601. Fluorine is more electronegative than chlorine, but the first electron-gain enthalpy of a gaseous chlorine atom is more negative than that of a gaseous fluorine atom under ordinary conditions.
Which explanation correctly reconciles these observations?
ⓐ. A more negative chlorine electron-gain enthalpy implies stronger chlorine electron attraction in every chemical bond.
ⓑ. Electronegativity concerns bonds, whereas electron-gain enthalpy concerns gaseous atoms; fluorine's compact shell increases electron repulsion on addition.
ⓒ. The electron-gain enthalpy of fluorine is positive because its small radius completely prevents electron addition.
ⓓ. Fluorine is more electronegative only in ionic substances, whereas chlorine is more electronegative in covalent bonds.
Correct Answer: Electronegativity concerns bonds, whereas electron-gain enthalpy concerns gaseous atoms; fluorine's compact shell increases electron repulsion on addition.
Explanation: \(\textbf{Distinguish the two properties:}\)
Electronegativity describes the tendency of an atom in a chemical bond to attract the shared electron density.
Electron-gain enthalpy is the enthalpy change when an isolated gaseous atom gains an electron.
These definitions refer to different physical situations.
Their periodic trends need not agree in every comparison.
\(\textbf{Examine fluorine's electronic structure:}\)
Fluorine is a second-period element.
Its valence electrons occupy compact \(2s\) and \(2p\) orbitals.
An added electron enters the relatively small \(2p\) region.
The existing electron density produces considerable electron–electron repulsion in that compact space.
\(\textbf{Examine chlorine's electronic structure:}\)
Chlorine is a third-period element.
Its valence electrons occupy the larger \(3s\) and \(3p\) regions.
The incoming electron enters a more spatially extended \(3p\) orbital.
The repulsive effect associated with accommodating the additional electron is less severe than in the unusually compact fluorine valence region.
\(\textbf{Interpret electron-gain enthalpy:}\)
The net electron-gain enthalpy reflects both attraction of the incoming electron by the nucleus and repulsion involving existing electrons.
The larger chlorine atom can therefore have a more negative first electron-gain enthalpy despite its lower electronegativity.
\(\textbf{Interpret electronegativity:}\)
Fluorine's high effective nuclear attraction and small atomic size make it exceptionally effective at attracting shared electron density in chemical bonds.
It remains the most electronegative element on the conventional Pauling scale.
\(\textbf{Reject the proposed identification:}\)
A single gas-phase electron-gain measurement does not directly determine a universal bonded-atom electronegativity ranking.
The two quantities are related to electron attraction but are not interchangeable.
Option B correctly explains the apparently contrasting periodic trends.
602. In a simplified Mulliken-type electronegativity model, define:
\[
\chi_M=\frac{I+A}{2}
\]
where \(I\) is the first ionization energy and \(A\) is the positive magnitude of the energy released when the corresponding neutral gaseous atom accepts one electron. All quantities are supplied in electronvolts per atom.
| Hypothetical atom | \(I\) (eV) | \(A\) (eV) |
|---|
| P | 9.0 | 1.0 |
| Q | 7.0 | 3.0 |
Which statement correctly evaluates their model electronegativities and the separate gas-phase processes?
ⓐ. P has greater \(\chi_M\) because its first ionization energy is higher; Q must therefore have the greater tendency to lose electrons.
ⓑ. Q has greater \(\chi_M\) because its electron affinity is higher; equal electronegativity would require both \(I\) and \(A\) to be equal individually.
ⓒ. P and Q have equal \(\chi_M\), although electron removal is easier from Q and the isolated electron-addition process releases more energy for Q.
ⓓ. P and Q have equal \(\chi_M\), so their ionic radii, electron configurations and chemical reactivities must also be identical.
Correct Answer: P and Q have equal \(\chi_M\), although electron removal is easier from Q and the isolated electron-addition process releases more energy for Q.
Explanation: \(\textbf{Identify the stipulated convention:}\)
The model defines electronegativity using the average of first ionization energy and positive electron-affinity magnitude.
The positive electron-affinity values here represent energy released during electron addition.
They should not be inserted as negative electron-gain enthalpies in the supplied formula.
\(\textbf{Calculate P's model electronegativity:}\)
\[
\chi_M(P)=\frac{I_P+A_P}{2}
\]
Substituting:
\[
\chi_M(P)=\frac{9.0+1.0}{2}
\]
\[
\chi_M(P)=5.0\ \mathrm{eV}
\]
\(\textbf{Calculate Q's model electronegativity:}\)
\[
\chi_M(Q)=\frac{I_Q+A_Q}{2}
\]
\[
\chi_M(Q)=\frac{7.0+3.0}{2}
\]
\[
\chi_M(Q)=5.0\ \mathrm{eV}
\]
\(\textbf{Compare the averages:}\)
\[
\chi_M(P)=\chi_M(Q)
\]
The two atoms have equal model electronegativities.
However, this equality arises from different component values.
\(\textbf{Compare the electron-removal processes:}\)
P requires \(9.0\ \mathrm{eV}\) for first ionization.
Q requires \(7.0\ \mathrm{eV}\).
Therefore, the specified first electron removal is energetically easier for Q.
\(\textbf{Compare electron attachment:}\)
Electron addition to P releases \(1.0\ \mathrm{eV}\).
Electron addition to Q releases \(3.0\ \mathrm{eV}\).
Thus, the specified isolated gas-phase electron-attachment process is more exothermic for Q.
\(\textbf{Interpret the apparent paradox:}\)
An arithmetic average can be equal even when its individual terms differ.
The equality of this electronegativity measure does not establish equality of ionization energies or electron-gain enthalpies.
Nor does it uniquely determine chemical reactivity, which depends on additional electronic and environmental factors.
\(\textbf{Final answer:}\)
\[
\boxed{\chi_M(P)=\chi_M(Q)=5.0\ \mathrm{eV}}
\]
603. A student uses a Born–Haber cycle to calculate the lattice formation enthalpy of \(\mathrm{NaCl(s)}\):
\[
\mathrm{Na(s)+\tfrac12Cl_2(g)\rightarrow NaCl(s)}
\]
The student is supplied with:
| Quantity | Enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| Na sublimation | \(+108\) |
| Na first ionization | \(+496\) |
| Cl₂ bond dissociation | \(+242\) |
| Cl first electron-gain enthalpy | \(-349\) |
| NaCl standard formation | \(-411\) |
The student writes:
\[
-411=108+496+242-349+U_f
\]
Which correction and lattice formation enthalpy are consistent with the balanced formation reaction?
ⓐ. Replace \(242\) with \(121\), because one mole of chloride ions requires only one mole of chlorine atoms obtained from half a mole of chlorine molecules; then \(U_f=-787\ \mathrm{kJ\,mol^{-1}}\).
ⓑ. Replace \(496\) with \(248\), because only half a mole of sodium is required; then \(U_f=-666\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. Replace \(-349\) with \(+349\), because electron addition is always endothermic; then \(U_f=-1606\ \mathrm{kJ\,mol^{-1}}\).
ⓓ. Reverse the sign of \(-411\), because lattice formation and standard formation are both exothermic; then \(U_f=-86\ \mathrm{kJ\,mol^{-1}}\).
Correct Answer: Replace \(242\) with \(121\), because one mole of chloride ions requires only one mole of chlorine atoms obtained from half a mole of chlorine molecules; then \(U_f=-787\ \mathrm{kJ\,mol^{-1}}\).
Explanation: \(\textbf{Identify the student's stoichiometric mistake:}\)
The formation reaction produces one mole of sodium chloride.
Each formula unit contains one sodium cation and one chloride anion.
Therefore, one mole of chlorine atoms is required.
The elemental reactant supplies these atoms from half a mole of \(\mathrm{Cl_2}\).
\(\textbf{Determine the correct chlorine atomization contribution:}\)
The dissociation of one mole of chlorine molecules is:
\[
\mathrm{Cl_2(g)\rightarrow2Cl(g)}
\]
\[
\Delta H=+242\ \mathrm{kJ\,mol^{-1}}
\]
For the required half-mole process:
\[
\mathrm{\tfrac12Cl_2(g)\rightarrow Cl(g)}
\]
\[
\Delta H_{\mathrm{Cl}}=\frac{242}{2}
\]
\[
\Delta H_{\mathrm{Cl}}=+121\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Check sodium sublimation:}\)
One mole of sodium atoms is required.
\[
\mathrm{Na(s)\rightarrow Na(g)}
\]
The full \(+108\ \mathrm{kJ\,mol^{-1}}\) contribution is therefore correct.
\(\textbf{Check sodium ionization:}\)
One sodium atom loses one electron per formula unit.
The full first ionization enthalpy is required.
\[
\Delta H_{\mathrm{IE}}=+496\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Check chloride formation:}\)
One chlorine atom gains one electron.
The supplied electron-gain enthalpy is negative.
\[
\Delta H_{\mathrm{EA}}=-349\ \mathrm{kJ\,mol^{-1}}
\]
Its sign must not be reversed.
\(\textbf{Construct the corrected cycle:}\)
\[
\Delta_fH^\circ=108+496+121-349+U_f
\]
Substituting:
\[
-411=376+U_f
\]
Rearranging:
\[
U_f=-411-376
\]
\[
U_f=-787\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the correction:}\)
The student's error was not a lattice sign error.
It was the use of a full mole of chlorine dissociation when the balanced formation equation requires only half a mole.
Correcting the stoichiometric coefficient restores the proper Hess cycle.
\(\textbf{Final answer:}\)
\[
\boxed{U_f=-787\ \mathrm{kJ\,mol^{-1}}}
\]
604. Sodium chloride exists as an extended ionic crystal under ordinary conditions, but neutral gaseous \(\mathrm{NaCl}\) units can be considered in a high-temperature vapour. A student claims that both must have identical bonding descriptions because their empirical compositions are the same.
Which statement most accurately distinguishes the structural units without treating ionic and covalent character as mutually exclusive categories?
ⓐ. Crystalline \(\mathrm{NaCl}\) consists of permanently isolated neutral NaCl molecules joined by ordinary C–C-type covalent bonds.
ⓑ. Gaseous \(\mathrm{NaCl}\) must consist of isolated \(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\) ions at infinite separation because it has the same formula as the solid.
ⓒ. Solid NaCl has an extended ionic lattice; an isolated gaseous NaCl unit is a strongly polar bonded species despite identical empirical composition.
ⓓ. Formation of gaseous NaCl necessarily requires both sodium and chlorine to acquire complete octets through formation of a nonpolar homonuclear bond.
Correct Answer: Solid NaCl has an extended ionic lattice; an isolated gaseous NaCl unit is a strongly polar bonded species despite identical empirical composition.
Explanation: \(\textbf{Examine the crystalline structure:}\)
Sodium chloride forms an extended ionic lattice.
Each sodium ion is surrounded by neighbouring chloride ions, and vice versa, according to the crystal arrangement.
The empirical formula expresses the simplest ratio of sodium to chlorine.
It does not identify an independently separated neutral molecule within the ordinary crystal.
\(\textbf{Identify the collective electrostatic interaction:}\)
The lattice is stabilized by attractions among oppositely charged ions.
The total stabilization depends on interactions throughout the extended crystalline arrangement.
A single isolated nearest-neighbour pair does not represent the complete lattice energy.
\(\textbf{Consider the gaseous molecular unit:}\)
At sufficiently high temperature, neutral NaCl species can exist in the vapour.
An individual gaseous unit has one sodium atom and one chlorine atom connected through a strongly polar interaction.
Its bonding can contain substantial ionic character without requiring the two atoms to be infinitely separated ions.
\(\textbf{Distinguish the physical descriptions:}\)
The crystal contains an extended array of interacting ions.
The isolated gaseous unit contains one bonded sodium–chlorine pair.
Consequently, their physical states and aggregation structures are different despite the same elemental ratio.
\(\textbf{Identify the formula misconception:}\)
An empirical formula specifies relative atom numbers.
It does not always specify the complete structural unit or whether the substance is molecular.
This distinction is especially important for ionic solids.
\(\textbf{Connect ionic and covalent character:}\)
The polarity of a heteronuclear bond can be discussed in terms of unequal electronic distribution.
The bonding description need not be forced into a completely ionic or completely covalent category in every state.
Option C correctly distinguishes the crystal from the molecular vapour without introducing a false absolute boundary.
605. Consider the two ionic solids \(\mathrm{LiCl}\) and \(\mathrm{CuCl}\). For the purpose of this comparison, assume their monovalent cations have sufficiently similar effective sizes that differences in outer electronic configuration are important.
The lithium cation has a noble-gas-like closed-shell configuration. The copper(I) cation has a filled \(3d^{10}\) subshell and is conventionally described as possessing a pseudo-noble-gas configuration. Both interact with chloride ions, which have polarizable electron clouds.
Which statement correctly applies the extended form of Fajan's rule?
ⓐ. \(\mathrm{LiCl}\) must have greater covalent character because lithium has fewer total electrons than copper.
ⓑ. The filled \(d\) subshell of \(\mathrm{Cu^+}\) can increase polarization relative to a similar noble-gas cation, giving \(\mathrm{CuCl}\) greater covalency than \(\mathrm{LiCl}\).
ⓒ. \(\mathrm{CuCl}\) must be purely metallic because the copper cation possesses ten d electrons.
ⓓ. Both compounds must have exactly the same ionic character because their cations carry charge \(+1\) and their anions are identical.
Correct Answer: The filled \(d\) subshell of \(\mathrm{Cu^+}\) can increase polarization relative to a similar noble-gas cation, giving \(\mathrm{CuCl}\) greater covalency than \(\mathrm{LiCl}\).
Explanation: \(\textbf{Identify the common formal charges:}\)
Lithium chloride contains formally monovalent lithium and chloride ions.
Copper(I) chloride contains formally monovalent copper and chloride ions.
Their cation-charge magnitudes are therefore identical.
However, ionic charge alone does not determine the full polarizing behaviour of a cation.
\(\textbf{Examine lithium's electronic configuration:}\)
The lithium cation has the electronic configuration \(1s^2\).
It possesses a compact noble-gas-like electron arrangement.
This is a conventional stable cationic configuration.
\(\textbf{Examine copper(I):}\)
The copper(I) cation has a filled \(3d^{10}\) subshell.
Its electronic configuration differs substantially from that of a simple noble-gas-configured main-group cation.
Filled d electrons shield nuclear charge less effectively than might be inferred from electron number alone.
This can influence the cation's polarizing power.
\(\textbf{Apply the configuration-sensitive Fajan trend:}\)
For suitably comparable ionic charges and sizes, cations with pseudo-noble-gas configurations can exert greater polarizing influence than cations with noble-gas configurations.
Their electron-cloud interaction with a polarizable anion can consequently acquire greater covalent character.
This tendency favours greater covalent character in \(\mathrm{CuCl}\).
\(\textbf{Distinguish formal cation configuration from metallic bonding:}\)
A copper(I) cation is not the same as an atom in metallic copper.
The presence of a filled d subshell does not automatically convert copper(I) chloride into a metal.
The chloride compound must be discussed using its actual solid-state bonding environment.
\(\textbf{Recognise the comparison limit:}\)
Fajan's rule provides qualitative trends rather than exact percentages of covalent character.
Crystal structure and detailed electron distribution can also matter.
The question explicitly controls the main cation-size complication to isolate the electronic-configuration effect.
Option B correctly identifies the relevant principle.
606. Two hypothetical ionic solids AX and BY are formed from gaseous atoms. Their complete energetic contributions per mole of formula units are supplied below:
| Energy contribution | AX (\(\mathrm{kJ\,mol^{-1}}\)) | BY (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| All endothermic atomization and ionization stages combined | \(+850\) | \(+1100\) |
| All electron-gain stages combined | \(-300\) | \(-450\) |
| Lattice formation enthalpy | \(-900\) | \(-950\) |
Each column describes a balanced formation reaction from its respective elemental reference states. Which conclusion is correct?
ⓐ. BY has the more negative formation enthalpy because its lattice formation enthalpy has the greater magnitude.
ⓑ. AX and BY have equal formation enthalpies because the extra ionization cost in BY is exactly compensated by its electron-gain stages.
ⓒ. AX has a formation enthalpy of \(-350\ \mathrm{kJ\,mol^{-1}}\), while BY has \(-300\ \mathrm{kJ\,mol^{-1}}\); AX is more exothermic despite BY having the larger lattice-enthalpy magnitude.
ⓓ. AX has a formation enthalpy of \(+350\ \mathrm{kJ\,mol^{-1}}\), while BY has \(+300\ \mathrm{kJ\,mol^{-1}}\), because lattice formation enthalpies must be inserted as positive magnitudes.
Correct Answer: AX has a formation enthalpy of \(-350\ \mathrm{kJ\,mol^{-1}}\), while BY has \(-300\ \mathrm{kJ\,mol^{-1}}\); AX is more exothermic despite BY having the larger lattice-enthalpy magnitude.
Explanation: \(\textbf{Identify the complete thermochemical balance:}\)
The formation enthalpy is the sum of all required energetic stages.
It includes preparation of gaseous atoms and cations, formation of anions and lattice formation.
A comparison using only the lattice contribution would omit other essential terms.
\(\textbf{Calculate the formation enthalpy of AX:}\)
\[
\Delta_fH^\circ(AX)=850-300-900
\]
Combining the ion-formation contributions:
\[
850-300=550
\]
Therefore:
\[
\Delta_fH^\circ(AX)=550-900
\]
\[
\Delta_fH^\circ(AX)=-350\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the formation enthalpy of BY:}\)
\[
\Delta_fH^\circ(BY)=1100-450-950
\]
Combining the non-lattice contributions:
\[
1100-450=650
\]
Therefore:
\[
\Delta_fH^\circ(BY)=650-950
\]
\[
\Delta_fH^\circ(BY)=-300\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Compare the lattice contributions:}\)
The BY lattice formation enthalpy is more negative.
Its magnitude is:
\[
|U_f(BY)|=950\ \mathrm{kJ\,mol^{-1}}
\]
For AX:
\[
|U_f(AX)|=900\ \mathrm{kJ\,mol^{-1}}
\]
Thus, BY has the greater stipulated lattice stabilization.
\(\textbf{Compare the complete formation reactions:}\)
AX has the more negative overall formation enthalpy.
Its combined atomization and ionization costs are lower.
BY's greater lattice stabilization and more favourable electron-gain contribution do not fully compensate for its larger endothermic preparation cost.
\(\textbf{Interpret the general principle:}\)
Greater lattice stabilization is one favourable factor in ionic compound formation.
It does not by itself guarantee a more exothermic formation reaction from the elements.
The complete Hess-law balance determines the net result.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta_fH^\circ(AX)=-350;\quad\Delta_fH^\circ(BY)=-300\ \mathrm{kJ\,mol^{-1}}}
\]
607. A thermochemical cycle for forming one mole of a hypothetical crystalline salt \(\mathrm{MCl_3(s)}\) from its elements includes the following stages:
The metal starts as \(\mathrm{M(s)}\), while chlorine starts as \(\mathrm{Cl_2(g)}\). The solid contains \(\mathrm{M^{3+}}\) and \(\mathrm{Cl^-}\). The cycle uses the sublimation enthalpy of M, its first three successive ionization enthalpies, the full dissociation enthalpy of \(\mathrm{Cl_2}\), the first electron-gain enthalpy of Cl and the lattice formation enthalpy of \(\mathrm{MCl_3}\).
Which expression correctly represents the standard enthalpy of formation, where \(S\) is the metal sublimation enthalpy, \(I_1,I_2,I_3\) are successive ionization enthalpies, \(D\) is the full \(\mathrm{Cl_2}\) bond dissociation enthalpy, \(E\) is the signed chlorine electron-gain enthalpy and \(U_f\) is the signed lattice formation enthalpy?
ⓐ. \(\Delta_fH^\circ=S+I_1+I_2+I_3+\frac32D+3E+U_f\)
ⓑ. \(\Delta_fH^\circ=S+3I_1+\frac12D+E+U_f\)
ⓒ. \(\Delta_fH^\circ=3S+I_1+I_2+I_3+3D+3E-U_f\)
ⓓ. \(\Delta_fH^\circ=S+I_1+I_2+I_3+\frac32D+E-U_f\)
Correct Answer: \(\Delta_fH^\circ=S+I_1+I_2+I_3+\frac32D+3E+U_f\)
Explanation: \(\textbf{Establish the balanced formation reaction:}\)
The crystal contains one trivalent metal cation and three chloride anions per formula unit.
Its formation reaction is:
\[
\mathrm{M(s)+\frac32Cl_2(g)\rightarrow MCl_3(s)}
\]
The coefficients in the thermochemical cycle must match this reaction.
\(\textbf{Prepare the gaseous metal atom:}\)
One metal atom is required per formula unit.
Therefore, one metal sublimation contribution appears.
The coefficient of \(S\) is one.
\(\textbf{Produce the trivalent metal cation:}\)
The gaseous atom loses three electrons successively.
The required stages are the first, second and third ionizations.
The total ionization cost is:
\[
I_1+I_2+I_3
\]
Three times the first ionization enthalpy would be incorrect because successive electrons are removed from differently charged species.
\(\textbf{Produce three chlorine atoms:}\)
The formation equation contains three-halves of a mole of chlorine molecules.
Each chlorine molecule contains two chlorine atoms.
The required amount of atomic chlorine is therefore three moles per mole of salt.
The corresponding dissociation contribution is:
\[
\frac32D
\]
\(\textbf{Form three chloride ions:}\)
Each chlorine atom gains one electron.
Three chlorine atoms therefore undergo three first electron-gain processes.
The total signed contribution is:
\[
3E
\]
The sign of \(E\) must be retained as supplied.
\(\textbf{Form the ionic lattice:}\)
The prepared gaseous ions combine:
\[
\mathrm{M^{3+}(g)+3Cl^-(g)\rightarrow MCl_3(s)}
\]
The lattice formation enthalpy is \(U_f\).
Because it is already defined for this forward formation process, its sign should not be reversed.
\(\textbf{Combine the stages:}\)
Adding the stages reproduces the balanced formation reaction.
Therefore, the complete expression is:
\[
\Delta_fH^\circ=S+I_1+I_2+I_3+\frac32D+3E+U_f
\]
\(\textbf{Identify the two principal stoichiometric traps:}\)
A trivalent metal requires three successive ionizations, not three repetitions of the first ionization.
Three chloride ions require three electron-gain events and three chlorine atoms.
The chlorine atomization coefficient is consequently \(\frac32\), not \(\frac12\).
Option A correctly preserves both conditions and the lattice sign convention.
608. A proposed electron-transfer reaction between isolated gaseous atoms is:
\[
\mathrm{Na(g)+Cl(g)\rightarrow Na^+(g)+Cl^-(g)}
\]
Use the supplied first ionization enthalpy of sodium, \(+496\ \mathrm{kJ\,mol^{-1}}\), and chlorine electron-gain enthalpy, \(-349\ \mathrm{kJ\,mol^{-1}}\).
Assertion (A): This isolated gas-phase electron-transfer reaction is endothermic under the supplied data, even though formation of crystalline sodium chloride from suitable gaseous ions can be strongly exothermic.
Reason (R): The energy released by electron attachment to chlorine is smaller than the energy required to ionize sodium, while subsequent ionic lattice formation supplies an additional stabilizing contribution absent from the isolated electron-transfer equation.
Which option is correct?
ⓐ. A is true, but R is false.
ⓑ. Both A and R are true, but R does not explain A.
ⓒ. 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: \(\textbf{Identify the gas-phase electron-transfer stages:}\)
The specified process begins with isolated gaseous sodium and chlorine atoms.
Sodium loses one electron to form a gaseous cation.
Chlorine gains that electron to form a gaseous anion.
No ionic crystal is formed in this reaction.
\(\textbf{Evaluate sodium ionization:}\)
Removing the electron from sodium requires energy.
The contribution is positive:
\[
\Delta H_{\mathrm{IE}}=+496\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Evaluate chlorine electron attachment:}\)
Adding the electron to chlorine releases energy.
The supplied contribution is:
\[
\Delta H_{\mathrm{EA}}=-349\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the isolated transfer enthalpy:}\)
\[
\Delta H_{\mathrm{transfer}}=\Delta H_{\mathrm{IE}}+\Delta H_{\mathrm{EA}}
\]
\[
\Delta H_{\mathrm{transfer}}=496-349
\]
\[
\Delta H_{\mathrm{transfer}}=+147\ \mathrm{kJ\,mol^{-1}}
\]
The positive result establishes that the specified isolated transfer is endothermic.
\(\textbf{Examine subsequent lattice formation:}\)
When gaseous sodium and chloride ions combine into a crystalline ionic lattice, electrostatic stabilization releases substantial energy.
This lattice contribution is additional to the isolated atom-to-ion conversion.
It can more than compensate for the positive gas-phase electron-transfer enthalpy.
\(\textbf{Evaluate the assertion:}\)
The first part is supported by the positive calculated transfer enthalpy.
The second part correctly recognizes a separate lattice-stabilization process.
Assertion A is therefore true.
\(\textbf{Evaluate the reason:}\)
Reason R explicitly identifies the competing ionization and electron-attachment terms.
It also identifies the missing lattice contribution in the isolated gas-phase reaction.
It therefore explains the assertion.
\(\textbf{Interpret the octet misconception:}\)
Obtaining noble-gas-like ionic configurations does not guarantee that the isolated electron-transfer step releases energy.
The full energetic stability of an ionic solid depends on the complete formation process.
The correct option is D.
609. Four hypothetical covalent bonds are formed between atoms having the following electronegativities on the same conventional scale:
| Atom | Electronegativity |
|---|
| P | 2.0 |
| Q | 2.8 |
| R | 3.2 |
| S | 3.8 |
Consider the bonds P–R, P–Q, Q–S and R–S. Assume comparable conditions and use electronegativity difference only to compare their predicted qualitative bond polarity, without assigning sharp ionic–covalent thresholds.
Which statement is correct?
ⓐ. P–R and Q–S have equal electronegativity differences, but their electron-rich ends must be on P and Q, respectively.
ⓑ. P–R has the largest electronegativity difference, while R–S has the smallest; the electron-rich end of every listed bond is the more electronegative atom.
ⓒ. P–Q and R–S have equal electronegativity differences, and therefore must have identical experimental dipole moments regardless of bond length.
ⓓ. Q–S has the largest electronegativity difference because S is the most electronegative atom, independent of the atom bonded to it.
Correct Answer: P–R has the largest electronegativity difference, while R–S has the smallest; the electron-rich end of every listed bond is the more electronegative atom.
Explanation: \(\textbf{Identify the governing comparison:}\)
The qualitative polarity of a heteronuclear covalent bond is related to the difference in electronegativity between its bonded atoms.
The electron-rich end is generally associated with the more electronegative atom.
An electronegativity difference is a comparative guide, not an exact measurement of transferred charge.
\(\textbf{Compare P–R:}\)
\[
\Delta\chi_{\mathrm{P-R}}=3.2-2.0
\]
\[
\Delta\chi_{\mathrm{P-R}}=1.2
\]
The electron-rich end is R.
\(\textbf{Compare P–Q:}\)
\[
\Delta\chi_{\mathrm{P-Q}}=2.8-2.0
\]
\[
\Delta\chi_{\mathrm{P-Q}}=0.8
\]
The electron-rich end is Q.
\(\textbf{Compare Q–S:}\)
\[
\Delta\chi_{\mathrm{Q-S}}=3.8-2.8
\]
\[
\Delta\chi_{\mathrm{Q-S}}=1.0
\]
The electron-rich end is S.
\(\textbf{Compare R–S:}\)
\[
\Delta\chi_{\mathrm{R-S}}=3.8-3.2
\]
\[
\Delta\chi_{\mathrm{R-S}}=0.6
\]
The electron-rich end is S.
\(\textbf{Establish the qualitative ranking:}\)
The electronegativity differences decrease as:
\[
1.2\gt 1.0\gt 0.8\gt 0.6
\]
Thus, P–R has the largest difference and R–S has the smallest.
\(\textbf{Identify the dipole-moment limitation:}\)
Bond dipole moment depends on effective charge separation and bond length.
Equal electronegativity differences would not independently establish identical experimental dipole moments.
The actual electron distribution and bond geometry would also matter.
\(\textbf{Avoid the terminal-atom shortcut:}\)
The most electronegative atom S does not necessarily produce the largest bond polarity with every bonding partner.
Polarity depends on both atoms in the bond.
Option B correctly combines the difference ranking with the directions of electron-density displacement.
610. A hypothetical monatomic cation \(\mathrm{M^{2+}}\) can form the ionic solids \(\mathrm{MX_2}\) and \(\mathrm{MY_2}\). X and Y are monovalent anions with the following properties:
| Property | \(\mathrm{X^-}\) | \(\mathrm{Y^-}\) |
|---|
| Radius | \(130\ \mathrm{pm}\) | \(210\ \mathrm{pm}\) |
| Polarizability | Lower | Higher |
Assume that the cation radius is \(70\ \mathrm{pm}\), nearest-neighbour separations are approximated by the sums of ionic radii, and the solids have comparable lattice structures. Which statement correctly combines the electrostatic and polarization tendencies?
ⓐ. \(\mathrm{MY_2}\) is expected to have stronger nearest-neighbour electrostatic attraction and less covalent character because Y is larger.
ⓑ. The lattice-energy and Fajan trends must be identical because the two salts have the same empirical ionic charge ratio.
ⓒ. The larger Y ion is more polarizable, which necessarily implies a higher lattice-enthalpy magnitude despite its larger separation from the cation.
ⓓ. \(\mathrm{MX_2}\) has shorter cation–anion separation and stronger local electrostatics; larger Y is more polarizable, favouring more covalency in \(\mathrm{MY_2}\).
Correct Answer: \(\mathrm{MX_2}\) has shorter cation–anion separation and stronger local electrostatics; larger Y is more polarizable, favouring more covalency in \(\mathrm{MY_2}\).
Explanation: \(\textbf{Identify the controlled variables:}\)
Both compounds contain the same divalent cation.
Both anions have charge magnitude one.
The principal charge-product factor is therefore the same in the nearest-neighbour comparison.
Their ionic radii and polarizabilities differ.
\(\textbf{Determine the cation–X separation:}\)
Using the stipulated radius-sum approximation:
\[
r_{\mathrm{M-X}}=r_{\mathrm M}+r_{\mathrm X}
\]
\[
r_{\mathrm{M-X}}=70+130
\]
\[
r_{\mathrm{M-X}}=200\ \mathrm{pm}
\]
\(\textbf{Determine the cation–Y separation:}\)
\[
r_{\mathrm{M-Y}}=70+210
\]
\[
r_{\mathrm{M-Y}}=280\ \mathrm{pm}
\]
\(\textbf{Compare electrostatic interactions:}\)
For comparable ion charges, the magnitude of the idealized Coulomb attraction varies inversely with separation.
Therefore, the shorter M–X separation favours stronger nearest-neighbour electrostatic attraction.
The comparison is qualitative for the complete crystal because actual lattice enthalpy also depends on the extended structure.
\(\textbf{Compare polarization:}\)
The Y anion is larger and more polarizable.
Its electron cloud can be distorted more readily by the same divalent cation.
According to Fajan's rule, this favours greater polarization-induced covalent character in the M–Y interaction.
\(\textbf{Distinguish the tendencies:}\)
The smaller X ion favours closer electrostatic contact.
The larger Y ion favours greater electron-cloud distortion.
These are different criteria and need not select the same compound.
\(\textbf{Interpret the supplied numerical information:}\)
The radius data establish the comparative separations.
They do not determine exact lattice formation enthalpies.
The polarizability data establish a qualitative covalent-character tendency.
They do not determine an exact percentage of covalent character.
\(\textbf{Final answer:}\)
\[
\boxed{r_{\mathrm{M-X}}=200\ \mathrm{pm}\lt r_{\mathrm{M-Y}}=280\ \mathrm{pm}}
\]
611. Consider three representative bonds involving carbon:
Bond P is the C–H bond of ethyne. Bond Q is the C–H bond of ethene. Bond R is the C–H bond of ethane. In the conventional hybrid-orbital model, the bonded carbon atoms are respectively \(sp\)-, \(sp^2\)- and \(sp^3\)-hybridized. The question concerns the qualitative electronegativity of the carbon hybrid orbital, the resulting tendency to attract shared electron density, and the limitations of inferring complete molecular polarity from one bond.
Which statement correctly connects hybridization with electronegativity and bond polarity?
ⓐ. The effective electronegativity associated with the carbon bonding orbital generally increases with s character, giving the qualitative carbon-orbital order \(sp\gt sp^2\gt sp^3\); this does not independently determine the permanent dipole moments of the complete molecules.
ⓑ. The effective electronegativity of carbon decreases as s character increases because s orbitals place electrons farther from the carbon nucleus.
ⓒ. All three C–H bonds must possess identical polarity because carbon's elemental electronegativity has one value on a given conventional scale.
ⓓ. The ethyne C–H bond must be purely ionic because its carbon is \(sp\)-hybridized, while the corresponding ethane bond must be nonpolar because its carbon is \(sp^3\)-hybridized.
Correct Answer: The effective electronegativity associated with the carbon bonding orbital generally increases with s character, giving the qualitative carbon-orbital order \(sp\gt sp^2\gt sp^3\); this does not independently determine the permanent dipole moments of the complete molecules.
Explanation: \(\textbf{Identify the hybrid compositions:}\)
An ideal \(sp\) orbital contains \(50\%\) s character.
An ideal \(sp^2\) orbital contains approximately \(33\frac13\%\) s character.
An ideal \(sp^3\) orbital contains \(25\%\) s character.
Thus, the carbon bonding orbitals have different electronic compositions despite belonging to the same element.
\(\textbf{Interpret greater s character:}\)
An s orbital has substantial electron density close to the nucleus.
Greater s character generally draws hybrid-orbital electron density closer to the carbon centre.
Consequently, an \(sp\) carbon bonding orbital tends to attract shared electron density more strongly than an analogous \(sp^2\) or \(sp^3\) orbital.
\(\textbf{Establish the effective electronegativity tendency:}\)
The conventional qualitative order is:
\[
sp\gt sp^2\gt sp^3
\]
This refers to differences associated with carbon's bonding environment.
It does not mean that carbon becomes a different element when its hybridization changes.
\(\textbf{Connect orbital character with C–H bond polarity:}\)
Hydrogen's bonding interaction with carbon depends partly on the electron-attracting ability of the carbon orbital.
The degree and distribution of C–H bond polarity may therefore differ among the three environments.
However, these effects should not be equated with complete electron transfer.
All three connections remain covalent in the conventional treatment.
\(\textbf{Distinguish bond polarity from molecular polarity:}\)
Molecular dipole moment is a vector property of the complete molecular charge distribution.
Ethyne and ethane possess symmetrical overall structures.
Their individual C–H bond polarities do not automatically produce nonzero permanent molecular dipoles.
Ethene also requires consideration of all its bond directions and molecular symmetry.
\(\textbf{Identify the scope of the hybridization trend:}\)
The s-character argument supports a qualitative electronic comparison.
It does not supply exact numerical bond dipole moments without further structural and electronic data.
The actual molecular dipole requires the complete vector sum of the charge-distribution contributions.
Option A correctly combines the orbital-electronegativity trend with its molecular limitation.
612. Boron trifluoride forms an addition compound with ammonia:
\[
\mathrm{BF_3+NH_3\rightarrow F_3B\leftarrow NH_3}
\]
In isolated \(\mathrm{BF_3}\), boron has three B–F sigma bonds and an electron-deficient valence shell. In ammonia, nitrogen has three N–H sigma bonds and one lone pair. Assume the conventional Lewis and VSEPR descriptions.
Which statement correctly integrates the electron-pair donation, formal charges and changes in local geometry?
ⓐ. Ammonia supplies a proton to boron, producing a trigonal-planar boron centre and a planar nitrogen centre.
ⓑ. Boron donates an electron pair to nitrogen; both centres retain their original molecular shapes because coordinate bonds are not counted in VSEPR.
ⓒ. N donates its lone pair to B; both centres become four-coordinate and approximately tetrahedral, with Lewis formal charges B \(-1\) and N \(+1\).
ⓓ. Nitrogen transfers an electron completely to fluorine, converting all three B–F bonds into ionic bonds and eliminating the B–N connection.
Correct Answer: N donates its lone pair to B; both centres become four-coordinate and approximately tetrahedral, with Lewis formal charges B \(-1\) and N \(+1\).
Explanation: \(\textbf{Identify the Lewis acid and base:}\)
Isolated \(\mathrm{BF_3}\) contains three B–F sigma connections and an electron-deficient boron centre.
Boron has an available acceptor orbital in the conventional trigonal-planar description.
Ammonia nitrogen possesses a lone pair that can be donated to a suitable electron-pair acceptor.
Thus, boron trifluoride acts as the Lewis acid and ammonia acts as the Lewis base.
\(\textbf{Trace the electron-pair movement:}\)
The B–N coordinate bond is formed using the nitrogen lone pair.
No proton transfer is required for the stated addition reaction.
Once formed, the B–N connection is counted as a sigma-bonding domain at both atoms.
Its coordinate origin does not make it invisible to VSEPR.
\(\textbf{Reassess the boron environment:}\)
Boron initially has three bonding domains and an approximately trigonal-planar arrangement.
In the adduct, it has four bonding connections.
Its local sigma framework becomes approximately tetrahedral.
The added electron pair completes boron's conventional octet.
\(\textbf{Reassess the nitrogen environment:}\)
Ammonia initially contains three bonding domains and one lone-pair domain.
The resulting molecular shape is trigonal pyramidal.
After donation, nitrogen has four bonding domains and no lone pair.
Its four attached atoms adopt an approximately tetrahedral arrangement.
\(\textbf{Assign formal charges:}\)
Treating the newly formed B–N connection as an ordinary shared electron pair, boron has four bonds and no lone pairs.
Its formal charge is \(-1\).
Nitrogen also has four bonds and no lone pairs.
Its formal charge is \(+1\).
These formal charges sum to zero, as required for the neutral adduct.
\(\textbf{Distinguish formal charge from complete electron transfer:}\)
The B–N bond is covalent in the Lewis donor–acceptor description.
Formal-charge separation does not imply that an isolated \(\mathrm{B^-}\) ion and an isolated \(\mathrm{N^+}\) ion have been produced.
The adduct remains one connected molecular species.
\(\textbf{Integrate the structural change:}\)
Electron-pair donation changes the number of sigma-bonding directions at boron.
At nitrogen, it converts a lone-pair domain into a bonding domain.
The two centres consequently acquire approximately tetrahedral local arrangements for different electron-counting reasons.
Option C correctly combines these effects.
613. Two gaseous hydrogen halides have the following experimentally specified molecular dipole moments and bond lengths:
| Molecule | Dipole moment | Bond length |
|---|
| HCl | \(1.08\ \mathrm D\) | \(127\ \mathrm{pm}\) |
| HBr | \(0.82\ \mathrm D\) | \(141\ \mathrm{pm}\) |
Use \(1\ \mathrm D=3.336\times10^{-30}\ \mathrm{C\,m}\) and \(e=1.60\times10^{-19}\ \mathrm C\). Define percentage ionic character using the effective charge \(q=\mu/r\), expressed as a percentage of \(e\).
Which statement correctly compares the percentage ionic characters and interprets the experimental dipoles?
ⓐ. HCl has approximately \(17.7\%\) and HBr approximately \(12.1\%\) ionic character; the larger HCl dipole occurs despite its shorter bond length because its effective charge separation is greater.
ⓑ. HCl has approximately \(12.1\%\) and HBr approximately \(17.7\%\) ionic character; the longer HBr bond necessarily has the more ionic bond.
ⓒ. Both bonds have approximately \(100\%\) ionic character because their molecules possess nonzero permanent dipole moments.
ⓓ. HCl and HBr have identical effective charges because hydrogen has the same electronegativity in both molecules.
Correct Answer: HCl has approximately \(17.7\%\) and HBr approximately \(12.1\%\) ionic character; the larger HCl dipole occurs despite its shorter bond length because its effective charge separation is greater.
Explanation: \(\textbf{Identify the dipole relation:}\)
For a diatomic molecule, the magnitude of the permanent molecular dipole is associated with the separation of effective positive and negative charges.
The stipulated effective-charge model uses:
\[
\mu=qr
\]
Therefore:
\[
q=\frac{\mu}{r}
\]
The percentage ionic character is:
\[
\%\mathrm{ionic\ character}=\frac{q}{e}\times100
\]
\(\textbf{Convert the HCl dipole:}\)
\[
\mu_{\mathrm{HCl}}=1.08(3.336\times10^{-30})
\]
\[
\mu_{\mathrm{HCl}}=3.60288\times10^{-30}\ \mathrm{C\,m}
\]
\(\textbf{Convert the HCl bond length:}\)
\[
r_{\mathrm{HCl}}=127\times10^{-12}\ \mathrm m
\]
\[
r_{\mathrm{HCl}}=1.27\times10^{-10}\ \mathrm m
\]
\(\textbf{Calculate the HCl effective charge:}\)
\[
q_{\mathrm{HCl}}=\frac{3.60288\times10^{-30}}{1.27\times10^{-10}}
\]
\[
q_{\mathrm{HCl}}\approx2.837\times10^{-20}\ \mathrm C
\]
\(\textbf{Calculate the HCl ionic percentage:}\)
\[
P_{\mathrm{HCl}}=\frac{2.837\times10^{-20}}{1.60\times10^{-19}}\times100
\]
\[
P_{\mathrm{HCl}}\approx17.7\%
\]
\(\textbf{Calculate the HBr effective charge:}\)
\[
\mu_{\mathrm{HBr}}=0.82(3.336\times10^{-30})
\]
\[
\mu_{\mathrm{HBr}}=2.73552\times10^{-30}\ \mathrm{C\,m}
\]
\[
r_{\mathrm{HBr}}=141\times10^{-12}=1.41\times10^{-10}\ \mathrm m
\]
\[
q_{\mathrm{HBr}}=\frac{2.73552\times10^{-30}}{1.41\times10^{-10}}
\]
\[
q_{\mathrm{HBr}}\approx1.940\times10^{-20}\ \mathrm C
\]
\(\textbf{Calculate the HBr ionic percentage:}\)
\[
P_{\mathrm{HBr}}=\frac{1.940\times10^{-20}}{1.60\times10^{-19}}\times100
\]
\[
P_{\mathrm{HBr}}\approx12.1\%
\]
\(\textbf{Interpret the comparison:}\)
HBr has the longer bond, but its molecular dipole is smaller.
This establishes that the difference in effective charge separation outweighs the difference in bond length.
The greater electronegativity of chlorine relative to bromine is consistent with the greater effective bond polarity of HCl.
\(\textbf{Identify the model limitation:}\)
Percentage ionic character does not represent a literal fraction of HCl or HBr molecules that are separate gaseous ions.
It is a conventional measure derived from the observed dipole and an idealized full-charge reference.
\(\textbf{Final answer:}\)
\[
\boxed{P_{\mathrm{HCl}}\approx17.7\%;\qquad P_{\mathrm{HBr}}\approx12.1\%}
\]
614. Sulfur trioxide and sulfur dioxide contain polar sulfur–oxygen bonds. Isolated gaseous \(\mathrm{SO_3}\) is trigonal planar, whereas gaseous \(\mathrm{SO_2}\) is bent. Consider the following statements.
Statement I: The presence of polar S–O bonds does not require \(\mathrm{SO_3}\) to possess a nonzero resultant permanent molecular dipole.
Statement II: In \(\mathrm{SO_3}\), three equivalent bond-dipole contributions can cancel by trigonal-planar symmetry.
Statement III: In bent \(\mathrm{SO_2}\), two equivalent S–O bond-dipole contributions cannot cancel by being antiparallel because the O–S–O angle is less than \(180^\circ\).
Statement IV: The difference in the molecular dipoles proves that sulfur must have different oxidation states in the two compounds.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and III only
Explanation: \(\textbf{Evaluate Statement I:}\)
A bond dipole is associated with unequal electron distribution along a bond.
A molecular dipole is the vector resultant of the contributions of the entire molecule.
Individual polar bonds can therefore exist within a molecule whose resultant permanent dipole is zero.
Gaseous sulfur trioxide provides such an example.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
The three oxygen atoms of isolated \(\mathrm{SO_3}\) occupy equivalent trigonal-planar positions.
The three S–O bond directions are separated by approximately \(120^\circ\).
Equivalent radial bond-dipole vectors cancel in this symmetric arrangement.
Thus, the molecular dipole can be zero despite significant polarity in the individual connections.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
Sulfur dioxide has a bent nuclear framework.
The two S–O bond-dipole vectors have an angle smaller than \(180^\circ\) between them.
Their components along the molecular-angle bisector reinforce rather than cancel.
Therefore, the molecule possesses a nonzero permanent dipole.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
Oxidation state is an electron-allocation convention applied to the chemical formula.
Sulfur has oxidation state \(+6\) in \(\mathrm{SO_3}\) and \(+4\) in \(\mathrm{SO_2}\).
Although the oxidation states are indeed different, that fact is not proved by the difference in molecular dipole moments.
Dipole moment depends on the molecular charge distribution and geometry, not uniquely on the central oxidation-state assignment.
Statement IV is therefore incorrect as a causal inference.
\(\textbf{Integrate the distinction:}\)
Bond polarity, molecular geometry and oxidation state are separate considerations.
An observed molecular dipole cannot independently establish a central atom's oxidation state.
Statements I, II and III are correct.
615. A hypothetical neutral molecule has the connectivity \(\mathrm{X-A-X}\). Each A–X bond possesses a bond-dipole vector of magnitude \(\mu_b\). The molecule undergoes a structural change from a bent arrangement with bond angle \(104^\circ\) to an ideal linear arrangement, while the individual bond-dipole magnitudes remain unchanged.
Which statement correctly identifies the effect of the change on the molecular dipole moment?
ⓐ. The linear form must have dipole moment \(2\mu_b\), because its two polar bonds lie on the same straight line.
ⓑ. The dipole moment cannot change because the two covalent bonds and their individual polarities remain unchanged.
ⓒ. The bent form must have zero dipole moment because the two terminal atoms are identical.
ⓓ. The bent form has a nonzero resultant \(2\mu_b\cos52^\circ\), whereas the linear form has zero resultant due to cancellation of the two equal antiparallel bond dipoles.
Correct Answer: The bent form has a nonzero resultant \(2\mu_b\cos52^\circ\), whereas the linear form has zero resultant due to cancellation of the two equal antiparallel bond dipoles.
Explanation: \(\textbf{Identify the fixed bond property:}\)
The two A–X bond-dipole magnitudes are equal and remain unchanged.
Thus, the structural transformation does not alter the stipulated local bond polarities.
Only the relative directions of the vectors change.
\(\textbf{Analyse the bent form:}\)
The angle between the two equal bond-dipole vectors is \(104^\circ\).
Their resultant lies along the angle bisector.
For equal vectors separated by angle \(\theta\):
\[
\mu_{\mathrm{resultant}}=2\mu_b\cos\left(\frac{\theta}{2}\right)
\]
Substituting:
\[
\mu_{\mathrm{bent}}=2\mu_b\cos\left(\frac{104^\circ}{2}\right)
\]
\[
\mu_{\mathrm{bent}}=2\mu_b\cos52^\circ
\]
This value is nonzero.
\(\textbf{Analyse the linear form:}\)
The bond angle becomes \(180^\circ\).
The equal bond-dipole vectors now point in opposite directions.
Their resultant is therefore zero.
\[
\mu_{\mathrm{linear}}=2\mu_b\cos90^\circ
\]
\[
\mu_{\mathrm{linear}}=0
\]
\(\textbf{Interpret the change:}\)
The molecular dipole disappears without either individual bond becoming nonpolar.
Its disappearance is entirely attributable to the changed spatial arrangement of the two bond vectors.
\(\textbf{Connect the result to molecular structure:}\)
A measured dipole moment can provide evidence concerning molecular symmetry.
However, dipole data alone cannot determine every bond angle unless additional information about individual bond dipoles is supplied.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{\mathrm{bent}}=2\mu_b\cos52^\circ;\qquad\mu_{\mathrm{linear}}=0}
\]
616. Water and oxygen difluoride have the structures \(\mathrm{H-O-H}\) and \(\mathrm{F-O-F}\), respectively. Both oxygen centres have two bonding domains and two lone-pair domains. Assume each molecule has a bent arrangement and that the electron-rich end of an individual bond lies towards the more electronegative bonded atom.
Consider the following statements.
Statement I: Both molecules have the conventional central-oxygen VSEPR classification \(\mathrm{AX_2E_2}\).
Statement II: The O–H bond-dipole vectors point towards oxygen, whereas the O–F bond-dipole vectors point towards fluorine.
Statement III: Although both molecules are bent, the resultant contributions from their two equivalent bond dipoles point in opposite senses relative to the central oxygen and terminal-atom sides.
Statement IV: Because both molecules possess two central lone pairs, their experimentally measured molecular dipole moments must be equal.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. II, III and IV only
ⓒ. I, III and IV only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: \(\textbf{Compare the oxygen electron domains:}\)
In water, oxygen forms two O–H sigma bonds and retains two lone pairs.
In oxygen difluoride, oxygen forms two O–F sigma bonds and also retains two lone pairs.
Both consequently contain four central electron domains.
Their conventional VSEPR classification is \(\mathrm{AX_2E_2}\).
Statement I is correct.
\(\textbf{Determine the O–H polarity direction:}\)
Oxygen is more electronegative than hydrogen.
The electron-rich end of each O–H bond therefore lies towards the central oxygen.
The two O–H bond-dipole contributions combine along the angle bisector towards the oxygen side.
\(\textbf{Determine the O–F polarity direction:}\)
Fluorine is more electronegative than oxygen.
The electron-rich end of each O–F bond consequently lies towards the terminal fluorine.
The two O–F bond-dipole contributions combine along their angle bisector towards the fluorine side.
Statement II is correct.
\(\textbf{Compare the resultant directions:}\)
The two molecules possess bent nuclear arrangements.
However, the directions of their bond-polarity contributions are reversed relative to the central atom.
The O–H contributions point towards the central oxygen side.
The O–F contributions point towards the terminal fluorine side.
Statement III is correct.
\(\textbf{Consider the complete molecular dipoles:}\)
Central lone-pair electron density also contributes to the overall charge distributions.
The different terminal elements change the bond polarities and electronic environments.
The bond angles and electron distributions need not be identical.
Therefore, the same number of lone pairs does not establish identical measured molecular dipole moments.
Statement IV is incorrect.
\(\textbf{Integrate the comparison:}\)
VSEPR electron-domain classification determines a useful qualitative geometry.
Electronegativity determines the direction of local bond polarization.
Complete molecular dipole moment requires consideration of both geometry and total electron distribution.
Statements I, II and III are correct.
617. Ozone, \(\mathrm{O_3}\), has three oxygen atoms and two equivalent O–O connections in its resonance hybrid. A student argues that its permanent molecular dipole must be zero because the electronegativity difference across each O–O bond is zero.
Which statement best evaluates the argument?
ⓐ. The claim is false: bent ozone has different central and terminal electron environments and resonance-separated charges despite identical atoms.
ⓑ. The zero elemental electronegativity differences necessarily force a zero molecular dipole despite ozone's bent structure.
ⓒ. Resonance makes the complete electron distribution symmetric about the central atom, cancelling every possible dipole.
ⓓ. The dipole arises only because one terminal oxygen permanently carries a different charge from the other terminal oxygen.
Correct Answer: The claim is false: bent ozone has different central and terminal electron environments and resonance-separated charges despite identical atoms.
Explanation: \(\textbf{Identify the student's assumption:}\)
An isolated bond between two identical atoms has no electronegativity difference arising from elemental identity.
However, a polyatomic molecule's permanent dipole depends on its complete three-dimensional electron distribution.
It cannot always be determined by treating each connection as an isolated diatomic bond.
\(\textbf{Examine ozone's nuclear geometry:}\)
Ozone has a bent O–O–O framework.
The central oxygen and two terminal oxygens occupy different structural environments.
The molecule therefore does not possess the symmetry of a linear, equivalent-ended diatomic arrangement.
\(\textbf{Examine the resonance contributors:}\)
Conventional octet-satisfying ozone contributors contain one O=O and one O–O connection.
The central oxygen carries positive formal charge.
The singly bonded terminal oxygen carries negative formal charge.
The two equivalent contributors exchange which terminal position receives the localized negative formal-charge assignment.
\(\textbf{Interpret the resonance hybrid:}\)
The two terminal oxygen positions are equivalent to one another.
However, the central oxygen position remains different from either terminal position.
The resonance hybrid consequently has a nonuniform electronic distribution.
The equivalent negative-charge contributions on the bent terminal directions can produce a nonzero resultant.
\(\textbf{Distinguish formal charge from oxidation state:}\)
All three atoms are oxygen in an elemental allotrope.
Their conventional oxidation state is zero.
Nonzero formal charges in resonance drawings do not require different elemental oxidation states.
Nor do they establish the presence of independent oxide ions.
\(\textbf{Identify the general lesson:}\)
Electronegativity difference is useful for describing local bond polarity.
Molecular polarity depends more generally on the complete charge distribution and its symmetry.
Therefore, even a molecule containing only atoms of one element can possess a permanent dipole when its inequivalent atomic positions and electron distribution permit one.
Option A correctly explains the limitation of the student's argument.
618. Lithium oxide and beryllium oxide contain oxide ions in their conventional ionic descriptions. Consider the comparative effects of changing from monovalent \(\mathrm{Li^+}\) to divalent \(\mathrm{Be^{2+}}\).
Beryllium forms a smaller, more highly charged cation than lithium. Oxygen is represented as \(\mathrm{O^{2-}}\) in both formal ionic descriptions. Assume the comparison concerns qualitative polarizing power and covalent character, not exact measured lattice enthalpies or melting temperatures.
Which conclusion is most defensible?
ⓐ. The lithium cation must distort the oxide electron cloud more strongly because its lower positive charge produces less electrostatic repulsion.
ⓑ. Smaller \(\mathrm{Be^{2+}}\) polarizes oxide more strongly and favours covalency, but this alone does not rank melting temperatures.
ⓒ. Both substances must contain identical degrees of covalent character because their anions possess the same electronic configuration.
ⓓ. Beryllium oxide must consist exclusively of isolated neutral BeO molecules because any polarization converts an ionic lattice completely into a molecular solid.
Correct Answer: Smaller \(\mathrm{Be^{2+}}\) polarizes oxide more strongly and favours covalency, but this alone does not rank melting temperatures.
Explanation: \(\textbf{Identify the common anion:}\)
Both conventional ionic descriptions involve oxide ions.
The anion electron-cloud characteristics are therefore held broadly constant in the comparison.
The decisive change concerns the cation.
\(\textbf{Compare the cation charges:}\)
Lithium forms a monovalent cation.
Beryllium forms a divalent cation.
The greater positive charge on beryllium increases its ability to attract and distort neighbouring electron density.
\(\textbf{Compare the cation sizes:}\)
The beryllium cation is smaller than the lithium cation.
A smaller radius concentrates positive charge into a smaller region.
This further increases the polarizing power of \(\mathrm{Be^{2+}}\).
\(\textbf{Apply Fajan's rule:}\)
An anion's electron cloud can become distorted in the field of a strongly polarizing cation.
Greater distortion contributes covalent character to an interaction that may also be described using formal ionic charges.
The combined small size and higher charge of \(\mathrm{Be^{2+}}\) therefore favour greater covalent character in beryllium oxide.
\(\textbf{Separate polarization from melting behaviour:}\)
The melting temperature depends on the thermodynamics of the solid and liquid phases.
Extended crystal interactions, cohesive energies and entropy changes can all matter.
A trend towards greater covalent character does not by itself establish a lower melting temperature.
\(\textbf{Avoid the molecular-solid misconception:}\)
Partial covalent character does not mean that the substance must consist of discrete neutral molecules.
An extended solid may contain bonding interactions with both ionic and covalent characteristics.
\(\textbf{Integrate the conclusion:}\)
Fajan's rule predicts a strong polarization tendency for the beryllium ion.
The rule is not a universal numerical formula for lattice energy, melting temperature or molecular aggregation.
Option B correctly states the supported inference without extending it beyond the information provided.
619. Three hypothetical elements P, Q and R have electronegativities \(2.0\), \(2.8\) and \(3.6\), respectively. Two linear molecules are constructed with fixed, outward-directed bond-dipole contributions:
Molecule I has the connectivity P–Q–R, with P–Q bond-dipole magnitude \(0.80\ \mathrm D\) and Q–R magnitude \(1.10\ \mathrm D\). Molecule II has the connectivity Q–P–R, with P–Q magnitude \(0.80\ \mathrm D\) and P–R magnitude \(1.50\ \mathrm D\). Each individual bond-dipole vector points towards the more electronegative atom. Treat both molecules as collinear, with no additional dipole contributions.
Which statement correctly gives the two resultant molecular dipole magnitudes and explains the different addition patterns?
ⓐ. Molecule I has \(0.30\ \mathrm D\), and Molecule II has \(2.30\ \mathrm D\), because bond vectors always point towards the central atom.
ⓑ. Both molecules have \(1.90\ \mathrm D\), because they contain the same three elements and all their bonds are polar.
ⓒ. Molecule I has \(1.90\ \mathrm D\), while Molecule II has \(0.70\ \mathrm D\); the first arrangement reinforces its bond vectors, whereas the second produces opposing bond vectors.
ⓓ. Molecule I has \(0.30\ \mathrm D\), while Molecule II has \(0.70\ \mathrm D\); linearity always requires subtraction of bond-dipole magnitudes.
Correct Answer: Molecule I has \(1.90\ \mathrm D\), while Molecule II has \(0.70\ \mathrm D\); the first arrangement reinforces its bond vectors, whereas the second produces opposing bond vectors.
Explanation: \(\textbf{Determine the polarity directions:}\)
Electronegativity increases in the order:
\[
\chi_P\lt \chi_Q\lt \chi_R
\]
The individual bond-dipole vectors point towards the more electronegative bonded atoms under the convention specified.
Their directions must be assigned separately for each molecular connectivity.
\(\textbf{Analyse Molecule I:}\)
The nuclear sequence is P–Q–R.
The P–Q bond dipole points from P towards Q.
The Q–R bond dipole points from Q towards R.
Both vectors point in the same overall direction along the molecular axis.
\(\textbf{Calculate the first resultant:}\)
\[
\mu_I=\mu_{\mathrm{P-Q}}+\mu_{\mathrm{Q-R}}
\]
\[
\mu_I=0.80+1.10
\]
\[
\mu_I=1.90\ \mathrm D
\]
The resultant points towards the R end.
\(\textbf{Analyse Molecule II:}\)
The nuclear sequence is Q–P–R.
The P–Q dipole points from P towards Q, which is towards the left.
The P–R dipole points from P towards R, which is towards the right.
The vectors therefore oppose one another.
\(\textbf{Calculate the second resultant:}\)
Taking the R direction as positive:
\[
\mu_{II,\mathrm{signed}}=1.50-0.80
\]
\[
\mu_{II,\mathrm{signed}}=+0.70\ \mathrm D
\]
Hence:
\[
\mu_{II}=0.70\ \mathrm D
\]
Its resultant also points towards the R end.
\(\textbf{Interpret the connectivity effect:}\)
Both molecules contain the same three elemental identities.
However, their middle atoms differ.
The electron-attracting direction of an individual bond is determined by its two bonded atoms, while the molecular vector sum is determined by their spatial arrangement.
The first molecule reinforces the contributions and the second partially cancels them.
\(\textbf{Identify the key misconception:}\)
Linearity alone does not require bond dipoles to cancel.
Two collinear bond-dipole vectors can reinforce if they point in the same direction.
The familiar cancellation in a symmetric \(\mathrm{X-A-X}\) molecule also requires equal oppositely directed contributions.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_I=1.90\ \mathrm D;\qquad\mu_{II}=0.70\ \mathrm D}
\]
620. Two gaseous molecules contain the same central atom A and three identical terminal atoms X. Each A–X bond is polar, and each terminal atom is connected to A through one sigma bond. The experimentally inferred structures are:
Molecule P is trigonal planar with three equivalent A–X directions \(120^\circ\) apart. Molecule Q is trigonal pyramidal with three equivalent A–X directions directed away from one central lone-pair region. Assume both species have nonzero individual bond-dipole magnitudes.
Which statement correctly distinguishes the central electron-domain populations and the symmetry-based bond-dipole resultants?
ⓐ. P has three bond domains and no lone pair, permitting dipole cancellation; Q has three bonds and one lone pair, preventing equivalent cancellation.
ⓑ. Both molecules have zero molecular dipole because three equivalent bonds always cancel regardless of molecular shape.
ⓒ. P must have a larger molecular dipole because trigonal-planar geometry places the bond dipoles closer together than trigonal-pyramidal geometry.
ⓓ. Q must have only three total electron domains because a lone pair is not included in VSEPR counting.
Correct Answer: P has three bond domains and no lone pair, permitting dipole cancellation; Q has three bonds and one lone pair, preventing equivalent cancellation.
Explanation: \(\textbf{Examine the trigonal-planar case:}\)
Molecule P has three sigma-bonding domains around A.
Its three equivalent bond directions lie in one plane.
The directions are separated by \(120^\circ\).
Equal vectors along these directions sum to zero.
\(\textbf{Identify P's central electron-domain population:}\)
The supplied ideal trigonal-planar shape is consistent with an \(\mathrm{AX_3}\) central environment.
No central lone-pair domain is present in this stated model.
The electron-domain arrangement and the molecular arrangement are both trigonal planar.
\(\textbf{Examine the pyramidal case:}\)
Molecule Q has three bonded terminal atoms.
It also has a lone-pair region on the central atom.
Its electron-domain population is therefore four.
The corresponding VSEPR description is \(\mathrm{AX_3E}\).
\(\textbf{Analyse Q's bond-dipole vectors:}\)
The three bond vectors are symmetrically distributed around the pyramidal axis.
Their components perpendicular to this axis cancel.
However, their components along the axis reinforce.
Therefore, the three equivalent bond-dipole vectors have a nonzero resultant in the stipulated geometry.
\(\textbf{Distinguish bond-vector and full-molecule results:}\)
The complete molecular dipole may also contain contributions associated with nonbonding electron density.
Consequently, an exact molecular dipole magnitude cannot be calculated from geometry alone.
Nevertheless, P has the symmetry required for complete bond-vector cancellation, whereas Q does not.
\(\textbf{Integrate the electronic and spatial arguments:}\)
Both species possess three polar bonds.
Their electron-domain populations differ because only Q possesses a central lone pair.
This changes the molecular shape and the vector cancellation condition.
Option A correctly combines the VSEPR and dipole inferences.
