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201. Ammonia and nitrogen trifluoride both contain three bonding domains and one central lone pair. Their approximate bond angles are \(107^\circ\) and \(102.5^\circ\), respectively. Which explanation most appropriately relates the observed angular difference to the different terminal atoms?
ⓐ. Fluorine draws N–F bonding density farther from nitrogen than hydrogen does in N–H, changing electron-domain repulsions.
ⓑ. Nitrogen trifluoride contains two central nitrogen lone pairs, whereas ammonia contains one.
ⓒ. The N–F bonds in nitrogen trifluoride count as two VSEPR domains each because fluorine is highly electronegative.
ⓓ. Ammonia is tetrahedral in molecular shape, whereas nitrogen trifluoride is trigonal pyramidal.
Correct Answer: Fluorine draws N–F bonding density farther from nitrogen than hydrogen does in N–H, changing electron-domain repulsions.
Explanation: \(\textbf{Identify the common VSEPR framework:}\)
Both ammonia and nitrogen trifluoride contain three bonds to nitrogen.
Each nitrogen atom also possesses one lone pair.
Therefore, both molecules have four electron domains.
Their electron-domain arrangements are approximately tetrahedral.
Their molecular shapes are trigonal pyramidal.
\(\textbf{Compare the bonded atoms:}\)
Hydrogen is less electronegative than nitrogen.
Fluorine is more electronegative than nitrogen.
Therefore, the distribution of bonding electron density differs between N–H and N–F bonds.
In ammonia, the N–H bonding electron density is drawn relatively towards nitrogen.
In nitrogen trifluoride, the highly electronegative fluorine atoms attract the N–F bonding electron density away from nitrogen.
\(\textbf{Interpret the repulsion changes:}\)
Electron-pair repulsion depends on the spatial distribution of electron density.
Changing the terminal atom can alter how strongly bonding regions interact near the central atom.
Therefore, two molecules with the same number of bonding domains and lone pairs need not have identical bond angles.
The comparatively smaller angle in \(\mathrm{NF_3}\) is consistent with the altered electron-density distribution and associated repulsive interactions.
\(\textbf{Reject the incorrect explanations:}\)
Both molecules contain one central nitrogen lone pair.
A single N–F bond contributes one VSEPR bonding domain.
Neither molecule becomes tetrahedral in molecular shape merely because its electron-domain arrangement is tetrahedral.
\(\textbf{Interpret the broader principle:}\)
The usual lone pair–lone pair and lone pair–bond pair repulsion hierarchy provides a qualitative framework.
However, exact measured angles also depend on the identities and electronic properties of the bonded atoms.
Thus, the common \(\mathrm{AX_3E}\) classification does not require equal bond angles.
Option A correctly identifies the relevant electronic difference.
202. Hydrazine, \(\mathrm{N_2H_4}\), has the connectivity \(\mathrm{H_2N-NH_2}\), with a single N–N bond. Which statement correctly describes the local VSEPR arrangement around each nitrogen atom and distinguishes it from the overall molecular geometry?
ⓐ. Each nitrogen is trigonal planar because the molecule contains four hydrogen atoms.
ⓑ. One nitrogen is tetrahedral and the other is trigonal planar because only one N–N bond is present.
ⓒ. Each nitrogen is linear because its lone pair lies along the N–N bond.
ⓓ. Each nitrogen has four domains and a locally trigonal-pyramidal shape; that does not require the entire molecule to be planar.
Correct Answer: Each nitrogen has four domains and a locally trigonal-pyramidal shape; that does not require the entire molecule to be planar.
Explanation: \(\textbf{Examine the Lewis connectivity:}\)
Hydrazine contains an N–N single bond.
Each nitrogen is additionally bonded to two hydrogen atoms.
Nitrogen contributes five valence electrons in its neutral atomic state.
In the conventional Lewis structure, each nitrogen forms three single bonds and retains one lone pair.
\(\textbf{Count domains around the first nitrogen:}\)
Two N–H bonds and one N–N bond provide three bonding domains.
The lone pair contributes one additional domain.
Therefore, the total electron-domain count is four.
Four domains favour a tetrahedral electron-domain arrangement.
Only three domains contain bonded atoms.
The local arrangement of the two hydrogens and the other nitrogen is trigonal pyramidal.
\(\textbf{Examine the second nitrogen:}\)
The second nitrogen has the same local connectivity.
It also possesses three bonding domains and one lone pair.
Its local VSEPR description is consequently identical.
Both nitrogen centres therefore have trigonal-pyramidal local molecular arrangements.
\(\textbf{Distinguish local and whole-molecule descriptions:}\)
A local VSEPR arrangement specifies the directions of electron regions around one central atom.
It does not by itself establish the relative orientation of the two \(\mathrm{NH_2}\) groups around the N–N bond.
The overall molecular conformation also depends on the relative positions of the atoms attached to the two nitrogen centres.
Therefore, the entire molecule need not be planar.
\(\textbf{Identify the conceptual error:}\)
The number of hydrogen atoms in the molecular formula does not determine the local geometry of nitrogen.
Likewise, an N–N single bond does not make either nitrogen centre linear.
The electron-domain count must be established separately for each relevant atom.
Option D correctly identifies the local shapes without imposing an unsupported global arrangement.
203. Nitromethane has the connectivity \(\mathrm{CH_3-NO_2}\). In its conventional resonance contributors, nitrogen is bonded to the methyl carbon and both oxygen atoms, with one N=O double bond and one N–O single bond. Which statement correctly compares the local geometries around the methyl carbon and the nitrogen atom?
ⓐ. Both centres are tetrahedral because each is bonded to three other atoms.
ⓑ. Methyl carbon is trigonal planar, whereas nitrogen is tetrahedral because the N=O double bond contributes two geometrical domains.
ⓒ. Methyl carbon is tetrahedral, whereas nitrogen is trigonal planar because nitrogen has three bonding domains and no central lone pair.
ⓓ. Methyl carbon is tetrahedral, whereas nitrogen is trigonal pyramidal because nitrogen has one central lone pair.
Correct Answer: Methyl carbon is tetrahedral, whereas nitrogen is trigonal planar because nitrogen has three bonding domains and no central lone pair.
Explanation: \(\textbf{Examine the methyl carbon:}\)
The carbon of the methyl group forms three C–H single bonds and one C–N single bond.
Therefore, it has four bonding domains.
It has no lone pair in the conventional neutral molecular structure.
Four bonding domains favour a tetrahedral arrangement.
Thus, the local geometry around methyl carbon is tetrahedral.
\(\textbf{Examine the nitro nitrogen:}\)
Nitrogen is directly connected to carbon and two oxygen atoms.
The N–C bond contributes one domain.
Each N–O bonding region also contributes one domain.
The N=O double bond counts as one electron domain rather than two.
Therefore, nitrogen has three bonding domains.
In the conventional resonance descriptions, nitrogen has no central lone pair.
Its three bonding directions are arranged approximately trigonal planar.
\(\textbf{Consider the formal-charge interpretation:}\)
A conventional nitro-group resonance contributor assigns positive formal charge to nitrogen and negative formal charge to the singly bonded oxygen.
This formal-charge distribution does not imply that nitrogen must have a central lone pair.
The nitrogen centre instead has three bonding directions and an octet in the localised representation.
The resonance hybrid makes the two N–O bonds equivalent in the idealised nitro-group description.
\(\textbf{Distinguish the two central environments:}\)
The methyl carbon has four independent bonding directions.
The nitro nitrogen has three independent bonding directions.
Consequently, the same molecule contains atoms with different local VSEPR geometries.
The overall molecular formula does not imply one common geometry for every atom.
Neither does the existence of a double bond automatically create an additional bond direction.
Thus, option C correctly identifies both local arrangements.
204. Carbonic acid has the connectivity \(\mathrm{HO-C(=O)-OH}\). Consider its conventional neutral Lewis structure, in which carbon is bonded to three oxygen atoms and each hydroxyl oxygen is bonded to carbon and hydrogen. Which statement correctly describes the electron-domain arrangement around the carbon atom and the two hydroxyl oxygen atoms?
ⓐ. Carbon is tetrahedral, and each hydroxyl oxygen is trigonal planar because the molecule contains one C=O double bond.
ⓑ. Carbon is trigonal planar; each hydroxyl oxygen has four electron domains and a bent local arrangement of its two bonded atoms.
ⓒ. Carbon is trigonal pyramidal because the C=O bond leaves one lone pair on carbon; both hydroxyl oxygens are linear.
ⓓ. Carbon and all three oxygen atoms have identical three-domain electron arrangements because they belong to the same molecule.
Correct Answer: Carbon is trigonal planar; each hydroxyl oxygen has four electron domains and a bent local arrangement of its two bonded atoms.
Explanation: \(\textbf{Identify the carbon connectivity:}\)
The central carbon is bonded to three oxygen atoms.
Two C–O connections are single bonds to hydroxyl groups.
The third is represented as a C=O double bond.
Each connection counts as one bonding domain around carbon.
Therefore, carbon has three electron domains.
It has no lone pair in the conventional neutral Lewis structure.
Three domains favour a trigonal-planar arrangement.
\(\textbf{Examine a hydroxyl oxygen:}\)
Each hydroxyl oxygen is bonded to carbon and hydrogen.
It forms two single bonds and retains two lone pairs.
Therefore, it possesses two bonding domains and two nonbonding domains.
The total domain count is four.
Its electron-domain arrangement is approximately tetrahedral.
However, the two bonded atoms form a bent local arrangement about that oxygen.
\(\textbf{Examine the second hydroxyl oxygen:}\)
The second hydroxyl oxygen has the same local bonding description.
It also has two bonding domains and two lone-pair domains.
Its local bonded-atom arrangement is bent.
\(\textbf{Distinguish the carbonyl oxygen:}\)
The oxygen represented as C=O is bonded only to carbon.
It has nonbonding electron pairs, but there is only one bonded neighbouring atom.
A conventional two-bond molecular angle is therefore not defined at that terminal oxygen.
\(\textbf{Integrate the local geometries:}\)
Different atoms within the same molecule can possess different electron-domain arrangements.
Carbon has three domains, whereas each hydroxyl oxygen has four.
The double bond changes the bonding-electron distribution but still counts as one geometrical domain around carbon.
Thus, the carbon centre is trigonal planar while the hydroxyl oxygen centres have bent local bonded-atom arrangements.
Option B is correct.
205. An ideal octahedral \(\mathrm{AX_6}\) molecule has six equivalent A–X bonds. An investigator counts every distinct angle formed by choosing two bond directions at the central atom. Which statement correctly gives the number of \(90^\circ\) angles, the number of \(180^\circ\) angles, and the structural implication for the six bond directions?
ⓐ. Twelve \(90^\circ\) angles and three \(180^\circ\) angles; the directions form three opposite pairs.
ⓑ. Nine \(90^\circ\) angles and six \(180^\circ\) angles; the directions form six opposite pairs.
ⓒ. Twelve \(90^\circ\) angles and two \(180^\circ\) angles; the remaining pair has angle \(120^\circ\).
ⓓ. Six \(90^\circ\) angles and nine \(180^\circ\) angles; every bond has three opposite neighbours.
Correct Answer: Twelve \(90^\circ\) angles and three \(180^\circ\) angles; the directions form three opposite pairs.
Explanation: \(\textbf{Identify the octahedral arrangement:}\)
An ideal octahedron has six radial directions from its centre towards its vertices.
These directions can be grouped into three pairs of opposite positions.
The angle between the members of an opposite pair is \(180^\circ\).
Every other pair of distinct bond directions is perpendicular and forms a \(90^\circ\) angle.
\(\textbf{Count all distinct bond-direction pairs:}\)
The number of ways to select two bond directions from six is:
\[
N_{\mathrm{total}}=\frac{6(6-1)}{2}
\]
\[
N_{\mathrm{total}}=15
\]
\(\textbf{Count opposite pairs:}\)
Each bond direction has exactly one opposite partner.
The six directions therefore form three distinct opposite pairs.
\[
N_{180^\circ}=3
\]
\(\textbf{Count perpendicular pairs:}\)
Every remaining pair forms a \(90^\circ\) angle.
Therefore:
\[
N_{90^\circ}=N_{\mathrm{total}}-N_{180^\circ}
\]
\[
N_{90^\circ}=15-3
\]
\[
N_{90^\circ}=12
\]
\(\textbf{Verify by counting from each direction:}\)
Every bond direction has four perpendicular neighbours.
Counting over all six directions gives twenty-four directional incidences.
Each \(90^\circ\) pair is counted twice in this procedure.
Therefore:
\[
N_{90^\circ}=\frac{6(4)}{2}
\]
\[
N_{90^\circ}=12
\]
The independent count agrees with the previous result.
\(\textbf{Interpret the structural implication:}\)
The three opposite pairs provide the characteristic geometrical framework of an ideal octahedron.
No \(120^\circ\) bond-direction relationships occur in this arrangement.
The result is useful for analysing octahedral molecular geometries and the placement of lone pairs in six-domain VSEPR structures.
\(\textbf{Final answer:}\)
\[
\boxed{N_{90^\circ}=12;\quad N_{180^\circ}=3}
\]
206. Two neutral xenon oxyfluorides, \(\mathrm{XeOF_2}\) and \(\mathrm{XeO_2F_2}\), are represented by conventional Lewis structures containing Xe=O double bonds and Xe–F single bonds. Each terminal oxygen has two lone pairs, and each terminal fluorine has three lone pairs. Which statement correctly compares the central xenon electron-domain arrangements and molecular shapes?
ⓐ. \(\mathrm{XeOF_2}\) is trigonal planar, whereas \(\mathrm{XeO_2F_2}\) is tetrahedral.
ⓑ. \(\mathrm{XeOF_2}\) has five electron domains and a T-shaped molecular arrangement, whereas \(\mathrm{XeO_2F_2}\) has five electron domains and a seesaw molecular arrangement.
ⓒ. \(\mathrm{XeOF_2}\) has four electron domains and a trigonal-pyramidal arrangement, whereas \(\mathrm{XeO_2F_2}\) has six domains and a square-planar arrangement.
ⓓ. Both species have four electron domains and tetrahedral molecular arrangements because all Xe=O double bonds count as two bonding domains.
Correct Answer: \(\mathrm{XeOF_2}\) has five electron domains and a T-shaped molecular arrangement, whereas \(\mathrm{XeO_2F_2}\) has five electron domains and a seesaw molecular arrangement.
Explanation: \(\textbf{Count the valence electrons in XeOF₂:}\)
Xenon contributes eight valence electrons, oxygen contributes six, and two fluorine atoms contribute fourteen.
\[
N_v(\mathrm{XeOF_2})=8+6+2(7)
\]
\[
N_v(\mathrm{XeOF_2})=28
\]
The Xe=O double bond uses four electrons, and the two Xe–F single bonds use four more.
The oxygen lone pairs account for four electrons, while the fluorine lone pairs account for twelve.
The electrons remaining on xenon are:
\[
N_{\mathrm{Xe,nonbonding}}=28-8-4-12
\]
\[
N_{\mathrm{Xe,nonbonding}}=4
\]
These form two central lone pairs.
Xenon therefore has three bonding domains and two lone-pair domains.
Its VSEPR description is \(\mathrm{AX_3E_2}\).
Five electron domains favour a trigonal-bipyramidal arrangement.
With both lone pairs occupying equatorial positions, the three bonded atoms form a T-shaped arrangement.
\(\textbf{Count the valence electrons in XeO₂F₂:}\)
\[
N_v(\mathrm{XeO_2F_2})=8+2(6)+2(7)
\]
\[
N_v(\mathrm{XeO_2F_2})=34
\]
Two Xe=O double bonds use eight electrons, while two Xe–F bonds use four.
The terminal oxygen and fluorine lone pairs together account for twenty electrons.
Therefore:
\[
N_{\mathrm{Xe,nonbonding}}=34-12-20
\]
\[
N_{\mathrm{Xe,nonbonding}}=2
\]
These form one central lone pair.
Xenon consequently has four bonding domains and one lone-pair domain.
Its description is \(\mathrm{AX_4E}\).
The electron-domain arrangement is trigonal bipyramidal, while the arrangement of the four bonded atoms is seesaw.
\(\textbf{Interpret the difference:}\)
Both species possess five electron domains, but their bonding-domain and lone-pair distributions differ.
The Xe=O double bonds count as one geometrical domain each.
Therefore, identical electron-domain counts do not require identical molecular shapes.
Option B correctly identifies both arrangements.
207. An ideal octahedral molecule has six polar bonds directed along three mutually perpendicular axes. Each opposite pair contains two unequal bond-dipole magnitudes, as shown below.
All bond-dipole vectors point radially outward from the central atom. Assume that the six stated vectors are the only contributions. What is the resultant molecular dipole-moment magnitude?
| Opposite bond directions | First direction | Opposite direction |
|---|---|---|
| x-axis pair | \(2.40\ \mathrm{D}\) | \(1.80\ \mathrm{D}\) |
| y-axis pair | \(1.50\ \mathrm{D}\) | \(1.50\ \mathrm{D}\) |
| z-axis pair | \(1.20\ \mathrm{D}\) | \(2.00\ \mathrm{D}\) |
ⓐ. \(0.20\ \mathrm{D}\)
ⓑ. \(1.40\ \mathrm{D}\)
ⓒ. \(3.20\ \mathrm{D}\)
ⓓ. \(1.00\ \mathrm{D}\)
Correct Answer: \(1.00\ \mathrm{D}\)
Explanation: \(\textbf{Resolve the octahedral vectors into opposite pairs:}\)
An ideal octahedral arrangement contains three mutually perpendicular pairs of opposite bond directions.
Within each pair, the two bond-dipole vectors are antiparallel.
Therefore, their resultant magnitude is the difference between their magnitudes.
The three pair-resultants must subsequently be combined as perpendicular vectors.
\(\textbf{Determine the x-axis contribution:}\)
\[
\mu_x=2.40-1.80
\]
\[
\mu_x=+0.60\ \mathrm{D}
\]
The positive sign indicates that the resultant points towards the first direction of the x-axis pair.
\(\textbf{Determine the y-axis contribution:}\)
\[
\mu_y=1.50-1.50
\]
\[
\mu_y=0\ \mathrm{D}
\]
The two y-axis contributions cancel completely.
\(\textbf{Determine the z-axis contribution:}\)
Taking the first direction of the z-axis pair as positive:
\[
\mu_z=1.20-2.00
\]
\[
\mu_z=-0.80\ \mathrm{D}
\]
The negative sign indicates that the z-axis resultant points towards the opposite direction.
\(\textbf{Combine the perpendicular components:}\)
The molecular dipole-moment magnitude is:
\[
\mu_R=\sqrt{\mu_x^2+\mu_y^2+\mu_z^2}
\]
Substituting:
\[
\mu_R=\sqrt{(0.60)^2+0^2+(-0.80)^2}
\]
\[
\mu_R=\sqrt{0.36+0.64}
\]
\[
\mu_R=\sqrt{1.00}
\]
\[
\mu_R=1.00\ \mathrm{D}
\]
\(\textbf{Check the interpretation:}\)
The equal y-axis bond dipoles cancel.
The x-axis and z-axis pairs leave nonzero contributions in perpendicular directions.
These cannot be added as ordinary scalars.
Although the bond directions retain ideal octahedral geometry, the unequal opposing bond-dipole magnitudes remove complete cancellation.
The resulting molecule is polar under the stipulated model.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_R=1.00\ \mathrm{D}}
\]
208. A hypothetical molecule \(\mathrm{AX_3Y_2}\) has an ideal trigonal-bipyramidal geometry and no central lone pair. Every A–X bond has a dipole magnitude of \(1.20\ \mathrm{D}\), while every A–Y bond has a magnitude of \(1.80\ \mathrm{D}\). All bond-dipole vectors point radially outward. Consider three arrangements:
Arrangement P: Both Y atoms occupy the axial positions.
Arrangement Q: Both Y atoms occupy equatorial positions.
Arrangement R: One Y atom occupies an axial position and the other occupies an equatorial position.
Which sequence correctly gives their resultant molecular dipole-moment magnitudes?
ⓐ. \(0,\ 0.60,\ 0.60\sqrt2\ \mathrm{D}\)
ⓑ. \(0,\ 1.20,\ 1.20\sqrt2\ \mathrm{D}\)
ⓒ. \(0.60,\ 0,\ 0.60\sqrt2\ \mathrm{D}\)
ⓓ. \(0.60,\ 0.60,\ 0.60\ \mathrm{D}\)
Correct Answer: \(0,\ 0.60,\ 0.60\sqrt2\ \mathrm{D}\)
Explanation: \(\textbf{Establish the symmetrical reference:}\)
Five equal radially outward bond-dipole vectors in an ideal trigonal-bipyramidal arrangement sum to zero.
The two axial vectors cancel one another.
The three equal equatorial vectors also cancel.
Replacing an X bond by a Y bond changes the vector magnitude along the substituted direction.
\(\textbf{Calculate the substitution contribution:}\)
\[
\Delta\mu=\mu_Y-\mu_X
\]
\[
\Delta\mu=1.80-1.20
\]
\[
\Delta\mu=0.60\ \mathrm{D}
\]
Each substitution is therefore equivalent to adding a \(0.60\ \mathrm{D}\) vector along its radial direction to the original zero-resultant arrangement.
\(\textbf{Evaluate Arrangement P:}\)
Both substitutions occur at opposite axial positions.
The two additional vectors have equal magnitudes and opposite directions.
Therefore:
\[
\mu_P=0.60-0.60
\]
\[
\mu_P=0\ \mathrm{D}
\]
\(\textbf{Evaluate Arrangement Q:}\)
Both substitutions occur at equatorial positions.
The two additional vectors have an included angle of \(120^\circ\).
Their resultant is:
\[
\mu_Q=2(0.60)\cos\left(\frac{120^\circ}{2}\right)
\]
\[
\mu_Q=1.20\cos60^\circ
\]
\[
\mu_Q=0.60\ \mathrm{D}
\]
\(\textbf{Evaluate Arrangement R:}\)
One additional vector points axially, and the other points equatorially.
These directions are perpendicular.
Therefore:
\[
\mu_R=\sqrt{(0.60)^2+(0.60)^2}
\]
\[
\mu_R=0.60\sqrt2\ \mathrm{D}
\]
\(\textbf{Interpret the positional dependence:}\)
The three arrangements have the same molecular formula and the same individual bond-dipole magnitudes.
They differ only in the positions occupied by the Y atoms.
Opposite axial substitutions cancel, whereas equatorial and mixed substitutions leave different nonzero resultants.
Therefore, molecular polarity depends on substitution pattern as well as bond polarity.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_P=0;\quad\mu_Q=0.60\ \mathrm{D};\quad\mu_R=0.60\sqrt2\ \mathrm{D}}
\]
209. Phosphorus trichloride, \(\mathrm{PCl_3}\), and phosphoryl chloride, \(\mathrm{POCl_3}\), both contain three P–Cl bonds. In a conventional Lewis description of \(\mathrm{POCl_3}\), phosphorus additionally forms a P=O double bond. Which statement correctly compares their central-atom electron-domain arrangements and molecular shapes?
ⓐ. Both are trigonal pyramidal because each contains three P–Cl bonds.
ⓑ. \(\mathrm{PCl_3}\) is trigonal pyramidal with four electron domains, whereas \(\mathrm{POCl_3}\) is approximately tetrahedral with four bonding domains and no central lone pair.
ⓒ. \(\mathrm{PCl_3}\) is trigonal planar, whereas \(\mathrm{POCl_3}\) is trigonal bipyramidal because the P=O double bond contributes two domains.
ⓓ. Both are tetrahedral in molecular shape because each phosphorus atom has a complete octet.
Correct Answer: \(\mathrm{PCl_3}\) is trigonal pyramidal with four electron domains, whereas \(\mathrm{POCl_3}\) is approximately tetrahedral with four bonding domains and no central lone pair.
Explanation: \(\textbf{Examine phosphorus trichloride:}\)
Phosphorus contributes five valence electrons.
It forms three P–Cl single bonds and retains one lone pair.
The central phosphorus therefore possesses three bonding domains and one lone-pair domain.
The total electron-domain count is four.
Its electron-domain arrangement is approximately tetrahedral.
However, only three domains contain bonded chlorine atoms.
The molecular shape is consequently trigonal pyramidal.
\(\textbf{Examine phosphoryl chloride:}\)
The conventional Lewis description contains three P–Cl single bonds and one P=O double bond.
Each P–Cl bond counts as one electron domain.
The P=O double bond also counts as one geometrical domain.
Therefore, phosphorus has four bonding domains.
In this conventional representation, it has no central lone pair.
\(\textbf{Determine the second molecular arrangement:}\)
Four bonding domains favour an approximately tetrahedral arrangement of the four atoms directly attached to phosphorus.
The oxygen and chlorine atoms are not identical, so the actual bond angles need not equal the ideal tetrahedral angle exactly.
Nevertheless, the molecular arrangement around phosphorus is described as approximately tetrahedral.
\(\textbf{Identify the changed electronic condition:}\)
The two molecules contain the same number of P–Cl bonds.
However, \(\mathrm{POCl_3}\) contains an additional bonding region involving oxygen.
The phosphorus lone pair present in \(\mathrm{PCl_3}\) is absent from the stated Lewis description of \(\mathrm{POCl_3}\).
Thus, the total electron-domain count remains four while the distribution between bonding and nonbonding domains changes.
\(\textbf{Interpret the comparison:}\)
A molecular formula or the number of bonds to one particular element cannot independently determine molecular geometry.
All bonding regions around the central atom must be included.
Multiple bonds count as one domain for the initial VSEPR assignment.
Therefore, option B correctly distinguishes the molecular shapes.
210. The carbonate ion reacts with a proton to form bicarbonate:
\[
\mathrm{CO_3^{2-}+H^+\rightarrow HCO_3^-}
\]
In a conventional Lewis description, the proton becomes bonded to one oxygen atom, producing an O–H group. Which statement correctly identifies the geometrical consequence of protonation?
ⓐ. The carbonate carbon changes from trigonal planar to tetrahedral because protonation increases its number of electron domains.
ⓑ. The carbonate carbon becomes trigonal pyramidal because the incoming proton creates a lone pair on carbon.
ⓒ. The carbon centre remains trigonal planar, while the protonated oxygen acquires two bonding domains and two lone pairs, giving a bent local bonded-atom arrangement.
ⓓ. Every atom in bicarbonate becomes tetrahedrally coordinated because the ion contains one hydrogen atom.
Correct Answer: The carbon centre remains trigonal planar, while the protonated oxygen acquires two bonding domains and two lone pairs, giving a bent local bonded-atom arrangement.
Explanation: \(\textbf{Identify the original carbonate arrangement:}\)
Carbonate contains carbon bonded to three oxygen atoms.
Each C–O bonding region counts as one VSEPR domain.
The central carbon has no lone pair in the conventional Lewis representations.
Therefore, it possesses three bonding domains and adopts a trigonal-planar arrangement.
Resonance makes the three C–O bonds equivalent in the carbonate ion.
\(\textbf{Follow the site of protonation:}\)
The incoming proton becomes attached to an oxygen atom.
It does not form an additional direct bond to carbon.
Consequently, the number of atoms directly attached to carbon remains three.
The carbon atom therefore continues to have three bonding domains.
Its approximate local arrangement remains trigonal planar.
\(\textbf{Examine the protonated oxygen:}\)
In the O–H group, oxygen is bonded to carbon and hydrogen.
It possesses two bonding domains.
In the conventional neutral hydroxyl-oxygen description, it also retains two lone pairs.
The total number of electron domains around this oxygen is four.
Its electron-domain arrangement is approximately tetrahedral.
The two bonded atoms form a bent local arrangement.
\(\textbf{Distinguish changes in bonding from changes in molecular shape:}\)
Protonation alters the electron distribution of the carbonate-derived group.
The three C–O bonds are no longer all equivalent in the same way as those of carbonate.
However, the number of central carbon bonding directions does not change.
The geometrical consequence is therefore different at carbon and at the protonated oxygen.
\(\textbf{Identify the underlying principle:}\)
A reaction can alter the local electron-domain arrangement around one atom while preserving the geometry around another atom.
The position where the new bond forms is essential.
It would be incorrect to assume that adding hydrogen to a polyatomic ion necessarily increases the coordination number of its original central atom.
Option C is correct.
211. Water and oxygen difluoride, \(\mathrm{H_2O}\) and \(\mathrm{OF_2}\), both possess two bonding domains and two lone pairs around oxygen. Their observed bond angles are approximately \(104.5^\circ\) and \(103.2^\circ\), respectively. Fluorine is more electronegative than oxygen, whereas hydrogen is less electronegative than oxygen. Which explanation is most consistent with these facts?
ⓐ. \(\mathrm{OF_2}\) has only one oxygen lone pair, so its smaller angle results from fewer electron domains.
ⓑ. The two angles must be identical because both molecules have the \(\mathrm{AX_2E_2}\) description.
ⓒ. The smaller angle in \(\mathrm{OF_2}\) proves that the F–O–F bonding domains contain no electron density near oxygen.
ⓓ. Different terminal-atom electronegativities modify bonding electron density and domain repulsions, so the bond angles can differ.
Correct Answer: Different terminal-atom electronegativities modify bonding electron density and domain repulsions, so the bond angles can differ.
Explanation: \(\textbf{Identify the common central arrangement:}\)
Oxygen forms two bonds in both molecules.
It also retains two lone pairs in the conventional Lewis descriptions.
Therefore, both molecules contain four electron domains around oxygen.
Their electron-domain arrangements are approximately tetrahedral.
Their molecular shapes are bent.
\(\textbf{Examine bonding in water:}\)
Oxygen is more electronegative than hydrogen.
The O–H shared electron density is consequently attracted towards oxygen.
This influences the concentration of bonding electron density near the central atom.
The observed H–O–H bond angle is approximately \(104.5^\circ\).
\(\textbf{Examine bonding in oxygen difluoride:}\)
Fluorine is more electronegative than oxygen.
The O–F bonding electron density is consequently attracted more strongly towards the terminal fluorine atoms.
The distribution of electron density around the oxygen centre therefore differs from that in water.
This affects the repulsive interactions involving bonding regions and oxygen lone pairs.
The observed F–O–F angle is approximately \(103.2^\circ\).
\(\textbf{Interpret the angular difference:}\)
The two molecules possess the same number of bonding and nonbonding domains.
However, they do not possess identical electronic environments.
The difference in terminal-atom electronegativity changes the distribution of shared electron density.
Consequently, the angular repulsions are not required to produce identical observed bond angles.
\(\textbf{Avoid an overstrong inference:}\)
The exact numerical difference cannot be calculated solely from the simple VSEPR domain count.
Bond length and other electronic factors also contribute to the observed geometry.
The common \(\mathrm{AX_2E_2}\) classification correctly predicts bent structures but not a universal exact angle.
Therefore, option D is the appropriate interpretation.
212. Gaseous aluminium chloride can exist as monomeric \(\mathrm{AlCl_3}\), while aluminium chloride also forms the dimer \(\mathrm{Al_2Cl_6}\), containing two chlorine atoms that bridge the aluminium centres. In the dimer, each aluminium is bonded to two terminal chlorine atoms and interacts with two bridging chlorine atoms through Al–Cl bonding regions. Which statement correctly compares the coordination geometry around aluminium in these two structural descriptions?
ⓐ. Monomeric \(\mathrm{AlCl_3}\) is tetrahedral, whereas every aluminium in the dimer becomes trigonal planar.
ⓑ. Monomeric \(\mathrm{AlCl_3}\) is trigonal planar, whereas each aluminium in the dimer has four Al–Cl bonding directions and an approximately tetrahedral local arrangement.
ⓒ. Aluminium remains trigonal planar in both because a bridging chlorine cannot contribute a bonding domain.
ⓓ. Dimerisation produces a linear arrangement around each aluminium because the two bridging chlorines must lie opposite one another.
Correct Answer: Monomeric \(\mathrm{AlCl_3}\) is trigonal planar, whereas each aluminium in the dimer has four Al–Cl bonding directions and an approximately tetrahedral local arrangement.
Explanation: \(\textbf{Examine monomeric aluminium chloride:}\)
An isolated \(\mathrm{AlCl_3}\) molecule contains three Al–Cl bonding regions around aluminium.
The conventional Lewis structure has no central aluminium lone pair.
Therefore, aluminium has three bonding domains.
The molecule adopts a trigonal-planar arrangement.
The central aluminium atom is electron deficient in the conventional localised-bond description.
This electron deficiency helps explain its Lewis-acid behaviour.
\(\textbf{Examine the dimeric structure:}\)
The question specifies two terminal chlorine atoms and two bridging chlorine atoms associated with each aluminium.
Thus, each aluminium has four Al–Cl bonding directions in the dimer.
The bridging chlorine atoms participate in connecting the two aluminium centres.
Their involvement in bridging interactions does not remove those bonding regions from a local geometrical description.
\(\textbf{Determine the local arrangement:}\)
Four bonding directions around aluminium favour an approximately tetrahedral local arrangement.
The actual angles need not be ideal because terminal and bridging chlorine environments differ.
Nevertheless, the coordination number has increased from three in the monomer to four in the dimer.
\(\textbf{Interpret the structural change:}\)
The molecular formula \(\mathrm{AlCl_3}\) alone does not describe every form in which aluminium chloride can exist.
Aggregation changes the local bonding environment.
Consequently, the geometry of the isolated monomer cannot automatically be assigned to each aluminium in the dimer.
The bridging arrangement also illustrates how electron-deficient centres can achieve additional bonding interactions.
\(\textbf{Distinguish coordination from electron counting:}\)
VSEPR analysis must be applied to the actual structural unit being considered.
It would be incorrect to ignore the bridging chlorine interactions merely because they are absent from the simplest monomeric formula.
Therefore, option B correctly identifies the different local arrangements.
213. The dimerisation of nitrogen dioxide is represented by:
\[
\mathrm{2NO_2(g)\rightleftharpoons N_2O_4(g)}
\]
In a conventional Lewis description of \(\mathrm{N_2O_4}\), the two nitrogen atoms are joined by an N–N single bond, and each nitrogen is bonded to two oxygen atoms. Each nitrogen–oxygen group has resonance contributors involving an N=O double bond and an N–O single bond. Which statement correctly describes the local VSEPR arrangement around each nitrogen in \(\mathrm{N_2O_4}\)?
ⓐ. Each nitrogen has three bonding domains and no lone pair, giving an approximately trigonal-planar local arrangement.
ⓑ. Each nitrogen has two bonding domains and one lone pair, giving a bent local arrangement.
ⓒ. Each nitrogen has four bonding domains because the N=O double bond contributes two independent directions, giving a tetrahedral arrangement.
ⓓ. Each nitrogen has two bonding domains and an unpaired electron, giving the same electronic description as monomeric \(\mathrm{NO_2}\).
Correct Answer: Each nitrogen has three bonding domains and no lone pair, giving an approximately trigonal-planar local arrangement.
Explanation: \(\textbf{Examine the original nitrogen dioxide molecules:}\)
Neutral \(\mathrm{NO_2}\) is an odd-electron species.
Its valence-electron count is:
\[
N_v(\mathrm{NO_2})=5+2(6)
\]
\[
N_v(\mathrm{NO_2})=17
\]
The unpaired electron is important to the electronic character of the monomer.
\(\textbf{Consider the dimerisation:}\)
Two nitrogen dioxide units combine to form \(\mathrm{N_2O_4}\).
The dimer contains an N–N single bond.
Its conventional Lewis representation is closed shell.
Therefore, the local electron arrangement around each nitrogen must be examined using the dimer structure rather than assuming that each nitrogen retains the monomer's unpaired electron.
\(\textbf{Count the bonding directions around one nitrogen:}\)
Each nitrogen is bonded to the other nitrogen.
That N–N single bond contributes one VSEPR domain.
The nitrogen is also bonded to two oxygen atoms.
Those N–O bonding regions contribute two more domains.
A double bond represented in a resonance contributor still counts as only one bonding domain.
Thus:
\[
N_{\mathrm{bonding\ domains}}=1+2
\]
\[
N_{\mathrm{bonding\ domains}}=3
\]
There is no central nitrogen lone pair in the conventional dimeric Lewis description.
\(\textbf{Determine the local geometry:}\)
Three bonding domains favour a trigonal-planar local arrangement.
Each nitrogen is therefore approximately trigonal planar with respect to the atoms directly attached to it.
This does not require every atom in the entire dimer to lie in one plane.
\(\textbf{Interpret the change from monomer to dimer:}\)
Dimerisation establishes a new N–N bonding interaction.
It changes the electronic description from two odd-electron units to a closed-shell molecular species.
The local VSEPR arrangement follows from the three actual bond directions around each nitrogen.
Option A is correct.
214. A hypothetical square-pyramidal molecule contains four equivalent basal A–X bonds and one axial A–Y bond. The basal A–X bond-dipole vectors each have magnitude \(1.50\ \mathrm{D}\), while the axial A–Y vector has magnitude \(0.80\ \mathrm{D}\). One basal X atom is replaced by Z, producing an A–Z bond-dipole magnitude of \(2.10\ \mathrm{D}\). Assume ideal square-pyramidal geometry and all bond dipoles directed radially outward. What is the resultant magnitude after substitution, considering only the stated bond-dipole contributions?
ⓐ. \(0.20\ \mathrm{D}\)
ⓑ. \(0.80\ \mathrm{D}\)
ⓒ. \(1.40\ \mathrm{D}\)
ⓓ. \(1.00\ \mathrm{D}\)
Correct Answer: \(1.00\ \mathrm{D}\)
Explanation: \(\textbf{Identify the original square-pyramidal vector arrangement:}\)
Four equivalent basal bond-dipole vectors point towards the four corners of a square.
They form two pairs of opposite vectors.
Consequently, the original basal resultant is zero.
The axial A–Y bond-dipole vector points perpendicular to the basal plane.
Its magnitude is \(0.80\ \mathrm{D}\).
\(\textbf{Determine the effect of basal substitution:}\)
One A–X bond is replaced by an A–Z bond.
The original magnitude was:
\[
\mu_X=1.50\ \mathrm{D}
\]
The new magnitude is:
\[
\mu_Z=2.10\ \mathrm{D}
\]
The additional contribution along the substituted basal direction is:
\[
\Delta\mu_{\mathrm{basal}}=\mu_Z-\mu_X
\]
\[
\Delta\mu_{\mathrm{basal}}=2.10-1.50
\]
\[
\Delta\mu_{\mathrm{basal}}=0.60\ \mathrm{D}
\]
The remaining basal vectors still cancel their original contributions.
Therefore, the basal resultant has magnitude \(0.60\ \mathrm{D}\).
\(\textbf{Combine the basal and axial contributions:}\)
The basal resultant lies in the square plane.
The axial vector is perpendicular to that plane.
Thus, the two remaining contributions are perpendicular.
\[
\mu_R=\sqrt{\mu_{\mathrm{basal}}^2+\mu_{\mathrm{axial}}^2}
\]
Substituting:
\[
\mu_R=\sqrt{(0.60)^2+(0.80)^2}
\]
\[
\mu_R=\sqrt{0.36+0.64}
\]
\[
\mu_R=\sqrt{1.00}
\]
\[
\mu_R=1.00\ \mathrm{D}
\]
\(\textbf{Check the physical interpretation:}\)
Before substitution, the basal dipoles cancelled and the model predicted an axial resultant.
Substitution breaks the basal symmetry and introduces a new perpendicular contribution.
The resulting molecular dipole is therefore larger than the original axial contribution.
The answer is not obtained by adding all five dipole magnitudes as scalars.
The geometrical directions determine the actual resultant.
\(\textbf{Final answer:}\)
\[
\boxed{\mu_R=1.00\ \mathrm{D}}
\]
215. A hypothetical molecule has the formula \(\mathrm{AX_3}\). Electron diffraction establishes that one X–A–X bond angle is close to \(180^\circ\), while the other two X–A–X angles are close to \(90^\circ\). All three A–X bonds are present, and the molecule is not linear. Which conventional VSEPR description is most consistent with these observations?
ⓐ. \(\mathrm{AX_3}\), with three electron domains and trigonal-planar geometry.
ⓑ. \(\mathrm{AX_3E}\), with four electron domains and trigonal-pyramidal geometry.
ⓒ. \(\mathrm{AX_3E_2}\), with five electron domains and T-shaped molecular geometry.
ⓓ. \(\mathrm{AX_3E_3}\), with six electron domains and trigonal-planar molecular geometry.
Correct Answer: \(\mathrm{AX_3E_2}\), with five electron domains and T-shaped molecular geometry.
Explanation: \(\textbf{Interpret the measured angular pattern:}\)
The molecule contains three bonded X atoms.
Two of the bond directions are approximately opposite because one X–A–X angle is close to \(180^\circ\).
The third bond direction is approximately perpendicular to each of those opposite directions.
This is the defining angular pattern of a T-shaped arrangement.
\(\textbf{Examine a three-domain structure:}\)
An \(\mathrm{AX_3}\) molecule with no central lone pairs ordinarily adopts a trigonal-planar arrangement.
Its three equivalent ideal bond angles are \(120^\circ\).
It would not exhibit the specified combination of one near-\(180^\circ\) angle and two near-\(90^\circ\) angles.
\(\textbf{Examine a four-domain structure:}\)
An \(\mathrm{AX_3E}\) species has a tetrahedral electron-domain arrangement.
Its molecular shape is trigonal pyramidal.
The three bond directions are not arranged as one opposite pair and one perpendicular direction.
Therefore, it is inconsistent with the reported idealised angular pattern.
\(\textbf{Examine the five-domain possibility:}\)
An \(\mathrm{AX_3E_2}\) species has five electron domains.
Its electron-domain arrangement is trigonal bipyramidal.
The two lone pairs preferentially occupy equatorial positions.
The remaining bonding domains occupy two opposite axial positions and one equatorial position.
The resulting molecular shape is T-shaped.
This gives one approximately \(180^\circ\) bond angle and two approximately \(90^\circ\) angles.
\(\textbf{Evaluate the six-domain alternative:}\)
An \(\mathrm{AX_3E_3}\) species can also be T-shaped in a suitable octahedral lone-pair arrangement.
Therefore, the measured angular pattern alone does not uniquely establish whether the species has five or six electron domains.
Therefore, bond-angle observations by themselves do not distinguish all possible electron-domain arrangements that can give a T-shaped molecule.
The six-domain alternative, however, is described as trigonal planar, which is inconsistent with the near-\(90^\circ\) and near-\(180^\circ\) angles.
Only the five-domain T-shaped description in option C agrees with the stated observations and its accompanying geometry label.
The bond angles identify the molecular shape, but additional information may be needed to determine the number of lone pairs on the central atom.
216. A student is given two hypothetical species, P and Q, each having a central atom bonded to two identical terminal atoms. P possesses three electron domains, while Q possesses four electron domains. The additional domains beyond the two bonding regions are lone pairs. Which statement correctly identifies their electron-domain arrangements and the limitation of predicting exact bond angles?
ⓐ. P is bent with trigonal-planar electron geometry; Q is bent with tetrahedral electron geometry; exact angles remain undetermined.
ⓑ. P is linear because it has only two bonded atoms; Q is bent because it has four electron domains.
ⓒ. P is bent and Q is linear because two lone pairs always force opposing bond directions.
ⓓ. Both are bent within identical tetrahedral electron-domain arrangements because each has two terminal atoms.
Correct Answer: P is bent with trigonal-planar electron geometry; Q is bent with tetrahedral electron geometry; exact angles remain undetermined.
Explanation: \(\textbf{Analyse Species P:}\)
P has two bonding domains and three total electron domains.
Therefore, it contains one central lone-pair domain.
Its VSEPR description is \(\mathrm{AX_2E}\).
Three electron domains favour a trigonal-planar arrangement.
However, only two domains contain bonded terminal atoms.
Therefore, the molecular shape is bent.
\(\textbf{Analyse Species Q:}\)
Q has two bonding domains and four total electron domains.
Therefore, it contains two central lone-pair domains.
Its VSEPR description is \(\mathrm{AX_2E_2}\).
Four electron domains favour an approximately tetrahedral arrangement.
Only two domains contain bonded terminal atoms.
Its molecular shape is consequently also bent.
\(\textbf{Distinguish shared molecular shape from electron geometry:}\)
The two species are both bent.
However, their electron-domain arrangements differ.
P belongs to the three-domain family, while Q belongs to the four-domain family.
Therefore, the term bent does not uniquely identify the total number of electron domains.
\(\textbf{Consider bond angles:}\)
The ideal electron-domain angle for three equivalent planar domains is \(120^\circ\).
The ideal tetrahedral electron-domain angle is approximately \(109.5^\circ\).
However, actual bond angles may deviate from these ideal values because lone pairs and bonding regions exert different repulsive influences.
The identities of the central and terminal atoms also affect the electronic distribution.
Therefore, the domain count alone does not provide an exact measured bond angle.
Option A correctly distinguishes the electron-domain arrangements and the limits of angle prediction.
217. Consider the following statements regarding electron-domain geometry, molecular geometry and molecular polarity.
Statement I: Two molecules may have the same molecular shape while possessing different electron-domain arrangements.
Statement II: Two molecules may have the same electron-domain arrangement but different molecular shapes.
Statement III: Every molecule with a trigonal-pyramidal shape must have three lone pairs on its central atom.
Statement IV: In a molecule with equivalent peripheral atoms, a sufficiently symmetrical arrangement can produce zero permanent dipole moment even when individual bonds are polar.
Which statements are correct?
ⓐ. Statements I and III only
ⓑ. Statements II and III only
ⓒ. Statements I, II and III only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: \(\textbf{Evaluate Statement I:}\)
Statement I is correct.
Molecular shape describes the arrangement of the bonded atomic nuclei.
Different electron-domain arrangements can leave bonded atoms in the same broad molecular shape.
For example, an \(\mathrm{AX_2E}\) species and an \(\mathrm{AX_2E_2}\) species can both be bent.
Their total electron-domain counts are three and four, respectively.
Therefore, identical molecular-shape names do not require identical electron-domain geometries.
\(\textbf{Evaluate Statement II:}\)
Statement II is correct.
A four-domain arrangement can correspond to several molecular shapes depending on how many positions contain lone pairs.
An \(\mathrm{AX_4}\) species is tetrahedral in molecular shape.
An \(\mathrm{AX_3E}\) species is trigonal pyramidal.
An \(\mathrm{AX_2E_2}\) species is bent.
All three have approximately tetrahedral electron-domain arrangements.
Thus, a common electron-domain arrangement does not guarantee a common molecular shape.
\(\textbf{Evaluate Statement III:}\)
Statement III is incorrect.
A conventional trigonal-pyramidal \(\mathrm{AX_3E}\) molecule has three bonding domains and one central lone pair.
It does not require three central lone pairs.
Three lone pairs around a central atom would lead to a different total domain count and potentially a different geometry.
The statement confuses the three bonded atoms with the number of nonbonding pairs.
\(\textbf{Evaluate Statement IV:}\)
Statement IV is correct.
Molecular dipole moment is a vector quantity.
Equal polar bond-dipole contributions can cancel when their spatial arrangement has the appropriate symmetry.
Carbon tetrachloride provides an example of a symmetrical molecule with polar bonds and zero permanent molecular dipole moment.
\(\textbf{Overall conclusion:}\)
Statements I and II correctly distinguish molecular shape from electron-domain geometry.
Statement III incorrectly assigns the central lone-pair count for a trigonal-pyramidal species.
Statement IV correctly recognises vector cancellation through symmetry.
Therefore, Statements I, II and IV are correct.
218. An octahedral \(\mathrm{AX_6}\) structure is modified by replacing two X atoms with Y atoms. Assume all original A–X bond-dipole magnitudes are equal and that the A–Y bonds have a different but equal magnitude. Which assertion–reason pair correctly describes the possible effect of placing the two Y atoms either opposite or adjacent?
Assertion (A): Under an ideal bond-dipole model, the opposite-Y arrangement can have zero resultant molecular dipole, whereas the adjacent-Y arrangement can have a nonzero resultant.
Reason (R): The two changes in bond-dipole magnitude cancel as opposite vectors when Y atoms occupy opposite positions, but combine as perpendicular vectors when Y atoms occupy adjacent octahedral positions.
Choose the correct option.
ⓐ. Both A and R are true, and R correctly explains A.
ⓑ. Both A and R are true, but R does not explain A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: Both A and R are true, and R correctly explains A.
Explanation: An ideal octahedral structure contains three pairs of opposite bond directions.
If all six A–X bond-dipole vectors have equal magnitudes, they cancel through symmetry.
Replacing an X atom with Y changes the dipole magnitude along the substituted direction.
The effect can be analysed as an additional vector relative to the original symmetrical arrangement.
\(\textbf{Examine opposite substitution:}\)
When the two Y atoms occupy opposite positions, the additional contributions are equal in magnitude and antiparallel.
They cancel.
The four unchanged A–X contributions also cancel in opposite pairs.
Therefore, the opposite-Y arrangement has zero resultant under the stated ideal model.
\(\textbf{Examine adjacent substitution:}\)
When Y atoms occupy adjacent octahedral positions, their directions are perpendicular.
The two additional contributions therefore do not cancel.
Instead, they combine to give a nonzero resultant whenever the A–Y bond-dipole magnitude differs from the A–X magnitude.
\(\textbf{Evaluate the assertion:}\)
The assertion correctly distinguishes the symmetry-preserving opposite arrangement from the symmetry-breaking adjacent arrangement.
\(\textbf{Evaluate the reason:}\)
The reason correctly identifies the vector relationships responsible for the different results.
It also supplies the direct explanation for the assertion.
The distinction does not depend on the absolute bond-dipole magnitudes, provided the substitution changes their magnitudes as specified.
The statement would require reconsideration if substitution altered the geometry or introduced additional dipole contributions, but those effects are excluded here.
Therefore, both A and R are true, and R correctly explains A.
219. The \(\mathrm{NH_2^-}\), \(\mathrm{NH_3}\) and \(\mathrm{NH_4^+}\) species contain the same nitrogen centre but different numbers of attached hydrogen atoms. Which sequence correctly gives the conventional central nitrogen lone-pair counts and the corresponding local arrangements?
ⓐ. \(\mathrm{NH_2^-}\): one lone pair, trigonal planar; \(\mathrm{NH_3}\): two lone pairs, bent; \(\mathrm{NH_4^+}\): no lone pair, tetrahedral.
ⓑ. \(\mathrm{NH_2^-}\): two lone pairs, bent; \(\mathrm{NH_3}\): one lone pair, trigonal pyramidal; \(\mathrm{NH_4^+}\): no lone pair, tetrahedral.
ⓒ. \(\mathrm{NH_2^-}\): two lone pairs, linear; \(\mathrm{NH_3}\): one lone pair, trigonal planar; \(\mathrm{NH_4^+}\): no lone pair, square planar.
ⓓ. All three have one central lone pair because nitrogen contributes five valence electrons in every species.
Correct Answer: \(\mathrm{NH_2^-}\): two lone pairs, bent; \(\mathrm{NH_3}\): one lone pair, trigonal pyramidal; \(\mathrm{NH_4^+}\): no lone pair, tetrahedral.
Explanation: \(\textbf{Analyse the amide ion:}\)
Nitrogen contributes five valence electrons, while the two hydrogen atoms contribute two.
The negative ionic charge adds one electron.
\[
N_v(\mathrm{NH_2^-})=5+2+1
\]
\[
N_v(\mathrm{NH_2^-})=8
\]
Two N–H single bonds use four electrons.
The remaining four electrons form two central nitrogen lone pairs.
Therefore, nitrogen has two bonding domains and two lone-pair domains.
Its electron-domain arrangement is tetrahedral.
The two bonded hydrogen atoms form a bent local arrangement.
\(\textbf{Analyse ammonia:}\)
\[
N_v(\mathrm{NH_3})=5+3
\]
\[
N_v(\mathrm{NH_3})=8
\]
Three N–H bonds use six electrons.
The remaining two electrons form one central lone pair.
Nitrogen therefore has three bonding domains and one lone-pair domain.
Its molecular shape is trigonal pyramidal.
\(\textbf{Analyse ammonium:}\)
\[
N_v(\mathrm{NH_4^+})=5+4-1
\]
\[
N_v(\mathrm{NH_4^+})=8
\]
Four N–H bonds account for all eight valence electrons.
Nitrogen has four bonding domains and no lone pair.
The molecular shape is tetrahedral.
\(\textbf{Interpret the progression:}\)
All three species contain eight total valence electrons.
However, those electrons are distributed differently among bonding and nonbonding domains.
The number of N–H bonds increases while the central lone-pair count decreases.
The total number of electron domains around nitrogen remains four in each case.
Consequently, the electron-domain family remains approximately tetrahedral while the molecular shape changes.
This demonstrates why total valence-electron count alone cannot determine molecular shape without a valid Lewis structure.
Option B correctly gives all three lone-pair counts and local geometries.
220. The ions \(\mathrm{SO_3^{2-}}\) and \(\mathrm{SO_4^{2-}}\) both contain sulfur bonded to oxygen atoms. In their conventional resonance-based VSEPR descriptions, sulfite possesses a central sulfur lone pair, whereas sulfate does not. Which statement correctly compares their molecular shapes and identifies the change in electron-domain distribution?
ⓐ. Sulfite is trigonal planar and sulfate is tetrahedral; both have four bonding domains.
ⓑ. Sulfite is trigonal pyramidal and sulfate is square planar; the central lone pair in sulfite becomes two lone pairs in sulfate.
ⓒ. Both are tetrahedral in molecular shape because they contain four central electron domains.
ⓓ. Sulfite is trigonal pyramidal with three bonding domains and one lone pair; sulfate is tetrahedral with four bonding domains and no central lone pair.
Correct Answer: Sulfite is trigonal pyramidal with three bonding domains and one lone pair; sulfate is tetrahedral with four bonding domains and no central lone pair.
Explanation: \(\textbf{Examine the sulfite ion:}\)
Sulfur is directly bonded to three oxygen atoms in \(\mathrm{SO_3^{2-}}\).
The conventional resonance contributors may contain S–O single and double bonds.
However, each S–O bonding region counts as one geometrical domain.
Sulfur therefore has three bonding domains.
It also possesses one central lone pair.
The total electron-domain count is four.
The electron-domain arrangement is approximately tetrahedral.
Because only three domains contain bonded oxygen atoms, the molecular shape is trigonal pyramidal.
\(\textbf{Examine the sulfate ion:}\)
Sulfur is directly bonded to four oxygen atoms in \(\mathrm{SO_4^{2-}}\).
The four S–O bonding regions constitute four electron domains.
The conventional VSEPR description contains no central sulfur lone pair.
Consequently, the molecular arrangement of the four oxygen atoms is tetrahedral.
\(\textbf{Identify the common feature:}\)
Both sulfur centres possess four electron domains in their conventional descriptions.
The underlying electron-domain arrangement is therefore approximately tetrahedral in both cases.
\(\textbf{Identify the decisive difference:}\)
Sulfite contains three bonded oxygen atoms and one lone pair around sulfur.
Sulfate contains four bonded oxygen atoms and no central lone pair.
Therefore, their molecular shapes differ despite the common electron-domain count.
\(\textbf{Consider resonance correctly:}\)
Resonance affects the distribution of bonding electron density among the S–O connections.
It does not change the number of oxygen atoms attached to sulfur.
Nor does representing a multiple bond create an additional independent VSEPR direction.
The central lone-pair count and the number of bonded atoms remain the decisive features for shape assignment.
Thus, option D is correct.