Molecular Structure and Bonding Theory Questions and Answers

Q: The CORRECT order of ionic character (in percentage) among the following is

The order is HBr \(\lt\) HCl \(\lt\) NaCl.The percentage ionic character of a bond grows with the electronegativity difference between its atoms. On the Pauling scale H is 2.20, Br 2.96, Cl 3.16 and Na 0.93, giving differences of about 0.8 for H-Br, 1.0 for H-Cl and 2.2 for Na-Cl.So HBr is the least ionic of the three, HCl comes next, and NaCl, an ionic solid, is by far the most ionic.The orders that put NaCl first or in the middle place a metal halide below covalent hydrogen halides, and the order HCl \(\lt\) HBr reverses the effect of chlorine being more electronegative than bromine.

Q: The repulsive interaction between different types of electron pairs increases in the order of

The order of increasing repulsion is bond pair - bond pair < lone pair - bond pair < lone pair - lone pair.This is the core rule of VSEPR theory. A lone pair is held by only one nucleus, so its electron cloud is broader and sits closer to the central atom than a bonding pair, which is drawn out between two nuclei. Lone pairs therefore repel more strongly, and two lone pairs repel each other most of all.That is why the H-O-H angle in water (about 104.5°) and the H-N-H angle in ammonia (about 107°) are smaller than the tetrahedral 109.5° of methane. The other three options put bond pair - bond pair repulsion above a repulsion involving a lone pair, which reverses the rule.

Q: The correct order of F-M-F bond angle of compounds given in the options is (where M represents the central element)

The correct order is XeF2 > BF3 > [NF4]+ > OF2.XeF2: three lone pairs occupy the equatorial positions of a trigonal bipyramid, so the molecule is linear, 180°.BF3: three bond pairs and no lone pair, trigonal planar, 120°.[NF4]+: four bond pairs, regular tetrahedron, 109.5°.OF2: two bond pairs and two lone pairs, bent, about 103°, below the tetrahedral angle because lone pairs repel more strongly.Putting BF3 ahead of XeF2 ignores the linear geometry, and orders placing OF2 above [NF4]+ or BF3 ignore the compression caused by its two lone pairs.

Q: The species that has 'T' shape is

The T-shaped species is BrF3.Bromine has 7 valence electrons. Three form bonds to fluorine and the remaining four make two lone pairs, giving five electron pairs in a trigonal bipyramid. The two lone pairs take equatorial positions, where they have more room, leaving the three fluorines in a T: two axial and one equatorial, with F-Br-F angles a little under 90°.The other three have four electron pairs around the central atom, one of them a lone pair, and so are trigonal pyramidal: PCl3 (three P-Cl bonds and a lone pair), SOCl2 (S=O, two S-Cl and a lone pair on sulfur) and [IO3]− (three I-O bonds and a lone pair on iodine). None of them is T-shaped.

Q: The correct order of dipole moments for the given species is

The correct order is BF3 = NH4+ < NF3 < NH3.BF3 is trigonal planar and NH4+ is a regular tetrahedron. In both, the bond dipoles cancel exactly by symmetry, so both have zero dipole moment and are equal, which rules out every option that puts one below the other.NH3 and NF3 are both pyramidal with a lone pair on nitrogen. In NH3, nitrogen is more electronegative than hydrogen, so the N-H bond dipoles point towards N and add to the lone-pair moment, about 1.5 D in all. In NF3, fluorine is more electronegative than nitrogen, so the N-F bond dipoles point away from N and largely cancel the lone-pair moment, leaving only about 0.2 D. So NF3 < NH3, and the options placing NF3 last are wrong.

Q: Consider the following species:SOCl2, XeOF4, ClF3, ClF5, XeF5+, SO32-, XeF3+, SF4List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.List-IList-II(P) See-saw(1) one(Q) T-Shaped(2) two(R) Trigonal Planar(3) three(S) Square Pyramidal(4) four(5) zero

The correct match is P → 1; Q → 2; R → 5; S → 3.Count bond pairs and lone pairs on each central atom (VSEPR):SpeciesBond pairs, lone pairsShapeSOCl23, 1trigonal pyramidalSO32-3, 1trigonal pyramidalSF44, 1see-sawClF33, 2T-shapedXeF3+3, 2T-shapedClF55, 1square pyramidalXeF5+5, 1square pyramidalXeOF45, 1square pyramidalSo one species is see-saw (SF4), two are T-shaped (ClF3 and XeF3+), three are square pyramidal, and none is trigonal planar: SOCl2 and SO32- look like candidates but each has a lone pair on sulfur, which makes them pyramidal. Options giving trigonal planar a count of one or two make exactly that mistake.

Q: The correct order of ONO bond angle in the given species is

The correct order is \(\mathrm{NO_2^- \lt NO_3^- \lt NO_2 \lt NO_2^+}\).The angles follow from the electron domains around nitrogen:\(\mathrm{NO_{2}^{+}}\): two bonding domains and no lone pair on N, so it is linear, 180°.\(\mathrm{NO_{2}}\): one unpaired electron on N. A single electron repels the bonds less than a lone pair, so the angle opens to about 134°.\(\mathrm{NO_{3}^{-}}\): three equivalent bonding domains, trigonal planar, 120°.\(\mathrm{NO_{2}^{-}}\): a full lone pair on N compresses the angle below 120°, to about 115°.The first option is this order reversed. The third puts nitrate below nitrite, though nitrite's lone pair makes it the smaller. The last puts \(\mathrm{NO_{2}^{+}}\) below \(\mathrm{NO_{2}}\), but a linear ion has the largest possible angle.

Q: The correct increasing order of dipole moment of NF3, H2S, CHCl3 and NH3 molecules is

The increasing order is \(NF_{3} \lt H_{2}S \lt CHCl_{3} \lt NH_{3}\).Approximate dipole moments are \(NF_{3}\) 0.24 D, \(H_{2}S\) 0.95 D, \(CHCl_{3}\) 1.04 D and \(NH_{3}\) 1.47 D.The key contrast is between \(NH_{3}\) and \(NF_{3}\). In ammonia the lone pair's moment points the same way as the N-H bond moments (towards N), so they add. In \(NF_{3}\) the N-F bond moments point away from N, opposing the lone pair, so the net moment is small. Any order with \(NF_{3}\) above \(NH_{3}\) is therefore wrong.

Q: Choose the INCORRECT pair of MOLECULE and its SHAPE among the following:

The incorrect pair is \(BrF_{5}\) Trigonal bipyramidal.Bromine in \(BrF_{5}\) has five bond pairs and one lone pair, six electron pairs in an octahedral arrangement. With one position taken by the lone pair, the molecule is square pyramidal.The other pairs are right: \(SF_{4}\) (4 bond pairs, 1 lone pair) is see-saw; \(NH_{3}\) (3 and 1) is trigonal pyramidal; \(XeF_{4}\) (4 and 2) is square planar; \(ClF_{3}\) (3 and 2) is T-shaped.

Q: Which of the following has least mean multiple bond enthalpy (in kJ mol-1 ) at 298 K?

The least is C = C.Mean multiple bond enthalpies (kJ mol-1) from the standard table are about: C=C 611, C=N 615, C&equiv;N 891, N&equiv;N 946, C&equiv;O 1070.Triple bonds are much stronger than double bonds, which rules out the three triple-bonded options. Between the two double bonds, C=N is slightly stronger than C=C, so C=C has the lowest value.

11.

The CORRECT order of ionic character (in percentage) among the following is

A.NaCl \(\lt\) HCl \(\lt\) HBr
B.HCl \(\lt\) NaCl \(\lt\) HBr
C.HCl \(\lt\) HBr \(\lt\) NaCl
D.HBr \(\lt\) HCl \(\lt\) NaCl

Exam Year(s)

GATE Life Sciences - 2026,

Answer: D

The order is HBr \(\lt\) HCl \(\lt\) NaCl.

The percentage ionic character of a bond grows with the electronegativity difference between its atoms. On the Pauling scale H is 2.20, Br 2.96, Cl 3.16 and Na 0.93, giving differences of about 0.8 for H-Br, 1.0 for H-Cl and 2.2 for Na-Cl.

So HBr is the least ionic of the three, HCl comes next, and NaCl, an ionic solid, is by far the most ionic.

The orders that put NaCl first or in the middle place a metal halide below covalent hydrogen halides, and the order HCl \(\lt\) HBr reverses the effect of chlorine being more electronegative than bromine.

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12.

The repulsive interaction between different types of electron pairs increases in the order of

A.bond pair - bond pair < lone pair - bond pair < lone pair - lone pair
B.lone pair - lone pair < lone pair - bond pair < bond pair - bond pair
C.lone pair - bond pair < lone pair - lone pair < bond pair - bond pair
D.lone pair - lone pair < bond pair - bond pair < lone pair - bond pair

Exam Year(s)

IIT JAM - 2026,

Answer: A

The order of increasing repulsion is bond pair - bond pair < lone pair - bond pair < lone pair - lone pair.

This is the core rule of VSEPR theory. A lone pair is held by only one nucleus, so its electron cloud is broader and sits closer to the central atom than a bonding pair, which is drawn out between two nuclei. Lone pairs therefore repel more strongly, and two lone pairs repel each other most of all.

That is why the H-O-H angle in water (about 104.5°) and the H-N-H angle in ammonia (about 107°) are smaller than the tetrahedral 109.5° of methane. The other three options put bond pair - bond pair repulsion above a repulsion involving a lone pair, which reverses the rule.

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13.

The correct order of F-M-F bond angle of compounds given in the options is (where M represents the central element)

A.\(\text{XeF}_{2} > \text{BF}_{3} > \text{OF}_{2} > [\text{NF}_{4}]^{+}\)
B.\(\text{XeF}_{2} > \text{BF}_{3} > [\text{NF}_{4}]^{+} > \text{OF}_{2}\)
C.\(\text{BF}_{3} > \text{XeF}_{2} > [\text{NF}_{4}]^{+} > \text{OF}_{2}\)
D.\(\text{XeF}_{2} > \text{OF}_{2} > \text{BF}_{3} > [\text{NF}_{4}]^{+}\)

Exam Year(s)

IIT JAM - 2026,

Answer: B

The correct order is XeF₂ > BF₃ > [NF₄]⁺ > OF₂.

XeF₂: three lone pairs occupy the equatorial positions of a trigonal bipyramid, so the molecule is linear, 180°.
BF₃: three bond pairs and no lone pair, trigonal planar, 120°.
[NF₄]⁺: four bond pairs, regular tetrahedron, 109.5°.
OF₂: two bond pairs and two lone pairs, bent, about 103°, below the tetrahedral angle because lone pairs repel more strongly.

Putting BF₃ ahead of XeF₂ ignores the linear geometry, and orders placing OF₂ above [NF₄]⁺ or BF₃ ignore the compression caused by its two lone pairs.

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14.

The species that has 'T' shape is

A.\(\text{PCl}_{3}\)
B.\(\text{SOCl}_{2}\)
C.\(\text{BrF}_{3}\)
D.\([\text{IO}_{3}]^{-}\)

Exam Year(s)

IIT JAM - 2026,

Answer: C

The T-shaped species is BrF₃.

Bromine has 7 valence electrons. Three form bonds to fluorine and the remaining four make two lone pairs, giving five electron pairs in a trigonal bipyramid. The two lone pairs take equatorial positions, where they have more room, leaving the three fluorines in a T: two axial and one equatorial, with F-Br-F angles a little under 90°.

The other three have four electron pairs around the central atom, one of them a lone pair, and so are trigonal pyramidal: PCl₃ (three P-Cl bonds and a lone pair), SOCl₂ (S=O, two S-Cl and a lone pair on sulfur) and [IO₃]⁻ (three I-O bonds and a lone pair on iodine). None of them is T-shaped.

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15.

The correct order of dipole moments for the given species is

A.\(\text{BF}_{3} = \text{NH}_{4}^{+} < \text{NF}_{3} < \text{NH}_{3}\)
B.\(\text{BF}_{3} < \text{NH}_{4}^{+} < \text{NF}_{3} < \text{NH}_{3}\)
C.\(\text{NH}_{4}^{+} < \text{BF}_{3} < \text{NH}_{3} < \text{NF}_{3}\)
D.\(\text{BF}_{3} < \text{NH}_{4}^{+} < \text{NH}_{3} < \text{NF}_{3}\)

Exam Year(s)

JEE Advanced - 2026,

Answer: A

The correct order is BF₃ = NH₄⁺ < NF₃ < NH₃.

BF₃ is trigonal planar and NH₄⁺ is a regular tetrahedron. In both, the bond dipoles cancel exactly by symmetry, so both have zero dipole moment and are equal, which rules out every option that puts one below the other.

NH₃ and NF₃ are both pyramidal with a lone pair on nitrogen. In NH₃, nitrogen is more electronegative than hydrogen, so the N-H bond dipoles point towards N and add to the lone-pair moment, about 1.5 D in all. In NF₃, fluorine is more electronegative than nitrogen, so the N-F bond dipoles point away from N and largely cancel the lone-pair moment, leaving only about 0.2 D. So NF₃ < NH₃, and the options placing NF₃ last are wrong.

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16.

Consider the following species:

SOCl₂, XeOF₄, ClF₃, ClF₅, XeF₅⁺, SO₃^2-, XeF₃⁺, SF₄

List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List-I	List-II
(P) See-saw	(1) one
(Q) T-Shaped	(2) two
(R) Trigonal Planar	(3) three
(S) Square Pyramidal	(4) four
	(5) zero

A.P → 1; Q → 2; R → 5; S → 3
B.P → 5; Q → 4; R → 2; S → 3
C.P → 3; Q → 2; R → 1; S → 4
D.P → 1; Q → 3; R → 5; S → 4

Exam Year(s)

JEE Advanced - 2026,

Answer: A

The correct match is P → 1; Q → 2; R → 5; S → 3.

Count bond pairs and lone pairs on each central atom (VSEPR):

Species	Bond pairs, lone pairs	Shape
SOCl₂	3, 1	trigonal pyramidal
SO₃^2-	3, 1	trigonal pyramidal
SF₄	4, 1	see-saw
ClF₃	3, 2	T-shaped
XeF₃⁺	3, 2	T-shaped
ClF₅	5, 1	square pyramidal
XeF₅⁺	5, 1	square pyramidal
XeOF₄	5, 1	square pyramidal

So one species is see-saw (SF₄), two are T-shaped (ClF₃ and XeF₃⁺), three are square pyramidal, and none is trigonal planar: SOCl₂ and SO₃^2- look like candidates but each has a lone pair on sulfur, which makes them pyramidal. Options giving trigonal planar a count of one or two make exactly that mistake.

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17.

The correct order of ONO bond angle in the given species is

A.\(\mathrm{NO_2^+ \lt NO_2 \lt NO_3^- \lt NO_2^-}\)
B.\(\mathrm{NO_2^- \lt NO_3^- \lt NO_2 \lt NO_2^+}\)
C.\(\mathrm{NO_3^- \lt NO_2^- \lt NO_2 \lt NO_2^+}\)
D.\(\mathrm{NO_2^- \lt NO_3^- \lt NO_2^+ \lt NO_2}\)

Exam Year(s)

JEE Advanced - 2026,

Answer: B

The correct order is \(\mathrm{NO_2^- \lt NO_3^- \lt NO_2 \lt NO_2^+}\).

The angles follow from the electron domains around nitrogen:

\(\mathrm{NO_{2}^{+}}\): two bonding domains and no lone pair on N, so it is linear, 180°.
\(\mathrm{NO_{2}}\): one unpaired electron on N. A single electron repels the bonds less than a lone pair, so the angle opens to about 134°.
\(\mathrm{NO_{3}^{-}}\): three equivalent bonding domains, trigonal planar, 120°.
\(\mathrm{NO_{2}^{-}}\): a full lone pair on N compresses the angle below 120°, to about 115°.

The first option is this order reversed. The third puts nitrate below nitrite, though nitrite's lone pair makes it the smaller. The last puts \(\mathrm{NO_{2}^{+}}\) below \(\mathrm{NO_{2}}\), but a linear ion has the largest possible angle.

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18.

The correct increasing order of dipole moment of NF₃, H₂S, CHCl₃ and NH₃ molecules is

A.\(NF_{3} \lt H_{2}S \lt CHCl_{3} \lt NH_{3}\)
B.\(NH_{3} \lt H_{2}S \lt CHCl_{3} \lt NF_{3}\)
C.\(NF_{3} \lt CHCl_{3} \lt H_{2}S \lt NH_{3}\)
D.\(NH_{3} \lt CHCl_{3} \lt H_{2}S \lt NF_{3}\)
E.\(CHCl_{3} \lt H_{2}S \lt NF_{3} \lt NH_{3}\)

Exam Year(s)

KEAM - 2025,

Answer: A

The increasing order is \(NF_{3} \lt H_{2}S \lt CHCl_{3} \lt NH_{3}\).

Approximate dipole moments are \(NF_{3}\) 0.24 D, \(H_{2}S\) 0.95 D, \(CHCl_{3}\) 1.04 D and \(NH_{3}\) 1.47 D.

The key contrast is between \(NH_{3}\) and \(NF_{3}\). In ammonia the lone pair's moment points the same way as the N-H bond moments (towards N), so they add. In \(NF_{3}\) the N-F bond moments point away from N, opposing the lone pair, so the net moment is small. Any order with \(NF_{3}\) above \(NH_{3}\) is therefore wrong.

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19.

Choose the INCORRECT pair of MOLECULE and its SHAPE among the following:

A.\(SF_{4}\) Seesaw
B.\(BrF_{5}\) Trigonal bipyramidal
C.\(NH_{3}\) Trigonal pyramidal
D.\(XeF_{4}\) Square planar
E.\(ClF_{3}\) T-shape

Exam Year(s)

KEAM - 2025,

Answer: B

The incorrect pair is \(BrF_{5}\) Trigonal bipyramidal.

Bromine in \(BrF_{5}\) has five bond pairs and one lone pair, six electron pairs in an octahedral arrangement. With one position taken by the lone pair, the molecule is square pyramidal.

The other pairs are right: \(SF_{4}\) (4 bond pairs, 1 lone pair) is see-saw; \(NH_{3}\) (3 and 1) is trigonal pyramidal; \(XeF_{4}\) (4 and 2) is square planar; \(ClF_{3}\) (3 and 2) is T-shaped.

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20.

Which of the following has least mean multiple bond enthalpy (in kJ mol^-1 ) at 298 K?

A.\(N \equiv N\)
B.\(C \equiv N\)
C.C = C
D.\(C \equiv O\)
E.C = N

Exam Year(s)

KEAM - 2025,

Answer: C

The least is C = C.

Mean multiple bond enthalpies (kJ mol^-1) from the standard table are about: C=C 611, C=N 615, C≡N 891, N≡N 946, C≡O 1070.

Triple bonds are much stronger than double bonds, which rules out the three triple-bonded options. Between the two double bonds, C=N is slightly stronger than C=C, so C=C has the lowest value.

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