401. An idealized oxygen molecule is placed in a specified electronically excited configuration by promoting one electron from a singly occupied antibonding \(\pi_{2p}^{*}\) orbital into a higher-energy, initially empty antibonding \(\sigma_{2p}^{*}\) orbital. Assume that no other occupancies change and that the promoted electron retains its spin.
Which statement correctly predicts what can and cannot be inferred from the elementary bond-order formula?
ⓐ. Bond order increases by one because the electron moves into a higher-energy molecular orbital.
ⓑ. Bond order decreases by one because every electronic excitation removes a bonding electron.
ⓒ. Bond order remains 2 because bonding and antibonding populations are unchanged; the excited state nevertheless has higher electronic energy.
ⓓ. Bond order remains 2, so the excited and ground-state configurations must have identical orbital energies and identical chemical stability.
Correct Answer: Bond order remains 2 because bonding and antibonding populations are unchanged; the excited state nevertheless has higher electronic energy.
Explanation: \(\textbf{Identify the ground-state antibonding occupancy:}\)
Ground-state O₂ contains two unpaired electrons in the degenerate antibonding \(\pi_{2p}^{*}\) orbitals.
Its calculated bond order is 2.
The higher-energy antibonding \(\sigma_{2p}^{*}\) orbital is initially empty.
\(\textbf{Track the promoted electron:}\)
An electron is removed from one antibonding pi orbital.
It is placed in a different antibonding orbital.
Therefore, the total number of antibonding electrons is unchanged.
The bonding-electron population also remains unchanged.
\(\textbf{Apply the bond-order relation:}\)
The MO bond-order formula is:
\[
BO=\frac{N_b-N_a}{2}
\]
Neither \(N_b\) nor \(N_a\) changes in the stipulated promotion.
Therefore, the calculated bond order remains 2.
\(\textbf{Examine the energy change:}\)
The destination \(\sigma_{2p}^{*}\) orbital is specified to be higher in energy than the starting \(\pi_{2p}^{*}\) orbital.
Moving an electron into the higher-energy orbital requires excitation energy.
Consequently, the new configuration is not energetically equivalent to the original configuration merely because their bond orders agree.
\(\textbf{Interpret the spin arrangement:}\)
One antibonding pi orbital remains singly occupied.
The newly occupied antibonding sigma orbital also contains one electron.
Under the stipulated spin-preserving change, two unpaired electrons remain.
\(\textbf{Recognise the limitation of bond order:}\)
The bond-order formula measures the difference between bonding and antibonding electron populations.
It does not distinguish the detailed energies of all occupied orbitals within those two categories.
Two configurations can therefore have equal bond orders but different electronic energies.
Equal bond order alone does not establish identical equilibrium bond lengths or dissociation enthalpies for distinct electronic states.
Option C correctly identifies this limitation.
402. Ozone contains two equivalent O–O connections with a simple resonance-average bond multiplicity of \(1.5\). The superoxide ion, \(\mathrm{O_2^-}\), has an MO bond order of \(1.5\) in the conventional molecular-orbital model. Consider the following statements.
Statement I: The value \(1.5\) in ozone arises from distributing localized single- and double-bond descriptions across equivalent connections, whereas the value in superoxide is calculated from bonding and antibonding molecular-orbital occupancies.
Statement II: Since both values are \(1.5\), the O–O bond dissociation enthalpies of ozone and superoxide must be identical.
Statement III: The equality of these values does not require ozone and superoxide to have identical molecular geometries or magnetic properties.
Statement IV: Ozone and superoxide must have the same number of unpaired electrons because any fractional bond order implies one unpaired electron.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: \(\textbf{Evaluate Statement I:}\)
Ozone is a triatomic molecule with two equivalent oxygen–oxygen connections.
Its conventional resonance contributors each contain one single and one double O–O bond.
The average bond multiplicity is represented as \(1.5\).
Superoxide is a diatomic ion.
Its MO bond order is calculated using the difference between bonding and antibonding electron populations.
The identical numerical values therefore arise from different descriptions of bonding.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Bond dissociation enthalpy depends on the complete energetic difference between a specified initial species and its dissociation products.
Ozone and superoxide have different numbers of atoms, different charges and different electronic structures.
An O–O connection in ozone is part of a triatomic bonding framework.
Superoxide contains one diatomic O–O connection.
Their equal bond-order values therefore cannot establish identical dissociation enthalpies.
Statement II is incorrect.
\(\textbf{Evaluate Statement III:}\)
Ozone has a bent molecular structure.
Superoxide is diatomic and therefore has a linear nuclear arrangement.
Their magnetic properties must also be determined from their electronic configurations rather than from the bond-order value alone.
Thus, equal bond order does not force equal geometry or magnetism.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
Fractional bond order does not universally imply a specific number of unpaired electrons.
Superoxide has one unpaired electron in the conventional MO description.
Ozone is a closed-shell, diamagnetic molecule in its ordinary ground state.
Therefore, their magnetic behaviours differ despite the equal numerical bond-order values.
Statement IV is incorrect.
\(\textbf{Overall conclusion:}\)
Bond order is useful for comparing related bonding situations, but its interpretation depends on the model and chemical environment.
An equal numerical bond-order value does not replace a complete electron-configuration or thermochemical analysis.
Statements I and III are correct.
403. Consider neutral nitrogen and its singly charged molecular ions:
\[
\mathrm{N_2^+,\quad N_2,\quad N_2^-}
\]
Use the conventional MO ordering for nitrogen. Assume that removing an electron from \(\mathrm{N_2}\) affects the occupied bonding \(\sigma_{2p_z}\) orbital, while adding an electron places it in an initially empty antibonding \(\pi_{2p}^{*}\) orbital. Which conclusion correctly compares the cation and anion with neutral nitrogen?
ⓐ. The two ions both have bond order \(2.5\) and one unpaired electron, but they reach this result through different changes in bonding and antibonding populations.
ⓑ. \(\mathrm{N_2^+}\) has bond order \(3.5\), whereas \(\mathrm{N_2^-}\) has bond order \(2.5\), because positive charge always increases bond order.
ⓒ. Both ions retain bond order 3 because their nitrogen nuclei are unchanged.
ⓓ. Both ions have bond order \(2.5\) and are diamagnetic because nitrogen contains an even number of protons.
Correct Answer: The two ions both have bond order \(2.5\) and one unpaired electron, but they reach this result through different changes in bonding and antibonding populations.
Explanation: \(\textbf{Establish the neutral reference:}\)
Neutral nitrogen has ten valence electrons.
In the conventional MO configuration, eight occupy bonding orbitals and two occupy antibonding orbitals.
\[
BO(\mathrm{N_2})=\frac{8-2}{2}
\]
\[
BO(\mathrm{N_2})=3
\]
All occupied orbitals are paired.
Neutral N₂ is consequently diamagnetic.
\(\textbf{Form the nitrogen cation:}\)
The highest occupied molecular orbital is bonding \(\sigma_{2p_z}\).
Removing one electron decreases the bonding population from eight to seven.
The antibonding population remains two.
\[
BO(\mathrm{N_2^+})=\frac{7-2}{2}
\]
\[
BO(\mathrm{N_2^+})=2.5
\]
The formerly paired bonding orbital becomes singly occupied.
Therefore, the cation contains one unpaired electron.
\(\textbf{Form the nitrogen anion:}\)
The additional electron enters an available antibonding \(\pi_{2p}^{*}\) orbital.
The bonding population remains eight.
The antibonding population increases from two to three.
\[
BO(\mathrm{N_2^-})=\frac{8-3}{2}
\]
\[
BO(\mathrm{N_2^-})=2.5
\]
The added antibonding electron is unpaired.
Therefore, the anion also has one unpaired electron.
\(\textbf{Compare the electronic mechanisms:}\)
Cation formation weakens net bonding by removing a stabilizing electron.
Anion formation weakens net bonding by adding a destabilizing electron.
Both changes reduce bond order by one-half unit.
The corresponding ions consequently have equal calculated bond orders but different orbital occupancies.
\(\textbf{Interpret the bond-length consequence:}\)
Both ions are expected to have weaker net bonding than neutral N₂ within the simple comparable-species trend.
Their equilibrium N–N distances would therefore generally be expected to exceed that of neutral N₂.
However, equal calculated bond order does not prove that the two ionic bond lengths or dissociation enthalpies are exactly equal.
Those properties depend on the detailed electronic energies and charge states.
\(\textbf{Final answer:}\)
\[
\boxed{\begin{aligned}
BO(\mathrm{N_2^+})&=BO(\mathrm{N_2^-})=2.5\\
n_u(\mathrm{N_2^+})&=n_u(\mathrm{N_2^-})=1
\end{aligned}}
\]
404. A researcher examines three hypothetical electronic states of the same homonuclear diatomic species X₂. Their nuclear charges are identical, and all three are described using the same elementary set of bonding and antibonding molecular orbitals.
| State | Bonding electrons | Antibonding electrons | Measured equilibrium distance | Measured dissociation enthalpy |
|---|
| P | 8 | 4 | \(120\ \mathrm{pm}\) | \(300\ \mathrm{kJ\,mol^{-1}}\) |
| Q | 8 | 5 | \(132\ \mathrm{pm}\) | \(225\ \mathrm{kJ\,mol^{-1}}\) |
| R | 7 | 4 | \(126\ \mathrm{pm}\) | \(245\ \mathrm{kJ\,mol^{-1}}\) |
The dissociation enthalpies refer to the specified states and their corresponding stated dissociation products. Which inference is supported by the complete data?
ⓐ. Q and R have different calculated bond orders, so their measured dissociation enthalpies must be identical.
ⓑ. P has the lowest calculated bond order because its internuclear separation is smallest.
ⓒ. Q and R both have bond order \(1.5\) but differ in observed bond length and dissociation enthalpy; bond order alone is insufficient.
ⓓ. The data contradict molecular-orbital theory because equal bond order requires identical bond lengths for all states of one diatomic species.
Correct Answer: Q and R both have bond order \(1.5\) but differ in observed bond length and dissociation enthalpy; bond order alone is insufficient.
Explanation: \(\textbf{Calculate the bond order of P:}\)
\[
BO_P=\frac{N_b-N_a}{2}
\]
Substituting:
\[
BO_P=\frac{8-4}{2}
\]
\[
BO_P=2
\]
\(\textbf{Calculate the bond order of Q:}\)
\[
BO_Q=\frac{8-5}{2}
\]
\[
BO_Q=1.5
\]
\(\textbf{Calculate the bond order of R:}\)
\[
BO_R=\frac{7-4}{2}
\]
\[
BO_R=1.5
\]
Therefore, Q and R have equal calculated bond orders.
\(\textbf{Compare their electron populations:}\)
Q has eight bonding and five antibonding electrons.
R has seven bonding and four antibonding electrons.
Their population differences are equal, but their total electron populations and detailed occupancies are not identical.
The bond-order formula retains only the difference between bonding and antibonding populations.
\(\textbf{Compare the measured bond lengths:}\)
\[
r_Q=132\ \mathrm{pm}
\]
\[
r_R=126\ \mathrm{pm}
\]
Thus:
\[
r_Q-r_R=6\ \mathrm{pm}
\]
\(\textbf{Compare the dissociation enthalpies:}\)
\[
D_Q=225\ \mathrm{kJ\,mol^{-1}}
\]
\[
D_R=245\ \mathrm{kJ\,mol^{-1}}
\]
Therefore:
\[
D_R-D_Q=20\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the overall trend:}\)
P has the greatest bond order, the shortest measured equilibrium distance and the greatest measured dissociation enthalpy.
That observation is consistent with the usual bond-order trend for related bonding situations.
However, Q and R demonstrate that equal calculated bond order does not force equality of detailed molecular properties.
The actual energetic distribution of electrons and the relevant dissociation reference states remain important.
The bond-order formula is a useful descriptor, not a complete calculation of the potential-energy curve.
\(\textbf{Evaluate the supported conclusion:}\)
The measured values for Q and R establish different bond lengths and dissociation enthalpies despite equal bond orders.
These differences do not contradict the stated MO electron counts.
They demonstrate a limitation of interpreting the bond-order number as an exact thermochemical or structural prediction.
Option C is correct.
405. The molecular-orbital model predicts bond order 1 for both \(\mathrm{Li_2}\) and \(\mathrm{F_2}\). Their measured bond lengths and dissociation enthalpies differ substantially. Consider the following statements.
Statement I: Equal calculated MO bond orders do not require equal bond lengths when the bonded elements have different atomic sizes.
Statement II: Equal calculated MO bond orders require equal bond dissociation enthalpies because the number of net bonding electron pairs is the only energetic factor.
Statement III: In the elementary MO descriptions, \(\mathrm{Li_2}\) obtains its positive valence bond order through occupancy of a bonding \(\sigma_{2s}\) orbital, whereas \(\mathrm{F_2}\) has additional \(2p\)-derived bonding and antibonding occupancies.
Statement IV: Both molecules are necessarily paramagnetic because a bond order of 1 implies two unpaired electrons.
Which statements are correct?
ⓐ. I and II only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I and III only
Correct Answer: I and III only
Explanation: \(\textbf{Evaluate Statement I:}\)
Lithium and fluorine have different atomic sizes and valence-orbital distributions.
Their equilibrium internuclear distances are therefore not determined solely by the numerical bond-order value.
A bond order of 1 in two different chemical systems does not require identical bond lengths.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Bond dissociation enthalpy depends on the complete energy difference between the bonded molecule and its dissociation products.
Atomic orbital energies, overlap, nuclear charge and repulsive contributions affect this difference.
The MO bond-order formula summarizes the difference between bonding and antibonding electron populations.
It does not provide an exact dissociation enthalpy.
Therefore, Statement II is incorrect.
\(\textbf{Evaluate Statement III:}\)
Each lithium atom contributes one valence electron.
The two electrons in neutral \(\mathrm{Li_2}\) occupy the bonding \(\sigma_{2s}\) orbital.
Its calculated bond order is 1.
Fluorine has a much larger valence-electron population.
Its conventional MO configuration includes occupied \(2s\)-derived orbitals and several \(2p\)-derived bonding and antibonding orbitals.
The difference between the complete bonding and antibonding populations again gives bond order 1.
Thus, the equal values arise from different detailed electron configurations.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
A bond order of 1 does not imply a particular unpaired-electron count.
In neutral \(\mathrm{Li_2}\), the two valence bonding electrons are paired.
In neutral \(\mathrm{F_2}\), the conventional occupied molecular orbitals also contain paired electrons.
Both molecules are diamagnetic in the elementary model.
Statement IV is therefore incorrect.
\(\textbf{Integrate the comparison:}\)
Bond order, electron pairing, bond length and bond dissociation enthalpy describe different aspects of molecular bonding.
They are related, but none should be treated as an exact substitute for the others.
The equal bond-order values of \(\mathrm{Li_2}\) and \(\mathrm{F_2}\) do not imply equivalent electronic structures or identical physical properties.
Statements I and III are correct.
406. A hypothetical metal M forms an ionic chloride, \(\mathrm{MCl_2(s)}\), by the reaction:
\[
\mathrm{M(s)+Cl_2(g)\rightarrow MCl_2(s)}
\]
The following thermochemical data are supplied:
| Process or quantity | Enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| \(\mathrm{M(s)\rightarrow M(g)}\) | \(+150\) |
| First ionization enthalpy of M | \(+600\) |
| Second ionization enthalpy of M | Unknown |
| \(\mathrm{Cl_2(g)\rightarrow2Cl(g)}\) | \(+240\) |
| \(\mathrm{Cl(g)+e^-\rightarrow Cl^-(g)}\) | \(-350\) |
| Lattice formation enthalpy of \(\mathrm{MCl_2}\) | \(-2100\) |
| Standard formation enthalpy of \(\mathrm{MCl_2(s)}\) | \(-660\) |
Assuming the data are internally consistent, what is the second ionization enthalpy of M, and why must it appear in the Born–Haber cycle?
ⓐ. \(1150\ \mathrm{kJ\,mol^{-1}}\); formation of \(\mathrm{M^{2+}}\) requires removal of a second electron from \(\mathrm{M^+(g)}\).
ⓑ. \(1750\ \mathrm{kJ\,mol^{-1}}\); both ionization enthalpies describe removal of electrons directly from neutral M.
ⓒ. \(450\ \mathrm{kJ\,mol^{-1}}\); the second ionization stage is offset by the second chlorine atom.
ⓓ. \(1150\ \mathrm{kJ\,mol^{-1}}\); it represents removal of an electron from gaseous chloride rather than from \(\mathrm{M^+}\).
Correct Answer: \(1150\ \mathrm{kJ\,mol^{-1}}\); formation of \(\mathrm{M^{2+}}\) requires removal of a second electron from \(\mathrm{M^+(g)}\).
Explanation: \(\textbf{Establish the required ionic products:}\)
The formula \(\mathrm{MCl_2}\) contains one \(\mathrm{M^{2+}}\) ion and two \(\mathrm{Cl^-}\) ions per formula unit.
Therefore, one metal atom must lose two electrons, while two chlorine atoms each gain one electron.
\(\textbf{Identify the successive ionization stages:}\)
\[
\mathrm{M(g)\rightarrow M^+(g)+e^-}
\]
\[
IE_1=+600\ \mathrm{kJ\,mol^{-1}}
\]
\[
\mathrm{M^+(g)\rightarrow M^{2+}(g)+e^-}
\]
The second ionization enthalpy corresponds to the second equation, not to another electron removal from neutral M.
\(\textbf{Calculate the chlorine contributions:}\)
One mole of \(\mathrm{Cl_2}\) provides two moles of gaseous chlorine atoms.
\[
\Delta H_{\mathrm{dissociation}}=+240\ \mathrm{kJ\,mol^{-1}}
\]
Both chlorine atoms accept electrons.
\[
\Delta H_{\mathrm{electron\ gain}}=2(-350)
\]
\[
\Delta H_{\mathrm{electron\ gain}}=-700\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Construct the Born–Haber balance:}\)
Let the second ionization enthalpy be \(x\).
\[
\Delta_fH^\circ=\Delta_{\mathrm{sub}}H+IE_1+x+D(\mathrm{Cl_2})+2\Delta_{\mathrm{eg}}H+\Delta_{\mathrm{latt,form}}H
\]
Substituting:
\[
-660=150+600+x+240-700-2100
\]
Combining known terms:
\[
-660=x-1810
\]
\[
x=-660+1810
\]
\[
x=1150\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Check the physical meaning:}\)
The calculated second ionization enthalpy is positive, as expected for electron removal from a gaseous cation.
Both ionization stages are energetically costly.
However, the large exothermic lattice-formation contribution helps compensate for those costs.
Omitting the second ionization stage would make the electron-accounting pathway inconsistent with the divalent metal ion in the crystal.
\(\textbf{Final answer:}\)
\[
\boxed{IE_2(\mathrm M)=+1150\ \mathrm{kJ\,mol^{-1}}}
\]
407. A monovalent metal M forms two hypothetical ionic solids, MX and MY, from the reactions:
\[
\mathrm{M(s)+\frac12X_2(g)\rightarrow MX(s)}
\]
\[
\mathrm{M(s)+\frac12Y_2(g)\rightarrow MY(s)}
\]
The metal sublimation enthalpy is \(+100\ \mathrm{kJ\,mol^{-1}}\), and its first ionization enthalpy is \(+500\ \mathrm{kJ\,mol^{-1}}\). The remaining data are:
| Contribution (\(\mathrm{kJ\,mol^{-1}}\)) | MX | MY |
|---|
| Half of the halogen-molecule dissociation enthalpy | \(+120\) | \(+80\) |
| Electron-gain enthalpy of the halogen atom | \(-320\) | \(-260\) |
| Lattice formation enthalpy | \(-720\) | \(-760\) |
Which conclusion correctly compares their formation enthalpies and explains why the less negative electron-gain enthalpy of Y does not prevent MY from having the more exothermic formation?
ⓐ. MX is more exothermic by \(20\ \mathrm{kJ\,mol^{-1}}\), because X has the more negative electron-gain enthalpy.
ⓑ. MY is more exothermic by \(100\ \mathrm{kJ\,mol^{-1}}\), because only the difference in lattice formation enthalpies matters.
ⓒ. Both formation enthalpies are equal because the favourable lattice and electron-gain differences cancel.
ⓓ. MY is more exothermic by \(20\ \mathrm{kJ\,mol^{-1}}\); its smaller atomization cost and more exothermic lattice formation together outweigh its less favourable electron-gain contribution.
Correct Answer: MY is more exothermic by \(20\ \mathrm{kJ\,mol^{-1}}\); its smaller atomization cost and more exothermic lattice formation together outweigh its less favourable electron-gain contribution.
Explanation: \(\textbf{Identify the common contributions:}\)
Both compounds contain the same monovalent metal.
Each formula unit therefore requires sublimation of one metal atom and its first ionization.
Their common energetic cost is:
\[
H_{\mathrm{metal}}=100+500
\]
\[
H_{\mathrm{metal}}=600\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate formation of MX:}\)
\[
\Delta_fH^\circ(\mathrm{MX})=H_{\mathrm{metal}}+\frac12D(\mathrm{X_2})+\Delta_{\mathrm{eg}}H(\mathrm X)+U_{\mathrm{form}}(\mathrm{MX})
\]
Substituting:
\[
\Delta_fH^\circ(\mathrm{MX})=600+120-320-720
\]
\[
\Delta_fH^\circ(\mathrm{MX})=-320\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate formation of MY:}\)
\[
\Delta_fH^\circ(\mathrm{MY})=H_{\mathrm{metal}}+\frac12D(\mathrm{Y_2})+\Delta_{\mathrm{eg}}H(\mathrm Y)+U_{\mathrm{form}}(\mathrm{MY})
\]
Substituting:
\[
\Delta_fH^\circ(\mathrm{MY})=600+80-260-760
\]
\[
\Delta_fH^\circ(\mathrm{MY})=-340\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the resulting enthalpies:}\)
\[
\Delta_fH^\circ(\mathrm{MY})-\Delta_fH^\circ(\mathrm{MX})=-340-(-320)
\]
\[
\Delta_fH^\circ(\mathrm{MY})-\Delta_fH^\circ(\mathrm{MX})=-20\ \mathrm{kJ\,mol^{-1}}
\]
Thus, MY has the more negative formation enthalpy.
\(\textbf{Resolve the competing contributions:}\)
The electron-gain process for Y releases \(60\ \mathrm{kJ\,mol^{-1}}\) less energy than that for X.
This disfavors MY formation.
However, preparing gaseous Y atoms costs \(40\ \mathrm{kJ\,mol^{-1}}\) less energy.
The MY lattice also releases \(40\ \mathrm{kJ\,mol^{-1}}\) more energy.
The combined advantage is:
\[
40+40-60=20\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the comparison:}\)
Electron-gain enthalpy is only one stage of ionic-solid formation.
A more negative electron-gain enthalpy does not independently guarantee a more negative formation enthalpy.
The complete Born–Haber balance determines the thermochemical comparison.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta_fH^\circ(\mathrm{MX})=-320,\quad\Delta_fH^\circ(\mathrm{MY})=-340\ \mathrm{kJ\,mol^{-1}}}
\]
408. The formation of an ionic oxide, \(\mathrm{MO(s)}\), is represented by:
\[
\mathrm{M(s)+\frac12O_2(g)\rightarrow MO(s)}
\]
The metal forms \(\mathrm{M^{2+}}\), and oxygen forms \(\mathrm{O^{2-}}\). A hypothetical complete Born–Haber data set is:
| Stage | Enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| \(\mathrm{M(s)\rightarrow M(g)}\) | \(+150\) |
| \(IE_1(\mathrm M)\) | \(+750\) |
| \(IE_2(\mathrm M)\) | \(+1450\) |
| \(\mathrm{\frac12O_2(g)\rightarrow O(g)}\) | \(+250\) |
| \(\mathrm{O(g)+e^-\rightarrow O^-(g)}\) | \(-140\) |
| \(\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)}\) | \(+780\) |
| \(\mathrm{M^{2+}(g)+O^{2-}(g)\rightarrow MO(s)}\) | \(-3800\) |
Which statement correctly gives the oxide formation enthalpy and explains why the positive second electron-gain enthalpy does not prohibit ionic-oxide formation?
ⓐ. \(-2120\ \mathrm{kJ\,mol^{-1}}\); the second electron-gain enthalpy must be treated as exothermic because the product is an anion.
ⓑ. \(-560\ \mathrm{kJ\,mol^{-1}}\); formation of isolated \(\mathrm{O^{2-}(g)}\) is energetically costly, but the lattice-formation contribution compensates sufficiently.
ⓒ. \(+560\ \mathrm{kJ\,mol^{-1}}\); a positive second electron-gain enthalpy necessarily makes the overall ionic formation endothermic.
ⓓ. \(-1340\ \mathrm{kJ\,mol^{-1}}\); the second electron-gain stage must be omitted because oxygen has already achieved an octet after accepting the first electron.
Correct Answer: \(-560\ \mathrm{kJ\,mol^{-1}}\); formation of isolated \(\mathrm{O^{2-}(g)}\) is energetically costly, but the lattice-formation contribution compensates sufficiently.
Explanation: \(\textbf{Identify the two-electron transfer:}\)
The product contains \(\mathrm{M^{2+}}\) and \(\mathrm{O^{2-}}\).
Therefore, the metal must lose two electrons, and oxygen must accept two.
Both successive ionization enthalpies and both electron-gain enthalpies are required.
\(\textbf{Calculate the cation-formation cost:}\)
\[
H_{\mathrm{cation}}=\Delta_{\mathrm{sub}}H+IE_1+IE_2
\]
Substituting:
\[
H_{\mathrm{cation}}=150+750+1450
\]
\[
H_{\mathrm{cation}}=2350\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the oxygen-atom preparation cost:}\)
\[
\mathrm{\frac12O_2(g)\rightarrow O(g)}
\]
\[
H_{\mathrm{atom}}=+250\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Combine the oxygen electron-gain stages:}\)
\[
H_{\mathrm{oxygen\ gain}}=(-140)+(+780)
\]
\[
H_{\mathrm{oxygen\ gain}}=+640\ \mathrm{kJ\,mol^{-1}}
\]
The net formation of gaseous \(\mathrm{O^{2-}}\) from a gaseous oxygen atom is endothermic under the supplied conditions.
\(\textbf{Construct the complete energy balance:}\)
\[
\Delta_fH^\circ=H_{\mathrm{cation}}+H_{\mathrm{atom}}+H_{\mathrm{oxygen\ gain}}+U_{\mathrm{form}}
\]
Substituting:
\[
\Delta_fH^\circ=2350+250+640-3800
\]
\[
\Delta_fH^\circ=3240-3800
\]
\[
\Delta_fH^\circ=-560\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Explain the positive second electron-gain enthalpy:}\)
The incoming second electron is added to an already negatively charged gaseous ion.
Electron–electron repulsion makes this electron addition energetically costly.
The positive enthalpy does not mean that oxide ions cannot exist within a strongly stabilizing ionic lattice.
\(\textbf{Interpret the complete process:}\)
The lattice releases \(3800\ \mathrm{kJ\,mol^{-1}}\) in the stipulated cycle.
This is sufficient to compensate for the net cost of preparing the gaseous ions.
The overall formation enthalpy is consequently negative.
Thus, the energetics of an isolated gaseous ion must not be confused with the net energetics of forming an ionic crystal.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta_fH^\circ(\mathrm{MO})=-560\ \mathrm{kJ\,mol^{-1}}}
\]
409. A hypothetical crystalline mixed halide, \(\mathrm{CaFCl(s)}\), contains \(\mathrm{Ca^{2+}}\), \(\mathrm{F^-}\) and \(\mathrm{Cl^-}\). Its formation reaction is:
\[
\mathrm{Ca(s)+\frac12F_2(g)+\frac12Cl_2(g)\rightarrow CaFCl(s)}
\]
The following hypothetical enthalpy data are given in \(\mathrm{kJ\,mol^{-1}}\):
| Quantity | Value |
|---|
| Calcium sublimation enthalpy | \(+180\) |
| Sum of first and second calcium ionization enthalpies | \(+1750\) |
| F–F dissociation enthalpy | \(+160\) |
| Cl–Cl dissociation enthalpy | \(+242\) |
| Fluorine electron-gain enthalpy | \(-330\) |
| Chlorine electron-gain enthalpy | \(-350\) |
| Formation enthalpy of solid CaFCl | \(-700\) |
What is the lattice formation enthalpy of CaFCl, and which stoichiometric feature distinguishes its Born–Haber cycle from that of a dihalide containing two identical halide ions?
ⓐ. \(-2151\ \mathrm{kJ\,mol^{-1}}\); one gaseous fluoride ion and one gaseous chloride ion are formed, requiring half a mole of each diatomic halogen.
ⓑ. \(-2341\ \mathrm{kJ\,mol^{-1}}\); one full mole of each halogen molecule must be dissociated.
ⓒ. \(-1471\ \mathrm{kJ\,mol^{-1}}\); only the electron-gain enthalpy of the more electronegative fluorine contributes.
ⓓ. \(-2151\ \mathrm{kJ\,mol^{-1}}\); both electrons lost by calcium must be transferred to the same fluorine atom.
Correct Answer: \(-2151\ \mathrm{kJ\,mol^{-1}}\); one gaseous fluoride ion and one gaseous chloride ion are formed, requiring half a mole of each diatomic halogen.
Explanation: \(\textbf{Establish the ionic formula:}\)
Calcium forms a doubly charged cation.
Fluoride and chloride each carry charge \(-1\).
Therefore, one \(\mathrm{Ca^{2+}}\), one \(\mathrm{F^-}\) and one \(\mathrm{Cl^-}\) give an electrically neutral formula unit.
The two electrons lost by calcium are accepted by two different halogen atoms.
\(\textbf{Calculate the calcium preparation cost:}\)
\[
H_{\mathrm{Ca}}=\Delta_{\mathrm{sub}}H+IE_1+IE_2
\]
Substituting:
\[
H_{\mathrm{Ca}}=180+1750
\]
\[
H_{\mathrm{Ca}}=1930\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate both halogen atomization contributions:}\)
One gaseous fluorine atom requires half a mole of \(\mathrm{F_2}\).
\[
H_{\mathrm F}=\frac12(160)=80\ \mathrm{kJ\,mol^{-1}}
\]
One gaseous chlorine atom requires half a mole of \(\mathrm{Cl_2}\).
\[
H_{\mathrm{Cl}}=\frac12(242)=121\ \mathrm{kJ\,mol^{-1}}
\]
The combined halogen preparation cost is:
\[
H_{\mathrm{halogen}}=80+121=201\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Combine the electron-gain contributions:}\)
\[
H_{\mathrm{electron\ gain}}=(-330)+(-350)
\]
\[
H_{\mathrm{electron\ gain}}=-680\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Apply the formation-enthalpy balance:}\)
Let the lattice formation enthalpy be \(U_{\mathrm{form}}\).
\[
\Delta_fH^\circ=H_{\mathrm{Ca}}+H_{\mathrm{halogen}}+H_{\mathrm{electron\ gain}}+U_{\mathrm{form}}
\]
Substituting:
\[
-700=1930+201-680+U_{\mathrm{form}}
\]
\[
-700=1451+U_{\mathrm{form}}
\]
\[
U_{\mathrm{form}}=-700-1451
\]
\[
U_{\mathrm{form}}=-2151\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Check the lattice reference process:}\)
The corresponding lattice formation reaction is:
\[
\mathrm{Ca^{2+}(g)+F^-(g)+Cl^-(g)\rightarrow CaFCl(s)}
\]
The formation enthalpy is negative because the specified gaseous ions combine into the crystal.
The reverse lattice dissociation enthalpy is \(+2151\ \mathrm{kJ\,mol^{-1}}\).
Unlike a dihalide containing two identical halide ions, the cycle requires separate fluorine and chlorine atomization and electron-gain contributions.
\(\textbf{Final answer:}\)
\[
\boxed{U_{\mathrm{form}}(\mathrm{CaFCl})=-2151\ \mathrm{kJ\,mol^{-1}}}
\]
410. A hypothetical ionic solid has the formula \(\mathrm{M_3N_2}\), where M forms \(\mathrm{M^{2+}}\) and nitrogen forms \(\mathrm{N^{3-}}\). It is produced formally from the elements:
\[
\mathrm{3M(s)+N_2(g)\rightarrow M_3N_2(s)}
\]
Consider the following proposed features of its Born–Haber cycle.
I. Three first-ionization and three second-ionization stages are required for the three metal atoms.
II. Only half a mole of \(\mathrm{N_2}\) needs to dissociate per mole of \(\mathrm{M_3N_2}\).
III. Each nitrogen atom undergoes three successive electron-gain stages, so six electrons are transferred in total.
IV. Six separate lattice formation enthalpies must be included because six electrons are transferred.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. II, III and IV only
ⓒ. I and III only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: \(\textbf{Establish charge neutrality:}\)
Three doubly charged metal cations give a total positive charge of \(+6\).
Two triply charged nitride ions give a total negative charge of \(-6\).
Therefore, the ionic formula is electrically neutral.
\(\textbf{Evaluate Statement I:}\)
Each metal atom must lose two electrons.
The first electron is removed from neutral gaseous M.
The second is removed from gaseous \(\mathrm{M^+}\).
For three metal atoms, both successive ionization stages occur three times.
Thus, Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
The compound contains two nitrogen atoms per formula unit.
One complete \(\mathrm{N_2}\) molecule supplies these atoms.
Therefore, one mole of \(\mathrm{N_2}\), not half a mole, must undergo dissociation per mole of product.
Statement II is incorrect.
\(\textbf{Evaluate Statement III:}\)
Each nitrogen atom accepts three electrons to become \(\mathrm{N^{3-}}\).
There are two nitrogen atoms per formula unit.
The complete electron-transfer requirement is consequently six electrons.
The Born–Haber accounting must include the first, second and third electron-gain stages for each nitrogen atom.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
The lattice formation enthalpy is defined for forming one mole of the specified crystalline compound from its constituent gaseous ions.
The relevant reaction is:
\[
\mathrm{3M^{2+}(g)+2N^{3-}(g)\rightarrow M_3N_2(s)}
\]
This represents one mole of formula units.
The number of transferred electrons does not determine how many times the molar lattice formation enthalpy is included.
Statement IV is incorrect.
\(\textbf{Integrate the stoichiometric requirements:}\)
The electron-transfer stages must be multiplied according to the numbers and charges of ions formed.
The molecular nitrogen dissociation coefficient follows the number of nitrogen atoms required.
The lattice formation term refers to the complete crystal formula unit.
Statements I and III are therefore correct.
411. Two hypothetical ion pairs, P and Q, are compared using the elementary Coulomb model. Assume point charges in the same medium and ignore short-range repulsive contributions.
| Property | P | Q |
|---|
| Cation charge | \(+e\) | \(+2e\) |
| Anion charge | \(-e\) | \(-2e\) |
| Interionic separation | \(200\ \mathrm{pm}\) | \(400\ \mathrm{pm}\) |
Which statement correctly compares the magnitudes of their electrostatic potential energies and attractive forces at the specified separations?
ⓐ. Q has twice the electrostatic potential-energy magnitude of P, but their attractive-force magnitudes are equal.
ⓑ. Q has four times the potential-energy magnitude and twice the attractive force of P.
ⓒ. P and Q have equal potential-energy magnitudes, but Q has twice the attractive force.
ⓓ. Q has half the potential-energy magnitude and one-quarter the attractive force because its interionic separation is doubled.
Correct Answer: Q has twice the electrostatic potential-energy magnitude of P, but their attractive-force magnitudes are equal.
Explanation: \(\textbf{Distinguish two electrostatic relations:}\)
Electrostatic potential energy and electrostatic force have different dependencies on separation.
For opposite point charges:
\[
|U|=\frac{k|q_+q_-|}{r}
\]
The attractive-force magnitude is:
\[
F=\frac{k|q_+q_-|}{r^2}
\]
The same Coulomb constant applies to both ion pairs.
\(\textbf{Compare the charge products:}\)
For P:
\[
|q_+q_-|_P=e^2
\]
For Q:
\[
|q_+q_-|_Q=(2e)(2e)=4e^2
\]
Therefore, Q has four times the charge-product magnitude.
\(\textbf{Compare their separations:}\)
\[
\frac{r_Q}{r_P}=\frac{400}{200}=2
\]
Thus, Q has twice the interionic separation.
\(\textbf{Calculate the potential-energy ratio:}\)
\[
\frac{|U_Q|}{|U_P|}=\frac{4e^2/(400)}{e^2/(200)}
\]
\[
\frac{|U_Q|}{|U_P|}=4\times\frac{200}{400}
\]
\[
\frac{|U_Q|}{|U_P|}=2
\]
\(\textbf{Calculate the attractive-force ratio:}\)
\[
\frac{F_Q}{F_P}=\frac{4e^2/(400)^2}{e^2/(200)^2}
\]
\[
\frac{F_Q}{F_P}=4\left(\frac{200}{400}\right)^2
\]
\[
\frac{F_Q}{F_P}=4\left(\frac12\right)^2
\]
\[
\frac{F_Q}{F_P}=1
\]
\(\textbf{Interpret the distinction:}\)
The greater charge product in Q increases both interaction magnitudes.
However, doubling the separation offsets the fourfold charge increase differently for force and potential energy.
The inverse-square dependence makes the forces equal.
The inverse-distance dependence leaves Q with twice the potential-energy magnitude.
These results concern isolated ion pairs under the stipulated model.
They are not exact predictions of molar lattice enthalpies, which depend on interactions throughout a crystal and its structure.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{|U_Q|}{|U_P|}=2;\qquad\frac{F_Q}{F_P}=1}
\]
412. Two hypothetical ionic solids, MX and MY, consist of the same monovalent cation \(\mathrm{M^+}\) and different monovalent anions \(\mathrm{X^-}\) and \(\mathrm{Y^-}\). Their lattice formation enthalpies are:
\[
\mathrm{M^+(g)+X^-(g)\rightarrow MX(s)}\qquad \Delta H=-690\ \mathrm{kJ\,mol^{-1}}
\]
\[
\mathrm{M^+(g)+Y^-(g)\rightarrow MY(s)}\qquad \Delta H=-840\ \mathrm{kJ\,mol^{-1}}
\]
Consider the formal gas-ion exchange process:
\[
\mathrm{MX(s)+Y^-(g)\rightarrow MY(s)+X^-(g)}
\]
Which statement correctly determines the reaction enthalpy and explains why halogen atomization and electron-gain stages are unnecessary for this particular cycle?
ⓐ. \(-1530\ \mathrm{kJ\,mol^{-1}}\); both lattice formation enthalpies must be added without reversing either process.
ⓑ. \(+150\ \mathrm{kJ\,mol^{-1}}\); MY has the more exothermic lattice formation and must therefore require energy to form.
ⓒ. \(-150\ \mathrm{kJ\,mol^{-1}}\); MX is first dissociated into gaseous ions, and the supplied gaseous \(\mathrm{Y^-}\) then forms MY without any additional anion-formation stage.
ⓓ. \(-150\ \mathrm{kJ\,mol^{-1}}\); the reaction requires converting gaseous Y atoms into \(\mathrm{Y^-}\) before any lattice interaction can occur.
Correct Answer: \(-150\ \mathrm{kJ\,mol^{-1}}\); MX is first dissociated into gaseous ions, and the supplied gaseous \(\mathrm{Y^-}\) then forms MY without any additional anion-formation stage.
Explanation: \(\textbf{Identify the required initial and final states:}\)
The reactants are crystalline MX and gaseous \(\mathrm{Y^-}\).
The products are crystalline MY and gaseous \(\mathrm{X^-}\).
Both anions are already specified as gaseous ions when present outside the solids.
\(\textbf{Reverse the MX lattice formation reaction:}\)
The first required step is:
\[
\mathrm{MX(s)\rightarrow M^+(g)+X^-(g)}
\]
Reversing the lattice formation reaction reverses the enthalpy sign.
\[
\Delta H_1=+690\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Form the MY lattice:}\)
The next step is:
\[
\mathrm{M^+(g)+Y^-(g)\rightarrow MY(s)}
\]
Its enthalpy is supplied directly.
\[
\Delta H_2=-840\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Combine the steps using Hess's law:}\)
Adding the two equations cancels gaseous \(\mathrm{M^+}\).
The resulting equation is the required exchange reaction.
\[
\Delta H_{\mathrm{exchange}}=\Delta H_1+\Delta H_2
\]
Substituting:
\[
\Delta H_{\mathrm{exchange}}=690-840
\]
\[
\Delta H_{\mathrm{exchange}}=-150\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Identify the stages that are not required:}\)
No neutral halogen atom or diatomic halogen molecule appears in the target reaction.
Therefore, no halogen dissociation or electron-gain stage is needed.
Adding such stages without their appropriate reverse processes would describe a different thermochemical pathway.
\(\textbf{Interpret the energetic comparison:}\)
Dissociating MX requires \(690\ \mathrm{kJ\,mol^{-1}}\).
Forming MY releases \(840\ \mathrm{kJ\,mol^{-1}}\).
The lattice-formation release exceeds the dissociation cost by \(150\ \mathrm{kJ\,mol^{-1}}\).
The net exchange is consequently exothermic under the stated gaseous-ion conditions.
This result must not be confused with the enthalpy of an exchange reaction occurring in aqueous solution.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{exchange}}=-150\ \mathrm{kJ\,mol^{-1}}}
\]
413. Four statements are made about the formation of ionic solids and their Born–Haber cycles.
I. Removal of an electron from an isolated gaseous neutral atom requires an input of energy in the conventional first-ionization process.
II. The first electron-gain enthalpy of every element must be negative because electrons are attracted towards positively charged nuclei.
III. The second electron-gain enthalpy for forming \(\mathrm{O^{2-}(g)}\) from \(\mathrm{O^-(g)}\) is positive because an electron is added to an already negatively charged ion.
IV. A negative lattice formation enthalpy guarantees that the standard formation enthalpy of the crystalline salt from its elements is negative.
V. For formation of one mole of \(\mathrm{MCl_2}\) from \(\mathrm{M(s)}\) and \(\mathrm{Cl_2(g)}\), a complete mole of chlorine molecules must be dissociated.
Which statements are correct?
ⓐ. I, II, III and V only
ⓑ. I, III and V only
ⓒ. I, III and IV only
ⓓ. II, IV and V only
Correct Answer: I, III and V only
Explanation: \(\textbf{Evaluate Statement I:}\)
The first ionization process removes an electron from a neutral gaseous atom.
Separating a bound electron from the attraction of its nucleus requires energy.
The first ionization enthalpy is therefore positive.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Electron-gain enthalpy measures the enthalpy change when an electron is added to an isolated gaseous atom.
For many nonmetals, the first electron-gain enthalpy is negative.
However, this is not universal.
The result depends on the electronic configuration and repulsive interactions within the atom.
Statement II is incorrect.
\(\textbf{Evaluate Statement III:}\)
The second electron-gain step for oxygen is:
\[
\mathrm{O^-(g)+e^-\rightarrow O^{2-}(g)}
\]
The incoming electron is added to an already negatively charged species.
Electrostatic repulsion makes the process energetically unfavourable.
The second electron-gain enthalpy is positive.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
Lattice formation from oppositely charged gaseous ions is exothermic.
However, preparing those gaseous ions from the elements may require substantial energy.
The overall formation enthalpy is the sum of all relevant stages.
A negative lattice formation term alone therefore cannot determine the sign of the total formation enthalpy.
Statement IV is incorrect.
\(\textbf{Evaluate Statement V:}\)
One formula unit of \(\mathrm{MCl_2}\) contains two chlorine atoms.
One molecule of \(\mathrm{Cl_2}\) provides precisely two chlorine atoms upon dissociation.
Therefore, formation of one mole of \(\mathrm{MCl_2}\) requires dissociation of one mole of \(\mathrm{Cl_2}\).
Statement V is correct.
\(\textbf{Overall conclusion:}\)
The correct statements separately concern ionization, successive electron gain and stoichiometric halogen dissociation.
Statements II and IV incorrectly generalize one energetic contribution to every element or to the complete formation process.
Thus, Statements I, III and V are correct.
414. Two hypothetical monovalent ionic solids, MX and NY, are dissolved separately in sufficient water. The following molar enthalpies refer to the stated processes:
| Contribution | MX | NY |
|---|
| Lattice dissociation into gaseous ions (\(\mathrm{kJ\,mol^{-1}}\)) | \(+750\) | \(+900\) |
| Combined hydration enthalpy of the constituent gaseous ions (\(\mathrm{kJ\,mol^{-1}}\)) | \(-820\) | \(-800\) |
Assume that the specified hydration steps convert each gaseous ion into its corresponding aqueous ion. Which statement correctly compares the enthalpies of dissolution and avoids an unsupported inference about solubility?
ⓐ. Both dissolutions are endothermic because lattice dissociation always requires energy.
ⓑ. MX dissolves endothermically by \(70\ \mathrm{kJ\,mol^{-1}}\), whereas NY dissolves exothermically by \(100\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. MX dissolves exothermically by \(70\ \mathrm{kJ\,mol^{-1}}\), while NY dissolves endothermically by \(100\ \mathrm{kJ\,mol^{-1}}\); these enthalpies alone do not establish their relative equilibrium solubilities.
ⓓ. NY must be more soluble because its lattice dissociation enthalpy is larger, irrespective of its hydration enthalpy and the entropy of dissolution.
Correct Answer: MX dissolves exothermically by \(70\ \mathrm{kJ\,mol^{-1}}\), while NY dissolves endothermically by \(100\ \mathrm{kJ\,mol^{-1}}\); these enthalpies alone do not establish their relative equilibrium solubilities.
Explanation: \(\textbf{Distinguish dissolution from lattice dissociation:}\)
Lattice dissociation produces gaseous ions from an ionic solid.
Dissolution in water produces aqueous ions.
The two processes have different final states and therefore different enthalpy changes.
\(\textbf{Construct the dissolution pathway:}\)
For each salt, imagine first separating the solid into gaseous ions.
Then hydrate the gaseous ions to form the final aqueous ions.
Hess's law gives:
\[
\Delta_{\mathrm{sol}}H=\Delta_{\mathrm{latt,diss}}H+\Delta_{\mathrm{hyd}}H
\]
\(\textbf{Calculate the MX dissolution enthalpy:}\)
\[
\Delta_{\mathrm{sol}}H(\mathrm{MX})=750+(-820)
\]
\[
\Delta_{\mathrm{sol}}H(\mathrm{MX})=-70\ \mathrm{kJ\,mol^{-1}}
\]
Thus, dissolution of MX is exothermic under the stipulated conditions.
\(\textbf{Calculate the NY dissolution enthalpy:}\)
\[
\Delta_{\mathrm{sol}}H(\mathrm{NY})=900+(-800)
\]
\[
\Delta_{\mathrm{sol}}H(\mathrm{NY})=+100\ \mathrm{kJ\,mol^{-1}}
\]
Thus, dissolution of NY is endothermic.
\(\textbf{Interpret the competing contributions:}\)
NY requires more energy for lattice dissociation than MX.
Its hydration process also releases less energy.
Consequently, hydration fails to compensate fully for its lattice-dissociation cost.
MX shows the opposite balance, giving an exothermic dissolution.
\(\textbf{Distinguish enthalpy from solubility:}\)
The sign of dissolution enthalpy alone does not establish whether dissolution is thermodynamically favourable under specified conditions.
Entropy and temperature also influence the Gibbs energy of dissolution.
Equilibrium solubility therefore cannot be ranked solely from the two enthalpy values.
A stronger lattice also does not, by itself, determine the complete aqueous dissolution behaviour.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta_{\mathrm{sol}}H(\mathrm{MX})=-70,\quad\Delta_{\mathrm{sol}}H(\mathrm{NY})=+100\ \mathrm{kJ\,mol^{-1}}}
\]
415. A hypothetical ionic bromide, \(\mathrm{MBr(s)}\), forms according to:
\[
\mathrm{M(s)+\frac12Br_2(g)\rightarrow MBr(s)}
\]
The supplied enthalpy contributions are:
| Contribution | Value (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Metal sublimation | \(+100\) |
| First metal ionization | \(+500\) |
| \(\mathrm{Br_2(g)\rightarrow2Br(g)}\) | \(+200\) |
| \(\mathrm{Br(g)+e^-\rightarrow Br^-(g)}\) | \(-300\) |
| Lattice formation | \(-700\) |
A student calculates the formation enthalpy as \(-200\ \mathrm{kJ\,mol^{-1}}\) by summing the correct metal, electron-gain and lattice contributions but making exactly one mistake in the halogen-atomization coefficient. Which statement correctly identifies the error and the formation enthalpy?
ⓐ. The student used the full Br–Br dissociation enthalpy instead of half its value; the correct formation enthalpy is \(-300\ \mathrm{kJ\,mol^{-1}}\).
ⓑ. The student used half the Br–Br dissociation enthalpy instead of its full value; the correct formation enthalpy is \(-100\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. The student should have multiplied the bromine electron-gain enthalpy by two; the correct formation enthalpy is \(-500\ \mathrm{kJ\,mol^{-1}}\).
ⓓ. The student should have reversed the lattice-formation sign; the correct formation enthalpy is \(+1100\ \mathrm{kJ\,mol^{-1}}\).
Correct Answer: The student used the full Br–Br dissociation enthalpy instead of half its value; the correct formation enthalpy is \(-300\ \mathrm{kJ\,mol^{-1}}\).
Explanation: \(\textbf{Determine the required amount of bromine atoms:}\)
One formula unit of MBr contains one bromine atom.
One \(\mathrm{Br_2}\) molecule contains two bromine atoms.
Therefore, formation of one mole of MBr requires half a mole of gaseous \(\mathrm{Br_2}\).
\(\textbf{Calculate the correct atomization contribution:}\)
\[
\mathrm{\frac12Br_2(g)\rightarrow Br(g)}
\]
\[
H_{\mathrm{Br\ atomization}}=\frac12D(\mathrm{Br_2})
\]
Substituting:
\[
H_{\mathrm{Br\ atomization}}=\frac12(200)
\]
\[
H_{\mathrm{Br\ atomization}}=+100\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Construct the formation balance:}\)
\[
\Delta_fH^\circ=\Delta_{\mathrm{sub}}H+IE_1+\frac12D(\mathrm{Br_2})+\Delta_{\mathrm{eg}}H+U_{\mathrm{form}}
\]
Substituting:
\[
\Delta_fH^\circ=100+500+100-300-700
\]
\[
\Delta_fH^\circ=-300\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Reconstruct the student's reported value:}\)
The student used the complete bromine dissociation enthalpy.
The resulting calculation would be:
\[
\Delta_fH^\circ_{\mathrm{student}}=100+500+200-300-700
\]
\[
\Delta_fH^\circ_{\mathrm{student}}=-200\ \mathrm{kJ\,mol^{-1}}
\]
This agrees with the erroneous value stated in the question.
\(\textbf{Identify the exact source of the discrepancy:}\)
\[
\Delta_fH^\circ_{\mathrm{student}}-\Delta_fH^\circ_{\mathrm{correct}}=-200-(-300)
\]
\[
\Delta H_{\mathrm{error}}=+100\ \mathrm{kJ\,mol^{-1}}
\]
The error equals the extra half-mole bromine dissociation enthalpy included unnecessarily.
\(\textbf{Interpret the reaction stoichiometry:}\)
A Born–Haber cycle must reproduce the specified overall formation reaction after all stages are added.
Using the full dissociation of \(\mathrm{Br_2}\) would produce two moles of bromine atoms while the reaction needs only one.
Thus, even correctly signed thermochemical terms give an incorrect answer when their coefficients are wrong.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta_fH^\circ(\mathrm{MBr})=-300\ \mathrm{kJ\,mol^{-1}}}
\]
416. A hypothetical metal M can form two distinct crystalline chlorides, MCl and \(\mathrm{MCl_2}\), corresponding to \(\mathrm{M^+}\) and \(\mathrm{M^{2+}}\). Their formation reactions are defined relative to the same elemental reference states. For MCl, the lattice formation enthalpy is \(-790\ \mathrm{kJ\,mol^{-1}}\), and its standard formation enthalpy is \(-400\ \mathrm{kJ\,mol^{-1}}\). The additional stages required when forming \(\mathrm{MCl_2}\) instead of MCl are:
| Additional or changed contribution | Value (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Second ionization of M | \(+900\) |
| Additional half-mole of \(\mathrm{Cl_2}\) dissociated | \(+120\) |
| Electron gain by the additional chlorine atom | \(-350\) |
Assume all other corresponding contributions are unchanged. What condition on the lattice formation enthalpy \(U_2\) of \(\mathrm{MCl_2}\) makes its standard formation enthalpy more negative than that of MCl?
ⓐ. \(U_2\lt -670\ \mathrm{kJ\,mol^{-1}}\)
ⓑ. \(U_2\lt -790\ \mathrm{kJ\,mol^{-1}}\)
ⓒ. \(U_2\lt -1460\ \mathrm{kJ\,mol^{-1}}\)
ⓓ. \(U_2\lt -1690\ \mathrm{kJ\,mol^{-1}}\)
Correct Answer: \(U_2\lt -1460\ \mathrm{kJ\,mol^{-1}}\)
Explanation: \(\textbf{Establish the comparison criterion:}\)
The question compares the molar formation enthalpies of two different chloride formula units from their respective elements.
For \(\mathrm{MCl_2}\) to have the more negative formation enthalpy:
\[
\Delta_fH^\circ(\mathrm{MCl_2})\lt \Delta_fH^\circ(\mathrm{MCl})
\]
The known reference value is:
\[
\Delta_fH^\circ(\mathrm{MCl})=-400\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Identify the additional ionization cost:}\)
MCl contains \(\mathrm{M^+}\), whereas \(\mathrm{MCl_2}\) contains \(\mathrm{M^{2+}}\).
Formation of the divalent metal ion requires the second ionization stage.
\[
\Delta H_{\mathrm{additional\ ionization}}=+900\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Identify the additional chlorine contribution:}\)
The second chloride ion requires one more gaseous chlorine atom.
\[
\Delta H_{\mathrm{atomization}}=+120\ \mathrm{kJ\,mol^{-1}}
\]
That atom accepts one electron.
\[
\Delta H_{\mathrm{electron\ gain}}=-350\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the additional non-lattice cost:}\)
\[
\Delta H_{\mathrm{additional}}=900+120-350
\]
\[
\Delta H_{\mathrm{additional}}=+670\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Express the formation-enthalpy difference:}\)
The lattice contribution changes from \(-790\) to \(U_2\).
\[
\Delta_fH^\circ(\mathrm{MCl_2})-\Delta_fH^\circ(\mathrm{MCl})=670+U_2-(-790)
\]
\[
\Delta_fH^\circ(\mathrm{MCl_2})-\Delta_fH^\circ(\mathrm{MCl})=1460+U_2
\]
\(\textbf{Apply the more-exothermic condition:}\)
The required difference must be negative.
\[
1460+U_2\lt 0
\]
\[
U_2\lt -1460\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the threshold:}\)
The second ionization cost is not compensated merely because the lattice formation enthalpy of \(\mathrm{MCl_2}\) is somewhat more negative.
Its additional lattice stabilization must outweigh the entire extra non-lattice cost of \(670\ \mathrm{kJ\,mol^{-1}}\).
At exactly \(-1460\ \mathrm{kJ\,mol^{-1}}\), the two formation enthalpies would be equal.
Only a more negative value satisfies the question.
\(\textbf{Final answer:}\)
\[
\boxed{U_2\lt -1460\ \mathrm{kJ\,mol^{-1}}}
\]
417. Sodium forms two chemically distinct oxygen-containing ionic solids, \(\mathrm{Na_2O}\) and \(\mathrm{Na_2O_2}\). Their conventional constituent ions are oxide and peroxide, respectively.
In sodium oxide, the oxygen-containing ion is monatomic \(\mathrm{O^{2-}}\). In sodium peroxide, the oxygen-containing ion is the intact diatomic peroxide ion, \(\mathrm{O_2^{2-}}\), containing an O–O bond. Both solids contain sodium cations.
Which statement correctly compares their ionic formulas, oxygen oxidation states and appropriate gaseous-ion products in their lattice dissociation equations?
ⓐ. \(\mathrm{Na_2O}\) yields \(2\mathrm{Na^+}+\mathrm{O^{2-}}\), while \(\mathrm{Na_2O_2}\) yields \(2\mathrm{Na^+}+\mathrm{O_2^{2-}}\); oxygen has states \(-2\) and \(-1\).
ⓑ. Both lattices dissociate into \(\mathrm{Na^+}\) and \(\mathrm{O^{2-}}\), but sodium peroxide gives two oxide ions per unit.
ⓒ. \(\mathrm{Na_2O_2}\) releases \(\mathrm{2Na^+(g)+2O^-(g)}\), since the O–O unit separates during lattice dissociation.
ⓓ. Oxygen is \(-1\) in both solids, while both lattices release the same monatomic gaseous oxygen ion.
Correct Answer: \(\mathrm{Na_2O}\) yields \(2\mathrm{Na^+}+\mathrm{O^{2-}}\), while \(\mathrm{Na_2O_2}\) yields \(2\mathrm{Na^+}+\mathrm{O_2^{2-}}\); oxygen has states \(-2\) and \(-1\).
Explanation: \(\textbf{Identify the sodium ions:}\)
Sodium forms the monovalent cation \(\mathrm{Na^+}\).
Two such ions contribute a total charge of \(+2\).
Therefore, the oxygen-containing anionic constituent must carry total charge \(-2\) in each stated formula unit.
\(\textbf{Examine sodium oxide:}\)
Sodium oxide contains the monatomic oxide ion \(\mathrm{O^{2-}}\).
Its lattice dissociation equation is:
\[
\mathrm{Na_2O(s)\rightarrow2Na^+(g)+O^{2-}(g)}
\]
Each oxygen atom has oxidation state \(-2\).
\(\textbf{Examine sodium peroxide:}\)
Sodium peroxide contains one diatomic peroxide ion per formula unit.
Its lattice dissociation equation is:
\[
\mathrm{Na_2O_2(s)\rightarrow2Na^+(g)+O_2^{2-}(g)}
\]
The peroxide ion contains two oxygen atoms sharing a total charge of \(-2\).
Each oxygen atom consequently has oxidation state \(-1\).
\(\textbf{Distinguish the anionic structures:}\)
The oxide ion is monatomic.
The peroxide ion is a bonded diatomic species.
Therefore, their gaseous constituent-ion descriptions are not interchangeable.
\(\textbf{Identify the lattice-enthalpy implication:}\)
Lattice dissociation refers to separation into the specified gaseous constituent ions.
It does not automatically require cleavage of covalent bonds inside a polyatomic constituent ion.
Consequently, the peroxide O–O bond is retained in the formal lattice dissociation process.
\(\textbf{Interpret the oxygen oxidation states:}\)
The oxidation state is assigned per oxygen atom, not simply from the total charge of the oxygen-containing ion.
A diatomic peroxide ion carrying charge \(-2\) does not contain two separate oxide ions.
Option A correctly combines charge balance, ionic composition and lattice reference states.
418. Four changes are made separately to hypothetical ionic systems. Assume comparable crystal structures and no changes in ionic charges unless explicitly stated. Match each change in List I with the most appropriate combined electrostatic and polarization inference in List II.
| List I | Changed condition | List II | Expected inference |
|---|
| P | The cation becomes smaller at unchanged charge; the anion is unchanged. | 1 | Weaker idealized electrostatic stabilization but greater anion polarizability |
| Q | The anion becomes larger at unchanged charge; the cation is unchanged. | 2 | Greater cation polarizing power and generally stronger electrostatic attraction at reduced separation |
| R | The cation charge magnitude increases at unchanged cation size; the anion is unchanged. | 3 | Greater cation polarizing power and greater electrostatic attraction arising from increased ionic charge product |
| S | The ionic charges and representative separation are unchanged, but the solid is formally separated into gaseous constituent ions. | 4 | Energy is required for lattice dissociation; the process reverses the lattice formation step |
Which matching is correct?
ⓐ. P–2, Q–1, R–3, S–4
ⓑ. P–1, Q–2, R–4, S–3
ⓒ. P–3, Q–1, R–2, S–4
ⓓ. P–2, Q–3, R–1, S–4
Correct Answer: P–2, Q–1, R–3, S–4
Explanation: \(\textbf{Evaluate Change P:}\)
A smaller cation generally has a more concentrated positive charge at the same charge magnitude.
Its polarizing power consequently increases.
The representative distance between cation and anion also decreases when other relevant conditions remain comparable.
This favours stronger idealized electrostatic attraction.
Therefore, P matches 2.
\(\textbf{Evaluate Change Q:}\)
A larger anion has a more extended electron cloud.
Its electron density is more readily distorted by a neighbouring cation.
The anion is consequently more polarizable.
However, the increased ionic size generally increases the representative interionic separation.
At unchanged charge product, this weakens idealized electrostatic attraction.
Therefore, Q matches 1.
\(\textbf{Evaluate Change R:}\)
Increasing cation charge magnitude while holding its size unchanged increases charge density.
The cation becomes more strongly polarizing.
The electrostatic charge product also increases.
At unchanged separation, the magnitude of Coulombic attraction increases.
Therefore, R matches 3.
\(\textbf{Evaluate Change S:}\)
Lattice formation combines gaseous constituent ions into an ionic crystal.
The reverse process separates the crystal into those gaseous ions.
The lattice dissociation enthalpy is positive under the usual thermochemical definition.
It has the opposite sign to the corresponding lattice formation enthalpy.
Therefore, S matches 4.
\(\textbf{Distinguish electrostatics from polarization:}\)
Stronger electrostatic attraction and greater covalent character are not interchangeable descriptions.
Electrostatic attraction depends on ionic charges and separation.
Covalent character can increase when an electron cloud is significantly polarized.
Some changes may influence both properties, but they need not change in the same direction.
\(\textbf{Apply the comparison limitation:}\)
The predicted lattice-stabilization trends assume comparable ionic structures.
They are not exact numerical lattice enthalpies derived from a single isolated ion pair.
Option A correctly matches all four changes.
419. The gaseous electron-transfer reaction
\[
\mathrm{Na(g)+Cl(g)\rightarrow Na^+(g)+Cl^-(g)}
\]
is compared with the formation of crystalline sodium chloride from its elements:
\[
\mathrm{Na(s)+\frac12Cl_2(g)\rightarrow NaCl(s)}
\]
Use the following approximate thermochemical data:
| Quantity | Enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| First ionization enthalpy of Na | \(+496\) |
| Electron-gain enthalpy of Cl | \(-349\) |
| Sodium sublimation | \(+108\) |
| Half of chlorine dissociation | \(+121\) |
| NaCl lattice formation | \(-787\) |
Which statement correctly compares the two reactions and explains why their enthalpy signs differ?
ⓐ. Both reactions are exothermic because chlorine has negative electron-gain enthalpy.
ⓑ. Gaseous electron transfer is exothermic by \(147\ \mathrm{kJ\,mol^{-1}}\), whereas crystalline sodium chloride formation is endothermic by \(411\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. Gaseous electron transfer is endothermic by \(147\ \mathrm{kJ\,mol^{-1}}\), while formation of crystalline NaCl from its elements is exothermic by \(411\ \mathrm{kJ\,mol^{-1}}\); the lattice formation contribution compensates for the other energetic costs.
ⓓ. Both reactions have enthalpy \(-787\ \mathrm{kJ\,mol^{-1}}\) because lattice formation and gaseous electron transfer describe the same process.
Correct Answer: Gaseous electron transfer is endothermic by \(147\ \mathrm{kJ\,mol^{-1}}\), while formation of crystalline NaCl from its elements is exothermic by \(411\ \mathrm{kJ\,mol^{-1}}\); the lattice formation contribution compensates for the other energetic costs.
Explanation: \(\textbf{Evaluate the isolated gaseous electron transfer:}\)
Sodium loses an electron, requiring its first ionization enthalpy.
Chlorine accepts an electron, releasing energy according to its electron-gain enthalpy.
The combined change is:
\[
\Delta H_{\mathrm{gas\ transfer}}=IE_1(\mathrm{Na})+\Delta_{\mathrm{eg}}H(\mathrm{Cl})
\]
Substituting:
\[
\Delta H_{\mathrm{gas\ transfer}}=496-349
\]
\[
\Delta H_{\mathrm{gas\ transfer}}=+147\ \mathrm{kJ\,mol^{-1}}
\]
The isolated gaseous electron transfer is therefore endothermic.
\(\textbf{Prepare the gaseous atoms from the elements:}\)
Sodium must be sublimed:
\[
\mathrm{Na(s)\rightarrow Na(g)}
\]
\[
\Delta H_{\mathrm{Na\ preparation}}=+108\ \mathrm{kJ\,mol^{-1}}
\]
Half a mole of chlorine molecules must be dissociated:
\[
\mathrm{\frac12Cl_2(g)\rightarrow Cl(g)}
\]
\[
\Delta H_{\mathrm{Cl\ preparation}}=+121\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Include crystal formation:}\)
The gaseous sodium and chloride ions combine into the ionic lattice.
\[
\mathrm{Na^+(g)+Cl^-(g)\rightarrow NaCl(s)}
\]
\[
\Delta H_{\mathrm{lattice}}=-787\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the standard formation enthalpy:}\)
\[
\Delta_fH^\circ=108+121+496-349-787
\]
\[
\Delta_fH^\circ=108+121+147-787
\]
\[
\Delta_fH^\circ=376-787
\]
\[
\Delta_fH^\circ=-411\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the opposite signs:}\)
The isolated electron-transfer reaction does not include crystal formation.
Its energetic balance contains only the ionization and electron-gain stages.
The complete formation reaction additionally includes atom preparation and the strongly exothermic lattice step.
The lattice energy released more than compensates for the preceding net costs.
\(\textbf{Recognise the physical distinction:}\)
The favourable formation of an ionic solid cannot be explained by claiming that electron transfer between isolated gaseous atoms is always exothermic.
The state and identity of the products are essential to the enthalpy calculation.
Thus, option C correctly distinguishes the two processes.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{gas\ transfer}}=+147,\quad\Delta_fH^\circ(\mathrm{NaCl})=-411\ \mathrm{kJ\,mol^{-1}}}
\]
420. Two hypothetical ionic compounds, AX and \(\mathrm{BX_2}\), contain the same monovalent anion \(\mathrm{X^-}\) but different metal cations, \(\mathrm{A^+}\) and \(\mathrm{B^{2+}}\), respectively. Assume that the cationic radii and relevant structural conditions are comparable, so that the effect of ionic charge can be examined independently. Consider the following argument:
“The higher charge on \(\mathrm{B^{2+}}\) favours stronger electrostatic lattice stabilization in \(\mathrm{BX_2}\). Therefore, formation of \(\mathrm{BX_2}\) from its constituent elements must always be more exothermic than formation of AX. Furthermore, the stronger electrostatic attraction proves that B–X bonding has greater ionic character than A–X bonding.”
Which option correctly identifies the two independent defects in this argument?
ⓐ. Lattice stabilization alone determines formation enthalpy, although greater cation charge can increase anion polarization and covalent character.
ⓑ. The second ionization enthalpy alone determines formation enthalpy, while greater cation charge suppresses anion polarization.
ⓒ. Formation enthalpy involves all energetic steps, but stronger lattice attraction necessarily makes the B–X interaction more ionic.
ⓓ. Formation enthalpy involves all energetic steps, and greater anion polarization can make the B–X interaction more covalent.
Correct Answer: Formation enthalpy involves all energetic steps, and greater anion polarization can make the B–X interaction more covalent.
Explanation: \(\textbf{Examine the electrostatic lattice interaction:}\)
The electrostatic attraction between oppositely charged ions depends on their charge magnitudes and separation.
For a simplified ionic interaction:
\[
|E_{\mathrm{electrostatic}}|\propto\frac{|z_+z_-|}{r}
\]
Here, \(z_+\) and \(z_-\) represent ionic charge numbers, and \(r\) represents interionic separation.
Under the stated comparable-size and structural assumptions, replacing \(\mathrm{A^+}\) with \(\mathrm{B^{2+}}\) increases the magnitude of the ionic-charge product.
This generally favours stronger electrostatic lattice stabilization.
However, this trend alone cannot establish the relative standard enthalpies of formation of AX and \(\mathrm{BX_2}\).
\(\textbf{Identify the first defect — incomplete formation-enthalpy reasoning:}\)
The standard enthalpy of formation of an ionic solid depends on the combined energetic contributions of its formation process.
These include atomization or sublimation, bond dissociation where applicable, ionization, electron gain and lattice formation.
Formation of gaseous \(\mathrm{B^{2+}}\) requires two successive electron removals:
\[
\mathrm{B(g)\rightarrow B^+(g)+e^-}
\]
\[
\mathrm{B^+(g)\rightarrow B^{2+}(g)+e^-}
\]
Both ionization steps require energy.
In particular, the second ionization enthalpy introduces an additional positive energetic contribution that is absent when forming \(\mathrm{A^+}\).
Furthermore, formation of one formula unit of \(\mathrm{BX_2}\) requires two anions, whereas AX requires only one.
Consequently, the two compounds involve different stoichiometric contributions from the relevant Born–Haber steps.
A greater lattice-stabilization magnitude does not necessarily make the overall formation enthalpy more negative.
The relative formation enthalpies cannot be determined without considering the complete energy balances.
\(\textbf{Identify the second defect — confusing lattice attraction with ionic character:}\)
According to Fajan's rule, the polarizing power of a cation generally increases with its positive charge and decreases with increasing ionic radius.
Because the cationic radii are assumed comparable, \(\mathrm{B^{2+}}\) has greater polarizing power than \(\mathrm{A^+}\).
The more highly charged cation can distort the electron cloud of the common anion \(\mathrm{X^-}\) more strongly.
Greater anion polarization increases covalent character in an otherwise predominantly ionic interaction.
Therefore, stronger electrostatic lattice attraction does not establish that the B–X interaction is more purely ionic.
Under the stated assumptions, the greater polarizing power of \(\mathrm{B^{2+}}\) instead favours greater covalent character.
\(\textbf{Combine the two conclusions:}\)
The first defect is the assumption that lattice stabilization alone determines the overall enthalpy of formation.
The second defect is the assumption that stronger electrostatic attraction necessarily implies greater ionic character.
These are independent errors because lattice energetics and ionic polarization describe different aspects of bonding.
Hence, the correct assessment must account for all formation-energy contributions and recognize that greater cation-induced polarization can increase covalent character.
Therefore, option D is correct.
