must begin with the complete valence configurations.
\(\textbf{Interpret the broader relevance:}\)
The number of unpaired electrons affects the paramagnetic behaviour of isolated species.
However, it does not independently determine the covalency of every compound containing the element.
The comparison therefore separates ground-state orbital occupancy from more general chemical bonding predictions.
P forms an \(np^1\) ion with one unpaired electron, while Q forms an \(np^4\) ion with two.
208. Three hypothetical p-block elements X, Y and Z lie in the same period and have the following properties.
| Element | Valence-shell configuration | General character of representative oxide |
|---|
| X | \(ns^2np^1\) | Amphoteric |
| Y | \(ns^2np^2\) | Acidic |
| Z | \(ns^2np^3\) | Acidic |
Assume these configurations and oxide classifications correspond to aluminium, silicon and phosphorus in Period 3. Which conclusion is justified?
ⓐ. Increasing oxide acidity proves that the number of occupied electron shells decreases from X to Z.
ⓑ. Oxides become less basic as metallic character decreases, although valence-electron count increases.
ⓒ. X must have the highest electronegativity because its oxide is amphoteric.
ⓓ. All three elements must possess identical ionization enthalpies because their outermost electrons occupy \(n=3\).
Correct Answer: Oxides become less basic as metallic character decreases, although valence-electron count increases.
Explanation: \(\textbf{Identify the three elements:}\)
The \(ns^2np^1\) configuration corresponds to Group 13.
In Period 3, the relevant element is aluminium.
The \(ns^2np^2\) configuration identifies Group 14 silicon.
The \(ns^2np^3\) configuration identifies Group 15 phosphorus.
Their valence-electron counts are three, four and five, respectively.
\(\textbf{Interpret the oxide classifications:}\)
Aluminium oxide exhibits amphoteric behaviour.
Silicon dioxide is an acidic oxide.
Representative phosphorus oxides are also acidic.
The supplied classifications are therefore consistent with the familiar change in oxide behaviour across this part of Period 3.
\(\textbf{Connect oxide behaviour with metallic character:}\)
Aluminium is metallic.
Silicon is a metalloid.
Phosphorus is non-metallic.
The broad decrease in metallic character is associated with increased attraction for electrons and increasingly non-metallic bonding tendencies.
These changes help explain why representative oxides become more acidic across the sequence.
\(\textbf{Reject the shell-number interpretation:}\)
All three atoms have outermost electrons occupying the third principal shell.
Therefore, increasing acidity is not caused by a decrease in principal quantum number.
The number of valence electrons increases across the sequence, but this does not require the oxide basicity to increase.
\(\textbf{Interpret the ionization and electronegativity trends:}\)
First ionization enthalpy generally increases across this part of the period, although the exact values depend on subshell occupancy.
Electronegativity also generally increases.
Thus, aluminium is not expected to have the greatest electronegativity of the three.
The correct conclusion must connect electronic configuration and chemical character without confusing electron count with metallicity.
The oxide classifications are consistent with increasing non-metallic character from aluminium to phosphorus.
209. A table of first ionization enthalpies contains the values shown below.
| Element | First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| B | 801 |
| C | 1086 |
| N | 1402 |
| O | 1314 |
| F | 1681 |
| Ne | 2081 |
A student draws a straight line through the first and last values and assumes every intermediate element must lie on or above that line because nuclear charge increases. Which observation most directly invalidates the proposed rule?
ⓐ. The ionization enthalpy of nitrogen exceeds that of oxygen, although oxygen has the greater nuclear charge.
ⓑ. Fluorine has a lower ionization enthalpy than neon, although fluorine is highly electronegative.
ⓒ. Carbon has a higher ionization enthalpy than boron, although both belong to the p-block.
ⓓ. Neon has a filled valence shell and a high first ionization enthalpy.
Correct Answer: The ionization enthalpy of nitrogen exceeds that of oxygen, although oxygen has the greater nuclear charge.
Explanation: \(\textbf{Identify the proposed model:}\)
The student assumes that first ionization enthalpy must rise smoothly and uniformly with increasing atomic number.
A general increase is indeed expected across a period because effective nuclear attraction tends to increase.
However, the precise values also depend on orbital energies and electron pairing.
Therefore, a simple straight-line model need not represent every local comparison correctly.
\(\textbf{Examine the nitrogen–oxygen pair:}\)
Nitrogen has the configuration \(1s^22s^22p^3\).
Its \(2p\) subshell is half-filled.
The three \(p\) electrons occupy separate orbitals and produce a relatively favourable arrangement.
Oxygen has the configuration \(1s^22s^22p^4\).
One of its \(2p\) orbitals contains a pair of electrons.
The additional repulsion makes removal of one electron comparatively easier.
\(\textbf{Interpret the measured values:}\)
Nitrogen's first ionization enthalpy is \(1402\,\mathrm{kJ\,mol^{-1}}\).
Oxygen's is \(1314\,\mathrm{kJ\,mol^{-1}}\).
Therefore:
\[
IE_1(\mathrm{O})\lt IE_1(\mathrm{N})
\]
This local decrease occurs despite oxygen having the greater nuclear charge.
\(\textbf{Distinguish a trend from a strict mathematical rule:}\)
A broad periodic trend may contain local exceptions.
Such exceptions do not invalidate the underlying role of increasing effective nuclear charge.
They show that other electronic factors can modify the observed values.
The remaining listed comparisons are broadly consistent with increasing ionization enthalpy or describe the high value of neon.
The nitrogen–oxygen comparison directly demonstrates that the student's strict monotonic assumption is unjustified.
The correct interpretation preserves the general trend while recognizing the effect of electron-pair repulsion.
210. Consider the following Assertion and Reason.
Assertion (A): Within a given period, the p-block element with the greatest number of valence electrons need not be the element with the strongest ordinary tendency to gain an electron.
Reason (R): A noble gas generally has a completed valence shell, so addition of an electron to its isolated gaseous atom requires occupation of a higher-energy available orbital.
Choose the correct option.
ⓐ. Both A and R are true, but R does not correctly explain A.
ⓑ. A is true, but R is false.
ⓒ. A is false, but R is true.
ⓓ. Both A and R are true, and R correctly explains A.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: \(\textbf{Examine the assertion:}\)
Across an ordinary p-block period, the number of valence electrons increases towards Group 18.
Group 17 elements possess seven valence electrons.
Group 18 elements, except helium in its special first-shell configuration, possess eight valence electrons.
The greater electron count of the noble gas does not imply a greater tendency to accept an additional electron.
Therefore, the assertion is true.
\(\textbf{Examine the halogen electron-gain process:}\)
A halogen has the general configuration \(ns^2np^5\).
Addition of one electron completes its \(p\) subshell.
The resulting \(ns^2np^6\) configuration is a closed-shell arrangement.
This electron-gain process is often strongly exothermic for the common halogens.
\(\textbf{Examine the noble-gas electron-gain process:}\)
A noble gas already possesses a completed outermost shell.
Its occupied valence \(s\) and \(p\) subshells cannot accommodate another electron.
The added electron must occupy a higher-energy available orbital.
The process is therefore generally energetically unfavourable for an isolated gaseous noble-gas atom.
The reason is true.
\(\textbf{Connect the reason with the assertion:}\)
The different electron-gain behaviour follows from the initial electronic configurations.
The halogen is one electron short of completing its valence shell.
The noble gas has already completed that shell.
Therefore, the noble gas's greater valence-electron count does not produce a stronger ordinary tendency towards electron gain.
The reason directly explains why the assertion is correct.
This comparison also illustrates why periodic chemical behaviour cannot be predicted from valence-electron count alone without considering subshell occupancy.
211. Two equimolar samples of neutral gaseous atoms X and Y have atomic numbers 16 and 17, respectively. They are separately supplied with exactly one mole of electrons per mole of atoms in idealized electron-attachment processes. Assume the resulting mononegative species can be examined as isolated gaseous ions. Which prediction correctly compares the resulting electronic configurations and the additional electrons needed to reach the argon configuration?
ⓐ. Both \(\mathrm{X^-}\) and \(\mathrm{Y^-}\) possess the argon configuration, so neither needs another electron.
ⓑ. \(\mathrm{X^-}\) contains 17 electrons and \(\mathrm{Y^-}\) contains 18; only the latter has the argon configuration.
ⓒ. \(\mathrm{X^-}\) has eighteen electrons, while \(\mathrm{Y^-}\) has seventeen; only Y requires another electron.
ⓓ. \(\mathrm{X^-}\) and \(\mathrm{Y^-}\) each require two additional electrons because both are p-block elements.
Correct Answer: \(\mathrm{X^-}\) contains 17 electrons and \(\mathrm{Y^-}\) contains 18; only the latter has the argon configuration.
Explanation: \(\textbf{Identify the neutral atoms:}\)
Atomic number 16 corresponds to sulphur.
Its neutral atom contains sixteen electrons and has the configuration \([\mathrm{Ne}]\,3s^23p^4\).
Atomic number 17 corresponds to chlorine.
Its neutral atom contains seventeen electrons and has the configuration \([\mathrm{Ne}]\,3s^23p^5\).
\(\textbf{Apply the stipulated electron attachment:}\)
Each neutral atom gains one electron.
The sulphur-derived mononegative species therefore contains:
\[
N_e(\mathrm{S^-})=16+1=17
\]
Its outermost configuration is \(3s^23p^5\).
This is not yet the argon configuration.
The chlorine-derived species contains:
\[
N_e(\mathrm{Cl^-})=17+1=18
\]
Its outermost configuration is \(3s^23p^6\).
This is identical to argon's closed-shell configuration.
\(\textbf{Determine the remaining electron requirement:}\)
The sulphur-derived ion would need one additional electron to reach eighteen electrons.
The chlorine-derived ion already contains eighteen electrons.
Thus, their further electron requirements differ even though both underwent the same one-electron attachment process.
\(\textbf{Interpret the group relationship:}\)
Sulphur belongs to Group 16 and has six valence electrons as a neutral atom.
Chlorine belongs to Group 17 and has seven.
The tendency of Group 16 elements to attain a closed shell by acquiring two electrons differs from the corresponding one-electron requirement of halogens.
However, electron counting alone does not prove that every isolated gaseous ion is energetically stable.
In particular, the second attachment of an electron to a negatively charged ion may be strongly opposed by electrostatic repulsion.
The correct comparison therefore distinguishes electronic completion from the thermodynamics of electron addition.
212. A comparison is made between two members of the same p-block group, X and Y, with Y lying below X. The following observations are supplied.
The covalent radius of Y is only slightly greater than that of X, even though Y has one additional principal electron shell. The first ionization enthalpy of Y is also slightly greater than that of X. Both atoms possess the general outermost configuration \(ns^2np^2\). Their chemical behaviour shows an increased relative importance of the +2 oxidation state for Y.
Which explanation best combines the irregular atomic-property trends with the oxidation-state observation?
ⓐ. The observations prove that Y has fewer valence electrons than X and therefore must belong to Group 13.
ⓑ. The observations are impossible because first ionization enthalpy must always decrease when the principal quantum number increases.
ⓒ. Inner-subshell shielding affects size and ionization trends, while increased inert-pair character favours lower oxidation states.
ⓓ. The increased stability of the +2 state proves that Y possesses only two electrons in its neutral valence shell.
Correct Answer: Inner-subshell shielding affects size and ionization trends, while increased inert-pair character favours lower oxidation states.
Explanation: \(\textbf{Identify the common group framework:}\)
Both atoms have the general valence-shell configuration \(ns^2np^2\).
They therefore belong to Group 14 and possess four valence electrons.
The lower oxidation state of +2 does not mean that the neutral atom contains only two valence electrons.
It means that only part of its valence-electron complement is formally involved in the corresponding bonding description.
\(\textbf{Interpret the radius observation:}\)
Moving down a group introduces an additional principal shell.
This generally increases atomic size.
However, inner subshells do not always shield the increasing nuclear charge equally effectively.
Filled \(d\) and \(f\) subshells can provide relatively poor screening.
Consequently, increased effective nuclear attraction may partly counteract the expansion expected from the new shell.
This can produce an unexpectedly small increase in covalent radius.
\(\textbf{Interpret the ionization-enthalpy observation:}\)
The outermost electron of Y occupies a higher principal shell than that of X.
A simple down-group argument would therefore predict easier electron removal.
Nevertheless, incomplete shielding and changes in orbital energies can modify that expectation.
A slight increase in first ionization enthalpy is therefore possible.
\(\textbf{Connect the lower oxidation state:}\)
For a Group 14 element, the +2 oxidation state can correspond formally to retention of the outermost \(ns^2\) pair.
The +4 state involves participation of that pair as well.
In heavier p-block elements, the \(ns^2\) pair becomes comparatively less favourable for bonding participation.
This inert-pair effect contributes to increasing relative stability of the lower oxidation state.
\(\textbf{Reconcile the observations:}\)
The small radius increase, ionization-enthalpy irregularity and lower oxidation-state preference do not all result from one simple numerical rule.
They reflect related changes in shielding, orbital energies and valence-electron participation.
The observations are compatible with heavy Group 14 behaviour and do not require a change in group membership.
The correct explanation therefore combines incomplete shielding with the increasing importance of the inert-pair effect.
213. A hypothetical Group 13 element M can form compounds in oxidation states +1 and +3. Its first three successive ionization enthalpies are \(600\), \(1800\) and \(2800\,\mathrm{kJ\,mol^{-1}}\), respectively. Two alternative compound-formation processes are considered:
| Process | Energy released after forming gaseous metal cations |
|---|
| Formation of an M(I) compound from \(\mathrm{M^+(g)}\) | \(1200\,\mathrm{kJ\,mol^{-1}}\) |
| Formation of an M(III) compound from \(\mathrm{M^{3+}(g)}\) | \(5400\,\mathrm{kJ\,mol^{-1}}\) |
The energy-release values include all remaining contributions required to produce the respective final compounds from the specified gaseous cations. Both processes begin with one mole of neutral gaseous M atoms. Which conclusion is correct?
ⓐ. The M(I) compound is favoured by \(800\,\mathrm{kJ\,mol^{-1}}\) because only the first ionization enthalpy determines oxidation-state stability.
ⓑ. The M(III) compound is favoured by \(400\,\mathrm{kJ\,mol^{-1}}\) because its additional stabilization outweighs the second and third ionization enthalpies.
ⓒ. The M(I) compound is favoured by \(400\,\mathrm{kJ\,mol^{-1}}\) because the additional stabilization of M(III) does not compensate for the additional ionization energy.
ⓓ. Both compounds are equally favoured because formation of each begins with the same neutral metal atom.
Correct Answer: The M(I) compound is favoured by \(400\,\mathrm{kJ\,mol^{-1}}\) because the additional stabilization of M(III) does not compensate for the additional ionization energy.
Explanation: \(\textbf{Identify the oxidation-state alternatives:}\)
Group 13 elements have the general valence-shell configuration \(ns^2np^1\).
Formation of M(I) corresponds formally to removal of one valence electron.
Formation of M(III) involves removal of three valence electrons.
The relative stability of compounds containing these oxidation states cannot be determined from the group number alone.
It depends on the energetic balance of electron removal and subsequent compound formation.
\(\textbf{Calculate the M(I) pathway:}\)
The first ionization process requires:
\[
IE_1=+600\,\mathrm{kJ\,mol^{-1}}
\]
The remaining compound-formation steps release:
\[
\Delta H_{\mathrm{stabilization,I}}=-1200\,\mathrm{kJ\,mol^{-1}}
\]
The total change is:
\[
\Delta H_{\mathrm I}=600-1200
\]
\[
\Delta H_{\mathrm I}=-600\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the M(III) pathway:}\)
Formation of the tripositive gaseous ion requires three successive ionizations.
\[
\Delta H_{\mathrm{ionization,III}}=600+1800+2800
\]
\[
=5200\,\mathrm{kJ\,mol^{-1}}
\]
The subsequent stabilization contribution is:
\[
\Delta H_{\mathrm{stabilization,III}}=-5400\,\mathrm{kJ\,mol^{-1}}
\]
Therefore:
\[
\Delta H_{\mathrm{III}}=5200-5400
\]
\[
\Delta H_{\mathrm{III}}=-200\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the complete pathways:}\)
\[
\Delta H_{\mathrm{III}}-\Delta H_{\mathrm I}=-200-(-600)
\]
\[
=+400\,\mathrm{kJ\,mol^{-1}}
\]
The M(I) pathway has the more negative enthalpy change.
It is therefore more exothermic by \(400\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Interpret the chemistry:}\)
The M(III) compound receives considerably greater stabilization after cation formation.
However, the additional energy required for the second and third ionizations outweighs this advantage.
This illustrates why strong bonding can stabilize a higher oxidation state without guaranteeing its energetic preference.
It also explains why the inert-pair effect must be interpreted alongside compound-formation energies rather than as an absolute prohibition against higher oxidation states.
The conclusion concerns the supplied enthalpy model; a complete equilibrium comparison would also require appropriate Gibbs-energy information.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm I}=-600,\quad\Delta H_{\mathrm{III}}=-200\ \mathrm{kJ\,mol^{-1}};\quad\text{M(I) is favoured enthalpically by }400\ \mathrm{kJ\,mol^{-1}}}
\]
214. Boron trichloride and nitrogen trichloride both contain their central elements in oxidation state +3. Nevertheless, isolated \(\mathrm{BCl_3}\) is an electron-pair acceptor, whereas \(\mathrm{NCl_3}\) contains a central-atom lone pair. Which statement correctly connects these differences with their valence-electron configurations?
ⓐ. Boron forms trigonal-planar electron-deficient \(\mathrm{BCl_3}\); nitrogen forms pyramidal \(\mathrm{NCl_3}\) with a lone pair.
ⓑ. Boron and nitrogen each have three valence electrons, but nitrogen acquires two additional electrons from chlorine before bonding.
ⓒ. Both trichlorides have trigonal-planar geometries because the oxidation state of the central element is +3.
ⓓ. Nitrogen trichloride has no central-atom lone pair because its nitrogen oxidation state is positive, while boron trichloride contains one lone pair on boron.
Correct Answer: Boron forms trigonal-planar electron-deficient \(\mathrm{BCl_3}\); nitrogen forms pyramidal \(\mathrm{NCl_3}\) with a lone pair.
Explanation: \(\textbf{Establish the central-atom configurations:}\)
Boron belongs to Group 13 and has the valence-shell configuration \(2s^22p^1\).
It therefore possesses three valence electrons.
Nitrogen belongs to Group 15 and has the configuration \(2s^22p^3\).
It possesses five valence electrons.
Thus, the atoms have different electron-pair arrangements available for ordinary bonding.
\(\textbf{Examine boron trichloride:}\)
Boron forms three B–Cl sigma bonds.
In the conventional Lewis structure, the central boron atom has three bonding pairs and no lone pair.
The molecule is trigonal planar.
Only six electrons are represented around boron through the three bonds.
The electron-deficient central atom can accept an electron pair from a suitable donor.
Therefore, \(\mathrm{BCl_3}\) behaves as a Lewis acid.
\(\textbf{Examine nitrogen trichloride:}\)
Nitrogen also forms three bonds with chlorine.
However, two of nitrogen's five valence electrons remain as a lone pair.
The central atom consequently has three bonding domains and one lone-pair domain.
Its ordinary molecular geometry is trigonal pyramidal.
The presence of a lone pair gives nitrogen an electron-pair-donating possibility, although actual Lewis basicity depends on its bonding environment.
\(\textbf{Compare oxidation states:}\)
Chlorine is more electronegative than both boron and nitrogen.
It is formally assigned oxidation state \(-1\) in the two trichlorides.
The central atom therefore has oxidation state +3 in each molecule.
However, identical oxidation states do not imply identical lone-pair arrangements or Lewis acid–base behaviour.
The decisive distinction is the total valence-electron count and the resulting electronic structure.
Thus, the comparison illustrates how Group 13 electron deficiency differs from the lone-pair chemistry of Group 15.
215. Carbon dioxide and sulphur dioxide both contain a central p-block element in oxidation state +4 and are classified as acidic oxides. However, isolated \(\mathrm{CO_2}\) is non-polar, whereas isolated \(\mathrm{SO_2}\) is polar. Which explanation correctly accounts for this difference?
ⓐ. The oxidation state +4 makes carbon dioxide linear but requires sulphur dioxide to contain sulphur in oxidation state +2.
ⓑ. Carbon dioxide is non-polar because its C–O bonds are non-polar, whereas sulphur dioxide contains polar S–O bonds.
ⓒ. Both molecules possess permanent dipole moments because each contains two polar bonds.
ⓓ. Carbon dioxide is linear and its bond dipoles cancel, whereas sulphur dioxide is bent and its bond dipoles have a non-zero resultant.
Correct Answer: Carbon dioxide is linear and its bond dipoles cancel, whereas sulphur dioxide is bent and its bond dipoles have a non-zero resultant.
Explanation: \(\textbf{Determine the oxidation states:}\)
Oxygen is assigned oxidation state \(-2\) in both ordinary dioxides.
Each molecule contains two oxygen atoms.
Therefore, the central atom has oxidation state +4 in both compounds.
This formal oxidation-state equality does not establish identical molecular geometries.
\(\textbf{Examine the carbon dioxide structure:}\)
Carbon has four valence electrons.
The conventional molecular structure of \(\mathrm{CO_2}\) contains two C=O double bonds.
The carbon centre has two bonding directions and no lone pair in the ordinary Lewis description.
Its molecular geometry is linear.
The two C–O bonds are polar.
However, their dipole vectors have equal magnitudes and opposite directions.
They therefore cancel, making the molecule non-polar.
\(\textbf{Examine the sulphur dioxide structure:}\)
Sulphur belongs to Group 16 and has six valence electrons.
In \(\mathrm{SO_2}\), its bonding arrangement includes a non-bonding electron domain on the central sulphur atom in the conventional electron-pair description.
The molecule adopts a bent geometry.
The polar S–O bond dipoles do not point in exactly opposite directions.
Their resultant is consequently non-zero.
Therefore, sulphur dioxide possesses a permanent molecular dipole moment.
\(\textbf{Connect the structural and chemical descriptions:}\)
Both compounds are acidic oxides, but acidic character does not require identical molecular polarity.
Similarly, the +4 oxidation state does not fix the coordination geometry.
The first element of Group 14 and the Period 3 Group 16 element have different valence-electron arrangements.
These differences affect bonding directions and molecular shape.
The correct explanation therefore requires distinguishing oxidation state, oxide classification and molecular dipole moment.
216. Radon belongs to Group 18 and has a completed \(6s^26p^6\) outermost configuration. Unlike the lighter familiar noble gases, radon is radioactive. Which conclusion correctly distinguishes its nuclear behaviour from its general electronic chemical character?
ⓐ. Radioactivity means radon has an incomplete \(6p\) subshell and therefore cannot be classified as a noble gas.
ⓑ. Radon must exhibit a positive oxidation state during every radioactive decay because nuclear changes involve oxidation.
ⓒ. Radon can be radioactive despite a full valence shell; nuclear decay and chemical inertness are distinct.
ⓓ. The completed outermost shell prevents radon nuclei from undergoing radioactive decay.
Correct Answer: Radon can be radioactive despite a full valence shell; nuclear decay and chemical inertness are distinct.
Explanation: \(\textbf{Identify radon's periodic position:}\)
Radon has atomic number 86 and belongs to Group 18.
Its ground-state outermost configuration is \(6s^26p^6\).
The outermost shell is therefore complete in the usual noble-gas sense.
This configuration contributes to radon's relatively low ordinary chemical reactivity.
\(\textbf{Examine radioactive instability:}\)
Radioactivity is associated with instability of the atomic nucleus.
It concerns changes involving nuclear particles and nuclear energy states.
The presence of a completed outer electron shell does not guarantee nuclear stability.
Therefore, an atom can be chemically unreactive while possessing an unstable nucleus.
\(\textbf{Distinguish chemical oxidation from nuclear transformation:}\)
Oxidation state is a formal description of electron distribution in chemical species.
Radioactive decay does not require the initial atom to undergo an ordinary chemical oxidation reaction.
A nuclear transformation can change the identity of the element independently of its prior bonding environment.
Thus, radon's radioactivity does not prove that its original valence shell is incomplete.
\(\textbf{Interpret the Group 18 consequence:}\)
Noble-gas membership is primarily associated with electronic configuration and chemical behaviour.
Nuclear stability is a separate property.
Helium, neon and argon are familiar examples of noble gases with stable naturally occurring isotopes.
Radon's radioactive character does not remove it from Group 18.
The correct conclusion preserves the distinction between electronic shell completion and nuclear stability.
217. The approximate bond dissociation enthalpies shown below refer to gas-phase species.
| Bond | Bond dissociation enthalpy \((\mathrm{kJ\,mol^{-1}})\) |
|---|
| O–O single bond | 146 |
| S–S single bond | 266 |
| O=O double bond in \(\mathrm{O_2}\) | 498 |
Oxygen and sulphur belong to Group 16. Which conclusion is consistent with the data and their characteristic elemental bonding?
ⓐ. Oxygen must show greater single-bond catenation than sulphur because the O=O double bond is strong.
ⓑ. Compact lone-pair repulsions weaken O–O single bonds, favouring sulphur's greater single-bonded catenation.
ⓒ. The stronger S–S single bond proves that molecular sulphur normally consists exclusively of S=S double bonds.
ⓓ. Oxygen and sulphur must have identical catenation tendencies because both have six valence electrons.
Correct Answer: Compact lone-pair repulsions weaken O–O single bonds, favouring sulphur's greater single-bonded catenation.
Explanation: \(\textbf{Distinguish single and multiple bonding:}\)
Oxygen and sulphur both have the general Group 16 valence configuration \(ns^2np^4\).
However, their atomic sizes and orbital overlaps differ.
A tendency to form a strong multiple bond does not automatically imply a tendency to form strong element–element single bonds.
\(\textbf{Examine oxygen single bonding:}\)
Oxygen is a small second-period element.
When two oxygen atoms form an ordinary O–O single bond, their non-bonding electron pairs lie relatively close together.
Repulsions involving these compact electron distributions weaken the single bond.
The supplied O–O single-bond enthalpy is only \(146\,\mathrm{kJ\,mol^{-1}}\).
\(\textbf{Examine sulphur single bonding:}\)
Sulphur atoms are larger.
The corresponding S–S single bond experiences a different balance of orbital overlap and lone-pair repulsion.
Its supplied bond dissociation enthalpy is \(266\,\mathrm{kJ\,mol^{-1}}\).
This stronger single bond supports the formation of cyclic and chain-like sulphur structures.
Ordinary \(\mathrm{S_8}\) molecules provide a familiar example of single-bonded sulphur rings.
\(\textbf{Examine molecular oxygen:}\)
The O=O double bond in \(\mathrm{O_2}\) is considerably stronger than an ordinary O–O single bond.
Effective overlap between compact oxygen \(2p\) orbitals contributes to strong multiple bonding.
Therefore, oxygen can favour a strongly bonded diatomic molecule while exhibiting comparatively weak single-bond catenation.
\(\textbf{Connect the competing tendencies:}\)
Catenation depends substantially on the stability of repeated element–element linkages.
Multiple-bond formation depends on a different orbital-overlap pattern.
It would be incorrect to infer oxygen's single-bond catenation tendency from the strength of its double bond.
The comparison illustrates how first-member behaviour can arise from the competing effects of compact-orbital overlap and lone-pair repulsion.
218. Consider three experimentally established comparisons between fluorine and chlorine.
| Property | Fluorine | Chlorine |
|---|
| First ionization enthalpy \((\mathrm{kJ\,mol^{-1}})\) | 1681 | 1251 |
| First electron-gain enthalpy \((\mathrm{kJ\,mol^{-1}})\) | -328 | -349 |
| X–X bond dissociation enthalpy \((\mathrm{kJ\,mol^{-1}})\) | 158 | 243 |
Which combination of explanations correctly accounts for all three comparisons?
ⓐ. Fluorine's compact shell raises ionization enthalpy; \(2p\) crowding makes its electron gain less exothermic and F–F bonding weaker than chlorine's.
ⓑ. Fluorine's greater ionization enthalpy reflects more occupied electron shells; chlorine's more negative electron-gain enthalpy reflects a completed \(3p^6\) shell before attachment; fluorine's weak bond reflects greater dispersion forces.
ⓒ. Fluorine's greater ionization enthalpy and weaker F–F bond both prove that its valence electrons are more weakly attracted to its nucleus than chlorine's.
ⓓ. Chlorine's more negative electron-gain enthalpy means chlorine must have greater electronegativity and aqueous oxidizing strength than fluorine, regardless of the other data.
Correct Answer: Fluorine's compact shell raises ionization enthalpy; \(2p\) crowding makes its electron gain less exothermic and F–F bonding weaker than chlorine's.
Explanation: \(\textbf{Interpret the ionization-enthalpy comparison:}\)
Fluorine has the configuration \(1s^22s^22p^5\).
Chlorine has the outermost configuration \(3s^23p^5\).
Fluorine's valence electrons occupy the smaller second principal shell.
They experience strong nuclear attraction within a relatively compact atomic structure.
Consequently, removing an electron from fluorine requires more energy than from chlorine.
\(\textbf{Interpret the electron-gain-enthalpy exception:}\)
Both atoms can gain one electron to complete their valence \(p\) subshell.
However, the receiving \(2p\) subshell of fluorine is exceptionally compact.
An incoming electron experiences appreciable repulsion from electrons already occupying that region.
Chlorine's \(3p\) subshell is larger and more diffuse.
The incoming electron experiences less severe crowding.
This contributes to chlorine's more negative first electron-gain enthalpy.
\(\textbf{Interpret the molecular bond-strength exception:}\)
Each fluorine atom possesses three lone pairs in the ordinary electron-pair description of \(\mathrm{F_2}\).
When two very small fluorine atoms form a bond, their non-bonding electron clouds approach closely.
Lone-pair repulsion consequently reduces the net strength of the F–F bond.
The larger chlorine atoms experience less severe non-bonding electron-pair crowding in \(\mathrm{Cl_2}\).
The Cl–Cl bond can therefore be stronger than the F–F bond.
\(\textbf{Reconcile the three properties:}\)
Ionization enthalpy concerns electron removal from an isolated atom.
Electron-gain enthalpy concerns addition of an electron to an isolated atom.
Bond dissociation enthalpy concerns cleavage of a bond within a molecule.
Different aspects of electronic structure control these processes.
Fluorine's small size can favour strong nuclear attraction while simultaneously increasing electron–electron repulsion in particular situations.
Thus, the three observations are compatible rather than contradictory.
The data must not be collapsed into a single unsupported rule about atomic size or electronegativity.
219. The first ionization enthalpies of xenon and argon are approximately \(1170\) and \(1521\,\mathrm{kJ\,mol^{-1}}\), respectively. A beam containing photons of wavelength \(90.0\,\mathrm{nm}\) is directed separately at isolated gaseous atoms of the two elements. Assume one-photon ionization from ground-state atoms and neglect secondary collisions. Given \(h=6.626\times10^{-34}\,\mathrm{J\,s}\), \(c=3.00\times10^8\,\mathrm{m\,s^{-1}}\) and \(N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}\), which conclusion is correct?
ⓐ. Neither gas can undergo one-photon ionization because both possess closed valence shells.
ⓑ. Both gases can undergo one-photon ionization because the photons carry more than \(1521\,\mathrm{kJ\,mol^{-1}}\).
ⓒ. Argon can undergo one-photon ionization, but xenon cannot, because argon has the greater ionization enthalpy.
ⓓ. Xenon can undergo one-photon ionization, but argon cannot, because the photon energy exceeds xenon's threshold but not argon's.
Correct Answer: Xenon can undergo one-photon ionization, but argon cannot, because the photon energy exceeds xenon's threshold but not argon's.
Explanation: \(\textbf{Identify the ionization requirement:}\)
The first ionization enthalpy is the energy required to remove one electron from an isolated gaseous atom in its ground state.
For one-photon ionization, the photon must supply at least the required energy.
A closed valence shell does not make ionization impossible when sufficiently energetic radiation is supplied.
\(\textbf{Calculate the energy of one photon:}\)
The photon-energy relation is:
\[
E_{\mathrm{photon}}=\frac{hc}{\lambda}
\]
The wavelength is:
\[
\lambda=90.0\,\mathrm{nm}=90.0\times10^{-9}\,\mathrm{m}
\]
Substitution gives:
\[
E_{\mathrm{photon}}=\frac{(6.626\times10^{-34})(3.00\times10^8)}{90.0\times10^{-9}}
\]
\[
E_{\mathrm{photon}}=2.209\times10^{-18}\,\mathrm{J}
\]
\(\textbf{Convert to molar photon energy:}\)
\[
E_{\mathrm{molar}}=E_{\mathrm{photon}}N_A
\]
\[
=(2.209\times10^{-18})(6.022\times10^{23})
\]
\[
=1.330\times10^6\,\mathrm{J\,mol^{-1}}
\]
\[
E_{\mathrm{molar}}=1330\,\mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the two ionization thresholds:}\)
For xenon:
\[
1330\gt 1170\,\mathrm{kJ\,mol^{-1}}
\]
The photon energy exceeds the first ionization requirement.
One-photon ionization of xenon is therefore energetically permitted.
For argon:
\[
1330\lt 1521\,\mathrm{kJ\,mol^{-1}}
\]
The photon energy is below the first ionization requirement.
One-photon ionization of ground-state argon is therefore not energetically permitted under the stipulated conditions.
\(\textbf{Explain the periodic difference:}\)
Xenon has a larger atomic size and more extensive shielding than argon.
Its outermost electrons are generally less strongly bound.
Consequently, xenon's first ionization enthalpy is lower.
The difference in thresholds allows the same radiation to ionize xenon selectively.
This does not mean that every xenon atom must absorb a photon or that ionization occurs with unit probability.
It establishes energetic possibility under the stated one-photon model.
\(\textbf{Final answer:}\)
\[
E_{\mathrm{molar}}=1330\,\mathrm{kJ\,mol^{-1}};\quad\mathrm{Xe:\ permitted,\ Ar:\ not\ permitted}
\]
220. Antimony forms oxides containing Sb(III) and Sb(V). Consider \(\mathrm{Sb_2O_3}\) and \(\mathrm{Sb_2O_5}\), both under conditions where their characteristic acid–base behaviour can be assessed. Which conclusion correctly combines their oxidation states with the influence of oxidation state on oxide acidity?
ⓐ. Both oxides must have identical basicity because they contain the same Group 15 element.
ⓑ. \(\mathrm{Sb_2O_3}\) is amphoteric; higher oxidation state in \(\mathrm{Sb_2O_5}\) favours acidic character.
ⓒ. \(\mathrm{Sb_2O_5}\) must be more basic because Sb(V) has a greater positive oxidation state.
ⓓ. \(\mathrm{Sb_2O_3}\) contains antimony in oxidation state +5, while \(\mathrm{Sb_2O_5}\) contains antimony in oxidation state +3.
Correct Answer: \(\mathrm{Sb_2O_3}\) is amphoteric; higher oxidation state in \(\mathrm{Sb_2O_5}\) favours acidic character.
Explanation: \(\textbf{Identify antimony's periodic character:}\)
Antimony belongs to Group 15.
It occupies an intermediate position between the strongly non-metallic upper members and metallic bismuth.
Its oxide chemistry consequently cannot be predicted by assuming exclusively acidic or exclusively basic behaviour.
\(\textbf{Determine the oxidation state in the trioxide:}\)
Oxygen has oxidation state \(-2\).
For neutral \(\mathrm{Sb_2O_3}\):
\[
2x+3(-2)=0
\]
\[
x=+3
\]
Thus, antimony is present in oxidation state +3.
Antimony(III) oxide exhibits amphoteric behaviour under suitable conditions.
\(\textbf{Determine the oxidation state in the pentoxide:}\)
For \(\mathrm{Sb_2O_5}\):
\[
2y+5(-2)=0
\]
\[
y=+5
\]
Thus, antimony is present in oxidation state +5.
\(\textbf{Interpret the acidity change:}\)
For many p-block elements, higher positive oxidation states are associated with more pronounced acidic behaviour in their oxides.
The higher oxidation state can increase the electron-withdrawing influence of the central element within the oxide framework.
This changes the relative tendency of the compound to exhibit acidic rather than basic reactions.
Consequently, antimony(V) oxide is more acidic in character than antimony(III) oxide.
\(\textbf{Connect the observation with periodic trends:}\)
The same element can form oxides with different acid–base behaviour when its oxidation state changes.
Therefore, metallic or non-metallic classification of the parent element is not sufficient by itself.
Oxidation state, bonding and actual reaction conditions also matter.
The trend should not be interpreted as proving that every higher oxide of every element is exclusively acidic.
For the specified antimony oxides, the correct comparison is amphoteric Sb(III) oxide and more acidic Sb(V) oxide.
