Molecular Structure and Bonding Theory Questions and Answers

Q: In which of the following compounds there is an expanded octet around the central atom?

The expanded octet is in H2SO4.In the usual Lewis structure of sulphuric acid, sulphur forms two S=O double bonds and two S–OH single bonds, giving it 12 electrons in its valence shell. Sulphur, a period 3 element, can exceed the octet.In SCl2, sulphur has two bond pairs and two lone pairs (8 electrons); NH3 and PCl3 have exactly an octet around N and P; and NO2 is an odd-electron molecule with fewer than eight electrons around nitrogen, an incomplete octet rather than an expanded one.

Q: Which of the following molecule has tetrahedral geometry?

The tetrahedral species is NH4+.In the ammonium ion, nitrogen has four bond pairs and no lone pair, so it is sp3 hybridised and regular tetrahedral with bond angles of 109.5°.The others have different shapes: SF6 is octahedral, PCl5 trigonal bipyramidal, BF3 trigonal planar and BeCl2 linear.

Q: In which one of the following compounds, there is complete octet of central atom?

The central atom has a complete octet in \(\mathrm{SCl_{2}}\).Sulphur has six valence electrons. In \(\mathrm{SCl_{2}}\) it forms two S-Cl bonds and keeps two lone pairs: 2 bond pairs and 2 lone pairs give 8 electrons around sulphur, a full octet.The others fall short. Boron in \(\mathrm{BF_{3}}\) and aluminium in \(\mathrm{AlCl_{3}}\) (printed AlCL3) have only three bond pairs, 6 electrons. Beryllium in \(\mathrm{BeH_{2}}\) has two bond pairs, 4 electrons. LiCl is essentially ionic, and \(\mathrm{Li^{+}}\) has the two-electron (helium-like) configuration, not an octet.

Q: Which one of the following molecule/ion has square planar shape?

The square planar species is \(\mathrm{XeF_{4}}\).Xenon in \(\mathrm{XeF_{4}}\) has 4 bond pairs and 2 lone pairs, six electron pairs in all, arranged octahedrally. The two lone pairs occupy opposite (trans) positions to keep apart, leaving the four fluorines at the corners of a square in one plane.\(\mathrm{SF_{4}}\) has 4 bond pairs and 1 lone pair, which gives a see-saw shape. \(\mathrm{NH_{4}^{+}}\), \(\mathrm{CH_{4}}\) and \(\mathrm{CH_{2}Cl_{2}}\) each have four bond pairs and no lone pair on the central atom, so all three are tetrahedral.

Q: Which one of the following molecules is nonpolar?

The nonpolar molecule is \(\mathrm{CO_{2}}\).Each C=O bond in carbon dioxide is polar, because oxygen is more electronegative than carbon. But the molecule is linear, O=C=O, so the two equal bond dipoles point in exactly opposite directions and cancel. The net dipole moment is zero.Water is bent and has a net dipole moment (about 1.85 D). Methanol and HCl are polar too: in each, the bond dipoles do not cancel. NaCl is an ionic compound, the extreme opposite of nonpolar; as a gaseous ion pair it has a very large dipole moment.

Q: Which of the following statement is incorrect about bond order?

The incorrect statement is that bond enthalpy decreases with increase in bond order.A higher bond order means more electron pairs shared between the atoms, which pulls them closer and holds them more strongly. So bond enthalpy increases and bond length decreases as bond order rises: compare N\(\equiv\)N (946 kJ/mol) with O=O (498 kJ/mol).The other statements are correct, including that isoelectronic species such as N2, CO and NO+ have identical bond orders.

Q: Which of the following molecules have dsp2 hybridisation?(i) \([Ni(CN)_{4}]^{2-}\) (ii) \(BrF_{5}\) (iii)\([Co(NH_{3})_{6}]^{3+}\) (iv) \([CrF_{6}]^{3-}\) (v) \([Pt(Cl)_{4}]^{2-}\)

The dsp2 species are (i) and (v).\([Ni(CN)_{4}]^{2-}\) and \([PtCl_{4}]^{2-}\) are square planar d8 complexes: strong-field CN−, and the large crystal-field splitting of a 5d metal for Pt, pair the d electrons and free one inner d orbital, giving dsp2 hybridisation.The others are not dsp2: BrF5 is square pyramidal (sp3d2), \([Co(NH_{3})_{6}]^{3+}\) is octahedral (d2sp3), and \([CrF_{6}]^{3-}\) is octahedral (d2sp3).

Q: Which of the following statement is INCORRECT?

The incorrect statement is The dipole moment of NF3 is much greater than that in NH3.It is the other way round: NH3 is about 1.47 D and NF3 only about 0.23 D. In NH3 the N-H bond dipoles point toward nitrogen, the same way as the lone-pair moment, so they add. In NF3 the N-F bond dipoles point away from nitrogen, against the lone-pair moment, so they largely cancel.The other statements are true: BF3 is trigonal planar and non-polar; CO and NO+ are isoelectronic (14 electrons) with bond order 3; resonance makes the two O-O bonds in ozone equal; and the carbonate ion has three equivalent canonical forms.

Q: In which of the following liquid inter molecular hydrogen bonding does not exist?

Intermolecular hydrogen bonding is absent in Diethylether.A hydrogen bond needs a hydrogen attached to N, O or F on one molecule and a lone pair on another. Diethyl ether, C2H5OC2H5, has an oxygen with lone pairs but no O-H hydrogen, so its molecules cannot hydrogen-bond to one another; this is why it boils at only 35 °C.Ethanoic acid (O-H, it even forms dimers), ethanol and phenol (O-H) and ethylamine (N-H) all have such hydrogens.

Q: Which one of the following molecules contains two 'sigma' bonds and two 'pi' bonds?

The molecule with two sigma and two pi bonds is CO2.Carbon dioxide is O=C=O. Each double bond is one sigma bond plus one pi bond, so the molecule has 2 σ and 2 π bonds.O2 (O=O) has 1 σ and 1 π; N2 (N&equiv;N) and CO have 1 σ and 2 π each; and C2H2 (H–C&equiv;C–H) has 3 σ (two C–H and one C–C) and 2 π.

21.

In which of the following compounds there is an expanded octet around the central atom?

A.\(\mathrm{SCl_{2}}\)
B.\(\mathrm{NO_{2}}\)
C.\(\mathrm{NH_{3}}\)
D.\(\mathrm{PCl_{3}}\)
E.\(\mathrm{H_{2}SO_{4}}\)

Exam Year(s)

KEAM - 2025,

Answer: E

The expanded octet is in H₂SO₄.

In the usual Lewis structure of sulphuric acid, sulphur forms two S=O double bonds and two S–OH single bonds, giving it 12 electrons in its valence shell. Sulphur, a period 3 element, can exceed the octet.

In SCl₂, sulphur has two bond pairs and two lone pairs (8 electrons); NH₃ and PCl₃ have exactly an octet around N and P; and NO₂ is an odd-electron molecule with fewer than eight electrons around nitrogen, an incomplete octet rather than an expanded one.

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22.

Which of the following molecule has tetrahedral geometry?

A.\(\mathrm{SF_{6}}\)
B.\(\mathrm{PCl_{5}}\)
C.\(\mathrm{BF_{3}}\)
D.\(\mathrm{BeCl_{2}}\)
E.\(\mathrm{NH_{4}^{+}}\)

Exam Year(s)

KEAM - 2025,

Answer: E

The tetrahedral species is NH₄⁺.

In the ammonium ion, nitrogen has four bond pairs and no lone pair, so it is sp³ hybridised and regular tetrahedral with bond angles of 109.5°.

The others have different shapes: SF₆ is octahedral, PCl₅ trigonal bipyramidal, BF₃ trigonal planar and BeCl₂ linear.

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23.

In which one of the following compounds, there is complete octet of central atom?

A.\(\mathrm{BF_{3}}\)
B.\(\mathrm{BeH_{2}}\)
C.\(\mathrm{SCl_{2}}\)
D.\(\mathrm{AlCL_{3}}\)
E.\(\mathrm{LiCl}\)

Exam Year(s)

KEAM - 2025,

Answer: C

The central atom has a complete octet in \(\mathrm{SCl_{2}}\).

Sulphur has six valence electrons. In \(\mathrm{SCl_{2}}\) it forms two S-Cl bonds and keeps two lone pairs: 2 bond pairs and 2 lone pairs give 8 electrons around sulphur, a full octet.

The others fall short. Boron in \(\mathrm{BF_{3}}\) and aluminium in \(\mathrm{AlCl_{3}}\) (printed AlCL₃) have only three bond pairs, 6 electrons. Beryllium in \(\mathrm{BeH_{2}}\) has two bond pairs, 4 electrons. LiCl is essentially ionic, and \(\mathrm{Li^{+}}\) has the two-electron (helium-like) configuration, not an octet.

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24.

Which one of the following molecule/ion has square planar shape?

A.\(\mathrm{SF_{4}}\)
B.\(\mathrm{NH_{4}^{+}}\)
C.\(\mathrm{CH_{2}Cl_{2}}\)
D.\(\mathrm{CH_{4}}\)
E.\(\mathrm{XeF_{4}}\)

Exam Year(s)

KEAM - 2025,

Answer: E

The square planar species is \(\mathrm{XeF_{4}}\).

Xenon in \(\mathrm{XeF_{4}}\) has 4 bond pairs and 2 lone pairs, six electron pairs in all, arranged octahedrally. The two lone pairs occupy opposite (trans) positions to keep apart, leaving the four fluorines at the corners of a square in one plane.

\(\mathrm{SF_{4}}\) has 4 bond pairs and 1 lone pair, which gives a see-saw shape. \(\mathrm{NH_{4}^{+}}\), \(\mathrm{CH_{4}}\) and \(\mathrm{CH_{2}Cl_{2}}\) each have four bond pairs and no lone pair on the central atom, so all three are tetrahedral.

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25.

Which one of the following molecules is nonpolar?

A.\(\mathrm{CO_{2}}\)
B.\(\mathrm{H_{2}O}\)
C.\(\mathrm{CH_{3}OH}\)
D.\(\mathrm{HCl}\)
E.\(\mathrm{NaCl}\)

Exam Year(s)

KEAM - 2025,

Answer: A

The nonpolar molecule is \(\mathrm{CO_{2}}\).

Each C=O bond in carbon dioxide is polar, because oxygen is more electronegative than carbon. But the molecule is linear, O=C=O, so the two equal bond dipoles point in exactly opposite directions and cancel. The net dipole moment is zero.

Water is bent and has a net dipole moment (about 1.85 D). Methanol and HCl are polar too: in each, the bond dipoles do not cancel. NaCl is an ionic compound, the extreme opposite of nonpolar; as a gaseous ion pair it has a very large dipole moment.

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26.

Which of the following statement is incorrect about bond order?

A.Bond order is the number of bonds between the two atoms in a molecule.
B.Isoelectronic molecules and ions have identical bond orders.
C.Bond enthalpy decreases with increase in bond order.
D.Bond length decreases with increase in bond order.
E.Bond enthalpy increases with increase in bond order.

Exam Year(s)

KEAM - 2025,

Answer: C

The incorrect statement is that bond enthalpy decreases with increase in bond order.

A higher bond order means more electron pairs shared between the atoms, which pulls them closer and holds them more strongly. So bond enthalpy increases and bond length decreases as bond order rises: compare N\(\equiv\)N (946 kJ/mol) with O=O (498 kJ/mol).

The other statements are correct, including that isoelectronic species such as N₂, CO and NO⁺ have identical bond orders.

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27.

Which of the following molecules have dsp² hybridisation?

(i) \([Ni(CN)_{4}]^{2-}\) (ii) \(BrF_{5}\) (iii)\([Co(NH_{3})_{6}]^{3+}\) (iv) \([CrF_{6}]^{3-}\) (v) \([Pt(Cl)_{4}]^{2-}\)

A.(i) and (ii)
B.(i) and (iii)
C.(ii) and (iii)
D.(i) and (v)
E.(iii) and (v)

Exam Year(s)

KEAM - 2025,

Answer: D

The dsp² species are (i) and (v).

\([Ni(CN)_{4}]^{2-}\) and \([PtCl_{4}]^{2-}\) are square planar d⁸ complexes: strong-field CN⁻, and the large crystal-field splitting of a 5d metal for Pt, pair the d electrons and free one inner d orbital, giving dsp² hybridisation.

The others are not dsp²: BrF₅ is square pyramidal (sp³d²), \([Co(NH_{3})_{6}]^{3+}\) is octahedral (d²sp³), and \([CrF_{6}]^{3-}\) is octahedral (d²sp³).

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28.

Which of the following statement is INCORRECT?

A.The dipole moment of \(\mathrm{BF_{3}}\) is zero.
B.The bond order of CO molecule is the same as the bond order in \(\mathrm{NO^{+}}\) ion.
C.In ozone molecule, the two O-O bond lengths are equal.
D.The dipole moment of \(NF_{3}\) is much greater than that in \(NH_{3}\)
E.Carbonate ion has three canonical forms.

Exam Year(s)

KEAM - 2025,

Answer: D

The incorrect statement is The dipole moment of NF₃ is much greater than that in NH₃.

It is the other way round: NH₃ is about 1.47 D and NF₃ only about 0.23 D. In NH₃ the N-H bond dipoles point toward nitrogen, the same way as the lone-pair moment, so they add. In NF₃ the N-F bond dipoles point away from nitrogen, against the lone-pair moment, so they largely cancel.

The other statements are true: BF₃ is trigonal planar and non-polar; CO and NO⁺ are isoelectronic (14 electrons) with bond order 3; resonance makes the two O-O bonds in ozone equal; and the carbonate ion has three equivalent canonical forms.

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29.

In which of the following liquid inter molecular hydrogen bonding does not exist?

A.\(CH_{3}COOH\)
B.\(\mathrm{C_{2}H_{5}OH}\)
C.Phenol
D.Diethylether
E.Ethylamine

Exam Year(s)

KEAM - 2025,

Answer: D

Intermolecular hydrogen bonding is absent in Diethylether.

A hydrogen bond needs a hydrogen attached to N, O or F on one molecule and a lone pair on another. Diethyl ether, C₂H₅OC₂H₅, has an oxygen with lone pairs but no O-H hydrogen, so its molecules cannot hydrogen-bond to one another; this is why it boils at only 35 °C.

Ethanoic acid (O-H, it even forms dimers), ethanol and phenol (O-H) and ethylamine (N-H) all have such hydrogens.

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30.

Which one of the following molecules contains two 'sigma' bonds and two 'pi' bonds?

A.\(\mathrm{O_{2}}\)
B.\(\mathrm{N_{2}}\)
C.\(\mathrm{C_{2}H_{2}}\)
D.\(\mathrm{CO_{2}}\)
E.CO

Exam Year(s)

KEAM - 2025,

Answer: D

The molecule with two sigma and two pi bonds is CO₂.

Carbon dioxide is O=C=O. Each double bond is one sigma bond plus one pi bond, so the molecule has 2 σ and 2 π bonds.

O₂ (O=O) has 1 σ and 1 π; N₂ (N≡N) and CO have 1 σ and 2 π each; and C₂H₂ (H–C≡C–H) has 3 σ (two C–H and one C–C) and 2 π.

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