Molecular Structure and Bonding Theory Questions and Answers

Q: The correct increasing order of dipole moment of the following molecules is

The increasing order is NF3 \(\lt\) CHCl3 \(\lt\) NH3 \(\lt\) H2O.The dipole moments are about 0.24 D for NF3, 1.04 D for CHCl3, 1.47 D for NH3 and 1.85 D for H2O.NF3 is low because the lone pair on nitrogen points opposite to the N-F bond dipoles and partly cancels them, whereas in NH3 the lone pair and the N-H bond dipoles point the same way. Water, with two strongly polar O-H bonds and a bent shape, has the largest moment.Every other option puts NH3 above H2O or misplaces NF3.

Q: The number of bonding pairs and lone pairs of electrons in BrF5 molecule are respectively

BrF5 has 5 bonding pairs and 1 lone pair.Bromine has 7 valence electrons. Five form bonds with the five fluorine atoms, leaving two electrons as one lone pair. Six electron pairs take an octahedral arrangement, and with one position occupied by the lone pair the molecule is square pyramidal.Four bonding pairs would leave room for only four fluorine atoms, so the options with 4 or 3 bond pairs cannot describe BrF5.

Q: Which of the following bond has highest mean bond enthalpy?

The highest mean bond enthalpy is that of C&equiv;O.The bond in carbon monoxide is the strongest known chemical bond between two atoms, about 1072 kJ mol-1. N&equiv;N is about 946 kJ mol-1, C&equiv;C about 839, O=O about 498 and N=N about 418.N&equiv;N is the usual trap because nitrogen is so unreactive, but C&equiv;O, with the same number of electrons and some extra ionic character, is stronger still.

Q: The correct increasing order of average bond length for C=C, N=O, C-C, N-O is

The increasing order is \(\mathrm{N{=}O \lt C{=}C \lt N{-}O \lt C{-}C}\).Double bonds are shorter than single bonds, and bonds between smaller atoms are shorter. The textbook's average bond lengths are N=O 122 pm, C=C 133 pm, N-O 136 pm and C-C 154 pm.Both double bonds come before both single bonds, and within each pair the bond involving nitrogen and oxygen, the smaller atoms, is the shorter. The other options put a single bond before a double bond somewhere.

Q: Which of the following set of molecules are having zero dipole moment?

The set with zero dipole moment is H2, CO2 and BF3.H2 has a non-polar bond. CO2 is linear, so its two C=O dipoles cancel. BF3 is trigonal planar, so its three B-F dipoles cancel.Every other set contains a polar molecule: HI and HCl are polar diatomics, and NH3 is pyramidal with a net dipole. CCl4 and CH4 are non-polar, but they appear only alongside a polar molecule. A molecule can have polar bonds and still a zero dipole moment if its shape is symmetric, as CO2, BF3 and CCl4 show.

Q: The bond order of oxygen molecules, superoxide ion and peroxide is/are :

The bond orders are 2, 1.5, 1.By molecular orbital theory, bond order is half the difference between bonding and antibonding electrons. Adding electrons to the antibonding \(\pi^{*}\) orbitals lowers it:\[ \begin{aligned} \mathrm{O_{2}}: \tfrac{1}{2}(10 - 6) &= 2 \\ \mathrm{O_{2}^{-}}: \tfrac{1}{2}(10 - 7) &= 1.5 \\ \mathrm{O_{2}^{2-}}: \tfrac{1}{2}(10 - 8) &= 1 \end{aligned} \]The other options give the same three values in the wrong order; bond order falls from oxygen to superoxide to peroxide.

Q: Which of the following molecule does not have zero dipole moment?

The molecule with a non-zero dipole moment is nitrogen trifluoride.NF3 is pyramidal, with a lone pair on nitrogen, so the three N-F bond dipoles do not cancel and the molecule has a net dipole moment (small, about 0.24 D).The others are symmetrical and their bond dipoles cancel: BF3 is trigonal planar, CCl4 and CH4 are tetrahedral, and BeF2 is linear, so each has zero dipole moment.

Q: One of the most widely used functions for the van der Waals potential is the Lennard-Jones expressed as :

The Lennard-Jones function is \(E_{LJ} = \sum (A_{ik}/r_{ik}^{12} - C_{ik}/r_{ik}^{6})\).The Lennard-Jones 12-6 potential has two terms. The positive r−12 term is the steep short-range repulsion that arises when electron clouds overlap. The negative r−6 term is the London dispersion attraction, which falls off as the sixth power of the distance. In force fields the sum runs over all nonbonded atom pairs, with Aik and Cik constants for each pair.Option A puts the powers on the wrong terms, making the attraction short-range. Options C and D add the two terms, which would make the interaction purely repulsive and leave no van der Waals minimum.

Q: The total number of lone pairs for the ion I3- is :

The triiodide ion carries 9 lone pairs in all.I3- has 3 × 7 + 1 = 22 valence electrons. Two I-I bonds use 4 of them, leaving 18 electrons, or 9 lone pairs. The central iodine has five electron pairs (two bonding, three lone pairs) arranged trigonal bipyramidally, with the three lone pairs equatorial, which is why the ion is linear. Each terminal iodine carries three lone pairs.So the total is 3 + 3 + 3 = 9. The answer 3 counts only the central atom's lone pairs, and 6 counts only the terminal atoms'.

Q: Which of the following diatomic molecules is paramagnetic ?

The paramagnetic molecule is B2.For the light diatomics up to N2, s-p mixing places the π2p orbitals below σ2p. B2 has 6 valence electrons: σ2s2 σ*2s2 π2p2. The two π electrons occupy the two degenerate π2p orbitals singly (Hund's rule), so B2 has two unpaired electrons and is paramagnetic.C2 (π2p4) and N2 (π2p4 σ2p2) have all electrons paired. F2 fills π*2p4, also leaving no unpaired electron, so all three are diamagnetic.

51.

The correct increasing order of dipole moment of the following molecules is

A.\(\mathrm{NF_3 \lt CHCl_3 \lt NH_3 \lt H_2O}\)
B.\(\mathrm{NF_3 \lt CHCl_3 \lt H_2O \lt NH_3}\)
C.\(\mathrm{H_2O \lt NF_3 \lt CHCl_3 \lt NH_3}\)
D.\(\mathrm{CHCl_3 \lt NF_3 \lt H_2O \lt NH_3}\)
E.\(\mathrm{NH_3 \lt NF_3 \lt CHCl_3 \lt H_2O}\)

Exam Year(s)

KEAM - 2026,

Answer: A

The increasing order is NF₃ \(\lt\) CHCl₃ \(\lt\) NH₃ \(\lt\) H₂O.

The dipole moments are about 0.24 D for NF₃, 1.04 D for CHCl₃, 1.47 D for NH₃ and 1.85 D for H₂O.

NF₃ is low because the lone pair on nitrogen points opposite to the N-F bond dipoles and partly cancels them, whereas in NH₃ the lone pair and the N-H bond dipoles point the same way. Water, with two strongly polar O-H bonds and a bent shape, has the largest moment.

Every other option puts NH₃ above H₂O or misplaces NF₃.

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52.

The number of bonding pairs and lone pairs of electrons in BrF₅ molecule are respectively

A.4 and 2
B.3 and 2
C.4 and 1
D.5 and 1
E.3 and 1

Exam Year(s)

KEAM - 2026,

Answer: D

BrF₅ has 5 bonding pairs and 1 lone pair.

Bromine has 7 valence electrons. Five form bonds with the five fluorine atoms, leaving two electrons as one lone pair. Six electron pairs take an octahedral arrangement, and with one position occupied by the lone pair the molecule is square pyramidal.

Four bonding pairs would leave room for only four fluorine atoms, so the options with 4 or 3 bond pairs cannot describe BrF₅.

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53.

Which of the following bond has highest mean bond enthalpy?

A.N=N
B.O=O
C.N≡N
D.C≡C
E.C≡O

Exam Year(s)

KEAM - 2026,

Answer: E

The highest mean bond enthalpy is that of C≡O.

The bond in carbon monoxide is the strongest known chemical bond between two atoms, about 1072 kJ mol^-1. N≡N is about 946 kJ mol^-1, C≡C about 839, O=O about 498 and N=N about 418.

N≡N is the usual trap because nitrogen is so unreactive, but C≡O, with the same number of electrons and some extra ionic character, is stronger still.

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54.

The correct increasing order of average bond length for C=C, N=O, C-C, N-O is

A.\(\mathrm{C{=}C \lt N{=}O \lt C{-}C \lt N{-}O}\)
B.\(\mathrm{N{-}O \lt N{=}O \lt C{-}C \lt C{=}C}\)
C.\(\mathrm{C{=}C \lt C{-}C \lt N{=}O \lt N{-}O}\)
D.\(\mathrm{C{-}C \lt N{=}O \lt C{=}C \lt N{-}O}\)
E.\(\mathrm{N{=}O \lt C{=}C \lt N{-}O \lt C{-}C}\)

Exam Year(s)

KEAM - 2026,

Answer: E

The increasing order is \(\mathrm{N{=}O \lt C{=}C \lt N{-}O \lt C{-}C}\).

Double bonds are shorter than single bonds, and bonds between smaller atoms are shorter. The textbook's average bond lengths are N=O 122 pm, C=C 133 pm, N-O 136 pm and C-C 154 pm.

Both double bonds come before both single bonds, and within each pair the bond involving nitrogen and oxygen, the smaller atoms, is the shorter. The other options put a single bond before a double bond somewhere.

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55.

Which of the following set of molecules are having zero dipole moment?

A.HI, \(\mathrm{CO_{2}}\) and \(\mathrm{CCl_{4}}\)
B.\(\mathrm{H_{2}}\), \(\mathrm{CO_{2}}\) and \(\mathrm{BF_{3}}\)
C.\(\mathrm{NH_{3}}\), \(\mathrm{CO_{2}}\) and \(\mathrm{CCl_{4}}\)
D.HI, \(\mathrm{CH_{4}}\) and \(\mathrm{CCl_{4}}\)
E.HI, HCl and \(\mathrm{CCl_{4}}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The set with zero dipole moment is H₂, CO₂ and BF₃.

H₂ has a non-polar bond. CO₂ is linear, so its two C=O dipoles cancel. BF₃ is trigonal planar, so its three B-F dipoles cancel.

Every other set contains a polar molecule: HI and HCl are polar diatomics, and NH₃ is pyramidal with a net dipole. CCl₄ and CH₄ are non-polar, but they appear only alongside a polar molecule. A molecule can have polar bonds and still a zero dipole moment if its shape is symmetric, as CO₂, BF₃ and CCl₄ show.

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56.

The bond order of oxygen molecules, superoxide ion and peroxide is/are :

A.1.5, 1, 2
B.2, 1, 1.5
C.2, 1.5, 1
D.1, 1.5, 2

Exam Year(s)

MPPSC Forest Service - 2022,

Answer: C

The bond orders are 2, 1.5, 1.

By molecular orbital theory, bond order is half the difference between bonding and antibonding electrons. Adding electrons to the antibonding \(\pi^{*}\) orbitals lowers it:

\[ \begin{aligned} \mathrm{O_{2}}: \tfrac{1}{2}(10 - 6) &= 2 \\ \mathrm{O_{2}^{-}}: \tfrac{1}{2}(10 - 7) &= 1.5 \\ \mathrm{O_{2}^{2-}}: \tfrac{1}{2}(10 - 8) &= 1 \end{aligned} \]

The other options give the same three values in the wrong order; bond order falls from oxygen to superoxide to peroxide.

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57.

Which of the following molecule does not have zero dipole moment?

A.Nitrogen trifluoride
B.Boron trifluoride
C.Carbon tetrachloride
D.Beryllium difluoride
E.Methane

Exam Year(s)

KEAM - 2026,

Answer: A

The molecule with a non-zero dipole moment is nitrogen trifluoride.

NF₃ is pyramidal, with a lone pair on nitrogen, so the three N-F bond dipoles do not cancel and the molecule has a net dipole moment (small, about 0.24 D).

The others are symmetrical and their bond dipoles cancel: BF₃ is trigonal planar, CCl₄ and CH₄ are tetrahedral, and BeF₂ is linear, so each has zero dipole moment.

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58.

One of the most widely used functions for the van der Waals potential is the Lennard-Jones expressed as :

A.\(E_{LJ} = \sum_{\text{nonbonded pairs}}\)\(\left(\frac{A_{ik}}{r_{ik}^{6}} - \frac{C_{ik}}{r_{ik}^{12}}\right)\)
B.\(E_{LJ} = \sum_{\text{nonbonded pairs}}\)\(\left(\frac{A_{ik}}{r_{ik}^{12}} - \frac{C_{ik}}{r_{ik}^{6}}\right)\)
C.\(E_{LJ} = \sum_{\text{nonbonded pairs}}\)\(\left(\frac{A_{ik}}{r_{ik}^{12}} + \frac{C_{ik}}{r_{ik}^{6}}\right)\)
D.\(E_{LJ} = \sum_{\text{nonbonded pairs}}\)\(\left(\frac{A_{ik}}{r_{ik}^{6}} + \frac{C_{ik}}{r_{ik}^{12}}\right)\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The Lennard-Jones function is \(E_{LJ} = \sum (A_{ik}/r_{ik}^{12} - C_{ik}/r_{ik}^{6})\).

The Lennard-Jones 12-6 potential has two terms. The positive r⁻¹² term is the steep short-range repulsion that arises when electron clouds overlap. The negative r⁻⁶ term is the London dispersion attraction, which falls off as the sixth power of the distance. In force fields the sum runs over all nonbonded atom pairs, with A_ik and C_ik constants for each pair.

Option A puts the powers on the wrong terms, making the attraction short-range. Options C and D add the two terms, which would make the interaction purely repulsive and leave no van der Waals minimum.

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59.

The total number of lone pairs for the ion I₃^- is :

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The triiodide ion carries 9 lone pairs in all.

I₃^- has 3 × 7 + 1 = 22 valence electrons. Two I-I bonds use 4 of them, leaving 18 electrons, or 9 lone pairs. The central iodine has five electron pairs (two bonding, three lone pairs) arranged trigonal bipyramidally, with the three lone pairs equatorial, which is why the ion is linear. Each terminal iodine carries three lone pairs.

So the total is 3 + 3 + 3 = 9. The answer 3 counts only the central atom's lone pairs, and 6 counts only the terminal atoms'.

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60.

Which of the following diatomic molecules is paramagnetic ?

A.\(\mathrm{B_{2}}\)
B.\(\mathrm{C_{2}}\)
C.\(\mathrm{N_{2}}\)
D.\(\mathrm{F_{2}}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The paramagnetic molecule is B₂.

For the light diatomics up to N₂, s-p mixing places the π2p orbitals below σ2p. B₂ has 6 valence electrons: σ2s² σ*2s² π2p². The two π electrons occupy the two degenerate π2p orbitals singly (Hund's rule), so B₂ has two unpaired electrons and is paramagnetic.

C₂ (π2p⁴) and N₂ (π2p⁴ σ2p²) have all electrons paired. F₂ fills π*2p⁴, also leaving no unpaired electron, so all three are diamagnetic.

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