101. A series of carbon–halogen bonds is considered. The electronegativities of carbon, chlorine and fluorine are \(2.5\), \(3.2\) and \(4.0\), respectively. The C–F bond is shorter than the corresponding C–Cl bond in comparable molecular environments. Which conclusion is justified?
ⓐ. C–Cl has the greater electronegativity difference, while C–F is the shorter bond.
ⓑ. C–F has the greater electronegativity difference, so its molecular dipole contribution must be exactly twice that of C–Cl.
ⓒ. C–F has the greater electronegativity difference and the shorter bond, but the relative dipole magnitudes require additional information.
ⓓ. C–Cl has the greater electronegativity difference and the longer bond, so it must have the greater dipole magnitude.
Correct Answer: C–F has the greater electronegativity difference and the shorter bond, but the relative dipole magnitudes require additional information.
Explanation: \(\textbf{Compare the bond electronegativity differences:}\)
For C–F:
\[
\Delta\chi_{\mathrm{C-F}}=4.0-2.5
\]
\[
\Delta\chi_{\mathrm{C-F}}=1.5
\]
For C–Cl:
\[
\Delta\chi_{\mathrm{C-Cl}}=3.2-2.5
\]
\[
\Delta\chi_{\mathrm{C-Cl}}=0.7
\]
Therefore, the electronegativity difference is greater for C–F.
\(\textbf{Identify the polarity direction:}\)
Fluorine and chlorine are both more electronegative than carbon.
Consequently, carbon is the relatively positive end of either bond.
The halogen is the relatively negative end.
The greater difference for C–F indicates a stronger tendency towards unequal electron sharing.
\(\textbf{Include the bond-length information:}\)
The question specifies that C–F is shorter than the corresponding C–Cl bond.
A shorter bond provides a smaller charge-separation distance.
However, the dipole-moment magnitude depends on both effective charge and separation.
\[
\mu=qr
\]
The electronegativity values alone do not provide exact effective charges.
Therefore, the numerical dipole contributions cannot be calculated from the given differences and qualitative bond-length ordering.
\(\textbf{Avoid the proportionality misconception:}\)
The electronegativity difference of C–F is not exactly twice that of C–Cl.
More importantly, even a simple numerical ratio of electronegativity differences would not establish the same ratio of dipole moments.
The electronic distributions of the two bonds are not determined by electronegativity difference alone.
Bond length and the molecular environment must also be considered.
Thus, option C correctly identifies the stronger electronegativity-based polarity tendency of C–F without making an unsupported quantitative dipole claim.
102. Consider the two molecular species \(\mathrm{H_2O}\) and \(\mathrm{OF_2}\). The electronegativity values are:
\[
\chi_{\mathrm H}=2.2,\qquad\chi_{\mathrm O}=3.5,\qquad\chi_{\mathrm F}=4.0
\]
Both molecules contain two single bonds involving oxygen. Which statement correctly compares oxygen's relative partial-charge character in the individual bonds?
ⓐ. Oxygen is relatively negative in O–H bonds but relatively positive in O–F bonds.
ⓑ. Oxygen is relatively positive in both O–H and O–F bonds.
ⓒ. Oxygen is relatively negative in both O–H and O–F bonds.
ⓓ. Oxygen is relatively positive in O–H bonds but relatively negative in O–F bonds.
Correct Answer: Oxygen is relatively negative in O–H bonds but relatively positive in O–F bonds.
Explanation: \(\textbf{Examine oxygen bonded to hydrogen:}\)
Oxygen is more electronegative than hydrogen.
\[
\chi_{\mathrm O}=3.5\gt 2.2=\chi_{\mathrm H}
\]
Therefore, oxygen attracts the shared O–H bonding electron density more strongly.
In the O–H bonds of water, oxygen has relative partial negative character.
Hydrogen has relative partial positive character.
\(\textbf{Examine oxygen bonded to fluorine:}\)
Fluorine is more electronegative than oxygen.
\[
\chi_{\mathrm F}=4.0\gt 3.5=\chi_{\mathrm O}
\]
Therefore, fluorine attracts the shared O–F bonding electron density more strongly.
In oxygen difluoride, oxygen has relative partial positive character compared with fluorine.
\(\textbf{Interpret the change in polarity direction:}\)
The same element can be the more electronegative member of one bond and the less electronegative member of another.
Oxygen lies between hydrogen and fluorine in the supplied electronegativity ordering.
Consequently, the relative partial-charge direction changes when oxygen is bonded to fluorine instead of hydrogen.
\(\textbf{Distinguish bond polarity from formal charge:}\)
The conventional Lewis structures of both molecules contain two single bonds and two lone pairs on oxygen.
Oxygen consequently has formal charge zero in either neutral molecule.
The existence of relative partial charges does not contradict this formal-charge assignment.
Formal charge is calculated by assigning half of the bonding electrons to each bonded atom.
Partial-charge character reflects the unequal distribution of actual electron density.
\(\textbf{Extend the inference carefully:}\)
The direction of each bond's polarisation can be established using electronegativity.
The exact molecular dipole moment also depends on the three-dimensional arrangement and the complete electron distribution.
Therefore, bond-polarity direction should not be confused with a complete quantitative description of molecular polarity.
Option A correctly identifies the reversal of oxygen's relative partial-charge character.
103. Consider the following statements about electronegativity trends.
Statement I: Electronegativity generally increases from left to right across a period because effective nuclear attraction increases and atomic size generally decreases.
Statement II: Electronegativity generally decreases down a group because the valence shell becomes more distant from the nucleus and shielding increases.
Statement III: The first electron-gain enthalpy must always become more negative in exactly the same order as electronegativity increases.
Statement IV: Electronegativity is a relative property of bonded atoms, so it should not be identified directly with an isolated-atom electron-addition enthalpy.
Which statements are correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
Statement I is correct.
Across a period, the nuclear charge increases while the additional electrons generally enter the same principal shell.
The effective nuclear attraction experienced by valence electrons consequently increases.
Atomic size generally decreases.
These changes increase the tendency of a bonded atom to attract shared electron density.
Therefore, electronegativity generally increases across a period.
\(\textbf{Evaluate Statement II:}\)
Statement II is correct.
Down a group, additional electron shells are introduced.
The valence electrons become more distant from the nucleus.
Inner-shell shielding also increases.
These effects generally reduce the attraction exerted on shared bonding electron density.
Consequently, electronegativity usually decreases down a group.
\(\textbf{Evaluate Statement III:}\)
Statement III is incorrect.
Electron-gain enthalpy depends on the enthalpy change accompanying electron addition to an isolated gaseous atom.
Electronegativity refers to electron attraction by a bonded atom.
The trends may be related, but they are not required to follow an identical order in every case.
Fluorine is more electronegative than chlorine, even though chlorine has a more negative first electron-gain enthalpy.
Therefore, a universal identical ordering is not justified.
\(\textbf{Evaluate Statement IV:}\)
Statement IV is correct.
The two properties represent different physical concepts.
A numerical electronegativity value is not simply the electron-gain enthalpy of the isolated element expressed in different units.
Bonding environment, relative electron attraction and the adopted electronegativity scale are relevant to its interpretation.
\(\textbf{Overall conclusion:}\)
Statements I and II correctly describe the general periodic trends and their physical basis.
Statement III incorrectly converts a broad trend into an exception-free equivalence.
Statement IV correctly identifies the conceptual distinction.
Hence, Statements I, II and IV are correct.
104. An ionic compound contains a small, highly charged cation and a large anion with a diffuse electron cloud. According to Fajan's rule, which interpretation of their interaction is most appropriate?
ⓐ. The cation is highly polarizable, and the anion has low polarizing power; covalent character necessarily decreases.
ⓑ. The cation has high polarizing power, and the anion has high polarizability; covalent character increases.
ⓒ. The cation has high polarizability, and the anion has high polarizing power; covalent character increases.
ⓓ. The cation has low polarizing power, and the anion has low polarizability; covalent character necessarily increases.
Correct Answer: The cation has high polarizing power, and the anion has high polarizability; covalent character increases.
Explanation: Fajan's rule relates the distortion of an anion's electron cloud to the development of covalent character in a bond that may initially be described as ionic.
The positively charged cation produces an electric field that can distort the electron distribution of a neighbouring anion.
The ability of the cation to cause such distortion is called its polarizing power.
A smaller cation generally has greater polarizing power because its positive charge is concentrated within a smaller region.
A higher positive ionic charge also generally increases polarizing power.
The anion's tendency to undergo electron-cloud distortion is described by its polarizability.
A larger anion usually has a more diffuse electron cloud.
Its outer electrons are less tightly held and can be distorted more readily.
Consequently, a large anion is generally more polarizable than a small anion of comparable charge.
\(\textbf{Connect the two properties:}\)
A small, highly charged cation can strongly distort the electron cloud of a large, polarizable anion.
This distortion increases the electron density between the two ions and introduces additional covalent character.
Therefore, the bond becomes less accurately represented by a perfectly ionic electron-distribution model.
Polarizing power and polarizability are not interchangeable terms.
The first concerns the ability to distort another ion's electron cloud.
The second concerns the ease with which an ion's own electron cloud is distorted.
The compound may still be described as an ionic substance overall while exhibiting appreciable covalent character.
Thus, option B correctly identifies the controlling properties and their bonding consequence.
105. Consider the chlorides \(\mathrm{LiCl}\), \(\mathrm{NaCl}\) and \(\mathrm{KCl}\). The cations are monovalent, and the anion is identical in all three compounds. According to Fajan's rule, which order of decreasing covalent character is expected?
ⓐ. \(\mathrm{KCl\gt NaCl\gt LiCl}\)
ⓑ. \(\mathrm{NaCl\gt LiCl\gt KCl}\)
ⓒ. \(\mathrm{LiCl\gt NaCl\gt KCl}\)
ⓓ. \(\mathrm{LiCl\gt KCl\gt NaCl}\)
Correct Answer: \(\mathrm{LiCl\gt NaCl\gt KCl}\)
Explanation: All three compounds contain cations carrying a charge of \(+1\).
They also contain the same chloride anion.
Therefore, the anion's charge and its intrinsic polarizability remain approximately unchanged across the series.
The principal changing factor is the size of the metal cation.
Lithium, sodium and potassium belong to Group 1.
The ionic radius increases down the group.
\[
r(\mathrm{Li^+})\lt r(\mathrm{Na^+})\lt r(\mathrm{K^+})
\]
A smaller cation produces a more concentrated electric field near the anion.
Consequently, the polarizing power decreases from lithium to potassium.
\[
\text{Polarizing power: }\mathrm{Li^+\gt Na^+\gt K^+}
\]
\(\textbf{Connect polarization with bonding character:}\)
The lithium ion can distort the chloride electron cloud more effectively than the larger sodium and potassium ions.
Greater electron-cloud distortion produces greater sharing of electron density between the ions.
According to Fajan's rule, this increases covalent character.
Therefore, LiCl is expected to have the greatest covalent character among the three compounds.
KCl is expected to have the least.
The comparison does not imply that LiCl is composed entirely of neutral covalent molecules under every physical condition.
Ionic and covalent character can coexist.
It also does not imply that greater covalent character necessarily produces a smaller magnitude of every lattice-energy contribution.
Covalent character describes electron-density distortion, whereas lattice enthalpy describes an energetic change involving the entire ionic solid.
The correct decreasing covalent-character order is \(\mathrm{LiCl\gt NaCl\gt KCl}\).
106. Consider the compounds \(\mathrm{AlF_3}\), \(\mathrm{AlCl_3}\), \(\mathrm{AlBr_3}\) and \(\mathrm{AlI_3}\). Assume the usual qualitative Fajan's-rule comparison for the same \(\mathrm{Al^{3+}}\) cation. Which statement correctly predicts the effect of changing the halide ion from fluoride to iodide?
ⓐ. Covalent character generally increases because anion polarizability increases down the halogen group.
ⓑ. Covalent character generally decreases because anion polarizability increases down the halogen group.
ⓒ. Covalent character generally increases because the charge on the halide ion increases down the group.
ⓓ. Covalent character generally decreases because the polarizing power of the aluminium ion decreases.
Correct Answer: Covalent character generally increases because anion polarizability increases down the halogen group.
Explanation: The same aluminium cation is present throughout the comparison.
Therefore, its formal ionic charge remains \(+3\).
The principal changing factor is the identity of the halide anion.
Fluoride, chloride, bromide and iodide ions all carry a single negative charge.
Their ionic radii increase down Group 17.
\[
r(\mathrm{F^-})\lt r(\mathrm{Cl^-})\lt r(\mathrm{Br^-})\lt r(\mathrm{I^-})
\]
The larger halide ions possess more extended electron clouds.
Their outer electrons can be distorted more readily by a neighbouring cation.
Consequently, their polarizability generally increases down the group.
\(\textbf{Apply Fajan's rule:}\)
The \(\mathrm{Al^{3+}}\) ion possesses a relatively high positive charge and substantial polarizing power.
When it interacts with a more polarizable anion, greater distortion of the anion's electron cloud becomes possible.
Greater polarization introduces additional covalent character into the bonding.
Thus, the expected covalent-character trend is:
\[
\mathrm{AlI_3\gt AlBr_3\gt AlCl_3\gt AlF_3}
\]
The comparison does not arise from an increase in the formal negative charge of the halide ions.
All four halide ions remain monovalent.
Nor does the aluminium cation's formal charge decrease as the halide is changed.
The trend arises principally from increasing anion size and polarizability.
Actual solid-state structures and physical properties may also depend on other factors.
Therefore, a Fajan's-rule ranking should not be treated as a complete prediction of every compound property.
Option A correctly identifies both the trend and its underlying reason.
107. Four hypothetical compounds contain the cations and anions described in the table. Assume that the stated ions are comparable in their electronic-shell characteristics apart from the listed size and charge differences.
| Compound | Cation charge | Cation radius | Anion charge | Anion radius |
|---|
| P | \(+1\) | \(110\ \mathrm{pm}\) | \(-1\) | \(130\ \mathrm{pm}\) |
| Q | \(+2\) | \(70\ \mathrm{pm}\) | \(-1\) | \(130\ \mathrm{pm}\) |
| R | \(+2\) | \(70\ \mathrm{pm}\) | \(-1\) | \(200\ \mathrm{pm}\) |
| S | \(+1\) | \(110\ \mathrm{pm}\) | \(-1\) | \(200\ \mathrm{pm}\) |
Using qualitative Fajan's-rule trends without assuming an exact quantitative polarization formula, which conclusion is justified?
ⓐ. Q has the greatest covalent character, while S has the least; R and P cannot be compared.
ⓑ. S has the greatest covalent character, while Q has the least; P and R cannot be compared.
ⓒ. R has the greatest covalent character, while P has the least; Q must exceed S.
ⓓ. R has the greatest covalent character, while P has the least; Q and S cannot be ranked uniquely.
Correct Answer: R has the greatest covalent character, while P has the least; Q and S cannot be ranked uniquely.
Explanation: \(\textbf{Identify the cation-polarizing factors:}\)
A smaller cation generally has greater polarizing power.
A higher positive ionic charge also increases polarizing power.
Compounds Q and R contain cations with charge \(+2\) and radius \(70\ \mathrm{pm}\).
Compounds P and S contain cations with charge \(+1\) and radius \(110\ \mathrm{pm}\).
Therefore, the cations in Q and R are stronger polarizers under the stated assumptions.
\(\textbf{Identify the anion-polarizability factors:}\)
All four anions carry charge \(-1\).
Their sizes differ.
The anions in R and S have radius \(200\ \mathrm{pm}\), whereas those in P and Q have radius \(130\ \mathrm{pm}\).
A larger anion generally has a more easily distorted electron cloud.
Therefore, the anions in R and S are more polarizable.
\(\textbf{Identify the strongest combined polarization case:}\)
Compound R combines the smaller, more highly charged cation with the larger, more polarizable anion.
Both relevant trends favour stronger polarization.
Consequently, R is expected to possess the greatest covalent character.
\(\textbf{Identify the weakest combined polarization case:}\)
Compound P combines the larger, less highly charged cation with the smaller, less polarizable anion.
Both trends favour weaker polarization.
Therefore, P is expected to have the least covalent character.
\(\textbf{Examine the competing middle cases:}\)
Compound Q has a stronger polarizing cation but a less polarizable anion.
Compound S has a weaker polarizing cation but a more polarizable anion.
These factors oppose one another.
Without a justified quantitative model or additional experimental information, the two compounds cannot be ranked uniquely.
\(\textbf{Interpret the conclusion:}\)
The available information establishes a strongest and weakest case without establishing a complete ordering.
Assigning Q above S merely because its cation has greater charge would ignore the competing anion-polarizability difference.
Assigning S above Q solely because it has the larger anion would ignore the stronger cation in Q.
Thus, option D is the defensible partial ranking.
108. Sodium chloride and silver chloride are compared using Fajan's rule. Sodium ions possess a noble-gas electronic configuration, whereas silver ions possess a filled \(d^{10}\) electronic configuration. Which statement most appropriately explains why AgCl has appreciably greater covalent character than NaCl?
ⓐ. Silver ions have a larger positive formal charge than sodium ions.
ⓑ. Silver chloride contains doubly negative chloride ions, unlike sodium chloride.
ⓒ. The \(d^{10}\) configuration of \(\mathrm{Ag^+}\) provides relatively poor shielding, contributing to stronger polarization of chloride.
ⓓ. Sodium ions share more electron density with chloride because their valence shells contain no d electrons.
Correct Answer: The \(d^{10}\) configuration of \(\mathrm{Ag^+}\) provides relatively poor shielding, contributing to stronger polarization of chloride.
Explanation: Both sodium chloride and silver chloride contain monovalent cations and chloride anions.
The formal ionic charges are therefore the same in the two compounds.
Sodium chloride contains \(\mathrm{Na^+}\), whose electronic configuration is:
\[
\mathrm{Na^+}:[Ne]
\]
Silver chloride contains \(\mathrm{Ag^+}\), whose electronic configuration is:
\[
\mathrm{Ag^+}:[Kr]4d^{10}
\]
The electronic configuration of a cation can influence its ability to polarize an anion.
Filled \(d\) subshells generally provide less effective shielding of nuclear charge than corresponding \(s\) and \(p\) electronic shells.
Consequently, cations with filled \(d^{10}\) configurations can exert appreciable polarizing effects.
\(\textbf{Connect electronic configuration with covalent character:}\)
The chloride ion has a relatively polarizable electron cloud.
The polarizing influence of \(\mathrm{Ag^+}\) can distort this cloud and increase the concentration of electron density between silver and chlorine.
This produces appreciable covalent character in AgCl.
Sodium chloride, by comparison, is more closely described by an ionic bonding model.
The distinction cannot be explained by assigning different formal charges to sodium and silver.
Both cations carry charge \(+1\).
Likewise, chloride remains singly negative in both compounds.
The comparison illustrates a qualification to simple cation-radius and charge rules.
Electronic configuration and shielding can affect polarizing power in addition to nominal ionic size and charge.
Therefore, option C correctly identifies the electronic explanation for the greater covalent character of silver chloride.
109. Consider the lithium halides LiF and LiI. Assume that their electrostatic lattice interactions can be compared qualitatively using the same monovalent cation and similar ionic structural considerations. Which statement correctly distinguishes the expected lattice-stabilisation and covalent-character trends?
ⓐ. LiI has stronger electrostatic lattice stabilisation and greater covalent character because iodide is larger.
ⓑ. LiF has stronger electrostatic lattice stabilisation, while LiI has greater covalent character because iodide is more polarizable.
ⓒ. LiF has stronger electrostatic lattice stabilisation and greater covalent character because fluoride is smaller.
ⓓ. LiI has stronger electrostatic lattice stabilisation, while LiF has greater covalent character because fluoride is less polarizable.
Correct Answer: LiF has stronger electrostatic lattice stabilisation, while LiI has greater covalent character because iodide is more polarizable.
Explanation: \(\textbf{Compare the ionic charges:}\)
Both lithium fluoride and lithium iodide contain the same monovalent lithium cation.
Their halide anions also carry the same charge magnitude.
Therefore, the ionic-charge products are identical in the two compounds.
\(\textbf{Compare electrostatic lattice stabilisation:}\)
The fluoride ion is smaller than the iodide ion.
Consequently, the representative separation between lithium and fluoride ions is smaller than that between lithium and iodide ions in comparable ionic arrangements.
Shorter separation generally favours stronger electrostatic attraction.
Therefore, LiF is expected to have the greater electrostatic lattice-stabilisation magnitude.
\(\textbf{Compare anion polarizability:}\)
The iodide ion is substantially larger than the fluoride ion.
Its electron cloud is more extended and can be distorted more readily.
Therefore, iodide is more polarizable than fluoride.
\(\textbf{Apply Fajan's rule:}\)
The lithium cation is small and possesses appreciable polarizing power.
It can distort the larger iodide electron cloud more readily than the compact fluoride electron cloud.
Greater polarization introduces additional covalent character.
Consequently, LiI exhibits greater covalent character than LiF.
\(\textbf{Interpret the opposite trends:}\)
The smaller fluoride ion favours stronger idealised electrostatic lattice attraction.
The larger iodide ion favours greater electron-cloud distortion and covalent character.
These trends arise from different aspects of chemical bonding.
They need not rank the two compounds in the same order.
Lattice enthalpy is a thermochemical quantity associated with forming or separating a crystalline ionic arrangement.
Fajan's rule concerns the extent to which the electron cloud is distorted and the bonding acquires covalent character.
Thus, a compound can possess relatively strong electrostatic lattice interactions without possessing the greatest covalent character.
Option B correctly preserves both distinctions.
110. Two hypothetical salts contain the same large, polarizable monovalent anion X. Salt P contains a monovalent cation \(\mathrm{M^+}\), whereas salt Q contains a divalent cation \(\mathrm{N^{2+}}\). Assume that the two cations have comparable ionic radii and electronic-shell characteristics, and that suitable neutral ionic compositions are formed. Which conclusion is most appropriate when the cation charge is increased under these controlled conditions?
ⓐ. Electrostatic attraction becomes stronger, but the cation's polarizing power necessarily decreases.
ⓑ. Electrostatic attraction becomes weaker, while the cation's polarizing power necessarily increases.
ⓒ. Both electrostatic attraction and polarizing power decrease because the anion is unchanged.
ⓓ. Both electrostatic attraction and polarizing power generally increase, so lattice attraction and covalent character may strengthen simultaneously.
Correct Answer: Both electrostatic attraction and polarizing power generally increase, so lattice attraction and covalent character may strengthen simultaneously.
Explanation: \(\textbf{Examine the ionic-charge change:}\)
The anion carries charge \(-1\) in both compounds.
The cation charge increases from \(+1\) to \(+2\).
For an idealised ionic pair, the electrostatic attraction depends on the product of the ionic-charge magnitudes.
At comparable separation, a greater charge product favours stronger electrostatic attraction.
The charge-product magnitude increases from one to two.
Therefore, stronger electrostatic attraction is expected under the stated assumptions.
\(\textbf{Examine the polarizing power:}\)
Polarizing power is the ability of a cation to distort a neighbouring anion's electron cloud.
For cations of comparable size and electronic-shell character, increasing the positive charge generally increases polarizing power.
The more highly charged cation can produce a stronger electric field near the anion.
\(\textbf{Include the anion's response:}\)
The anion is specified to be large and polarizable.
Its electron cloud can therefore be distorted appreciably.
When the cation's polarizing power increases, the extent of anion polarization can increase.
According to Fajan's rule, greater polarization introduces additional covalent character into the bond.
\(\textbf{Resolve the apparent contradiction:}\)
Stronger electrostatic attraction and greater covalent character are not mutually exclusive descriptions.
The former concerns the attraction associated with ionic charges.
The latter concerns distortion and redistribution of electron density.
A stronger ionic field can contribute to both effects.
Therefore, the greater charge of the divalent cation does not require the bonding to become perfectly ionic.
It can produce stronger electrostatic interactions while also increasing covalent character.
\(\textbf{Apply the necessary qualification:}\)
The two salts have different empirical compositions because one contains a monovalent cation and the other a divalent cation.
Their actual lattice enthalpies cannot be ranked quantitatively without considering the complete ionic arrangements and other structural factors.
Nevertheless, the direction of the local electrostatic and polarizing effects follows from the controlled assumptions.
Thus, option D correctly integrates ionic attraction with Fajan's-rule polarization.
111. Two hypothetical molecules, P and Q, each contain two identical polar A–B bonds. The magnitude of each bond dipole is \(1.20\ \mathrm{D}\).
Molecule P has the linear arrangement B–A–B, with a B–A–B angle of \(180^\circ\). Molecule Q has the bent arrangement B–A–B, with a B–A–B angle of \(120^\circ\). In both molecules, B is more electronegative than A, and both A–B bond dipoles point from A towards B. The individual bond-dipole magnitudes are identical in P and Q.
Which pair correctly gives the resultant molecular dipole moments of P and Q, respectively?
ⓐ. \(0\ \mathrm{D}\) and \(2.40\ \mathrm{D}\)
ⓑ. \(1.20\ \mathrm{D}\) and \(1.20\ \mathrm{D}\)
ⓒ. \(0\ \mathrm{D}\) and \(1.20\ \mathrm{D}\)
ⓓ. \(0\ \mathrm{D}\) and \(2.08\ \mathrm{D}\)
Correct Answer: \(0\ \mathrm{D}\) and \(1.20\ \mathrm{D}\)
Explanation: \(\textbf{Identify the vector nature of dipole moment:}\)
A bond dipole has both magnitude and direction.
The molecular dipole moment is the vector resultant of the contributing bond dipoles, together with any other relevant contributions to the electron distribution.
The two A–B bond dipoles in each molecule have equal magnitudes.
However, their relative directions differ because the molecular geometries differ.
\(\textbf{Examine the linear molecule:}\)
In P, the two bond-dipole vectors point away from the central A atom along opposite directions.
Their included angle is \(180^\circ\).
Therefore:
\[
\mu_{\mathrm P}=\left|\mu_b-\mu_b\right|
\]
\[
\mu_{\mathrm P}=1.20-1.20
\]
\[
\mu_{\mathrm P}=0\ \mathrm{D}
\]
\(\textbf{Examine the bent molecule:}\)
In Q, the included angle between the two equal bond-dipole vectors is \(120^\circ\).
For two vectors of equal magnitude \(\mu_b\), separated by angle \(\theta\):
\[
\mu_{\mathrm{resultant}}=2\mu_b\cos\left(\frac{\theta}{2}\right)
\]
Substituting:
\[
\mu_{\mathrm Q}=2(1.20)\cos\left(\frac{120^\circ}{2}\right)
\]
\[
\mu_{\mathrm Q}=2.40\cos60^\circ
\]
\[
\mu_{\mathrm Q}=2.40\left(\frac12\right)
\]
\[
\mu_{\mathrm Q}=1.20\ \mathrm{D}
\]
\(\textbf{Interpret the structural change:}\)
Both molecules contain polar bonds of the same magnitude.
The linear arrangement gives complete vector cancellation.
The bent arrangement gives a nonzero resultant directed along the bisector of the B–A–B angle.
Thus, molecular polarity cannot be inferred from bond polarity alone.
The two molecules differ in overall dipole moment even though their individual bond-dipole magnitudes are identical.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{\mathrm P}=0\ \mathrm{D};\qquad\mu_{\mathrm Q}=1.20\ \mathrm{D}}
\]
112. The dipole moment of a hypothetical heteronuclear diatomic molecule AB is initially \(1.80\ \mathrm{D}\). In another hypothetical bonding condition, its effective separated charge becomes two-thirds of its original value, while its bond length becomes \(1.50\) times the original length. Assuming the simple dipole relation remains applicable, what is the new dipole-moment magnitude?
ⓐ. \(1.20\ \mathrm{D}\)
ⓑ. \(4.05\ \mathrm{D}\)
ⓒ. \(2.70\ \mathrm{D}\)
ⓓ. \(1.80\ \mathrm{D}\)
Correct Answer: \(1.80\ \mathrm{D}\)
Explanation: \(\textbf{Identify the governing relation:}\)
For a simple separation of equal and opposite charges, dipole-moment magnitude is given by:
\[
\mu=qr
\]
Here, \(q\) is the magnitude of the effective separated charge and \(r\) is the separation between the charge centres.
Both quantities affect the dipole moment.
\(\textbf{Represent the original condition:}\)
Let the original effective charge and separation be \(q_1\) and \(r_1\).
\[
\mu_1=q_1r_1
\]
The original dipole moment is:
\[
\mu_1=1.80\ \mathrm{D}
\]
\(\textbf{Represent the changed condition:}\)
The effective charge becomes two-thirds of its original magnitude.
\[
q_2=\frac23q_1
\]
The separation increases by a factor of \(1.50\).
\[
r_2=1.50r_1
\]
\(\textbf{Calculate the new dipole moment:}\)
\[
\mu_2=q_2r_2
\]
Substituting:
\[
\mu_2=\left(\frac23q_1\right)(1.50r_1)
\]
Rearranging:
\[
\mu_2=\left(\frac23\times1.50\right)q_1r_1
\]
\[
\mu_2=1.00\mu_1
\]
Therefore:
\[
\mu_2=1.00(1.80)
\]
\[
\mu_2=1.80\ \mathrm{D}
\]
\(\textbf{Check the competing effects:}\)
The reduction in effective separated charge tends to decrease dipole moment.
The increase in charge separation tends to increase dipole moment.
In this case, the two changes compensate exactly.
Therefore, the dipole moment remains unchanged despite changes in both underlying quantities.
This demonstrates why a longer bond need not possess a larger dipole moment when the effective charge also changes.
Likewise, a larger effective charge does not independently guarantee a larger dipole if the separation becomes sufficiently smaller.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_2=1.80\ \mathrm{D}}
\]
113. A hypothetical heteronuclear diatomic molecule has a bond length of \(100\ \mathrm{pm}\) and an observed dipole moment of \(1.80\ \mathrm{D}\). Estimate its percentage ionic character by comparing the observed dipole moment with that expected for complete separation of one electronic charge across the bond length.
Use:
\[
e=1.60\times10^{-19}\ \mathrm{C}
\]
\[
1\ \mathrm{D}=3.33\times10^{-30}\ \mathrm{C\,m}
\]
ⓐ. Approximately \(18.8\%\)
ⓑ. Approximately \(25.0\%\)
ⓒ. Approximately \(48.0\%\)
ⓓ. Approximately \(37.5\%\)
Correct Answer: Approximately \(37.5\%\)
Explanation: \(\textbf{Identify the comparison required:}\)
Percentage ionic character can be estimated by comparing the observed molecular dipole moment with the idealised dipole moment corresponding to complete separation of one elementary charge.
The idealised reference assumes charges \(+e\) and \(-e\) separated by the bond length.
\(\textbf{Convert the bond length into SI units:}\)
\[
r=100\ \mathrm{pm}
\]
Since:
\[
1\ \mathrm{pm}=10^{-12}\ \mathrm{m}
\]
Therefore:
\[
r=100\times10^{-12}
\]
\[
r=1.00\times10^{-10}\ \mathrm{m}
\]
\(\textbf{Calculate the idealised fully ionic dipole:}\)
\[
\mu_{\mathrm{ionic}}=er
\]
Substituting:
\[
\mu_{\mathrm{ionic}}=(1.60\times10^{-19})(1.00\times10^{-10})
\]
\[
\mu_{\mathrm{ionic}}=1.60\times10^{-29}\ \mathrm{C\,m}
\]
\(\textbf{Convert the reference dipole into debye:}\)
\[
\mu_{\mathrm{ionic}}=\frac{1.60\times10^{-29}}{3.33\times10^{-30}}\ \mathrm{D}
\]
\[
\mu_{\mathrm{ionic}}\approx4.805\ \mathrm{D}
\]
\(\textbf{Determine percentage ionic character:}\)
\[
\%\text{ ionic character}=\frac{\mu_{\mathrm{observed}}}{\mu_{\mathrm{ionic}}}\times100
\]
Substituting:
\[
\%\text{ ionic character}=\frac{1.80}{4.805}\times100
\]
\[
\%\text{ ionic character}\approx37.5\%
\]
\(\textbf{Interpret the result:}\)
The observed dipole moment is substantially smaller than the idealised full-charge-separation value.
Thus, the effective separated charge is smaller than one elementary charge.
The estimate indicates appreciable bond polarity without requiring the bond to be treated as completely ionic.
The percentage is based on a simplified charge-separation interpretation of dipole moment.
It is not obtained directly from electronegativity difference.
A bond can be strongly polar while retaining substantial covalent character.
\(\textbf{Final answer:}\)
\[
\boxed{\%\text{ ionic character}\approx37.5\%}
\]
114. Consider two hypothetical trigonal-planar molecules, \(\mathrm{BF_3}\) and \(\mathrm{BF_2Cl}\). Assume that each molecule has an ideal \(120^\circ\) angle between adjacent bonds. The B–F and B–Cl bond-dipole magnitudes are \(1.50\ \mathrm{D}\) and \(0.90\ \mathrm{D}\), respectively. All bond dipoles point away from boron towards the halogen. Ignore any additional contributions beyond these bond-dipole vectors.
Which statement correctly compares their resultant molecular dipole moments?
ⓐ. \(\mathrm{BF_3}\) has zero dipole; \(\mathrm{BF_2Cl}\) has \(0.60\ \mathrm{D}\) directed towards the fluorine side.
ⓑ. \(\mathrm{BF_3}\) has zero dipole; \(\mathrm{BF_2Cl}\) has \(2.40\ \mathrm{D}\) directed towards chlorine.
ⓒ. \(\mathrm{BF_3}\) has \(1.50\ \mathrm{D}\); \(\mathrm{BF_2Cl}\) has \(0.60\ \mathrm{D}\) directed towards chlorine.
ⓓ. \(\mathrm{BF_3}\) has \(4.50\ \mathrm{D}\); \(\mathrm{BF_2Cl}\) has \(3.90\ \mathrm{D}\) directed towards fluorine.
Correct Answer: \(\mathrm{BF_3}\) has zero dipole; \(\mathrm{BF_2Cl}\) has \(0.60\ \mathrm{D}\) directed towards the fluorine side.
Explanation: \(\textbf{Examine the symmetrical molecule:}\)
In ideal trigonal-planar \(\mathrm{BF_3}\), the three B–F bonds are equivalent.
Their bond-dipole vectors have equal magnitudes and are separated by \(120^\circ\).
Three equal vectors arranged symmetrically in this manner cancel.
Therefore:
\[
\mu(\mathrm{BF_3})=0
\]
The zero molecular dipole does not mean that the B–F bonds are nonpolar.
\(\textbf{Identify the change introduced by substitution:}\)
In \(\mathrm{BF_2Cl}\), one B–F bond is replaced by a B–Cl bond.
The molecule retains the stipulated trigonal-planar geometry.
However, the three bond-dipole magnitudes are no longer equal.
The two B–F vectors each have magnitude \(1.50\ \mathrm{D}\).
\(\textbf{Combine the two fluorine-directed vectors:}\)
The angle between them is \(120^\circ\).
Their resultant is:
\[
\mu_{\mathrm{FF}}=2(1.50)\cos\left(\frac{120^\circ}{2}\right)
\]
\[
\mu_{\mathrm{FF}}=3.00\cos60^\circ
\]
\[
\mu_{\mathrm{FF}}=1.50\ \mathrm{D}
\]
This resultant points along the bisector between the fluorine bonds.
In a trigonal-planar arrangement, that direction is opposite to the B–Cl bond direction.
\(\textbf{Include the chlorine bond dipole:}\)
The B–Cl bond dipole has magnitude:
\[
\mu_{\mathrm{B-Cl}}=0.90\ \mathrm{D}
\]
It points opposite to the combined fluorine-directed resultant.
Therefore:
\[
\mu_{\mathrm{BF_2Cl}}=1.50-0.90
\]
\[
\mu_{\mathrm{BF_2Cl}}=0.60\ \mathrm{D}
\]
The resultant points towards the fluorine side.
\(\textbf{Interpret the symmetry breaking:}\)
The original molecule possesses three equivalent bond-dipole vectors and complete cancellation.
Replacing one terminal atom breaks that equality.
A nonzero molecular dipole results even though the assumed bond angles are unchanged.
This distinction shows that molecular polarity depends on both geometry and the magnitudes of the constituent bond dipoles.
\(\textbf{Final answer:}\)
\[
\boxed{\mu(\mathrm{BF_3})=0;\quad\mu(\mathrm{BF_2Cl})=0.60\ \mathrm{D}\text{ towards the fluorine side}}
\]
115. Ammonia and nitrogen trifluoride are both trigonal-pyramidal molecules containing one lone pair on nitrogen. Experimentally, the dipole moment of ammonia is appreciably greater than that of nitrogen trifluoride, despite the high electronegativity of fluorine. Which explanation best accounts for this observation?
ⓐ. The N–F bonds are nonpolar, whereas N–H bonds are strongly ionic.
ⓑ. The nitrogen lone pair is absent in \(\mathrm{NF_3}\), removing its contribution.
ⓒ. The three N–F bond dipoles cancel completely because \(\mathrm{NF_3}\) is trigonal planar.
ⓓ. The resultant bond-dipole contribution and lone-pair contribution reinforce in \(\mathrm{NH_3}\) but substantially oppose in \(\mathrm{NF_3}\).
Correct Answer: The resultant bond-dipole contribution and lone-pair contribution reinforce in \(\mathrm{NH_3}\) but substantially oppose in \(\mathrm{NF_3}\).
Explanation: Both ammonia and nitrogen trifluoride contain three bonds around nitrogen and one nitrogen lone pair.
Their molecular shapes are trigonal pyramidal.
The presence of the lone pair makes the electron distribution different from that of a corresponding symmetrical trigonal-planar molecule.
In ammonia, nitrogen is more electronegative than hydrogen.
Each N–H bond dipole is directed towards nitrogen.
The resultant contribution of the three N–H bond dipoles points approximately towards the nitrogen side of the molecule.
The lone-pair electron distribution contributes to molecular polarity in broadly the same direction.
These contributions therefore reinforce each other.
In nitrogen trifluoride, fluorine is more electronegative than nitrogen.
The N–F bond dipoles are consequently directed towards the fluorine atoms.
Their resultant points towards the fluorine side, opposite to the direction associated with the nitrogen lone-pair region.
These contributions substantially oppose one another.
The resulting molecular dipole of \(\mathrm{NF_3}\) is therefore relatively small.
The comparison cannot be explained by saying that N–F bonds are nonpolar.
Nor can it be explained by assigning a trigonal-planar shape to \(\mathrm{NF_3}\).
Both molecules possess nitrogen lone pairs, and both contain polar bonds.
The important distinction is the vector relationship between the bond-dipole resultant and the contribution associated with the nonbonding electron distribution.
Therefore, the greater dipole moment of ammonia does not contradict the high electronegativity of fluorine.
116. Consider an ideal tetrahedral molecule \(\mathrm{CHCl_3}\). Assume that each C–Cl bond has a dipole moment of \(1.60\ \mathrm{D}\) directed from carbon towards chlorine, while the C–H bond has a dipole moment of \(0.40\ \mathrm{D}\) directed from hydrogen towards carbon. Treat these as the only contributions and assume ideal tetrahedral bond directions.
What is the resultant molecular dipole-moment magnitude?
ⓐ. \(1.20\ \mathrm{D}\)
ⓑ. \(2.00\ \mathrm{D}\)
ⓒ. \(4.80\ \mathrm{D}\)
ⓓ. \(5.20\ \mathrm{D}\)
Correct Answer: \(2.00\ \mathrm{D}\)
Explanation: \(\textbf{Use tetrahedral vector symmetry:}\)
In an ideal tetrahedron, four equal vectors directed from the centre towards the four vertices have a zero resultant.
Therefore, the vector sum of three such directions is equal and opposite to the fourth direction.
This property permits the three C–Cl bond-dipole vectors to be combined without resolving each into separate Cartesian components.
\(\textbf{Identify the three chlorine-directed contributions:}\)
Each C–Cl bond dipole has magnitude:
\[
\mu_{\mathrm{C-Cl}}=1.60\ \mathrm{D}
\]
The three chlorine bonds occupy three vertices of the tetrahedron.
Their resultant has the same magnitude as one C–Cl bond dipole.
It points opposite to the direction from carbon towards hydrogen.
Therefore:
\[
\mu_{\mathrm{3(C-Cl),resultant}}=1.60\ \mathrm{D}
\]
\(\textbf{Determine the C–H dipole direction:}\)
Carbon is more electronegative than hydrogen.
The C–H bond dipole therefore points from hydrogen towards carbon.
This direction is also opposite to the radial direction from carbon towards hydrogen.
Consequently, the C–H dipole and the resultant of the three C–Cl dipoles point in the same direction.
\(\textbf{Add the aligned contributions:}\)
The C–H bond-dipole magnitude is:
\[
\mu_{\mathrm{C-H}}=0.40\ \mathrm{D}
\]
Since the two resultant vectors are parallel:
\[
\mu_{\mathrm{CHCl_3}}=\mu_{\mathrm{3(C-Cl),resultant}}+\mu_{\mathrm{C-H}}
\]
\[
\mu_{\mathrm{CHCl_3}}=1.60+0.40
\]
\[
\mu_{\mathrm{CHCl_3}}=2.00\ \mathrm{D}
\]
\(\textbf{Interpret the molecular arrangement:}\)
The molecular dipole points towards the side containing the three chlorine atoms.
The three C–Cl dipoles must not be added as ordinary scalars.
Their directions reduce their vector resultant to the magnitude of a single C–Cl bond dipole under ideal tetrahedral symmetry.
The C–H dipole then reinforces that resultant.
This reasoning also explains why a molecule containing several identical polar bonds need not have a dipole equal to their scalar sum.
The calculated value applies to the idealised bond-dipole model specified in the question.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{\mathrm{CHCl_3}}=2.00\ \mathrm{D}}
\]
117. Consider the following assertion and reason.
Assertion (A): A molecule of \(\mathrm{CCl_4}\) has zero permanent molecular dipole moment despite containing four polar C–Cl bonds.
Reason (R): In its tetrahedral geometry, the four equivalent C–Cl bond-dipole vectors cancel one another.
Choose the correct option.
ⓐ. Both A and R are true, but R does not correctly explain A.
ⓑ. A is true, but R is false.
ⓒ. Both A and R are true, and R correctly explains A.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: The carbon–chlorine bond is polar because chlorine is more electronegative than carbon.
Its shared electron density is therefore displaced towards chlorine.
Consequently, each C–Cl bond contributes a bond dipole.
However, the presence of polar bonds does not automatically produce a nonzero molecular dipole.
Molecular dipole moment is a vector quantity.
The directions of the individual bond dipoles must be considered in the actual molecular geometry.
In \(\mathrm{CCl_4}\), carbon is tetrahedrally surrounded by four equivalent chlorine atoms.
The four C–Cl bond-dipole vectors have equal magnitudes and point towards the four vertices of a regular tetrahedron.
Their vector sum is zero.
Thus, the molecule has no permanent resultant dipole moment.
The assertion is therefore true.
The reason correctly identifies tetrahedral symmetry and vector cancellation as the explanation.
It is not sufficient to say that the bonds are covalent, because many covalent molecules possess nonzero dipole moments.
It is also incorrect to conclude that the C–Cl bonds individually lack polarity.
The molecule is nonpolar overall because its polar bond contributions cancel through symmetry.
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.
118. A hypothetical bent molecule AB₂ contains two identical polar A–B bonds. Each bond has a dipole-moment magnitude of \(1.00\ \mathrm{D}\). The measured molecular dipole moment is \(\sqrt{3}\ \mathrm{D}\). Assume that the measured resultant arises only from the two bond-dipole vectors, directed from A towards B.
What is the B–A–B bond angle?
ⓐ. \(60^\circ\)
ⓑ. \(90^\circ\)
ⓒ. \(120^\circ\)
ⓓ. \(150^\circ\)
Correct Answer: \(60^\circ\)
Explanation: \(\textbf{Identify the unknown:}\)
The individual bond-dipole magnitudes and their resultant are known.
The angle between the two bond-dipole vectors must be determined.
Because both vectors point from the central A atom towards the terminal B atoms, their included angle is the B–A–B bond angle.
\(\textbf{Write the vector-resultant relation:}\)
For two vectors of equal magnitude \(\mu_b\) separated by angle \(\theta\):
\[
\mu_{\mathrm{resultant}}=2\mu_b\cos\left(\frac{\theta}{2}\right)
\]
The supplied values are:
\[
\mu_{\mathrm{resultant}}=\sqrt3\ \mathrm{D}
\]
\[
\mu_b=1.00\ \mathrm{D}
\]
\(\textbf{Substitute the data:}\)
\[
\sqrt3=2(1.00)\cos\left(\frac{\theta}{2}\right)
\]
Therefore:
\[
\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt3}{2}
\]
\(\textbf{Determine the half-angle:}\)
For the physically relevant molecular bond-angle range:
\[
\frac{\theta}{2}=30^\circ
\]
Hence:
\[
\theta=2(30^\circ)
\]
\[
\theta=60^\circ
\]
\(\textbf{Check the result:}\)
Substituting the calculated angle:
\[
\mu_{\mathrm{resultant}}=2(1.00)\cos30^\circ
\]
\[
\mu_{\mathrm{resultant}}=2\left(\frac{\sqrt3}{2}\right)
\]
\[
\mu_{\mathrm{resultant}}=\sqrt3\ \mathrm{D}
\]
This agrees with the measured value.
The resultant is smaller than the scalar sum of \(2.00\ \mathrm{D}\), as expected for two nonparallel vectors.
It is also greater than zero because the vectors are not oppositely directed.
The calculation illustrates how molecular geometry can sometimes be inferred from dipole data when individual bond-dipole magnitudes and other contributions are specified.
\(\textbf{Final answer:}\)
\[
\boxed{\theta=60^\circ}
\]
119. A hypothetical bent molecule ABC contains two polar bonds, AB and BC. Their dipole-moment magnitudes are \(3.00\ \mathrm{D}\) and \(4.00\ \mathrm{D}\), respectively. The bond-dipole vectors are perpendicular in the stated molecular arrangement. Ignore other contributions to the molecular dipole.
A student obtains \(1.00\ \mathrm{D}\) by subtracting the two bond-dipole magnitudes. Which conclusion correctly evaluates the student's method and gives the actual resultant?
ⓐ. Subtraction is valid because the dipoles originate from different bonds; the resultant is \(1.00\ \mathrm{D}\).
ⓑ. Subtraction is invalid because the dipoles are perpendicular; the resultant is \(5.00\ \mathrm{D}\).
ⓒ. Subtraction is invalid because the dipoles are perpendicular; the resultant is \(7.00\ \mathrm{D}\).
ⓓ. Subtraction is valid only for a bent molecule; the resultant is \(5.00\ \mathrm{D}\).
Correct Answer: Subtraction is invalid because the dipoles are perpendicular; the resultant is \(5.00\ \mathrm{D}\).
Explanation: \(\textbf{Identify the vector relationship:}\)
A molecular dipole moment is the vector resultant of the contributing dipole moments.
The magnitudes cannot generally be added or subtracted without accounting for direction.
The question specifies that the two vectors are perpendicular.
Therefore, their included angle is \(90^\circ\).
\(\textbf{Write the general resultant relation:}\)
For vectors of magnitudes \(\mu_1\) and \(\mu_2\), separated by angle \(\theta\):
\[
\mu_R^2=\mu_1^2+\mu_2^2+2\mu_1\mu_2\cos\theta
\]
The supplied values are:
\[
\mu_1=3.00\ \mathrm{D}
\]
\[
\mu_2=4.00\ \mathrm{D}
\]
\[
\theta=90^\circ
\]
\(\textbf{Apply the perpendicular-vector condition:}\)
Since:
\[
\cos90^\circ=0
\]
The cross term vanishes.
\[
\mu_R^2=(3.00)^2+(4.00)^2
\]
\[
\mu_R^2=9.00+16.00
\]
\[
\mu_R^2=25.00
\]
Therefore:
\[
\mu_R=\sqrt{25.00}
\]
\[
\mu_R=5.00\ \mathrm{D}
\]
\(\textbf{Diagnose the student's error:}\)
The difference \(4.00-3.00=1.00\ \mathrm{D}\) would be appropriate for exactly oppositely directed vectors.
The scalar sum \(4.00+3.00=7.00\ \mathrm{D}\) would apply to parallel vectors pointing in the same direction.
Neither condition is satisfied here.
The correct result lies between the difference and the sum, as expected for perpendicular vectors.
The calculation demonstrates why geometry must accompany bond-dipole magnitudes in molecular-polarity questions.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_R=5.00\ \mathrm{D}}
\]
120. Two hypothetical linear molecules have the structures X–A–X and X–A–Y. In both, the central atom A is less electronegative than its terminal atoms. The individual bond-dipole magnitudes are:
For X–A–X, each A–X bond has a magnitude of \(1.80\ \mathrm{D}\).
For X–A–Y, the A–X bond has a magnitude of \(1.80\ \mathrm{D}\), while the A–Y bond has a magnitude of \(1.10\ \mathrm{D}\).
Assume that all bond dipoles point from A towards the relevant terminal atom. Which conclusion is correct?
ⓐ. Both molecules have zero resultant dipole because both are linear.
ⓑ. X–A–X has \(3.60\ \mathrm{D}\), while X–A–Y has \(2.90\ \mathrm{D}\).
ⓒ. X–A–X has zero resultant dipole, while X–A–Y has \(2.90\ \mathrm{D}\).
ⓓ. X–A–X has zero resultant dipole, while X–A–Y has \(0.70\ \mathrm{D}\).
Correct Answer: X–A–X has zero resultant dipole, while X–A–Y has \(0.70\ \mathrm{D}\).
Explanation: \(\textbf{Identify the effect of linear geometry:}\)
In a linear triatomic molecule with the central atom in the middle, the two bond directions are opposite.
Therefore, the two bond-dipole vectors are antiparallel under the stated direction convention.
However, antiparallel vectors cancel completely only when their magnitudes are equal.
\(\textbf{Evaluate X–A–X:}\)
The molecule contains two equivalent A–X bonds.
Their dipole magnitudes are both \(1.80\ \mathrm{D}\).
Since the vectors point in opposite directions:
\[
\mu_{\mathrm{XAX}}=\left|1.80-1.80\right|
\]
\[
\mu_{\mathrm{XAX}}=0\ \mathrm{D}
\]
The molecule is therefore nonpolar under the stipulated bond-dipole model.
\(\textbf{Evaluate X–A–Y:}\)
The A–X and A–Y bonds have unequal dipole magnitudes.
Their directions remain opposite because the molecule is linear.
The resultant magnitude is:
\[
\mu_{\mathrm{XAY}}=\left|1.80-1.10\right|
\]
\[
\mu_{\mathrm{XAY}}=0.70\ \mathrm{D}
\]
The resultant points towards X, corresponding to the larger bond-dipole vector.
\(\textbf{Interpret the changed terminal atom:}\)
Replacing one X with Y removes the equality of the opposing bond dipoles.
The molecular geometry remains linear, but the molecular dipole becomes nonzero.
Therefore, linearity alone does not ensure that a molecule is nonpolar.
Complete cancellation also requires the relevant dipole contributions to possess suitable magnitudes and symmetry.
The total dipole is not the scalar sum of the two individual bond dipoles.
The important distinction is between geometrical opposition and equal-magnitude cancellation.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_{\mathrm{XAX}}=0\ \mathrm{D};\quad\mu_{\mathrm{XAY}}=0.70\ \mathrm{D}}
\]
