301. Formaldehyde is converted into methanol by hydrogenation:
\[
\mathrm{H_2C=O+H_2\rightarrow CH_3OH}
\]
Use the conventional localized valence bond and hybridization descriptions of the reactant and product. Which statement correctly identifies the changes around carbon and oxygen?
ⓐ. Carbon remains \(sp^2\)-hybridized because the C–O sigma bond is retained, while oxygen changes from \(sp^3\) to \(sp\).
ⓑ. Carbon changes from \(sp^2\) to \(sp^3\), but oxygen retains a carbonyl-type \(sp^2\) arrangement because it has two lone pairs in both molecules.
ⓒ. Carbon changes from \(sp^2\) to \(sp\), while oxygen changes from \(sp^3\) to \(sp^2\).
ⓓ. Carbon changes from \(sp^2\) to \(sp^3\); oxygen changes from a carbonyl-type three-domain arrangement to a hydroxyl four-domain arrangement, while the C=O pi component is lost.
Correct Answer: Carbon changes from \(sp^2\) to \(sp^3\); oxygen changes from a carbonyl-type three-domain arrangement to a hydroxyl four-domain arrangement, while the C=O pi component is lost.
Explanation: \(\textbf{Examine the carbonyl carbon:}\)
In formaldehyde, carbon forms two C–H single bonds and one C=O double bond.
These are three distinct sigma-bonding directions.
Therefore, the carbon centre is conventionally \(sp^2\)-hybridized.
One unhybridized carbon p orbital participates in the pi component of the carbonyl double bond.
\(\textbf{Examine the carbonyl oxygen:}\)
Oxygen participates in the C=O double bond and possesses two lone pairs.
In the conventional hybrid-orbital description, the C–O sigma-bonding region and two lone-pair regions form a three-domain framework.
The remaining p orbital participates in carbonyl pi bonding.
This is described using an approximately \(sp^2\)-type oxygen environment.
\(\textbf{Identify the reaction's bond changes:}\)
Hydrogenation changes the C=O double bond into a C–O single bond.
One hydrogen atom becomes attached to carbon and the other to oxygen.
The original C–O sigma connection remains.
The carbonyl pi component is removed.
\(\textbf{Examine the methanol carbon:}\)
Carbon now has three C–H bonds and one C–O bond.
These provide four sigma-bonding directions.
The conventional carbon hybridization becomes \(sp^3\).
\(\textbf{Examine the methanol oxygen:}\)
The hydroxyl oxygen forms a C–O bond and an O–H bond.
It also retains two lone pairs.
It therefore has two bonding and two nonbonding domains.
Its four-domain environment is described using the conventional \(sp^3\) framework.
\(\textbf{Integrate the orbital changes:}\)
Both carbon and oxygen undergo changes in their local bonding environments.
The oxygen lone-pair count remains two, but the number of its sigma-bonding directions increases.
Consequently, an unchanged lone-pair count does not imply unchanged hybridization.
The carbonyl pi bond is removed while two new sigma bonds are formed.
Thus, option D correctly identifies both local orbital changes.
302. Cyanide ion is protonated at carbon in the following reaction:
\[
\mathrm{CN^-+H^+\rightarrow HCN}
\]
In the conventional Lewis representation of \(\mathrm{CN^-}\), carbon and nitrogen are connected by a triple bond, and each has one lone pair. The proton bonds to carbon using the electron pair initially assigned to carbon. Which statement correctly describes the changes?
ⓐ. Carbon changes from \(sp\) to \(sp^3\), and the C≡N triple bond becomes a single bond.
ⓑ. Carbon retains an approximately \(sp\)-hybridized framework; its lone pair becomes a C–H sigma-bonding pair, and the C≡N triple bond remains unchanged.
ⓒ. Nitrogen changes from \(sp\) to \(sp^2\) because protonation creates an additional C–N pi bond.
ⓓ. Carbon retains its negative formal charge after forming the C–H bond because the electron pair originally belonged to carbon.
Correct Answer: Carbon retains an approximately \(sp\)-hybridized framework; its lone pair becomes a C–H sigma-bonding pair, and the C≡N triple bond remains unchanged.
Explanation: \(\textbf{Establish the cyanide Lewis description:}\)
Carbon contributes four valence electrons, nitrogen contributes five, and the negative charge adds one.
\[
N_v(\mathrm{CN^-})=4+5+1
\]
\[
N_v(\mathrm{CN^-})=10
\]
The C≡N triple bond accounts for six electrons.
The remaining four electrons form one lone pair on carbon and one on nitrogen.
The conventional formal charge on carbon is:
\[
FC_{\mathrm C}=4-2-\frac62
\]
\[
FC_{\mathrm C}=-1
\]
Nitrogen has formal charge zero.
\(\textbf{Identify the proton-accepting pair:}\)
The problem specifies that protonation occurs at carbon.
The carbon lone pair forms the new C–H bonding interaction.
The electron pair changes from nonbonding to bonding in the conventional Lewis description.
\(\textbf{Examine the carbon orbital framework before protonation:}\)
Carbon participates in one C≡N bonding direction and possesses one lone pair.
The conventional two-domain framework is described as \(sp\).
The two unhybridized p orbitals participate in the two C≡N pi bonds.
\(\textbf{Examine the carbon environment after protonation:}\)
In HCN, carbon has one C–H sigma bond and one C≡N bonding connection.
These provide two sigma-bonding directions.
The conventional hybridization remains \(sp\).
Its two unhybridized p orbitals continue to support the two carbon–nitrogen pi interactions.
\(\textbf{Check the formal-charge change:}\)
The carbon lone pair is no longer assigned wholly to carbon.
Carbon forms four bonding pairs in the resulting H–C≡N representation and has no nonbonding electrons.
Its formal charge becomes zero.
The nitrogen lone pair remains present.
\(\textbf{Interpret the result:}\)
Protonation changes electron-pair occupancy and formal charge without necessarily changing the hybridization label.
The decisive condition is that the number of central electron-domain directions remains two.
The retained C≡N triple bond also preserves the two pi interactions.
Therefore, option B is correct.
303. Acetaldehyde, \(\mathrm{CH_3CHO}\), and oxirane, a three-membered cyclic ether, have the same molecular formula \(\mathrm{C_2H_4O}\).
Structure P: \(\mathrm{CH_3-C(=O)-H}\), containing a carbonyl double bond. Structure Q: A three-membered ring in which the two carbon atoms are bonded to each other and both are separately bonded to the oxygen atom; each carbon also carries two hydrogen atoms. All bonds in Q are single bonds.
Which statement correctly compares their sigma–pi bonding counts and the conventional hybridizations of the carbon centres?
ⓐ. P has six sigma bonds and one pi bond, with \(sp^3\) methyl carbon and \(sp^2\) carbonyl carbon; Q has seven sigma bonds and no pi bond, with both carbons \(sp^3\)-hybridized.
ⓑ. P and Q both have seven sigma bonds because they share the same molecular formula; only P has one pi bond.
ⓒ. P has five sigma bonds and two pi bonds; Q has seven sigma bonds and one pi bond arising from ring closure.
ⓓ. P has six sigma bonds and one pi bond with both carbons \(sp^2\)-hybridized; Q has six sigma bonds and no pi bond with both carbons \(sp^3\)-hybridized.
Correct Answer: P has six sigma bonds and one pi bond, with \(sp^3\) methyl carbon and \(sp^2\) carbonyl carbon; Q has seven sigma bonds and no pi bond, with both carbons \(sp^3\)-hybridized.
Explanation: \(\textbf{Identify the structural difference:}\)
The two compounds have identical molecular formulas.
However, their connectivities are different.
Acetaldehyde contains a carbonyl double bond.
Oxirane contains a three-membered ring with only single bonds.
Consequently, their sigma and pi bond counts need not be identical.
\(\textbf{Count the bonds in acetaldehyde:}\)
Acetaldehyde has four C–H bonds.
It also has one C–C single bond and one C=O double bond.
Every C–H bond contributes one sigma bond.
The C–C single bond contributes one sigma bond.
The C=O double bond contributes one sigma and one pi bond.
Therefore:
\[
N_{\sigma,P}=4+1+1
\]
\[
N_{\sigma,P}=6
\]
\[
N_{\pi,P}=1
\]
\(\textbf{Identify its carbon hybridizations:}\)
The methyl carbon has four sigma-bonding directions and is conventionally \(sp^3\)-hybridized.
The carbonyl carbon has three sigma-bonding directions and is conventionally \(sp^2\)-hybridized.
\(\textbf{Count the bonds in oxirane:}\)
The ring contains one C–C bond and two C–O bonds.
Four C–H bonds are also present.
All seven connections are single bonds.
Therefore:
\[
N_{\sigma,Q}=1+2+4
\]
\[
N_{\sigma,Q}=7
\]
\[
N_{\pi,Q}=0
\]
\(\textbf{Identify its carbon hybridizations:}\)
Each ring carbon forms two C–H bonds, one C–C bond and one C–O bond.
Each therefore has four sigma-bonding directions.
Both carbons are conventionally \(sp^3\)-hybridized.
The strained three-membered ring can distort ideal orbital geometry, but the conventional electron-domain classification remains four-coordinate.
\(\textbf{Interpret the isomer comparison:}\)
A ring can account for the degree of unsaturation of a molecular formula without containing a pi bond.
Therefore, equal molecular formulas do not guarantee equal sigma–pi bond counts.
The carbonyl double bond in P and the ring closure in Q represent different structural arrangements.
Thus, option A is correct.
304. Four conventional orbital descriptions are proposed for carbon atoms in different molecular environments:
| Centre | Stated sigma-bonding directions | Stated unhybridized p orbitals | Proposed hybridization |
|---|
| P | 2 | 2 | \(sp\) |
| Q | 3 | 1 | \(sp^2\) |
| R | 4 | 0 | \(sp^3\) |
| S | 3 | 2 | \(sp^2\) |
Assume that the listed sigma directions are accommodated by the corresponding hybrid orbitals and that the available valence basis contains one s and three p orbitals. Which assessment is correct?
ⓐ. Only P and R are internally consistent because \(sp^2\) hybridization cannot leave any p orbital unhybridized.
ⓑ. All four are consistent because hybridization can increase the total number of valence orbitals.
ⓒ. P, Q and R are consistent, whereas S violates orbital counting because three \(sp^2\) hybrids leave only one unhybridized p orbital.
ⓓ. P, Q and S are consistent, whereas R is impossible because a carbon atom must retain at least one unhybridized p orbital.
Correct Answer: P, Q and R are consistent, whereas S violates orbital counting because three \(sp^2\) hybrids leave only one unhybridized p orbital.
Explanation: \(\textbf{Establish the available orbital inventory:}\)
The specified carbon valence basis contains one s orbital and three p orbitals.
The total number of available atomic orbitals is four.
Hybridization redistributes those orbitals into hybrid and unhybridized sets.
It does not create additional orbitals.
\(\textbf{Evaluate Centre P:}\)
An \(sp\) set is formed from one s orbital and one p orbital.
It produces two hybrid orbitals.
The other two p orbitals remain unhybridized.
Therefore, two sigma-bonding directions and two residual p orbitals are consistent with the stated model.
P is valid.
\(\textbf{Evaluate Centre Q:}\)
An \(sp^2\) set uses one s and two p orbitals.
It produces three hybrid orbitals.
One p orbital remains unhybridized.
Therefore, three sigma-bonding directions and one residual p orbital are consistent.
Q is valid.
\(\textbf{Evaluate Centre R:}\)
An \(sp^3\) set combines one s and three p orbitals.
It produces four hybrid orbitals.
No p orbital remains unhybridized in the specified basis.
Therefore, four sigma-bonding directions and zero residual p orbitals are consistent.
R is valid.
\(\textbf{Evaluate Centre S:}\)
The proposed \(sp^2\) hybridization requires three hybrid orbitals.
These consume one s and two p orbitals.
Only one p orbital can remain unhybridized.
The claim of two remaining p orbitals would require five orbital directions in total: three hybrids plus two unhybridized p orbitals.
That exceeds the four orbitals originally available.
S is therefore internally inconsistent.
\(\textbf{Integrate the orbital-counting principle:}\)
The number of hybrid orbitals equals the number of atomic orbitals mixed.
The remaining orbitals must be subtracted from the original valence basis.
Any proposed description that violates this inventory fails even if its hybridization label appears plausible.
Thus, option C is correct.
305. A conjugated molecule has the connectivity \(\mathrm{CH_2=CH-CH=CH_2}\). Consider two hypothetical conformations around the central C–C single bond:
Conformation P: The four carbon atoms and their unhybridized p-orbital axes are arranged so that adjacent p orbitals can overlap continuously along the conjugated chain. Conformation Q: One terminal \(\mathrm{CH_2=CH}\) fragment is rotated by \(90^\circ\) about the central carbon–carbon single bond relative to the other fragment. The individual terminal C=C bonds remain intact, and the local bonding environments at the carbon atoms are assumed not to change substantially.
Which inference best distinguishes this twisting process from rotation about a carbon–carbon double bond?
ⓐ. In Q, both C=C pi bonds must disappear because the two fragments no longer lie in one plane.
ⓑ. Q can retain the two local C=C pi bonds while reducing continuous p-orbital overlap across the central C–C connection; the carbon atoms may retain their \(sp^2\) local descriptions.
ⓒ. Twisting the central single bond necessarily changes the middle carbon atoms from \(sp^2\) to \(sp^3\) because their p orbitals become perpendicular.
ⓓ. P and Q have identical continuous conjugation because a sigma bond guarantees uninterrupted sideways overlap between all four carbon p orbitals.
Correct Answer: Q can retain the two local C=C pi bonds while reducing continuous p-orbital overlap across the central C–C connection; the carbon atoms may retain their \(sp^2\) local descriptions.
Explanation: \(\textbf{Examine the original conjugated chain:}\)
The molecule contains two carbon–carbon double bonds separated by one carbon–carbon single bond.
Each carbon has three sigma-bonding directions.
Therefore, the four carbon atoms are conventionally \(sp^2\)-hybridized.
Each retains one unhybridized p orbital.
When these p orbitals are appropriately aligned, electron density can be delocalized across the conjugated chain.
\(\textbf{Identify the bond about which rotation occurs:}\)
The specified rotation occurs about the central C–C single bond.
That connection contains a sigma bond.
The rotation does not directly require breaking either terminal C=C double bond.
The central sigma interaction can remain present.
\(\textbf{Examine Conformation P:}\)
The four p-orbital axes have a suitable common orientation.
Continuous sideways overlap can occur along the carbon chain.
The conjugated arrangement is consequently supported.
\(\textbf{Examine Conformation Q:}\)
The two unsaturated fragments are rotated relative to one another.
Their p-orbital systems are no longer continuously aligned across the central single bond.
The effective orbital interaction responsible for delocalization between the two fragments is therefore reduced in the idealized perpendicular arrangement.
However, the p orbitals within each individual C=C bond can remain appropriately aligned with each other.
Thus, each local double bond can retain its pi component.
\(\textbf{Interpret hybridization:}\)
The stipulated local sigma-bonding directions at each carbon remain three.
Consequently, the local \(sp^2\) descriptions need not change merely because the two fragments adopt a different relative orientation.
\(\textbf{Distinguish the two rotation processes:}\)
Twisting directly about a C=C double bond disrupts the lateral overlap constituting that particular pi bond.
Twisting about the intervening C–C single bond can instead interrupt conjugation between otherwise intact pi bonds.
This is the central distinction tested by the two conformations.
Option B is correct.
306. In the conventional NCERT hybridization model, an isolated gaseous \(\mathrm{PCl_5}\) molecule is described using five \(sp^3d\) hybrid orbitals. A student states that all five P–Cl bonds must therefore be geometrically and energetically equivalent. Which conclusion correctly evaluates the student's claim?
ⓐ. The claim is correct because five hybrid orbitals necessarily form the vertices of a regular pentagon.
ⓑ. The claim is correct because hybridization prohibits the existence of different axial and equatorial bonding environments.
ⓒ. The claim is incorrect because an \(sp^3d\) description produces only four orbitals and one unhybridized d orbital.
ⓓ. The claim is incorrect: axial and equatorial positions in trigonal bipyramids differ in angle, and \(sp^3d\) alone does not ensure equivalent bonds.
Correct Answer: The claim is incorrect: axial and equatorial positions in trigonal bipyramids differ in angle, and \(sp^3d\) alone does not ensure equivalent bonds.
Explanation: \(\textbf{Identify the conventional orbital inventory:}\)
An \(sp^3d\) hybrid set is described using one s orbital, three p orbitals and one d orbital.
Five atomic orbitals are therefore involved in the conventional mixing scheme.
The resulting description contains five hybrid orbitals.
The student is correct about the number of hybrid orbitals but incorrect about their necessary geometrical equivalence.
\(\textbf{Examine the trigonal-bipyramidal arrangement:}\)
The arrangement contains three equatorial positions in one plane.
It also contains two axial positions perpendicular to that plane.
Adjacent equatorial directions are separated by \(120^\circ\).
Each axial direction forms \(90^\circ\) angles with all three equatorial directions.
The two axial directions are opposite one another.
\(\textbf{Compare the local bonding environments:}\)
An axial bonding region has three equatorial neighbours at \(90^\circ\).
An equatorial bonding region has only two axial neighbours at \(90^\circ\).
Therefore, axial and equatorial positions are not geometrically equivalent.
Their different angular environments can influence bonding interactions and bond lengths.
\(\textbf{Apply the comparison to PCl₅:}\)
The axial P–Cl bonds in gaseous \(\mathrm{PCl_5}\) are longer and weaker than the equatorial bonds.
This observation is compatible with the unequal geometrical environments.
It does not contradict the conventional five-domain hybridization description.
\(\textbf{Recognise the model's limitation:}\)
The \(sp^3d\) label is a conventional orbital model used to describe the five-coordinate arrangement in the NCERT treatment.
It should not be treated as independent experimental proof of a unique detailed d-orbital bonding mechanism.
The geometry and bonding properties must still be distinguished.
Thus, option D correctly rejects the claim of five identical bond environments.
307. In the conventional NCERT hybridization model, a five-domain trigonal-bipyramidal framework is associated with mixing one s, three p and one d orbital, whereas a six-domain octahedral framework is associated with mixing one s, three p and two d orbitals. Consider the following claims:
Statement I: The five-domain framework produces five hybrid orbitals, while the six-domain framework produces six.
Statement II: The average s-orbital contribution per hybrid, calculated across the complete set, is \(\frac15\) in the five-hybrid set and \(\frac16\) in the six-hybrid set.
Statement III: The five trigonal-bipyramidal directions must be energetically equivalent because their average s-orbital contribution is \(\frac15\).
Statement IV: Orbital counting within these conventional models does not, by itself, establish the exact experimental contribution of d orbitals to bonding in a particular molecule.
Which statements are correct?
ⓐ. I and III only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
The number of hybrid orbitals equals the number of atomic orbitals mixed.
A conventional \(sp^3d\) set combines one s, three p and one d orbital.
It therefore contains five hybrid orbitals.
An \(sp^3d^2\) set combines one s, three p and two d orbitals.
It consequently contains six hybrid orbitals.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Each complete hybrid set contains the contribution of one original s orbital.
When the total s contribution is averaged across five hybrids, the average fraction is:
\[
\overline{f_s}(sp^3d)=\frac15
\]
For six hybrids:
\[
\overline{f_s}(sp^3d^2)=\frac16
\]
These are averages over the complete respective sets.
They need not establish identical orbital compositions at geometrically inequivalent positions.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
A trigonal bipyramid contains two axial and three equatorial positions.
These positions have different angular environments.
Each axial direction is perpendicular to all three equatorial directions.
Each equatorial direction is perpendicular to only the two axial directions.
Therefore, the five directions are not geometrically equivalent.
An average orbital-composition fraction cannot prove that all corresponding bonding environments are energetically identical.
Statement III is incorrect.
\(\textbf{Evaluate Statement IV:}\)
The conventional \(sp^3d\) and \(sp^3d^2\) labels provide useful textbook descriptions of directional bonding arrangements.
However, a successful geometrical classification does not independently establish the precise physical participation of d orbitals in an actual molecular electronic structure.
A more detailed bonding description may require additional electronic-structure considerations.
Statement IV is correct.
\(\textbf{Overall conclusion:}\)
Orbital counting establishes the number of hybrids.
The averaged s contribution follows from the stipulated conventional orbital inventory.
Neither result proves equivalence of all bonding positions or an exact microscopic d-orbital contribution.
Thus, Statements I, II and IV are correct.
308. Isolated gaseous \(\mathrm{PCl_5}\) can accept a chloride ion to form the hexachlorophosphate ion:
\[
\mathrm{PCl_5+Cl^-\rightarrow PCl_6^-}
\]
Use conventional single-bond Lewis descriptions and the textbook hybridization models for five- and six-coordinate phosphorus. Which statement correctly combines the phosphorus oxidation state, formal charge and local geometry?
ⓐ. Phosphorus retains oxidation state \(+5\), its formal charge changes from \(0\) to \(-1\), and its conventional description changes from \(sp^3d\), trigonal bipyramidal, to \(sp^3d^2\), octahedral.
ⓑ. Phosphorus changes from oxidation state \(+5\) to \(+6\), while its formal charge remains zero.
ⓒ. Phosphorus retains oxidation state \(+5\) and formal charge zero because an incoming chloride ion cannot modify formal-charge assignments.
ⓓ. Phosphorus changes from \(sp^3d\) to \(sp^3\) because accepting an electron pair reduces the number of bonding directions.
Correct Answer: Phosphorus retains oxidation state \(+5\), its formal charge changes from \(0\) to \(-1\), and its conventional description changes from \(sp^3d\), trigonal bipyramidal, to \(sp^3d^2\), octahedral.
Explanation: \(\textbf{Determine the oxidation state in PCl₅:}\)
Chlorine is assigned oxidation state \(-1\).
For neutral phosphorus pentachloride:
\[
x+5(-1)=0
\]
\[
x=+5
\]
\(\textbf{Determine the oxidation state in PCl₆⁻:}\)
For hexachlorophosphate:
\[
x+6(-1)=-1
\]
\[
x=+5
\]
Thus, the formal oxidation state of phosphorus remains \(+5\).
\(\textbf{Compare the Lewis formal charges:}\)
In the conventional PCl₅ structure, phosphorus has five single bonds and no lone pair.
Its formal charge is:
\[
FC_{\mathrm P}=5-0-\frac{10}{2}
\]
\[
FC_{\mathrm P}=0
\]
In the conventional \(\mathrm{PCl_6^-}\) structure, phosphorus has six single bonds and no lone pair.
Its formal charge becomes:
\[
FC_{\mathrm P}=5-0-\frac{12}{2}
\]
\[
FC_{\mathrm P}=-1
\]
The six chlorine atoms have formal charge zero in this representation.
The total formal charge is therefore \(-1\), as required.
\(\textbf{Track the coordination change:}\)
Phosphorus in gaseous PCl₅ has five bonding domains.
Its molecular arrangement is trigonal bipyramidal.
The conventional hybridization label is \(sp^3d\).
After accepting the additional chloride, phosphorus has six bonding directions.
The product ion has an octahedral arrangement.
Its conventional textbook hybridization label is \(sp^3d^2\).
\(\textbf{Interpret the non-obvious distinction:}\)
The reaction increases the phosphorus coordination number.
It also changes the Lewis formal charge assigned to phosphorus.
However, it does not change the phosphorus oxidation state.
Formal charge assumes equal sharing of bonding electrons for bookkeeping, whereas oxidation state formally assigns the bonding electrons towards chlorine.
The two quantities therefore need not change together.
The conventional hybridization labels describe the corresponding directional arrangements without independently proving a specific detailed d-orbital bonding mechanism.
Option A correctly combines these distinctions.
309. Nitrous oxide, \(\mathrm{N_2O}\), is a linear molecule with N–N–O connectivity. Consider the following conventional resonance contributors:
Contributor P: \(\mathrm{N\equiv N^+-O^-}\), with a triple N–N bond and a single N–O bond. Contributor Q: \(\mathrm{^-N=N^+=O}\), with double N–N and N–O bonds. Assume that all atoms satisfy the octet rule in both contributors. Oxygen is more electronegative than nitrogen.
Which statement correctly evaluates their resonance relationship and their expected qualitative importance?
ⓐ. P and Q are structural isomers because the formal negative charge occupies different atoms.
ⓑ. Q must dominate because placing negative formal charge on nitrogen is always preferable to placing it on oxygen.
ⓒ. P and Q are valid resonance contributors with unchanged nuclear connectivity; P is generally favoured by placing the negative formal charge on the more electronegative oxygen atom.
ⓓ. P and Q must contribute equally because they contain the same number of atoms and the same total number of electrons.
Correct Answer: P and Q are valid resonance contributors with unchanged nuclear connectivity; P is generally favoured by placing the negative formal charge on the more electronegative oxygen atom.
Explanation: \(\textbf{Verify the electron count:}\)
Nitrous oxide contains two nitrogen atoms and one oxygen atom.
Its total number of valence electrons is:
\[
N_v=2(5)+6
\]
\[
N_v=16
\]
Both proposed contributors use the same sixteen valence electrons.
They also preserve the N–N–O atomic connectivity.
\(\textbf{Examine Contributor P:}\)
The terminal nitrogen forms a triple bond and retains one lone pair.
Its formal charge is zero.
The central nitrogen forms four bonds in total and has formal charge \(+1\).
The terminal oxygen forms one bond and possesses three lone pairs.
Its formal charge is \(-1\).
The net formal charge is zero, and the octets are satisfied.
\(\textbf{Examine Contributor Q:}\)
The terminal nitrogen forms a double bond and possesses two lone pairs.
Its formal charge is \(-1\).
The central nitrogen forms two double bonds and has formal charge \(+1\).
The terminal oxygen forms a double bond and possesses two lone pairs.
Its formal charge is zero.
The total formal charge is again zero.
The proposed arrangement also satisfies the octet rule.
\(\textbf{Apply the resonance criterion:}\)
The two contributors differ in the distribution of bonding and nonbonding electrons.
The nuclei remain in the same order and connectivity.
Therefore, P and Q are resonance contributors rather than structural isomers.
\(\textbf{Compare their expected contributions:}\)
Both contributors contain a positive and a negative formal charge.
However, P places the negative formal charge on oxygen.
Q places it on the terminal nitrogen.
Oxygen is more electronegative than nitrogen and generally accommodates negative formal charge more favourably when other important conditions are comparable.
Thus, P is the more favourable contributor on this conventional qualitative criterion.
\(\textbf{Interpret the resonance hybrid:}\)
The actual molecule is not an equal-time mixture of independently existing P and Q structures.
It is represented by a resonance hybrid.
Valid contributors need not have equal importance merely because they contain the same atoms and total number of electrons.
The relative distribution of formal charge provides one important criterion for judging their contributions.
Therefore, option C is correct.
310. The nitrate ion, \(\mathrm{NO_3^-}\), has three equivalent oxygen positions in its trigonal-planar structure. Its conventional resonance contributors each contain one N=O double bond and two N–O single bonds. Suppose a student uses these three equivalent contributors to infer the average bond multiplicity and the arrangement of nitrogen orbitals. Which statement is most appropriate?
ⓐ. Every N–O bond has permanent bond order 1, because the actual ion contains two single bonds in each contributor.
ⓑ. Every N–O bond has permanent bond order 2, because the double bond can occupy any oxygen position.
ⓒ. Nitrogen is \(sp^3\)-hybridized because the three oxygen atoms and the negative ionic charge create four electron domains.
ⓓ. The three N–O bonds are equivalent; the simple resonance-based average bond multiplicity is \(\frac43\), and nitrogen has a three-domain, approximately \(sp^2\)-type framework.
Correct Answer: The three N–O bonds are equivalent; the simple resonance-based average bond multiplicity is \(\frac43\), and nitrogen has a three-domain, approximately \(sp^2\)-type framework.
Explanation: \(\textbf{Identify the resonance contributors:}\)
A conventional nitrate contributor contains one N=O double bond and two N–O single bonds.
The position of the double bond can be assigned to any one of the three oxygen atoms.
These assignments generate three equivalent contributors.
The actual nitrate ion is represented by their resonance hybrid.
\(\textbf{Evaluate the bond equivalence:}\)
No oxygen position is distinguished in the symmetrical nitrate ion.
Therefore, the three N–O bonds are equivalent.
They do not consist of one permanently double bond and two permanently single bonds.
Their bonding character is intermediate between the individual localized single- and double-bond descriptions.
\(\textbf{Infer the simple resonance-average multiplicity:}\)
Each conventional contributor contains a total N–O bond-order sum of:
\[
2+1+1=4
\]
The three equivalent N–O connections share this conventional bond-order contribution in the resonance description.
Therefore:
\[
\overline{b}_{\mathrm{N-O}}=\frac{4}{3}
\]
\(\textbf{Identify the nitrogen electron domains:}\)
Nitrogen is directly connected to three oxygen atoms.
Each N–O bonding region counts as one geometrical domain.
Nitrogen has no central lone pair in the conventional octet-satisfying representations.
The total domain count is three.
The molecular arrangement is trigonal planar.
In the conventional hybrid-orbital model, a three-directional sigma framework is described using \(sp^2\) hybridization.
\(\textbf{Interpret the unhybridized orbital contribution:}\)
An \(sp^2\) description leaves one p orbital unhybridized on nitrogen.
The pi-electron distribution is not correctly described as a permanently localized N=O pi bond to one particular oxygen.
The resonance hybrid instead represents delocalization among the oxygen positions.
\(\textbf{Recognise the numerical limitation:}\)
The fraction \(\frac43\) is the simple average bond multiplicity inferred from the equivalent Lewis contributors.
It is not a claim that a fraction of a localized double bond appears on one bond at a particular instant.
Nor should every measured bond property be calculated by directly averaging separate single- and double-bond data.
Thus, option D correctly combines resonance equivalence, bond multiplicity and hybridization.
311. Two conventional Lewis contributors for the sulfite ion, \(\mathrm{SO_3^{2-}}\), are proposed.
Contributor P: Sulfur has three S–O single bonds and one lone pair. Each oxygen has three lone pairs. Contributor Q: Sulfur has one S=O double bond, two S–O single bonds and one lone pair. The doubly bonded oxygen has two lone pairs, while each singly bonded oxygen has three. In both contributors, sulfur is connected to the same three oxygen atoms.
Which evaluation correctly compares formal charges and the likely relative importance of these contributors within the conventional expanded-octet Lewis model?
ⓐ. P places formal charge \(+1\) on sulfur and \(-1\) on each oxygen; Q places formal charge zero on sulfur and the double-bonded oxygen, with \(-1\) on each singly bonded oxygen. Q reduces formal-charge separation.
ⓑ. P places formal charge \(-1\) on sulfur and zero on every oxygen; Q places \(+2\) on sulfur. Therefore, P must dominate.
ⓒ. Both contributors have identical formal-charge distributions because changing a single bond into a double bond never affects formal charge.
ⓓ. Q is invalid because any sulfur atom forming more than four shared electron pairs must violate the second-period octet restriction.
Correct Answer: P places formal charge \(+1\) on sulfur and \(-1\) on each oxygen; Q places formal charge zero on sulfur and the double-bonded oxygen, with \(-1\) on each singly bonded oxygen. Q reduces formal-charge separation.
Explanation: \(\textbf{Identify the ion's total charge:}\)
The sulfite ion has overall charge \(-2\).
Any proposed Lewis contributor must have formal charges that sum to \(-2\).
Both P and Q preserve the same sulfur–oxygen connectivity.
They differ in the distribution of bonding and nonbonding electrons.
\(\textbf{Calculate the formal charge in P:}\)
Sulfur possesses one lone pair and forms three single bonds.
Therefore:
\[
FC_{\mathrm S}=6-2-\frac62
\]
\[
FC_{\mathrm S}=+1
\]
Each singly bonded oxygen possesses three lone pairs.
For each oxygen:
\[
FC_{\mathrm O}=6-6-\frac22
\]
\[
FC_{\mathrm O}=-1
\]
The total is:
\[
(+1)+3(-1)=-2
\]
Therefore, P is consistent with the ionic charge.
\(\textbf{Calculate the formal charges in Q:}\)
Sulfur has one lone pair, one double bond and two single bonds.
The total number of bonding electrons around sulfur is eight.
Therefore:
\[
FC_{\mathrm S}=6-2-\frac82
\]
\[
FC_{\mathrm S}=0
\]
The doubly bonded oxygen has formal charge zero.
Each singly bonded oxygen has formal charge \(-1\).
The sum remains:
\[
0+0+2(-1)=-2
\]
\(\textbf{Compare charge separation:}\)
P contains a positively charged sulfur centre and three negatively charged oxygen centres.
Q eliminates the positive formal charge on sulfur and one negative oxygen formal charge.
Therefore, Q has less formal-charge separation.
\(\textbf{Interpret the conventional contributor preference:}\)
Reducing unnecessary formal-charge separation is an important criterion for evaluating Lewis contributors.
The conventional expanded-octet description permits the sulfur representation used in Q.
The second-period octet restriction that applies to atoms such as nitrogen does not transfer unchanged to this textbook sulfur representation.
The three oxygen positions are equivalent, so the double bond of Q may be placed at any one of them in corresponding resonance contributors.
The actual sulfite ion is represented by a resonance hybrid rather than one permanently localized contributor.
Option A correctly identifies the formal-charge distributions and their conventional qualitative comparison.
312. A hypothetical resonance-stabilized species has two valid Lewis contributors, P and Q, whose estimated energies relative to a common separated-atom reference are \(-350\) and \(-390\ \mathrm{kJ\,mol^{-1}}\), respectively. A model calculation assigns the actual resonance hybrid an energy of \(-475\ \mathrm{kJ\,mol^{-1}}\) relative to the same reference.
The vertical axis is energy, increasing upward. The separated-atom reference is at \(0\ \mathrm{kJ\,mol^{-1}}\). The level assigned to contributor P is at \(-350\ \mathrm{kJ\,mol^{-1}}\). The level assigned to contributor Q is at \(-390\ \mathrm{kJ\,mol^{-1}}\). The resonance-hybrid level is lowest, at \(-475\ \mathrm{kJ\,mol^{-1}}\). These levels are model estimates, not separately isolable molecular forms.
Using the most stable individual contributor as the comparison reference, what is the resonance stabilization energy in magnitude, and which interpretation is correct?
ⓐ. \(40\ \mathrm{kJ\,mol^{-1}}\); the hybrid alternates between P and Q at different times.
ⓑ. \(85\ \mathrm{kJ\,mol^{-1}}\); the hybrid is stabilized below the lowest-energy individual contributor in the stated model.
ⓒ. \(125\ \mathrm{kJ\,mol^{-1}}\); the hybrid must contain exactly half the electron density of each contributor.
ⓓ. \(475\ \mathrm{kJ\,mol^{-1}}\); resonance stabilization is always measured from the separated-atom reference.
Correct Answer: \(85\ \mathrm{kJ\,mol^{-1}}\); the hybrid is stabilized below the lowest-energy individual contributor in the stated model.
Explanation: \(\textbf{Identify the appropriate reference contributor:}\)
The two individual contributors have different estimated energies.
P lies at:
\[
E_P=-350\ \mathrm{kJ\,mol^{-1}}
\]
Q lies at:
\[
E_Q=-390\ \mathrm{kJ\,mol^{-1}}
\]
Q has the lower energy.
Therefore, Q is the more stable individual contributor according to the supplied model.
\(\textbf{Identify the resonance-hybrid energy:}\)
The actual hybrid is assigned:
\[
E_{\mathrm{hybrid}}=-475\ \mathrm{kJ\,mol^{-1}}
\]
This lies below both individual contributor levels.
\(\textbf{Calculate the stabilization relative to Q:}\)
The resonance stabilization energy in magnitude is the difference between the lowest-energy individual contributor and the hybrid.
\[
E_{\mathrm{stabilization}}=E_Q-E_{\mathrm{hybrid}}
\]
Substituting:
\[
E_{\mathrm{stabilization}}=(-390)-(-475)
\]
\[
E_{\mathrm{stabilization}}=85\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Distinguish the alternative energy differences:}\)
The difference between P and Q is:
\[
E_P-E_Q=(-350)-(-390)
\]
\[
E_P-E_Q=40\ \mathrm{kJ\,mol^{-1}}
\]
This compares two contributors rather than comparing the most stable contributor with the actual hybrid.
The energy difference between P and the hybrid would be \(125\ \mathrm{kJ\,mol^{-1}}\), but P is not the stipulated reference contributor.
\(\textbf{Interpret the physical meaning:}\)
The resonance hybrid is not a separate molecular form that physically alternates with P or Q.
The contributors are alternative electron-accounting representations of the same nuclear arrangement.
Their energies in this problem are theoretical model estimates rather than experimentally isolated energies of separate structures.
The lower hybrid energy represents the additional stabilization associated with the delocalized electronic description.
This does not imply equal contributions from P and Q.
The separated-atom reference establishes a common energy scale, but it is not the reference used for the requested resonance stabilization.
\(\textbf{Final answer:}\)
\[
\boxed{E_{\mathrm{stabilization}}=85\ \mathrm{kJ\,mol^{-1}}}
\]
313. Three pairs of structural representations are examined.
Pair P: Two Lewis representations of \(\mathrm{NO_2^-}\) differ only in which oxygen has the N=O double bond and which oxygen carries negative formal charge. Pair Q: \(\mathrm{CH_3-C(=O)-CH_3}\) and \(\mathrm{CH_2=C(OH)-CH_3}\) differ in the position of a hydrogen atom and the arrangement of double bonds. Pair R: \(\mathrm{H-C\equiv N}\) and \(\mathrm{H-N^+\equiv C^-}\) differ in which atom is directly bonded to hydrogen.
Which classification correctly distinguishes resonance from structural rearrangement?
ⓐ. P, Q and R are all resonance pairs because each pair contains the same elements.
ⓑ. P and Q are resonance pairs, while R is a pair of structural isomers.
ⓒ. Only P is resonance; Q involves keto–enol hydrogen migration and R changes atomic connectivity.
ⓓ. Only R is a resonance pair because its two representations have different formal-charge distributions.
Correct Answer: Only P is resonance; Q involves keto–enol hydrogen migration and R changes atomic connectivity.
Explanation: \(\textbf{Establish the resonance criterion:}\)
Valid resonance contributors represent alternative distributions of electrons for the same arrangement of atomic nuclei.
The positions and connectivity of the nuclei must remain unchanged.
Only the bonding and nonbonding electron assignments change.
Different formal-charge distributions are possible, but formal-charge redistribution alone is not sufficient if nuclear connectivity also changes.
\(\textbf{Evaluate Pair P:}\)
Both nitrite representations contain nitrogen directly bonded to the same two oxygen atoms.
One contributor places the N=O double bond on one side.
The other places it on the opposite side.
The associated negative formal-charge assignment shifts correspondingly.
No atom changes its position or identity in the nuclear framework.
Therefore, the two drawings are valid resonance contributors.
The actual nitrite ion is described by their resonance hybrid, with equivalent N–O bonds.
\(\textbf{Evaluate Pair Q:}\)
The first structure is a ketone.
The second is its enol form.
A hydrogen atom changes its attachment from carbon to oxygen.
The arrangement of double bonds also changes.
Because a hydrogen nucleus changes its bonding connectivity, the two structures cannot be resonance contributors.
They are related through keto–enol tautomerism.
This involves a chemical interconversion between distinct structural forms rather than alternative electron-only descriptions of one fixed nuclear arrangement.
\(\textbf{Evaluate Pair R:}\)
In HCN, hydrogen is directly bonded to carbon.
In HNC, hydrogen is directly bonded to nitrogen.
The central atom therefore differs between the two connectivities.
These are distinct structural isomers.
The formal-charge distribution in HNC does not transform the HCN and HNC structures into resonance contributors.
The location of the hydrogen nucleus is decisive.
\(\textbf{Integrate the classifications:}\)
Equal elemental composition is not sufficient for resonance.
Similarly, a changed formal-charge assignment is not sufficient when the nuclear connectivity changes.
The distinguishing test is whether the structures preserve the same atomic framework while differing only in electron distribution.
Only Pair P satisfies that condition.
Therefore, option C is correct.
314. Two hypothetical homonuclear diatomic molecules, P₂ and Q₂, have the following potential-energy characteristics. All energies are measured relative to the corresponding separated ground-state atoms.
| Property | P₂ | Q₂ |
|---|
| Equilibrium internuclear separation | \(80\ \mathrm{pm}\) | \(115\ \mathrm{pm}\) |
| Minimum potential energy | \(-420\ \mathrm{kJ\,mol^{-1}}\) | \(-510\ \mathrm{kJ\,mol^{-1}}\) |
| Energy at a specified distorted separation | \(-100\ \mathrm{kJ\,mol^{-1}}\) at \(60\ \mathrm{pm}\) | \(-300\ \mathrm{kJ\,mol^{-1}}\) at \(140\ \mathrm{pm}\) |
Assume that both molecules remain on their respective ground-state potential-energy curves during the specified distortions. Which conclusion correctly compares their bond lengths, dissociation energies and distortion-energy requirements?
ⓐ. P₂ has the shorter and stronger bond; distortion of P₂ requires \(320\ \mathrm{kJ\,mol^{-1}}\), while distortion of Q₂ requires \(210\ \mathrm{kJ\,mol^{-1}}\).
ⓑ. Q₂ has the shorter and stronger bond; distortion of P₂ requires \(520\ \mathrm{kJ\,mol^{-1}}\), while distortion of Q₂ requires \(810\ \mathrm{kJ\,mol^{-1}}\).
ⓒ. P₂ has the shorter bond but Q₂ has the greater dissociation energy; the specified distortions release \(320\) and \(210\ \mathrm{kJ\,mol^{-1}}\), respectively.
ⓓ. P₂ has the shorter bond but Q₂ has the greater dissociation energy; the specified distortions require \(320\) and \(210\ \mathrm{kJ\,mol^{-1}}\), respectively.
Correct Answer: P₂ has the shorter bond but Q₂ has the greater dissociation energy; the specified distortions require \(320\) and \(210\ \mathrm{kJ\,mol^{-1}}\), respectively.
Explanation: \(\textbf{Identify the equilibrium bond lengths:}\)
The equilibrium separation corresponds to the minimum of each potential-energy curve.
For P₂:
\[
r_P=80\ \mathrm{pm}
\]
For Q₂:
\[
r_Q=115\ \mathrm{pm}
\]
Therefore, P₂ has the shorter equilibrium bond.
\(\textbf{Compare the dissociation energies:}\)
Both curves use separated ground-state atoms as their zero-energy references.
The dissociation energy is consequently the energy difference between the minimum and zero.
For P₂:
\[
D_P=0-(-420)
\]
\[
D_P=420\ \mathrm{kJ\,mol^{-1}}
\]
For Q₂:
\[
D_Q=0-(-510)
\]
\[
D_Q=510\ \mathrm{kJ\,mol^{-1}}
\]
Therefore, Q₂ has the greater dissociation energy.
\(\textbf{Determine the compression cost for P₂:}\)
The energy changes from \(-420\) to \(-100\ \mathrm{kJ\,mol^{-1}}\).
\[
\Delta E_P=E_{\mathrm{distorted}}-E_{\mathrm{equilibrium}}
\]
\[
\Delta E_P=-100-(-420)
\]
\[
\Delta E_P=+320\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Determine the stretching cost for Q₂:}\)
The energy changes from \(-510\) to \(-300\ \mathrm{kJ\,mol^{-1}}\).
\[
\Delta E_Q=-300-(-510)
\]
\[
\Delta E_Q=+210\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Interpret the comparison:}\)
Both distortions raise the potential energy above the respective equilibrium minima.
Therefore, both require energy.
The shorter equilibrium bond is not necessarily the stronger bond when different chemical systems are compared.
Bond length and bond dissociation energy are related to the complete bonding environment rather than linked by an exception-free numerical rule.
The supplied curves establish their properties independently.
\(\textbf{Final answer:}\)
\[
\boxed{r_P\lt r_Q,\quad D_Q\gt D_P,\quad\Delta E_P=+320,\quad\Delta E_Q=+210\ \mathrm{kJ\,mol^{-1}}}
\]
315. Two atoms A and B each possess a half-filled p orbital. The internuclear axis is defined as the z-axis. Three relative orbital arrangements are proposed.
Arrangement P: Both atoms use \(p_z\) orbitals, with their axes directed along the A–B internuclear axis. Arrangement Q: Both atoms use \(p_x\) orbitals, whose axes are parallel to each other and perpendicular to the A–B axis. Arrangement R: Atom A uses a \(p_x\) orbital while atom B uses a \(p_y\) orbital; the two p-orbital axes are mutually perpendicular and both perpendicular to the internuclear z-axis. In all arrangements, electron spins permit ordinary covalent pairing.
Which classification follows from the symmetry and orientation of the overlapping orbitals?
ⓐ. P gives a sigma-type interaction, Q gives a pi-type interaction, and R gives zero net overlap in the ideal orthogonal arrangement.
ⓑ. P and Q both give sigma-type interactions, while R gives a pi-type interaction.
ⓒ. P gives a pi-type interaction, Q gives a sigma-type interaction, and R gives a stronger pi interaction than Q.
ⓓ. All three arrangements give equivalent interactions because every p orbital has the same shape.
Correct Answer: P gives a sigma-type interaction, Q gives a pi-type interaction, and R gives zero net overlap in the ideal orthogonal arrangement.
Explanation: The symmetry of the atomic orbitals must be considered relative to the internuclear axis.
The orientation of that axis is not determined by the orbital labels alone.
Here, the internuclear direction is explicitly defined as z.
\(\textbf{Examine Arrangement P:}\)
The two \(p_z\) orbitals point along the A–B axis.
Their lobes can overlap directly in the internuclear direction.
This produces an axially symmetrical bonding interaction of sigma type.
The electron density is associated with overlap along the bond axis.
\(\textbf{Examine Arrangement Q:}\)
The two \(p_x\) orbitals are perpendicular to the internuclear axis but parallel to each other.
They can overlap laterally on opposite sides of that axis.
This is the characteristic orientation for a pi-type interaction.
The corresponding electron density has a nodal plane containing the internuclear axis.
\(\textbf{Examine Arrangement R:}\)
The \(p_x\) and \(p_y\) orbital axes are mutually perpendicular.
Their positive and negative wavefunction regions do not provide the required matching symmetry for effective net overlap in the ideal arrangement.
The corresponding overlap integral vanishes by symmetry.
Therefore, R does not provide an ordinary sigma or pi bonding overlap between the specified orbital pair.
\(\textbf{Distinguish orbital occupancy from geometry:}\)
All three arrangements satisfy the stipulated opposite-spin pairing condition.
Nevertheless, their spatial overlap characteristics differ.
Compatible spin pairing alone does not make every orbital orientation equally favourable.
Likewise, equal orbital shapes do not imply equal overlap when their directions differ.
Thus, option A correctly combines orbital orientation and bonding symmetry.
316. In the conventional valence bond description of methane formation, carbon is first represented using a promoted valence configuration and then four equivalent \(sp^3\) hybrid orbitals. Consider the following statements.
Statement I: The ground-state carbon configuration \(2s^2\,2p^2\) contains only two unpaired valence electrons in the elementary orbital-box description.
Statement II: Promotion of one \(2s\) electron to an available \(2p\) orbital produces a model configuration with four unpaired valence electrons.
Statement III: The promotion step releases energy and is therefore, by itself, the cause of methane's stability.
Statement IV: Formation of four C–H bonding interactions can compensate for the promotion energy, and hybridization provides a model for their approximately tetrahedral equivalence.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
Carbon has atomic number six.
Its ground-state electronic configuration is \(1s^2\,2s^2\,2p^2\).
The two \(2s\) electrons are paired.
According to Hund's rule, the two \(2p\) electrons occupy separate degenerate p orbitals with parallel spins.
Therefore, the elementary orbital-box description contains two unpaired valence electrons.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
The conventional promoted configuration is represented as \(2s^1\,2p^3\).
The \(2s\) orbital then contains one electron.
Each of the three \(2p\) orbitals also contains one electron.
Thus, four unpaired valence electrons are available in the model.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
Promotion moves an electron to a higher-energy atomic orbital.
It requires energy rather than releasing energy by itself.
Therefore, the statement incorrectly treats an energetically costly preparatory step as an independent stabilising process.
Statement III is incorrect.
\(\textbf{Evaluate Statement IV:}\)
The promoted valence orbitals can be combined into four equivalent \(sp^3\) hybrid orbitals in the conventional model.
These hybrids point towards tetrahedral directions.
Each can overlap with a hydrogen \(1s\) orbital to form a C–H sigma bond.
The energy released through formation of the molecular bonding arrangement can more than compensate for the promotion cost.
Therefore, the complete molecule may be stable even though an individual preparatory step is endothermic.
Statement IV is correct.
\(\textbf{Overall conclusion:}\)
Promotion and hybridization describe different aspects of the elementary bonding model.
The stability of methane must be considered from the complete bonding process rather than from the promotion step alone.
Statements I, II and IV are correct.
317. A researcher studies two hypothetical covalent bonds using a simplified valence bond model.
| Condition | Bond P | Bond Q |
|---|
| Spatial overlap of the proposed bonding orbitals | Large | Moderate |
| Relative energies of the interacting atomic orbitals | Substantially different | Comparable |
| Electron spins | Compatible with pairing | Compatible with pairing |
A student concludes that P must have the stronger bond solely because its geometrical orbital overlap is greater. Which evaluation is most defensible?
ⓐ. P must have the stronger bond because the greater spatial overlap always outweighs the poorer orbital-energy match.
ⓑ. Q must have the stronger bond because closer orbital energies necessarily outweigh P's greater spatial overlap.
ⓒ. The ranking is indeterminate: overlap, energy matching, attraction and repulsion all affect the relative bond strengths.
ⓓ. Neither bond can be compared until the overlapping orbitals have exactly equal energies and identical spatial shapes.
Correct Answer: The ranking is indeterminate: overlap, energy matching, attraction and repulsion all affect the relative bond strengths.
Explanation: Valence bond theory relates covalent bond formation to interactions between suitable atomic orbitals.
Effective spatial overlap is an important condition for the development of a stabilising bonding interaction.
However, the final bond energy does not depend on geometrical overlap alone.
The relative energies of the interacting orbitals influence how effectively their electronic distributions can combine.
\(\textbf{Examine Bond P:}\)
P possesses the greater geometrical overlap.
That is a potentially favourable factor.
However, its atomic orbital energies are substantially different.
The extent to which this difference affects the resulting bond cannot be quantified from the information supplied.
\(\textbf{Examine Bond Q:}\)
Q possesses only moderate overlap.
Nevertheless, its interacting orbital energies are comparable.
This is another favourable condition for effective orbital interaction.
The supplied description does not quantify how its energetic advantage compares with P's greater spatial overlap.
\(\textbf{Consider the complete energy balance:}\)
The strength of a stable bond reflects the potential-energy minimum relative to an appropriate dissociation reference.
Attractive electron–nucleus interactions, electron–electron repulsions and nucleus–nucleus repulsion all contribute.
Therefore, no single qualitative feature can universally determine the bond-strength ordering.
\(\textbf{Identify the student's error:}\)
The student treats one favourable condition as sufficient to establish the complete energetic outcome.
The opposite extreme would also be incorrect: comparable orbital energies do not guarantee that Q must be stronger.
Exact equality of atomic orbital energies is not a universal prerequisite for heteronuclear covalent bonding.
Therefore, option C correctly identifies the limitation of the supplied evidence.
318. The carbon atoms in ethyne, ethene and ethane have conventional hybridizations \(sp\), \(sp^2\) and \(sp^3\), respectively. Suppose the three molecules are compared using the following structural features:
| Molecule | Carbon–carbon connection | Carbon sigma-bonding directions per atom |
|---|
| Ethyne | Triple bond | 2 |
| Ethene | Double bond | 3 |
| Ethane | Single bond | 4 |
Which statement correctly integrates the unhybridized p-orbital counts, carbon–carbon pi-bond counts and conventional s-character trends?
ⓐ. Ethyne has two unhybridized p orbitals per carbon and two C–C pi bonds; ethene has one of each; ethane has neither, while hybrid s character decreases in the sequence \(sp\gt sp^2\gt sp^3\).
ⓑ. Ethyne has three unhybridized p orbitals per carbon and three C–C pi bonds; ethene has two of each; ethane has one of each.
ⓒ. The unhybridized p-orbital count is the same in all three molecules because every carbon atom originally has three p orbitals.
ⓓ. Ethane has the greatest hybrid s character because its carbon atoms form the largest number of sigma bonds.
Correct Answer: Ethyne has two unhybridized p orbitals per carbon and two C–C pi bonds; ethene has one of each; ethane has neither, while hybrid s character decreases in the sequence \(sp\gt sp^2\gt sp^3\).
Explanation: \(\textbf{Analyse ethyne:}\)
Each ethyne carbon has two sigma-bonding directions.
It is conventionally \(sp\)-hybridized.
One s and one p orbital form two hybrid orbitals.
Two p orbitals remain unhybridized on each carbon.
Their sideways overlaps provide two pi components of the C≡C triple bond.
Each \(sp\) hybrid has \(50\%\) s character.
\(\textbf{Analyse ethene:}\)
Each ethene carbon has three sigma-bonding directions.
It is conventionally \(sp^2\)-hybridized.
One s and two p orbitals produce three hybrids.
One p orbital remains unhybridized on each carbon.
The remaining p orbitals provide the single pi component of the C=C double bond.
Each \(sp^2\) hybrid has approximately \(33\frac13\%\) s character.
\(\textbf{Analyse ethane:}\)
Each ethane carbon has four sigma-bonding directions.
It is conventionally \(sp^3\)-hybridized.
All three original valence p orbitals participate in the hybrid set.
No unhybridized p orbital remains in this simplified model.
The C–C connection is a single sigma bond with no accompanying pi component.
Each \(sp^3\) hybrid has \(25\%\) s character.
\(\textbf{Identify the combined trend:}\)
The number of sigma-bonding directions increases from two to four across the sequence.
The number of unhybridized p orbitals decreases from two to zero.
The number of C–C pi components similarly decreases from two to zero.
However, the s character per hybrid decreases as the number of hybrids increases.
Therefore, a larger number of sigma bonds around carbon does not imply greater s character in each hybrid.
Option A correctly integrates all three features.
319. In the conventional hybridization model, the carbon atom in carbon dioxide is \(sp\)-hybridized, whereas the carbon atom in carbonate, \(\mathrm{CO_3^{2-}}\), is \(sp^2\)-hybridized. Consider the hypothetical association:
\[
\mathrm{CO_2+O^{2-}\rightarrow CO_3^{2-}}
\]
Which statement correctly tracks the carbon-centred orbital changes and the role of resonance in the product?
ⓐ. Carbon changes from \(sp\) to \(sp^3\), because carbonate contains an additional oxide ion and an overall charge of \(-2\).
ⓑ. Carbon changes from \(sp^2\) to \(sp\), and two independent carbon p orbitals become available for additional pi bonds in carbonate.
ⓒ. Carbon retains \(sp\) hybridization, because both reactant and product contain carbon–oxygen multiple bonding.
ⓓ. Carbon changes \(sp\rightarrow sp^2\): two to three sigma directions, two to one unhybridized p orbitals; carbonate has three resonance-equivalent C–O bonds.
Correct Answer: Carbon changes \(sp\rightarrow sp^2\): two to three sigma directions, two to one unhybridized p orbitals; carbonate has three resonance-equivalent C–O bonds.
Explanation: \(\textbf{Examine the carbon dioxide structure:}\)
Carbon dioxide has the connectivity O=C=O.
Each C=O double bond provides one sigma-bonding direction.
Carbon consequently has two sigma-bonding directions.
The conventional hybridization is \(sp\).
Two unhybridized p orbitals remain on carbon.
They participate in the pi-bonding framework associated with the two C=O connections.
\(\textbf{Follow formation of carbonate:}\)
The product contains carbon directly bonded to three oxygen atoms.
The new carbon–oxygen connection introduces an additional sigma-bonding direction.
The carbon centre therefore has three bonding domains.
The conventional hybridization becomes \(sp^2\).
Only one carbon p orbital remains unhybridized.
\(\textbf{Analyse the resonance contribution:}\)
The carbonate ion has three equivalent oxygen positions.
Its conventional resonance contributors contain one C=O double bond and two C–O single bonds.
The identity of the double-bonded oxygen changes among the equivalent contributors.
The actual ion is represented by a resonance hybrid.
Its three C–O bonds are therefore equivalent rather than permanently assigned different bond multiplicities.
\(\textbf{Identify the orbital distinction:}\)
The two unhybridized p orbitals associated with the conventional \(sp\) description of CO₂ are reduced to one in carbonate's \(sp^2\) framework.
The product still possesses pi-electron delocalization, but the bonding is distributed among three oxygen positions.
\(\textbf{Interpret the geometrical change:}\)
Carbon dioxide is linear around carbon.
Carbonate is trigonal planar.
The change follows from the number of directly bonded oxygen atoms, not merely from the magnitude of the ionic charge.
Option D correctly connects sigma-domain counting, residual p orbitals, geometry and resonance.
320. A hypothetical carbon compound contains three chemically distinct carbon centres with the following local connections:
Centre P is bonded by four single bonds to four neighbouring atoms. Centre Q forms one double bond and two single bonds. Centre R forms one triple bond and one single bond. Each centre has no lone pair in the stated conventional Lewis representation.
An investigator wishes to assign hybridizations and count the number of unhybridized p orbitals remaining at each carbon. Which sequence is correct for P, Q and R?
ⓐ. \(sp^3/1;\ sp^2/2;\ sp/3\)
ⓑ. \(sp^3/0;\ sp^2/1;\ sp/2\)
ⓒ. \(sp^2/1;\ sp/2;\ sp^3/0\)
ⓓ. \(sp^3/0;\ sp^3/0;\ sp^2/1\)
Correct Answer: \(sp^3/0;\ sp^2/1;\ sp/2\)
Explanation: \(\textbf{Apply the sigma-domain counting principle:}\)
The conventional hybridization assignment depends on the number of sigma-bonding directions and lone-pair domains around the central atom.
A double bond contributes one sigma-bonding direction.
A triple bond also contributes one sigma-bonding direction.
The additional components of multiple bonds are pi interactions.
\(\textbf{Evaluate Centre P:}\)
Four single bonds provide four sigma-bonding directions.
Therefore, P is \(sp^3\)-hybridized.
One s and three p orbitals combine to form four hybrid orbitals.
No valence p orbital remains unhybridized.
\(\textbf{Evaluate Centre Q:}\)
One double bond and two single bonds provide three sigma-bonding directions.
Therefore, Q is \(sp^2\)-hybridized.
One s and two p orbitals form three hybrids.
One p orbital remains unhybridized.
That p orbital can participate in the pi component of the double bond.
\(\textbf{Evaluate Centre R:}\)
One triple bond and one single bond provide two sigma-bonding directions.
Therefore, R is \(sp\)-hybridized.
One s and one p orbital form two hybrids.
Two p orbitals remain unhybridized.
These can participate in the two pi components of the triple bond.
\(\textbf{Check orbital conservation:}\)
Each starting carbon centre has one valence s and three valence p orbitals available in the model.
The number of hybrids plus the number of remaining unhybridized p orbitals is four in every case.
The stated pairs therefore satisfy the orbital-counting requirement.
The correct sequence is P: \(sp^3/0\), Q: \(sp^2/1\), R: \(sp/2\).
Thus, option B is correct.
