501. Two hypothetical species contain three equivalent terminal fluorine atoms bonded to a central atom:
Species P is neutral \(\mathrm{ClF_3}\). Species Q is the hypothetical ion \(\mathrm{XeF_3^-}\), assumed to possess three Xe–F single bonds and to obey the conventional VSEPR electron-domain model. Both are to be treated as isolated species for this comparison.
Which statement correctly predicts their electron-domain counts and molecular shapes?
ⓐ. P has five electron domains and Q has six; both can have T-shaped arrangements of the three bonded fluorine atoms, despite their different underlying electron-domain geometries.
ⓑ. Both have five electron domains and must be trigonal planar because each contains three fluorine atoms.
ⓒ. P has six electron domains and is square pyramidal, whereas Q has five electron domains and is linear.
ⓓ. P has five electron domains and is T-shaped, whereas Q has six electron domains and must be trigonal planar because its three lone pairs occupy mutually perpendicular directions.
Correct Answer: P has five electron domains and Q has six; both can have T-shaped arrangements of the three bonded fluorine atoms, despite their different underlying electron-domain geometries.
Explanation: \(\textbf{Count electrons in chlorine trifluoride:}\)
\[
N_v(\mathrm{ClF_3})=7+3(7)
\]
\[
N_v(\mathrm{ClF_3})=28
\]
Three Cl–F single bonds use six electrons.
The terminal fluorine lone pairs use eighteen electrons.
Four electrons remain on chlorine as two lone pairs.
\(\textbf{Determine P's VSEPR arrangement:}\)
P has three bonding domains and two lone pairs.
Its classification is \(\mathrm{AX_3E_2}\).
Five electron domains give a trigonal-bipyramidal electron-domain arrangement.
The two lone pairs favour equatorial positions.
The remaining fluorine atoms occupy two axial and one equatorial position, producing a T-shaped molecular arrangement.
\(\textbf{Count electrons in the hypothetical xenon ion:}\)
\[
N_v(\mathrm{XeF_3^-})=8+3(7)+1
\]
\[
N_v(\mathrm{XeF_3^-})=30
\]
Three Xe–F single bonds use six electrons.
The fluorine lone pairs use eighteen electrons.
Six electrons remain on xenon as three lone pairs.
\(\textbf{Determine Q's domain geometry:}\)
Q has three bonding domains and three lone-pair domains.
Its VSEPR classification is \(\mathrm{AX_3E_3}\).
The six electron domains are treated using an octahedral arrangement.
\(\textbf{Place the three lone pairs:}\)
Within the ideal octahedral model, the three lone pairs preferentially occupy positions minimizing the strongest lone-pair repulsions.
A representative arrangement places two lone pairs opposite each other and the third perpendicular to them.
The remaining three bond directions then contain one opposite pair and one perpendicular direction.
Their molecular arrangement is T-shaped.
\(\textbf{Identify the significant inference:}\)
The two species have different central electron-domain counts.
Their electron-domain arrangements are therefore different.
Nevertheless, the positions of their bonded atoms can give the same qualitative T-shaped molecular geometry.
Thus, observing a T shape alone need not uniquely identify the total number of central electron domains.
Option A is correct.
502. A hypothetical neutral xenon molecule has the formula \(\mathrm{XeF_2Cl_2}\). Assume a square-planar \(\mathrm{AX_4E_2}\) arrangement with two equivalent Xe–F bonds and two equivalent Xe–Cl bonds.
Arrangement P: The two fluorine atoms occupy opposite positions of the square, and the two chlorine atoms occupy the remaining opposite positions. Arrangement Q: The two fluorine atoms occupy adjacent positions, while the chlorine atoms occupy the other two adjacent positions. Each Xe–F bond-dipole vector has magnitude \(1.50\ \mathrm D\), and each Xe–Cl vector has magnitude \(1.00\ \mathrm D\). All vectors point outward from xenon towards the terminal atoms. Neglect any additional asymmetric electron-density contributions.
Which statement correctly compares the predicted dipole magnitudes?
ⓐ. P has \(0.50\ \mathrm D\) and Q has zero, because opposite terminal atoms always reinforce one another.
ⓑ. P and Q both have zero dipole because any square-planar molecule is nonpolar.
ⓒ. P has \(2.50\ \mathrm D\) and Q has \(0.50\ \mathrm D\), because only the differences in bond-dipole magnitudes matter.
ⓓ. P has zero resultant and Q has approximately \(0.707\ \mathrm D\); the changed positions of the unlike ligands remove complete vector cancellation.
Correct Answer: P has zero resultant and Q has approximately \(0.707\ \mathrm D\); the changed positions of the unlike ligands remove complete vector cancellation.
Explanation: \(\textbf{Identify the square-planar geometry:}\)
The four bond directions lie in one plane.
Adjacent directions are perpendicular.
Opposite directions are antiparallel.
The two central lone pairs occupy positions perpendicular to the molecular plane in the ideal octahedral electron-domain arrangement.
\(\textbf{Analyse Arrangement P:}\)
The fluorine atoms occupy opposite positions.
Their equal outward-directed bond-dipole vectors cancel.
The chlorine atoms also occupy opposite positions.
Their equal bond-dipole vectors cancel separately.
Therefore:
\[
\mu_P=0
\]
\(\textbf{Analyse the adjacent arrangement:}\)
In Q, the two fluorine atoms occupy two adjacent directions.
The two chlorine atoms occupy the opposite pair of directions.
Choose perpendicular axes along the adjacent fluorine bond directions.
\(\textbf{Resolve the horizontal component:}\)
One Xe–F bond contributes \(+1.50\ \mathrm D\).
The opposite Xe–Cl bond contributes \(-1.00\ \mathrm D\).
\[
\mu_x=1.50-1.00
\]
\[
\mu_x=0.50\ \mathrm D
\]
\(\textbf{Resolve the vertical component within the square:}\)
The second Xe–F bond has an analogous opposite Xe–Cl contribution.
\[
\mu_y=1.50-1.00
\]
\[
\mu_y=0.50\ \mathrm D
\]
\(\textbf{Calculate Q's resultant:}\)
The two components are perpendicular.
\[
\mu_Q=\sqrt{\mu_x^2+\mu_y^2}
\]
Substituting:
\[
\mu_Q=\sqrt{(0.50)^2+(0.50)^2}
\]
\[
\mu_Q=\sqrt{0.50}
\]
\[
\mu_Q\approx0.707\ \mathrm D
\]
\(\textbf{Interpret the geometrical change:}\)
Both arrangements are square planar.
Both contain the same numbers of Xe–F and Xe–Cl bonds.
However, the positions of the different bonds determine whether opposite vectors cancel.
Consequently, molecular formula and central geometry alone do not establish the dipole magnitude when different substituents occupy nonequivalent arrangements.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_P=0;\qquad\mu_Q\approx0.707\ \mathrm D}
\]
503. Sulfur hexafluoride, \(\mathrm{SF_6}\), and the hexachlorophosphate ion, \(\mathrm{PCl_6^-}\), each contain six terminal halogen atoms surrounding a central atom. Use conventional single-bond Lewis electron counting and assign fluorine and chlorine oxidation state \(-1\).
Which statement correctly combines their valence-electron counts, VSEPR geometries, central formal charges and central oxidation states?
ⓐ. Both have \(48\) valence electrons, six bond domains, octahedral geometry and no central lone pairs; S has formal charge \(0\)/oxidation state \(+6\), P \(-1\)/\(+5\).
ⓑ. Both contain forty-eight valence electrons and six bonding domains; phosphorus must be square pyramidal because its ion carries a negative charge.
ⓒ. \(\mathrm{SF_6}\) contains forty-eight valence electrons, while \(\mathrm{PCl_6^-}\) contains forty-seven, so one must have an unpaired electron.
ⓓ. Both are tetrahedral because six terminal atoms can be arranged in four hybrid-orbital directions when the central atom has an expanded valence shell.
Correct Answer: Both have \(48\) valence electrons, six bond domains, octahedral geometry and no central lone pairs; S has formal charge \(0\)/oxidation state \(+6\), P \(-1\)/\(+5\).
Explanation: \(\textbf{Count the valence electrons of sulfur hexafluoride:}\)
Sulfur contributes six valence electrons.
Six fluorine atoms contribute forty-two.
\[
N_v(\mathrm{SF_6})=6+6(7)
\]
\[
N_v(\mathrm{SF_6})=48
\]
\(\textbf{Count the valence electrons of hexachlorophosphate:}\)
Phosphorus contributes five valence electrons.
Six chlorine atoms contribute forty-two.
The negative charge adds one electron.
\[
N_v(\mathrm{PCl_6^-})=5+6(7)+1
\]
\[
N_v(\mathrm{PCl_6^-})=48
\]
\(\textbf{Determine their central electron domains:}\)
Each conventional structure contains six central–halogen single bonds.
These bonds use twelve electrons.
The six terminal halogen atoms together possess thirty-six nonbonding electrons.
No central lone pair remains in either structure.
Both have six bonding domains and zero lone-pair domains.
The ideal VSEPR geometry is octahedral.
\(\textbf{Calculate sulfur's formal charge:}\)
Sulfur has no assigned nonbonding electrons and participates in six single bonds.
\[
FC_{\mathrm S}=6-0-\frac{12}{2}
\]
\[
FC_{\mathrm S}=0
\]
\(\textbf{Calculate phosphorus's formal charge:}\)
\[
FC_{\mathrm P}=5-0-\frac{12}{2}
\]
\[
FC_{\mathrm P}=-1
\]
\(\textbf{Determine the oxidation states:}\)
In neutral \(\mathrm{SF_6}\), the six fluorines contribute oxidation-state sum \(-6\).
Therefore, sulfur is \(+6\).
In \(\mathrm{PCl_6^-}\), the six chlorines contribute \(-6\), while the overall ion charge is \(-1\).
Therefore, phosphorus is \(+5\).
\(\textbf{Interpret the comparison:}\)
The two species have identical total valence-electron counts and electron-domain geometries.
However, their central atoms have different original valence-electron counts.
Their formal charges and oxidation states therefore differ.
A shared octahedral structure does not require identical central oxidation states or formal charges.
Option A correctly distinguishes these properties.
504. The two hypothetical isolated ions \(\mathrm{ICl_2^+}\) and \(\mathrm{ICl_2^-}\) have the same I–Cl–Cl atomic composition but different total charges. Assume conventional I–Cl single bonds and terminal chlorine octets.
Which statement correctly compares their iodine lone-pair populations, electron-domain arrangements, molecular shapes and iodine oxidation states?
ⓐ. Both ions contain three iodine lone pairs and are linear because the number of chlorine atoms is unchanged.
ⓑ. \(\mathrm{ICl_2^+}\) is linear with five domains, while \(\mathrm{ICl_2^-}\) is bent with four domains; iodine has oxidation state \(+1\) in both.
ⓒ. Both ions are bent because an iodine centre bearing two chlorine bonds necessarily has two lone pairs, regardless of ionic charge.
ⓓ. \(\mathrm{ICl_2^+}\) has two I lone pairs, four domains, a bent shape and I \(+3\); \(\mathrm{ICl_2^-}\) has three lone pairs, five domains, a linear shape and I \(+1\).
Correct Answer: \(\mathrm{ICl_2^+}\) has two I lone pairs, four domains, a bent shape and I \(+3\); \(\mathrm{ICl_2^-}\) has three lone pairs, five domains, a linear shape and I \(+1\).
Explanation: \(\textbf{Count electrons in the cation:}\)
Iodine contributes seven valence electrons.
The two chlorine atoms contribute fourteen.
The positive charge removes one electron.
\[
N_v(\mathrm{ICl_2^+})=7+2(7)-1
\]
\[
N_v(\mathrm{ICl_2^+})=20
\]
\(\textbf{Distribute the cation electrons:}\)
Two I–Cl single bonds require four electrons.
The terminal chlorine lone pairs require twelve.
Four electrons remain on iodine as two lone pairs.
The central iodine therefore has two bonding domains and two lone-pair domains.
Its VSEPR classification is \(\mathrm{AX_2E_2}\).
The electron-domain arrangement is approximately tetrahedral, and the molecular shape is bent.
\(\textbf{Count electrons in the anion:}\)
\[
N_v(\mathrm{ICl_2^-})=7+2(7)+1
\]
\[
N_v(\mathrm{ICl_2^-})=22
\]
The two I–Cl bonds and terminal chlorine lone pairs again account for sixteen electrons.
Six electrons remain on iodine.
These form three lone pairs.
\(\textbf{Determine the anion geometry:}\)
The central atom has two bonding domains and three lone-pair domains.
Its classification is \(\mathrm{AX_2E_3}\).
Five electron domains are arranged trigonal bipyramidally.
The three lone pairs favour equatorial positions.
The two chlorine atoms occupy axial positions, giving a linear molecular arrangement.
\(\textbf{Determine the oxidation states:}\)
Each chlorine is assigned oxidation state \(-1\).
For the cation:
\[
x-2=+1
\]
\[
x=+3
\]
For the anion:
\[
x-2=-1
\]
\[
x=+1
\]
\(\textbf{Interpret the charge dependence:}\)
Changing the ionic charge changes the total valence-electron count even though the terminal-atom composition remains fixed.
This changes the central lone-pair population and electron-domain arrangement.
Thus, molecular shape cannot be inferred from the number of bonded atoms alone.
Option D correctly identifies both electronic structures and the corresponding oxidation states.
505. Consider the idealized octahedral electron-domain arrangements of \(\mathrm{BrF_5}\) and \(\mathrm{XeF_4}\). Bromine pentafluoride has one central lone pair and five bonding domains. Xenon tetrafluoride has two central lone pairs and four bonding domains.
Count only interactions between a central lone-pair domain and a central bonding domain that are separated by \(90^\circ\). Treat all six electron-domain positions as ideal octahedral directions.
Which statement correctly compares these interaction counts and the resulting molecular shapes?
ⓐ. \(\mathrm{BrF_5}\) has four such interactions and is square pyramidal; \(\mathrm{XeF_4}\) has eight such interactions and is square planar, with its two lone pairs opposite one another.
ⓑ. \(\mathrm{BrF_5}\) has five such interactions and is trigonal bipyramidal; \(\mathrm{XeF_4}\) has four such interactions and is tetrahedral.
ⓒ. Both have eight such interactions because both contain six electron domains; both must be square planar.
ⓓ. \(\mathrm{BrF_5}\) has four such interactions and is square planar; \(\mathrm{XeF_4}\) has six and is square pyramidal.
Correct Answer: \(\mathrm{BrF_5}\) has four such interactions and is square pyramidal; \(\mathrm{XeF_4}\) has eight such interactions and is square planar, with its two lone pairs opposite one another.
Explanation: \(\textbf{Establish the octahedral arrangement:}\)
Six electron domains occupy the six directions towards the vertices of an ideal octahedron.
Each position has one opposite position at \(180^\circ\).
The remaining four positions lie at \(90^\circ\) to it.
This geometrical relationship permits the required interaction counts.
\(\textbf{Analyse bromine pentafluoride:}\)
The central bromine has five bonding domains and one lone pair.
The lone pair occupies one octahedral direction.
The bond opposite it lies at \(180^\circ\).
The other four bonds lie at \(90^\circ\).
Therefore, there are four stipulated \(90^\circ\) lone-pair–bonding-domain interactions.
\(\textbf{Determine the bromine pentafluoride shape:}\)
Five bonded fluorine atoms occupy all positions except the lone-pair position.
Their arrangement consists of a square base and one apical fluorine.
The molecular shape is square pyramidal.
\(\textbf{Analyse xenon tetrafluoride:}\)
The central xenon atom has four bonding domains and two lone pairs.
The two lone pairs are placed opposite one another in the preferred ideal arrangement.
This avoids a \(90^\circ\) lone-pair–lone-pair interaction.
The four fluorine atoms occupy the remaining square-planar positions.
\(\textbf{Count the xenon interactions:}\)
Each central lone pair is perpendicular to all four Xe–F bonding domains.
Therefore, each gives four stipulated \(90^\circ\) lone-pair–bonding-domain interactions.
There are two lone pairs.
\[
N_{90^\circ,\mathrm{LP-BP}}=2(4)
\]
\[
N_{90^\circ,\mathrm{LP-BP}}=8
\]
\(\textbf{Interpret the comparison:}\)
The two species possess the same total number of electron domains.
However, the distribution between bonding and lone-pair domains differs.
The lone-pair positions determine different arrangements of the bonded atoms.
The count of \(90^\circ\) lone-pair–bonding-domain interactions therefore differs as specified.
Option A is correct.
506. Consider an idealized octahedral molecule \(\mathrm{AX_3Y_3}\) with six equivalent geometrical bond directions labelled \(+x,-x,+y,-y,+z,-z\). All bond-dipole vectors point from A towards their terminal atoms. Each A–X bond contributes \(1.40\ \mathrm D\), while each A–Y bond contributes \(0.80\ \mathrm D\).
Arrangement P: X atoms occupy \(+x,+y,+z\), while Y atoms occupy \(-x,-y,-z\). Arrangement Q: X atoms occupy \(+x,-x,+z\), while Y atoms occupy \(-z,+y,-y\). The positions are ideal and all bond lengths and bond-dipole magnitudes are fixed.
Which statement correctly compares the resultant molecular dipole magnitudes?
ⓐ. Both arrangements have zero dipole because each is octahedral and contains three X and three Y atoms.
ⓑ. P has \(0.60\ \mathrm D\), while Q has approximately \(1.04\ \mathrm D\).
ⓒ. P has approximately \(1.04\ \mathrm D\), while Q has \(0.60\ \mathrm D\).
ⓓ. P has \(4.20\ \mathrm D\), while Q has \(2.40\ \mathrm D\).
Correct Answer: P has approximately \(1.04\ \mathrm D\), while Q has \(0.60\ \mathrm D\).
Explanation: \(\textbf{Identify the octahedral vector directions:}\)
The six bond directions occur as three opposite pairs.
Vectors along opposite directions have opposite signs.
The three coordinate axes are mutually perpendicular.
\(\textbf{Analyse Arrangement P:}\)
The three X bond dipoles point along the positive coordinate axes.
The three Y bond dipoles point along the corresponding negative axes.
Therefore, each coordinate axis has the same net contribution.
\(\textbf{Calculate P's x component:}\)
\[
\mu_{x,P}=1.40-0.80
\]
\[
\mu_{x,P}=0.60\ \mathrm D
\]
The same calculation applies to the other two directions:
\[
\mu_{y,P}=0.60\ \mathrm D
\]
\[
\mu_{z,P}=0.60\ \mathrm D
\]
\(\textbf{Calculate P's vector magnitude:}\)
\[
\mu_P=\sqrt{\mu_{x,P}^2+\mu_{y,P}^2+\mu_{z,P}^2}
\]
Substituting:
\[
\mu_P=\sqrt{3(0.60)^2}
\]
\[
\mu_P=0.60\sqrt3
\]
\[
\mu_P\approx1.039\ \mathrm D
\]
\(\textbf{Analyse Arrangement Q:}\)
The X atoms at \(+x\) and \(-x\) produce equal and opposite vectors.
Their contributions cancel.
The Y atoms at \(+y\) and \(-y\) also cancel.
Only the opposite pair along the z-axis contains unlike terminal atoms.
\(\textbf{Calculate Q's remaining component:}\)
\[
\mu_{z,Q}=1.40-0.80
\]
\[
\mu_{z,Q}=0.60\ \mathrm D
\]
All other components are zero.
Therefore:
\[
\mu_Q=0.60\ \mathrm D
\]
\(\textbf{Interpret the geometric distinction:}\)
Both arrangements contain six bonds and the same three-to-three substituent composition.
Both also have ideal octahedral bond directions.
Nevertheless, the distribution of unlike ligands among opposite pairs changes the degree of dipole cancellation.
The arrangement of substituents is therefore essential to the resultant.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_P=0.60\sqrt3\approx1.04\ \mathrm D;\qquad\mu_Q=0.60\ \mathrm D}
\]
507. A hypothetical T-shaped \(\mathrm{AX_2Y}\) molecule has three bonding domains and two equatorial lone pairs in an ideal trigonal-bipyramidal electron-domain arrangement. The bond-dipole magnitudes are \(1.30\ \mathrm D\) for A–X and \(0.50\ \mathrm D\) for A–Y. Consider two possible placements of the bonded atoms:
Arrangement P: Both X atoms occupy the two opposite axial positions, and Y occupies the remaining equatorial bonding position. Arrangement Q: One X occupies the positive axial position, Y occupies the negative axial position, and the second X occupies the equatorial bonding position. Axial and equatorial directions are mutually perpendicular. Treat the supplied bond-dipole magnitudes as fixed.
Which statement correctly compares their resultant dipole magnitudes?
ⓐ. P has \(0.80\ \mathrm D\), while Q has \(1.30\ \mathrm D\).
ⓑ. P has \(0.50\ \mathrm D\), while Q has approximately \(1.53\ \mathrm D\).
ⓒ. Both have \(0.50\ \mathrm D\), because all T-shaped molecules with the same formula possess identical dipoles.
ⓓ. P has zero dipole, while Q has \(2.60\ \mathrm D\), because the two X bonds cancel regardless of their positions.
Correct Answer: P has \(0.50\ \mathrm D\), while Q has approximately \(1.53\ \mathrm D\).
Explanation: \(\textbf{Establish the T-shaped geometry:}\)
An \(\mathrm{AX_3E_2}\) species has five electron domains.
The two lone pairs favour equatorial positions in the trigonal-bipyramidal arrangement.
The three bonded atoms occupy two opposite axial positions and one equatorial position.
The axial and equatorial bond directions are perpendicular.
\(\textbf{Examine Arrangement P:}\)
Both axial positions contain X.
Their equal bond-dipole vectors point in opposite directions.
Therefore, the axial contributions cancel.
The only remaining contribution is the equatorial A–Y dipole.
\[
\mu_P=\mu_Y
\]
\[
\mu_P=0.50\ \mathrm D
\]
\(\textbf{Examine Arrangement Q:}\)
One axial position contains X.
The opposite axial position contains Y.
Their bond-dipole magnitudes differ.
Therefore, the axial contributions do not cancel completely.
\(\textbf{Calculate the residual axial component:}\)
\[
\mu_{\mathrm{axial},Q}=\mu_X-\mu_Y
\]
Substituting:
\[
\mu_{\mathrm{axial},Q}=1.30-0.50
\]
\[
\mu_{\mathrm{axial},Q}=0.80\ \mathrm D
\]
\(\textbf{Identify the perpendicular component:}\)
The remaining X atom occupies the equatorial bonding position.
Its contribution is perpendicular to the axial resultant.
\[
\mu_{\mathrm{equatorial},Q}=1.30\ \mathrm D
\]
\(\textbf{Calculate Q's total dipole:}\)
\[
\mu_Q=\sqrt{\mu_{\mathrm{axial},Q}^2+\mu_{\mathrm{equatorial},Q}^2}
\]
Substituting:
\[
\mu_Q=\sqrt{(0.80)^2+(1.30)^2}
\]
\[
\mu_Q=\sqrt{0.64+1.69}
\]
\[
\mu_Q=\sqrt{2.33}
\]
\[
\mu_Q\approx1.526\ \mathrm D
\]
\(\textbf{Interpret the structural effect:}\)
Both arrangements have the same molecular formula, number of lone pairs and ideal T-shaped geometry.
The relative positions of the different terminal atoms nevertheless affect vector cancellation.
Therefore, these structural details can change molecular polarity without changing the central VSEPR classification.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_P=0.50\ \mathrm D;\qquad\mu_Q\approx1.53\ \mathrm D}
\]
508. Hydrogen peroxide has the connectivity \(\mathrm{H-O-O-H}\). Each oxygen atom forms two sigma bonds and possesses two lone pairs. Consider two hypothetical conformations in which the O–O and O–H bond lengths and both H–O–O bond angles remain unchanged, but the H–O–O–H dihedral angle differs.
Which statement correctly distinguishes the local VSEPR prediction from the possible change in overall molecular polarity?
ⓐ. Each oxygen remains \(\mathrm{AX_2E_2}\); O–O rotation reorients O–H dipoles and can change the total molecular dipole.
ⓑ. Changing the dihedral angle necessarily converts each oxygen from \(sp^3\) to \(sp\) hybridization because the molecule contains four atoms.
ⓒ. The molecular dipole must remain constant because the O–H bond lengths and H–O–O bond angles are unchanged; torsional orientation never affects vector addition.
ⓓ. Every hydrogen peroxide conformation must be planar because each oxygen possesses two lone pairs, and no bond rotation is possible without breaking an O–H bond.
Correct Answer: Each oxygen remains \(\mathrm{AX_2E_2}\); O–O rotation reorients O–H dipoles and can change the total molecular dipole.
Explanation: \(\textbf{Determine the local oxygen environments:}\)
Each oxygen in hydrogen peroxide forms one O–H bond and one O–O bond.
Each also possesses two lone pairs.
Therefore, each oxygen has four electron domains.
Its local VSEPR classification is \(\mathrm{AX_2E_2}\).
\(\textbf{Identify the local molecular geometry:}\)
The four electron domains have an approximately tetrahedral arrangement.
The two bonded atoms surrounding each oxygen define a bent local geometry.
This description applies to each oxygen separately.
\(\textbf{Examine the conformational change:}\)
The stipulated conformations have identical bond lengths and local H–O–O angles.
They differ only in the torsional orientation around the O–O connection.
Such a change alters the relative spatial positions of the two hydrogen atoms.
It does not require breaking either O–H covalent bond.
\(\textbf{Consider molecular dipole addition:}\)
The molecular dipole depends on the full three-dimensional distribution of molecular charge.
Its bond-dipole contributions must therefore be combined as vectors.
Changing the dihedral angle changes the relative directions of contributions associated with opposite ends of the molecule.
Their degree of cancellation can consequently change.
\(\textbf{Distinguish local and global geometry:}\)
The number of electron domains on either oxygen remains four.
The conventional local hybridization description consequently need not change during the rotation.
However, the global molecular conformation changes.
Local VSEPR classifications alone do not specify the complete torsional geometry of a molecule containing several bonded central atoms.
\(\textbf{Interpret the experimental implication:}\)
Different conformations can possess different overall dipole moments even when their bond lengths and local bond angles are similar.
An exact numerical dipole comparison would require further information about the charge distribution and dihedral angles.
Option A correctly identifies what remains invariant and what can change.
509. Two hypothetical central-atom hybrid-orbital schemes are proposed using the conventional NCERT hybridization model.
| Scheme | Atomic orbitals mixed | Electron-pair occupancy of the resulting hybrids |
|---|
| P | One s, three p and one d | Three bonding pairs and two lone pairs |
| Q | One s, three p and two d | Four bonding pairs and two lone pairs |
Assume that all hybrid orbitals within a scheme share the s-orbital contribution equally, and that the lone pairs occupy positions minimizing their mutual repulsion.
Which statement correctly compares the hybridization labels, ideal molecular shapes and s-character percentages?
ⓐ. P uses \(sp^3d\) hybridization and is T-shaped with \(20\%\) s character per hybrid; Q uses \(sp^3d^2\) hybridization and is square planar with \(16\frac23\%\) s character per hybrid.
ⓑ. P uses \(sp^3d\) hybridization and is trigonal planar with \(25\%\) s character; Q uses \(sp^3d^2\) hybridization and is tetrahedral with \(20\%\) s character.
ⓒ. P and Q both have square-planar molecular shapes because each contains two lone pairs; their s-character percentages are identical.
ⓓ. P uses \(sp^3d^2\) hybridization and is T-shaped, while Q uses \(sp^3d\) hybridization and is square planar; both have \(25\%\) s character.
Correct Answer: P uses \(sp^3d\) hybridization and is T-shaped with \(20\%\) s character per hybrid; Q uses \(sp^3d^2\) hybridization and is square planar with \(16\frac23\%\) s character per hybrid.
Explanation: \(\textbf{Determine the orbital count in P:}\)
Scheme P mixes one s, three p and one d orbital.
The number of hybrid orbitals equals the number of atomic orbitals mixed.
\[
N_{\mathrm{hybrid},P}=1+3+1=5
\]
The conventional designation is \(sp^3d\).
Five electron domains correspond to a trigonal-bipyramidal electron-domain arrangement.
\(\textbf{Determine the molecular shape of P:}\)
Three hybrid orbitals accommodate bonding pairs, while two accommodate lone pairs.
The two lone pairs preferentially occupy equatorial positions because this avoids an unfavourable \(90^\circ\) interaction between them.
The three remaining bonding directions form a T-shaped arrangement.
Therefore, P has a T-shaped molecular geometry rather than a trigonal-planar geometry.
\(\textbf{Calculate the s-character fraction in P:}\)
One s orbital contributes to five equivalent hybrid-orbital compositions in the stipulated idealized model.
\[
f_s(P)=\frac15
\]
\[
\%\mathrm{s}(P)=\frac15\times100=20\%
\]
\(\textbf{Determine the orbital count in Q:}\)
Scheme Q mixes one s, three p and two d orbitals.
\[
N_{\mathrm{hybrid},Q}=1+3+2=6
\]
The conventional designation is \(sp^3d^2\).
Six electron domains correspond to an octahedral electron-domain arrangement.
\(\textbf{Determine the molecular shape of Q:}\)
Four domains contain bonding pairs and two contain lone pairs.
The lone pairs preferentially occupy opposite octahedral positions to avoid a \(90^\circ\) lone-pair–lone-pair interaction.
The four bonded atoms consequently occupy a square plane.
Thus, Q is square planar despite possessing six electron domains.
\(\textbf{Calculate the s-character fraction in Q:}\)
\[
f_s(Q)=\frac16
\]
\[
\%\mathrm{s}(Q)=\frac16\times100
\]
\[
\%\mathrm{s}(Q)=16\frac23\%
\]
\(\textbf{Integrate the comparison:}\)
The same number of central lone pairs does not guarantee the same molecular shape.
The total number of electron domains and the preferred lone-pair positions must also be considered.
Likewise, mixing more orbitals reduces the equal-share s-character fraction under the stipulated model.
The \(sp^3d\) and \(sp^3d^2\) labels here are conventional textbook geometrical descriptions, not proof of a unique modern quantum-mechanical bonding mechanism.
\(\textbf{Final answer:}\)
\[
\begin{aligned}
P:&\ sp^3d,\ \text{T-shaped},\ 20\%\ \mathrm{s}\\
Q:&\ sp^3d^2,\ \text{square planar},\ 16\frac23\%\ \mathrm{s}
\end{aligned}
\]
510. A student proposes the following neutral or ionic species using conventional Lewis structures:
| Species | Proposed central-atom bonding |
|---|
| P | \(\mathrm{NF_5}\), containing five ordinary N–F single bonds |
| Q | \(\mathrm{PF_5}\), containing five ordinary P–F single bonds |
| R | \(\mathrm{NF_4^+}\), containing four ordinary N–F single bonds |
Assume the proposals are evaluated using the ordinary octet restriction for second-period nitrogen and conventional expanded-valence-shell descriptions for phosphorus.
Which assessment correctly distinguishes the three structures?
ⓐ. P, Q and R are all permitted because every atom can form as many single bonds as it has available neighbouring fluorine atoms.
ⓑ. P and Q are permitted \(sp^3d\) structures, but R is forbidden because a positively charged nitrogen cannot form four sigma bonds.
ⓒ. P and R are permitted tetrahedral structures, while Q is forbidden because phosphorus and nitrogen belong to the same group.
ⓓ. P violates nitrogen's second-period octet; Q is five-domain trigonal bipyramidal, whereas R is four-domain tetrahedral.
Correct Answer: P violates nitrogen's second-period octet; Q is five-domain trigonal bipyramidal, whereas R is four-domain tetrahedral.
Explanation: \(\textbf{Examine the proposed nitrogen pentafluoride structure:}\)
Five ordinary N–F single bonds would place five bonding pairs around the central nitrogen.
That corresponds to ten electrons in nitrogen's conventional valence-shell bonding description.
Second-period nitrogen cannot accommodate an expanded octet in an ordinary Lewis structure.
Therefore, the proposed neutral \(\mathrm{NF_5}\) structure is not accepted under the stated model.
\(\textbf{Examine phosphorus pentafluoride:}\)
Phosphorus belongs to the third period.
The conventional NEET bonding model permits a five-bond expanded-valence-shell description for \(\mathrm{PF_5}\).
Five bonding domains and no central lone pair give a trigonal-bipyramidal electron-domain and molecular arrangement.
The traditional hybridization label is \(sp^3d\).
\(\textbf{Count the electrons in tetrafluoronitrogen cation:}\)
Nitrogen contributes five valence electrons.
Four fluorine atoms contribute twenty-eight.
The positive charge removes one electron.
\[
N_v(\mathrm{NF_4^+})=5+4(7)-1
\]
\[
N_v(\mathrm{NF_4^+})=32
\]
Four N–F single bonds use eight electrons.
The fluorine lone pairs account for the remaining twenty-four.
No central nitrogen lone pair remains.
\(\textbf{Determine the cation geometry:}\)
Nitrogen has four bonding domains.
The central octet contains four shared electron pairs, corresponding to eight electrons.
The conventional local hybridization is \(sp^3\).
The molecular geometry is tetrahedral.
\(\textbf{Identify the decisive distinction:}\)
Nitrogen's second-period octet limitation prevents the proposed five-bond structure.
It does not prevent formation of a four-coordinate nitrogen cation.
Phosphorus can be assigned five bonding domains in the conventional expanded-valence-shell description.
Thus, charge, period and electron count must be evaluated together rather than selecting a structure solely from the group number.
Option D is correct.
511. The linear azide ion, \(\mathrm{N_3^-}\), can be represented by the following conventional octet-satisfying Lewis contributors:
Contributor P: \(\mathrm{^-N=N^+=N^-}\), containing two N=N double bonds. Contributor Q: \(\mathrm{N\equiv N^+-N^{2-}}\), containing one triple bond and one single bond. Contributor R: The mirror of Q, with the single and triple bonds interchanged between the two sides of the central nitrogen. In each contributor, the nitrogen nuclei occupy the same linear N–N–N framework.
Which statement correctly evaluates the formal-charge distributions, contributor preference and equivalence of the two N–N connections?
ⓐ. Q and R must be constitutional isomers because their single and triple bonds occur on opposite sides of the central atom.
ⓑ. P avoids a formal charge of \(-2\) on a terminal nitrogen and is conventionally an important contributor; Q and R are mirror-related resonance contributors, and the actual ion has equivalent N–N connections.
ⓒ. P is invalid because two double bonds necessarily require the central nitrogen to possess ten valence-shell electrons.
ⓓ. The actual azide ion must contain one permanently single and one permanently triple N–N bond because Q and R each contain those bond orders.
Correct Answer: P avoids a formal charge of \(-2\) on a terminal nitrogen and is conventionally an important contributor; Q and R are mirror-related resonance contributors, and the actual ion has equivalent N–N connections.
Explanation: \(\textbf{Establish the valence-electron count:}\)
Each nitrogen atom supplies five valence electrons.
The negative charge adds one electron.
\[
N_v(\mathrm{N_3^-})=3(5)+1
\]
\[
N_v(\mathrm{N_3^-})=16
\]
\(\textbf{Examine Contributor P:}\)
The central nitrogen forms two double bonds, giving four shared electron pairs and an octet.
Its formal charge is \(+1\).
Each terminal nitrogen forms one double bond and has two lone pairs.
Each terminal nitrogen therefore has formal charge \(-1\).
The three formal charges sum to \(-1\), matching the ionic charge.
\(\textbf{Examine Contributor Q:}\)
The triply bonded terminal nitrogen has one lone pair and formal charge zero.
The central nitrogen forms one triple bond and one single bond, giving formal charge \(+1\).
The singly bonded terminal nitrogen has three lone pairs and formal charge \(-2\).
Their sum is again \(-1\).
\(\textbf{Compare the contributors:}\)
Contributor R interchanges the terminal bonding patterns of Q.
All three contributors preserve the same N–N–N nuclear connectivity.
Therefore, they are resonance contributors rather than constitutional isomers.
\(\textbf{Identify the important formal-charge criterion:}\)
Contributor P distributes the negative formal charge over the two terminal nitrogens.
Q and R place formal charge \(-2\) on one terminal atom.
Avoiding large formal-charge magnitudes is an important conventional contributor-selection criterion.
P is therefore a particularly important resonance contributor.
\(\textbf{Infer the equivalence of the terminal connections:}\)
The two terminal nitrogen positions are chemically equivalent in the isolated ion.
The actual electronic distribution respects this equivalence.
Consequently, the two N–N connections do not remain permanently single and triple bonds on opposite sides.
Their bonding is represented by the resonance hybrid.
\(\textbf{Connect resonance with geometry:}\)
The central nitrogen has two bonding directions.
A double or triple bond contributes one sigma-bonding direction.
The linear framework is therefore compatible with the conventional two-domain description.
Option B correctly combines formal-charge reasoning, resonance and molecular geometry.
512. Two oxoanions are described using explicitly specified conventional expanded-valence-shell Lewis-resonance models:
For \(\mathrm{SO_4^{2-}}\), each selected contributor has two S=O double bonds and two S–O single bonds, with negative formal charges on the singly bonded oxygen atoms. For \(\mathrm{PO_4^{3-}}\), each selected contributor has one P=O double bond and three P–O single bonds, with negative formal charges on the singly bonded oxygen atoms. In both ions, the four terminal oxygen positions are geometrically equivalent. Count only the contributors defined by these patterns.
Which statement correctly gives the numbers of contributors, the simple resonance-averaged central–oxygen bond multiplicities and the central VSEPR geometries?
ⓐ. Sulfate has four contributors and average S–O multiplicity \(1.25\); phosphate has six contributors and average P–O multiplicity \(1.50\); both are square planar.
ⓑ. Both ions have four contributors and average bond multiplicity \(1.50\), because both contain four oxygen atoms.
ⓒ. Sulfate has six contributors and average S–O multiplicity \(1.50\); phosphate has four contributors and average P–O multiplicity \(1.25\); both have tetrahedral central bonding arrangements.
ⓓ. Sulfate has twelve contributors and average S–O multiplicity \(2.00\); phosphate has three contributors and average P–O multiplicity \(1.00\); both are trigonal pyramidal.
Correct Answer: Sulfate has six contributors and average S–O multiplicity \(1.50\); phosphate has four contributors and average P–O multiplicity \(1.25\); both have tetrahedral central bonding arrangements.
Explanation: \(\textbf{Interpret the stipulated contributor models:}\)
The question defines a particular set of conventional Lewis contributors for each oxoanion.
Their bond-order averages are bookkeeping results within these selected models.
They should not be confused with independently measured, exact quantum-mechanical bond orders.
\(\textbf{Count the sulfate contributors:}\)
Each sulfate contributor has two S=O bonds among four oxygen positions.
The number of ways to choose those positions is:
\[
N_{\mathrm{sulfate}}=\binom42
\]
\[
N_{\mathrm{sulfate}}=\frac{4!}{2!\,2!}
\]
\[
N_{\mathrm{sulfate}}=6
\]
\(\textbf{Calculate sulfate's average multiplicity:}\)
Each specified contributor has two double bonds and two single bonds.
The total formal bond multiplicity across four S–O connections is:
\[
2(2)+2(1)=6
\]
The simple average is:
\[
\overline b_{\mathrm{S-O}}=\frac64
\]
\[
\overline b_{\mathrm{S-O}}=1.50
\]
\(\textbf{Count the phosphate contributors:}\)
Each phosphate contributor contains one P=O bond.
There are four equivalent oxygen positions at which it can be placed.
\[
N_{\mathrm{phosphate}}=\binom41
\]
\[
N_{\mathrm{phosphate}}=4
\]
\(\textbf{Calculate phosphate's average multiplicity:}\)
Each specified contributor contains one double and three single P–O bonds.
\[
\overline b_{\mathrm{P-O}}=\frac{2+3(1)}{4}
\]
\[
\overline b_{\mathrm{P-O}}=\frac54
\]
\[
\overline b_{\mathrm{P-O}}=1.25
\]
\(\textbf{Determine the central geometries:}\)
Each ion contains four central–oxygen sigma-bonding directions.
Multiple bonds count as one bonding domain each in VSEPR.
Neither stipulated central structure requires a lone-pair domain.
Therefore, both central arrangements are tetrahedral.
\(\textbf{Interpret the structural equivalence:}\)
The equivalent oxygen positions permit redistribution of the specified multiple-bond assignments.
The resonance hybrids therefore avoid treating one particular central–oxygen connection as a permanently localized double bond.
The calculated averages reflect the selected conventional contributors rather than literal rapid alternation among structures.
\(\textbf{Final answer:}\)
\[
\begin{aligned}
\mathrm{SO_4^{2-}}:&\ 6\ \text{contributors},\ \overline b=1.50\\
\mathrm{PO_4^{3-}}:&\ 4\ \text{contributors},\ \overline b=1.25\\
&\text{Both central arrangements tetrahedral}
\end{aligned}
\]
513. Nitric acid has the connectivity \(\mathrm{HO-NO_2}\). In the conventional octet-satisfying Lewis description, the central nitrogen is attached to three oxygen atoms, one of which also carries hydrogen. On deprotonation, nitric acid forms the nitrate ion:
\[
\mathrm{HNO_3\rightarrow NO_3^-+H^+}
\]
Which statement correctly compares the resonance equivalence of the N–O bonds and the local nitrogen geometry before and after deprotonation?
ⓐ. In nitric acid, all three N–O bonds are equivalent because nitrogen has three oxygen neighbours; nitrate has two equivalent bonds and one distinct bond.
ⓑ. In nitric acid, the two oxygen atoms not bonded to hydrogen can exchange their N–O bonding assignments through resonance, while the N–OH connection remains distinguishable; in nitrate, all three oxygen positions become equivalent. The central nitrogen remains trigonal planar in both.
ⓒ. Nitric acid has tetrahedral nitrogen because the O–H bond is counted as a fourth central nitrogen electron domain; nitrate is linear because its negative charge adds a lone pair to nitrogen.
ⓓ. Deprotonation breaks an N–O covalent bond, reducing the number of nitrogen bonding domains from three to two.
Correct Answer: In nitric acid, the two oxygen atoms not bonded to hydrogen can exchange their N–O bonding assignments through resonance, while the N–OH connection remains distinguishable; in nitrate, all three oxygen positions become equivalent. The central nitrogen remains trigonal planar in both.
Explanation: \(\textbf{Identify the nitric acid connectivity:}\)
The central nitrogen is directly bonded to three oxygen atoms.
One oxygen is additionally bonded to hydrogen.
This hydroxyl oxygen therefore occupies a structurally different position from the other two oxygens.
\(\textbf{Examine the octet-satisfying Lewis contributors:}\)
Nitrogen is a second-period atom and must obey the ordinary octet restriction.
Its three N–O sigma bonds can be accompanied by one N=O pi component.
The conventional contributors place this pi component on either oxygen not bonded to hydrogen.
A negative formal charge appears on the remaining singly bonded non-hydroxyl oxygen, while nitrogen carries formal charge \(+1\).
\(\textbf{Identify the resonance equivalence in the acid:}\)
The two non-hydroxyl oxygens have equivalent positions.
Interchanging the double-bond assignment between them gives equivalent conventional resonance contributors.
The N–OH connection remains distinguishable because that oxygen retains its O–H bond.
Thus, all three N–O connections are not necessarily equivalent in nitric acid.
\(\textbf{Follow deprotonation:}\)
Removing the hydroxyl proton produces \(\mathrm{NO_3^-}\).
The oxygen that formerly carried hydrogen no longer has an O–H connection.
The three oxygen positions become equivalent in the isolated nitrate ion.
\(\textbf{Examine nitrate resonance:}\)
The conventional nitrate contributors distribute one N=O double-bond assignment among the three oxygen positions.
No one N–O bond remains permanently designated as the only double bond.
The actual ion is a resonance hybrid with three equivalent N–O connections.
\(\textbf{Compare the nitrogen geometries:}\)
Both nitric acid and nitrate have three central nitrogen–oxygen sigma-bonding directions.
Each multiple bond contributes one electron domain in the VSEPR model.
Neither description requires a central nitrogen lone pair.
Therefore, nitrogen has a trigonal-planar arrangement in both species.
\(\textbf{Identify the significant chemical change:}\)
Deprotonation changes the equivalence of the oxygen positions.
It does not change the number of nitrogen sigma-bonding directions.
This illustrates how resonance equivalence can change while central hybridization and local molecular geometry remain the same.
Option B is correct.
514. Ozone, \(\mathrm{O_3}\), has eighteen valence electrons and a bent O–O–O nuclear framework. Its conventional octet-satisfying resonance contributors are represented as \(\mathrm{O=O^+-O^-}\) and its mirror image, with the single and double bonds interchanged.
Consider the following statements.
Statement I: The two terminal oxygen positions are equivalent in the resonance hybrid even though either Lewis contributor depicts one single and one double O–O bond.
Statement II: The central oxygen can have formal charge \(+1\) in the conventional contributors even though ozone is an elemental substance whose oxygen oxidation states are zero.
Statement III: Equivalent terminal bonds require a zero molecular dipole moment regardless of whether the molecule is linear or bent.
Statement IV: The bent nuclear framework is consistent with two central bonding domains and one central lone-pair domain in the conventional electron-domain analysis.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
The two conventional ozone contributors have the same O–O–O nuclear connectivity.
They differ in the assignments of the single and double O–O bonds.
The terminal oxygen positions are interchangeable.
The actual resonance hybrid therefore has equivalent terminal O–O connections.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Formal charge allocates bonding electrons equally between the bonded atoms.
In the stipulated contributors, the central oxygen has three bonding pairs and one lone pair.
Its formal charge is \(+1\).
The singly bonded terminal oxygen has formal charge \(-1\).
However, oxidation state uses a different electron-allocation convention.
Ozone contains only oxygen atoms, so it is an elemental allotrope and its oxygen atoms have oxidation state zero under the conventional assignment.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
Equal bond-dipole contributions cancel completely only when their vector arrangement permits cancellation.
In a linear symmetrical arrangement, two equal opposite bond-dipole vectors can cancel.
A bent arrangement generally does not provide that complete cancellation.
Ozone is bent and has a nonzero permanent molecular dipole moment.
Therefore, equal bond lengths do not imply zero molecular polarity.
Statement III is incorrect.
\(\textbf{Evaluate Statement IV:}\)
The central oxygen forms two sigma-bonding connections.
It also possesses one lone pair in the conventional resonance contributors.
The total central electron-domain count is three.
The corresponding electron-domain arrangement is trigonal planar, but the molecular arrangement of the three oxygen nuclei is bent.
Statement IV is correct.
\(\textbf{Integrate the distinct descriptions:}\)
Formal charge, oxidation state, resonance equivalence and molecular dipole moment describe different aspects of the same species.
The resonance contributors explain equivalent O–O bonding.
VSEPR explains the bent nuclear framework.
Dipole moment depends on the full molecular charge distribution and geometry.
Statements I, II and IV are correct.
515. The thiocyanate ion, \(\mathrm{SCN^-}\), has the fixed nuclear connectivity S–C–N and contains sixteen valence electrons. Consider the following two octet-satisfying Lewis contributors:
\[
\mathrm{^-S-C\equiv N}
\]
\[
\mathrm{S=C=N^-}
\]
In each drawing, carbon is the central atom and all terminal lone pairs must be included when evaluating formal charges. Which statement correctly connects the location of formal charge, the geometry around carbon and the significance of these drawings?
ⓐ. The first contributor places formal charge \(-1\) on sulfur and the second places it on nitrogen; carbon has two bonding domains and a linear S–C–N arrangement in both, and the drawings represent resonance rather than atom rearrangement.
ⓑ. The first contributor places formal charge \(-1\) on nitrogen and the second places it on sulfur; carbon remains linear, but a triple bond cannot participate in any resonance description.
ⓒ. The first contributor places formal charge \(-1\) on sulfur and the second on nitrogen; carbon must become bent in the second drawing because two double bonds require two central lone pairs.
ⓓ. Both drawings place the formal charge on carbon; the different S–C and C–N bond multiplicities show that sulfur and nitrogen exchange nuclear positions between the contributors.
Correct Answer: The first contributor places formal charge \(-1\) on sulfur and the second places it on nitrogen; carbon has two bonding domains and a linear S–C–N arrangement in both, and the drawings represent resonance rather than atom rearrangement.
Explanation: \(\textbf{Check the valence-electron inventory:}\)
Sulfur contributes six valence electrons, carbon four and nitrogen five.
The single negative charge supplies one additional electron.
\[
N_v=6+4+5+1=16
\]
Both specified contributors use these sixteen electrons and obey the ordinary octet requirement for carbon and nitrogen.
\(\textbf{Evaluate the first contributor:}\)
In \(\mathrm{^-S-C\equiv N}\), sulfur has one single bond and three lone pairs.
Its formal charge follows from the standard electron-counting relation:
\[
FC=V-N_{\mathrm{nonbonding}}-\frac12N_{\mathrm{bonding}}
\]
\[
FC_{\mathrm S}=6-6-1=-1
\]
Nitrogen has a triple bond and one lone pair, so its formal charge is:
\[
FC_{\mathrm N}=5-2-3=0
\]
Carbon has four shared bonding pairs in total and formal charge zero.
\(\textbf{Evaluate the second contributor:}\)
In \(\mathrm{S=C=N^-}\), sulfur has two lone pairs and a double bond.
\[
FC_{\mathrm S}=6-4-2=0
\]
Nitrogen has two lone pairs and a double bond.
\[
FC_{\mathrm N}=5-4-2=-1
\]
Carbon again has formal charge zero, and the total formal charge remains \(-1\).
\(\textbf{Determine the central arrangement:}\)
The central carbon is bonded to two atoms in each contributor.
A double or triple bond is counted as one bonding domain for elementary VSEPR geometry.
Two central bonding domains and no central lone pair give a linear S–C–N arrangement.
The number of sigma connections does not change between the drawings.
\(\textbf{Interpret the resonance relationship:}\)
Only electron-pair placement and associated formal charges differ between the contributors.
The S–C–N nuclear framework remains unchanged.
The real ion is not a mixture of two species whose sulfur and nitrogen atoms interchange positions.
Resonance allows the electron distribution to extend across the connected atoms, while the two different terminal elements need not possess equal bond lengths.
Thus, the formal charges and common linear arrangement stated in option A are consistent.
516. Ethene contains one C=C double bond. In its usual planar geometry, an unhybridized p orbital on each carbon is parallel to the other and can overlap laterally.
Arrangement P: The two \(\mathrm{CH_2}\) groups lie in the same plane, and their unhybridized p orbitals are parallel. Arrangement Q: One \(\mathrm{CH_2}\) group is hypothetically rotated by \(90^\circ\) about the C–C internuclear axis, while the two carbon nuclei remain connected through the original sigma framework. The unhybridized p orbitals are then mutually perpendicular. No new atoms are attached and no C–H covalent bonds are broken.
Which statement correctly predicts the principal bonding consequence of the rotation?
ⓐ. Q must contain a stronger C=C pi bond because perpendicular p orbitals have maximum sideways overlap.
ⓑ. The C–C sigma bond remains; \(90^\circ\) rotation eliminates ordinary p–p pi overlap, explaining restricted alkene rotation.
ⓒ. Both sigma and pi bonding must remain unchanged because rotation around any bond is energetically free.
ⓓ. Q necessarily becomes ethane because changing the orientation of p orbitals automatically adds two hydrogen atoms.
Correct Answer: The C–C sigma bond remains; \(90^\circ\) rotation eliminates ordinary p–p pi overlap, explaining restricted alkene rotation.
Explanation: \(\textbf{Identify the original sigma framework:}\)
Each ethene carbon forms three sigma-bonding connections.
Two are C–H connections, and one is the C–C sigma bond.
The conventional local hybridization is \(sp^2\).
The sigma bond is oriented along the internuclear C–C axis.
\(\textbf{Identify the original pi interaction:}\)
Each carbon retains one unhybridized p orbital.
In planar ethene, these p orbitals are parallel.
Their lateral overlap produces the pi component of the C=C double bond.
\(\textbf{Examine the rotation:}\)
The stipulated rotation changes the relative orientations of the unhybridized p orbitals.
At a relative angle of \(90^\circ\), they are mutually perpendicular.
Their net lateral overlap vanishes by symmetry in the idealized arrangement.
Therefore, the original pi stabilization is lost.
\(\textbf{Distinguish the sigma contribution:}\)
The sigma connection lies along the carbon–carbon internuclear axis.
Rotation about this axis does not automatically eliminate the sigma-bonding direction.
The carbon atoms can therefore remain connected through the sigma framework in the stipulated hypothetical structure.
\(\textbf{Interpret the energy change:}\)
The planar arrangement benefits from both sigma and pi bonding.
A twisted structure that loses the pi interaction lacks that contribution to stabilization.
This helps explain why rotation about an ordinary C=C double bond requires substantially more energy than rotation about an isolated C–C single bond.
\(\textbf{Reject the chemical-composition error:}\)
Changing a torsional angle does not add hydrogen atoms.
Ethene therefore does not become ethane merely by rotating one \(\mathrm{CH_2}\) group.
Hydrogenation would require a separate chemical reaction.
\(\textbf{Integrate the orbital reasoning:}\)
The restricted rotation arises from the directionality of p-orbital overlap.
It cannot be inferred solely by counting the connected nuclei.
Option B correctly distinguishes the retained sigma connection from the disrupted pi interaction.
517. Carbon dioxide and ketene have the following structures:
\[
\mathrm{O=C=O}
\]
\[
\mathrm{H_2C=C=O}
\]
In both molecules, the central carbon forms two double bonds. Assume conventional localized sigma–pi bonding descriptions and approximately trigonal-planar arrangements around the terminal \(\mathrm{CH_2}\) carbon of ketene.
Which statement correctly compares their local hybridizations and the symmetry-based prediction of their molecular dipoles?
ⓐ. Both central carbons are \(sp^2\)-hybridized because a double bond always requires three hybrid orbitals; both molecules are necessarily nonpolar.
ⓑ. The central carbon of CO₂ is \(sp\)-hybridized, but the central carbon of ketene is \(sp^3\)-hybridized because ketene contains hydrogen atoms.
ⓒ. Both central carbons are \(sp\)-hybridized, and both molecules must have zero dipole moment because their two double bonds point in opposite directions.
ⓓ. Both central carbons are \(sp\); ketene's terminal \(\mathrm{CH_2}\) carbon is \(sp^2\), and its dipoles are not forced to cancel as in \(\mathrm{CO_2}\).
Correct Answer: Both central carbons are \(sp\); ketene's terminal \(\mathrm{CH_2}\) carbon is \(sp^2\), and its dipoles are not forced to cancel as in \(\mathrm{CO_2}\).
Explanation: \(\textbf{Examine the central carbon of carbon dioxide:}\)
The carbon atom forms one C–O sigma connection in each of two opposite directions.
It consequently has two sigma-bonding domains.
Its conventional hybridization is \(sp\).
Two unhybridized p orbitals remain available for the two pi components.
\(\textbf{Examine the central carbon of ketene:}\)
Ketene also contains a carbon atom forming two double bonds in opposite directions.
Each double bond contributes one sigma-bonding direction to this central carbon.
Its conventional local hybridization is therefore \(sp\).
Two residual p orbitals participate in the pi components of the cumulated double bonds.
\(\textbf{Determine the terminal carbon hybridization:}\)
The terminal \(\mathrm{CH_2}\) carbon of ketene forms two C–H sigma bonds and one C–C sigma bond.
It therefore possesses three sigma-bonding directions.
The conventional local hybridization is \(sp^2\).
One unhybridized p orbital participates in the adjacent C=C pi interaction.
\(\textbf{Compare the dipole cancellation conditions:}\)
Carbon dioxide has two equivalent oxygen termini and a linear nuclear arrangement.
The equal C=O bond-dipole contributions point in opposite directions.
They cancel in the ideal molecule.
Ketene has chemically different terminal environments: a \(\mathrm{CH_2}\) group at one end and an oxygen atom at the other.
Its charge distribution is therefore not constrained by the same terminal-exchange symmetry.
\(\textbf{Interpret the correct inference:}\)
The central-carbon geometry is similar in the two molecules.
However, identical central hybridization does not require identical overall molecular polarity.
The absence of symmetry-enforced cancellation in ketene permits a nonzero molecular dipole.
An exact magnitude would require further electronic information.
Option D correctly combines the local orbital frameworks and molecular symmetry.
518. Vinyl chloride and ethyl chloride have the structures:
\[
\mathrm{CH_2=CH-Cl}
\]
\[
\mathrm{CH_3CH_2-Cl}
\]
In vinyl chloride, a chlorine lone pair can interact with the adjacent carbon–carbon pi system. Consider the following statements.
Statement I: The carbon bonded directly to chlorine is \(sp^2\)-hybridized in vinyl chloride but \(sp^3\)-hybridized in ethyl chloride.
Statement II: A minor charge-separated resonance contributor of vinyl chloride can be represented as \(\mathrm{^-CH_2-CH=Cl^+}\), with the chlorine lone pair contributing to a C–Cl pi interaction.
Statement III: The possible resonance contribution in vinyl chloride means that its C–Cl sigma bond must disappear from the resonance hybrid.
Statement IV: Comparing the two C–Cl bond lengths requires considering their differing hybridization and electronic environments; one should not attribute every difference solely to chlorine's electronegativity.
Which statements are correct?
ⓐ. I, II and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
The chlorine-bearing carbon in vinyl chloride participates in a C=C double bond.
It has three sigma-bonding directions and an unhybridized p orbital.
Its conventional local hybridization is \(sp^2\).
The corresponding carbon in ethyl chloride forms four sigma bonds.
Its conventional hybridization is \(sp^3\).
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Chlorine in vinyl chloride possesses lone-pair electron density.
A suitable chlorine orbital can interact with the adjacent carbon–carbon pi system.
A charge-separated contributor can be drawn by shifting the original C=C pi electron pair towards the terminal carbon and forming a C=Cl pi contribution.
In the resulting representation, chlorine carries positive formal charge and the terminal carbon carries negative formal charge.
The original atomic connectivity remains unchanged.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
The C–Cl sigma connection is present in both the ordinary and charge-separated contributors.
Resonance changes the electron distribution associated with the pi system.
It does not require removal of the sigma bond connecting carbon and chlorine.
Statement III is incorrect.
\(\textbf{Evaluate Statement IV:}\)
The carbon directly bonded to chlorine has different hybridization in the two molecules.
Vinyl chloride also permits additional lone-pair–pi-system interaction that is absent from the corresponding saturated ethyl chloride framework.
These electronic differences can influence the C–Cl bond properties.
Thus, a bond-length comparison cannot be reduced to chlorine's electronegativity alone.
Statement IV is correct.
\(\textbf{Integrate the reasoning:}\)
Resonance requires suitable orbital alignment and unchanged nuclear connectivity.
Hybridization describes the local sigma framework.
Both concepts can contribute to the bonding differences between a vinylic and an alkyl C–Cl connection.
Statements I, II and IV are correct.
519. Ethene is hydrogenated to ethane according to:
\[
\mathrm{C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)}
\]
Use the following representative gaseous bond enthalpies:
| Bond | Bond enthalpy (\(\mathrm{kJ\,mol^{-1}}\)) |
|---|
| C=C | 610 |
| C–C | 348 |
| H–H | 436 |
| C–H | 413 |
Which statement correctly combines the approximate hydrogenation enthalpy with the changes in carbon hybridization and sigma–pi bond counts?
ⓐ. \(\Delta H\approx+128\ \mathrm{kJ\,mol^{-1}}\); both carbons change from \(sp^3\) to \(sp^2\), and the number of pi bonds increases.
ⓑ. \(\Delta H\approx-128\ \mathrm{kJ\,mol^{-1}}\); both carbons change from \(sp^2\) to \(sp^3\), the reactant side contains six sigma bonds and one pi bond, and the product contains seven sigma bonds and no pi bonds.
ⓒ. \(\Delta H\approx-541\ \mathrm{kJ\,mol^{-1}}\); both carbons remain \(sp^2\), and only the C=C pi bond is removed.
ⓓ. \(\Delta H\approx-128\ \mathrm{kJ\,mol^{-1}}\); the product retains one C=C pi bond because hydrogenation changes only the C–H bond count.
Correct Answer: \(\Delta H\approx-128\ \mathrm{kJ\,mol^{-1}}\); both carbons change from \(sp^2\) to \(sp^3\), the reactant side contains six sigma bonds and one pi bond, and the product contains seven sigma bonds and no pi bonds.
Explanation: \(\textbf{Identify the bonds changed by hydrogenation:}\)
Ethene contains one carbon–carbon double bond.
Hydrogenation converts that connection into a carbon–carbon single bond.
One H–H bond is also consumed.
Two new C–H bonds are formed.
\(\textbf{Determine the bonds broken:}\)
The bond-enthalpy method treats the existing C=C and H–H bonds as broken in the accounting cycle.
\[
H_{\mathrm{broken}}=D(\mathrm{C=C})+D(\mathrm{H-H})
\]
Substituting:
\[
H_{\mathrm{broken}}=610+436
\]
\[
H_{\mathrm{broken}}=1046\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Determine the bonds formed:}\)
The products contain the new C–C single bond and two additional C–H connections.
\[
H_{\mathrm{formed}}=D(\mathrm{C-C})+2D(\mathrm{C-H})
\]
Substituting:
\[
H_{\mathrm{formed}}=348+2(413)
\]
\[
H_{\mathrm{formed}}=348+826
\]
\[
H_{\mathrm{formed}}=1174\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate the approximate reaction enthalpy:}\)
\[
\Delta H\approx\sum D_{\mathrm{broken}}-\sum D_{\mathrm{formed}}
\]
\[
\Delta H\approx1046-1174
\]
\[
\Delta H\approx-128\ \mathrm{kJ\,mol^{-1}}
\]
The negative sign indicates an exothermic process.
\(\textbf{Count the reactant-side sigma and pi bonds:}\)
Ethene contains four C–H sigma bonds and one C–C sigma bond.
The C=C connection also contains one pi component.
Hydrogen contributes one additional H–H sigma bond.
Therefore:
\[
N_{\sigma,\mathrm{reactants}}=4+1+1=6
\]
\[
N_{\pi,\mathrm{reactants}}=1
\]
\(\textbf{Count the product bonds:}\)
Ethane contains six C–H sigma bonds and one C–C sigma bond.
\[
N_{\sigma,\mathrm{product}}=6+1=7
\]
\[
N_{\pi,\mathrm{product}}=0
\]
\(\textbf{Identify the hybridization change:}\)
Each ethene carbon initially has three sigma-bonding directions and is conventionally \(sp^2\)-hybridized.
Each ethane carbon has four sigma-bonding directions and is \(sp^3\)-hybridized.
The carbon p orbitals that formerly contributed to the C=C pi bond are no longer required for that interaction.
\(\textbf{Interpret the thermochemical calculation:}\)
The overall reaction enthalpy depends on all the bonds broken and formed.
It cannot be calculated correctly by considering only the disappearance of the pi bond while omitting the H–H and newly formed C–H connections.
\(\textbf{Final answer:}\)
\[
\begin{aligned}
\Delta H&\approx-128\ \mathrm{kJ\,mol^{-1}}\\
N_\sigma:&\ 6\rightarrow7\\
N_\pi:&\ 1\rightarrow0\\
\mathrm{C\ hybridization}:&\ sp^2\rightarrow sp^3
\end{aligned}
\]
520. The sulfite ion, \(\mathrm{SO_3^{2-}}\), has twenty-six valence electrons. Consider a conventional expanded-valence-shell resonance model in which each of three equivalent contributors contains one S=O double bond, two S–O single bonds and one central sulfur lone pair.
Which statement correctly combines the number of these contributors, their simple resonance-averaged S–O bond multiplicity, the central molecular shape and sulfur oxidation state?
ⓐ. Three contributors; average S–O multiplicity \(1\); trigonal-planar shape; sulfur oxidation state \(+6\).
ⓑ. Six contributors; average S–O multiplicity \(1.5\); tetrahedral molecular shape; sulfur oxidation state \(+4\).
ⓒ. Three contributors; average S–O multiplicity \(1\frac13\); trigonal-pyramidal shape; sulfur oxidation state \(+4\).
ⓓ. Four contributors; average S–O multiplicity \(1.25\); T-shaped molecular geometry; sulfur oxidation state \(+6\).
Correct Answer: Three contributors; average S–O multiplicity \(1\frac13\); trigonal-pyramidal shape; sulfur oxidation state \(+4\).
Explanation: \(\textbf{Confirm the total electron count:}\)
Sulfur contributes six valence electrons.
Three oxygen atoms contribute eighteen.
The ionic charge supplies two additional electrons.
\[
N_v=6+3(6)+2
\]
\[
N_v=26
\]
\(\textbf{Determine the specified resonance possibilities:}\)
Each stipulated contributor contains one S=O double bond.
The other two oxygen atoms are connected through S–O single bonds.
The double-bond assignment can be placed on any one of three oxygen positions.
Therefore, the specified model contains three equivalent contributors.
\(\textbf{Calculate the average bond multiplicity:}\)
Each contributor contains one double and two single S–O bonds.
The sum of formal bond multiplicities is:
\[
2+1+1=4
\]
The simple average across the three oxygen positions is:
\[
\overline b_{\mathrm{S-O}}=\frac43
\]
\[
\overline b_{\mathrm{S-O}}=1\frac13
\]
\(\textbf{Determine the electron-domain arrangement:}\)
Sulfur has three S–O sigma-bonding directions.
It also possesses one central lone pair.
A multiple bond contributes one VSEPR bonding domain.
Therefore, the central atom has four electron domains.
The electron-domain arrangement is approximately tetrahedral.
\(\textbf{Identify the molecular shape:}\)
Three domains contain bonded oxygen atoms.
The fourth contains a lone pair.
The arrangement of the oxygen nuclei is consequently trigonal pyramidal.
\(\textbf{Determine sulfur's oxidation state:}\)
Each oxygen is assigned oxidation state \(-2\).
Let sulfur have oxidation state \(x\).
\[
x+3(-2)=-2
\]
\[
x-6=-2
\]
\[
x=+4
\]
\(\textbf{Interpret the distinction:}\)
The conventional resonance contributors redistribute the formal multiple-bond assignment among equivalent oxygen positions.
They do not eliminate the central sulfur lone pair.
Thus, resonance-related bond equivalence can coexist with a trigonal-pyramidal molecular shape.
The calculated average multiplicity belongs to the specified Lewis model and should not be treated as an exact experimental bond order.
\(\textbf{Final answer:}\)
\[
\begin{aligned}
N_{\mathrm{contributors}}&=3\\
\overline b_{\mathrm{S-O}}&=\frac43\\
\mathrm{Shape}&=\text{trigonal pyramidal}\\
\mathrm{Oxidation\ state\ of\ S}&=+4
\end{aligned}
\]
