The shapes of the compoundsClF3, XeOF2, N3− and XeO3F2respectively, are:

The shapes are T-shape, T-shape, linear and trigonal bipyramidal.ClF3: three bonds and two lone pairs on Cl, five electron pairs in a trigonal bipyramid with the lone pairs equatorial, leaving a T shape.XeOF2: Xe forms one double bond to O and two bonds to F and keeps two lone pairs; three bonded positions and two equatorial lone pairs again give a T shape.N3−: 16 valence electrons, isoelectronic with CO2, so linear.XeO3F2: five bonded groups and no lone pair on Xe, so trigonal bipyramidal.Calling ClF3 or XeOF2 trigonal planar ignores their lone pairs, azide is not bent, and XeO3F2 has no lone pair to make it square pyramidal.

Molecular Structure and Bonding Theory Questions and Answers

Q: The order of polarity of NH3, NF3 and BF3 is

The order of polarity is BF3 < NF3 < NH3.BF3 is trigonal planar, so its three B-F bond dipoles cancel and its dipole moment is zero. NH3 and NF3 are both pyramidal with a lone pair on nitrogen. In NH3 the N-H bond dipoles point towards nitrogen, the same way as the lone-pair dipole, so they add: about 1.47 D. In NF3 the N-F bond dipoles point away from nitrogen, against the lone-pair dipole, so they largely cancel: about 0.23 D.Any order placing NF3 above NH3 forgets that opposition, and any order not putting BF3 first ignores its symmetry.

Q: 1sA and 1sB are the normalized eigenfunctions of two hydrogen atoms HA and HB, respectively. If \(S = \langle 1s_{A} | 1s_{B} \rangle\), the option that is ALWAYS CORRECT is

The statement that is always correct is \(0 \le S \le 1\).The 1s functions are real, positive everywhere and normalised. Their overlap integral is therefore a real number that cannot be negative, and by the Schwarz inequality it cannot exceed the product of the norms, 1. It equals 1 only when the two nuclei coincide and approaches 0 as they move far apart.\(S = 1\) holds only at zero separation and \(S = 0\) only at infinite separation, so neither is always true. The overlap of two real functions is real, so it is never an imaginary constant.

Q: Consider the figure given below, where M is a metal and L is a monodentate ligand. The σ-bonding ligand group orbital (LGO) having same symmetry with \(d_{z^{2}}\) orbital of M in the octahedral coordination geometry is

The ligand group orbital is \(\frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)\).In \(O_{h}\) the \(d_{z^{2}}\) orbital belongs to eg. Its LGO must copy the orbital's shape: large positive lobes along +z and -z, where \(\sigma_{1}\) and \(\sigma_{2}\) point, and the smaller negative collar in the xy plane, where \(\sigma_{3}\) to \(\sigma_{6}\) point. So \(\sigma_{1}\) and \(\sigma_{2}\) take the same positive weight, twice that of the four equatorial orbitals, which take a negative sign. The normalising factor is\[ \frac{1}{\sqrt{4 + 4 + 1 + 1 + 1 + 1}} = \frac{1}{\sqrt{12}} \]The all-positive combination \(\frac{1}{\sqrt{6}}(\sigma_{1} + \dots + \sigma_{6})\) is the a1g LGO that matches the 4s orbital. The \(2\sigma_{1} - 2\sigma_{2}\) option changes sign between +z and -z, which a gerade d orbital never does. The \(\frac{1}{\sqrt{6}}\) option with equal weights of opposite sign is not a symmetry-adapted orbital at all: it is a mixture of the a1g and eg combinations.

Q: The hybridization of atomic orbitals of sulphur in SF4 is

Sulphur in SF4 is sp\(^{3}\)d hybridised.Sulphur has six valence electrons. Four form bonds to fluorine and the remaining two make one lone pair, giving five electron pairs around the central atom. Five pairs need five hybrid orbitals, sp3d, arranged as a trigonal bipyramid; the lone pair occupies an equatorial position, so the molecule has a see-saw shape.sp3 would allow only four pairs (as in SCl2 with two lone pairs, or a tetrahedral molecule), dsp2 gives a square planar set of four, and sp3d2 six pairs, as in SF6.

Q: As predicted by MO theory, the bond order and magnetic nature of NO+ are

NO+ has bond order three and is diamagnetic.NO+ has 10 valence electrons (5 from N, 6 from O, minus 1), the same as N2 and CO. Filling the molecular orbitals:\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2} \\ &\pi_{2p}^{4}\,\sigma_{2p}^{2} \end{aligned} \]\[ \text{bond order} = \frac{8 - 2}{2} = 3 \]All electrons are paired, so the ion is diamagnetic. Neutral NO has one more electron, in a \(\pi^{*}\) orbital, giving bond order 2.5 and paramagnetism; options with bond order two or paramagnetism describe neither species correctly.

Q: Among the following, the one having smallest bond angle is

The smallest bond angle is in PH3, about 93°.The measured angles are roughly NH3 107°, NF3 102°, PF3 98° and PH3 93°.Going down the group from N to P, the central atom becomes larger and less electronegative. Its bonding pairs sit farther from the nucleus and repel each other less, and the bonds use orbitals of nearly pure p character, so the angle closes toward 90°. That puts both phosphorus compounds below both nitrogen compounds.Within the nitrogen pair, fluorine pulls the bonding pairs away from N, reducing their mutual repulsion, so NF3 has a smaller angle than NH3. Between the phosphorus compounds, PH3 is the smaller; PF3 is opened slightly by the bulk of the fluorine atoms and by their lone pairs interacting with phosphorus.

Q: The bond order in \(\mathrm{N_{2}^{2-}}\) species is

The bond order of \(\mathrm{N_{2}^{2-}}\) is 2.N2 has 10 valence electrons; the dianion has 12. Filling the molecular orbitals in the order used for N2:\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2p}^{4} \\ &\sigma_{2p}^{2}\,\pi_{2p}^{*2} \end{aligned} \]That is 8 bonding and 4 antibonding electrons, so\[ \text{bond order} = \frac{8 - 4}{2} = 2 \]3 is the bond order of neutral N2, and 2.5 is that of N2−, where only one electron has entered \(\pi^{*}\). Each extra electron goes into an antibonding orbital and lowers the bond order by half, so 3.5 is impossible here.

Q: A few species are given in Column I. Column II contains the hybrid orbitals used by the central atom of the species for bonding.The CORRECT match for the species to their central atom hybridization is(Given: Atomic numbers of B: 5; C: 6; O: 8; F: 9; P: 15; Cl: 17; I: 53)Column ISpeciesColumn IIHybrid orbitals used by the central atom for bondingi. I3−a. spii. PCl3b. sp2iii. BF3c. sp3iv. CO2d. sp3d

The correct match is i-d, ii-c, iii-b, iv-a.Count the electron domains (bond pairs plus lone pairs) on the central atom; that number fixes the hybridization.I3−: the central I has 2 bonds and 3 lone pairs, 5 domains, so sp3d (linear ion, lone pairs equatorial).PCl3: 3 bonds and 1 lone pair, 4 domains, so sp3 (pyramidal).BF3: 3 bonds and no lone pair, so sp2 (trigonal planar).CO2: two double bonds, 2 domains, so sp (linear).The other options give I3− sp (counting only its two bonds), or swap PCl3 with BF3, or give CO2 a trigonal hybrid; each fails one of these counts.

Let \(\phi_{x}^{C}\) and \(\phi_{z}^{C}\) denote the wavefunctions of the 2p_x and 2p_z orbitals of carbon, respectively, and \(\phi_{x}^{O}\) and \(\phi_{z}^{O}\) represent the wavefunctions of the 2p_x and 2p_z orbitals of oxygen, respectively. If \(c_{1}\) and \(c_{2}\) are constants used in linear combinations and the CO molecule is oriented along the z axis, then, according to molecular orbital theory, the π-bonding molecular orbital has a wavefunction given by

A.\(c_{1}\phi_{z}^{C} + c_{2}\phi_{x}^{O}\)
B.\(c_{1}\phi_{z}^{C} + c_{2}\phi_{z}^{O}\)
C.\(c_{1}\phi_{x}^{C} + c_{2}\phi_{z}^{O}\)
D.\(c_{1}\phi_{x}^{C} + c_{2}\phi_{x}^{O}\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: D

The π-bonding orbital is \(c_{1}\phi_{x}^{C} + c_{2}\phi_{x}^{O}\).

With the molecule along the z axis, a π orbital is built from p orbitals perpendicular to the bond, that is, 2p_x (or 2p_y), and only orbitals of the same symmetry about the axis can combine. So the π MO combines 2p_x on carbon with 2p_x on oxygen, in phase for the bonding combination.

2p_z on carbon with 2p_z on oxygen point along the bond and give a σ orbital, not a π orbital. A 2p_z with a 2p_x on the other atom has zero net overlap by symmetry: one is symmetric and the other antisymmetric under reflection through the yz plane, so the two mixed options describe no bond at all.

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The order of polarity of NH₃, NF₃ and BF₃ is

A.NH\(_{3}\) \(\lt\) NF\(_{3}\) \(\lt\) BF\(_{3}\)
B.BF\(_{3}\) \(\lt\) NF\(_{3}\) \(\lt\) NH\(_{3}\)
C.BF\(_{3}\) \(\lt\) NH\(_{3}\) \(\lt\) NF\(_{3}\)
D.NF\(_{3}\) \(\lt\) BF\(_{3}\) \(\lt\) NH\(_{3}\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: B

The order of polarity is BF₃ < NF₃ < NH₃.

BF₃ is trigonal planar, so its three B-F bond dipoles cancel and its dipole moment is zero. NH₃ and NF₃ are both pyramidal with a lone pair on nitrogen. In NH₃ the N-H bond dipoles point towards nitrogen, the same way as the lone-pair dipole, so they add: about 1.47 D. In NF₃ the N-F bond dipoles point away from nitrogen, against the lone-pair dipole, so they largely cancel: about 0.23 D.

Any order placing NF₃ above NH₃ forgets that opposition, and any order not putting BF₃ first ignores its symmetry.

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The shapes of the compounds

ClF₃, XeOF₂, N₃⁻ and XeO₃F₂

respectively, are:

A.T-shape, T-shape, linear and trigonal bipyramidal
B.trigonal planar, T-shape, V-shape and square pyramidal
C.T-shape, trigonal planar, linear and square pyramidal
D.trigonal planar, trigonal planar, V-shape and trigonal bipyramidal

Exam Year(s)

GATE Chemistry - 2021,

Answer: A

The shapes are T-shape, T-shape, linear and trigonal bipyramidal.

ClF₃: three bonds and two lone pairs on Cl, five electron pairs in a trigonal bipyramid with the lone pairs equatorial, leaving a T shape.
XeOF₂: Xe forms one double bond to O and two bonds to F and keeps two lone pairs; three bonded positions and two equatorial lone pairs again give a T shape.
N₃⁻: 16 valence electrons, isoelectronic with CO₂, so linear.
XeO₃F₂: five bonded groups and no lone pair on Xe, so trigonal bipyramidal.

Calling ClF₃ or XeOF₂ trigonal planar ignores their lone pairs, azide is not bent, and XeO₃F₂ has no lone pair to make it square pyramidal.

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1s_A and 1s_B are the normalized eigenfunctions of two hydrogen atoms H_A and H_B, respectively. If \(S = \langle 1s_{A} | 1s_{B} \rangle\), the option that is ALWAYS CORRECT is

A.\(S = 1\)
B.\(S = 0\)
C.\(S\) = imaginary constant
D.\(0 \le S \le 1\)

Exam Year(s)

GATE Chemistry - 2022,

Answer: D

The statement that is always correct is \(0 \le S \le 1\).

The 1s functions are real, positive everywhere and normalised. Their overlap integral is therefore a real number that cannot be negative, and by the Schwarz inequality it cannot exceed the product of the norms, 1. It equals 1 only when the two nuclei coincide and approaches 0 as they move far apart.

\(S = 1\) holds only at zero separation and \(S = 0\) only at infinite separation, so neither is always true. The overlap of two real functions is real, so it is never an imaginary constant.

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Consider the figure given below, where M is a metal and L is a monodentate ligand. The σ-bonding ligand group orbital (LGO) having same symmetry with \(d_{z^{2}}\) orbital of M in the octahedral coordination geometry is

Octahedral complex ML6: ligands L on the +z axis (sigma 1), -z axis (sigma 2), +x (sigma 3), -y (sigma 4), +y (sigma 5) and -x (sigma 6), with M at the centre and the x, y, z axes marked

A.\(\frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)\)
B.\(\frac{1}{\sqrt{12}}(2\sigma_1-2\sigma_2+\sigma_3-\sigma_4+\sigma_5-\sigma_6)\)
C.\(\frac{1}{\sqrt{6}}(\sigma_1+\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)\)
D.\(\frac{1}{\sqrt{6}}(\sigma_1+\sigma_2+\sigma_3+\sigma_4+\sigma_5+\sigma_6)\)

Exam Year(s)

GATE Chemistry - 2026,

Answer: A

The ligand group orbital is \(\frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)\).

In \(O_{h}\) the \(d_{z^{2}}\) orbital belongs to e_g. Its LGO must copy the orbital's shape: large positive lobes along +z and -z, where \(\sigma_{1}\) and \(\sigma_{2}\) point, and the smaller negative collar in the xy plane, where \(\sigma_{3}\) to \(\sigma_{6}\) point. So \(\sigma_{1}\) and \(\sigma_{2}\) take the same positive weight, twice that of the four equatorial orbitals, which take a negative sign. The normalising factor is

\[ \frac{1}{\sqrt{4 + 4 + 1 + 1 + 1 + 1}} = \frac{1}{\sqrt{12}} \]

The all-positive combination \(\frac{1}{\sqrt{6}}(\sigma_{1} + \dots + \sigma_{6})\) is the a_1g LGO that matches the 4s orbital. The \(2\sigma_{1} - 2\sigma_{2}\) option changes sign between +z and -z, which a gerade d orbital never does. The \(\frac{1}{\sqrt{6}}\) option with equal weights of opposite sign is not a symmetry-adapted orbital at all: it is a mixture of the a_1g and e_g combinations.

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The hybridization of atomic orbitals of sulphur in SF₄ is

A.dsp\(^{2}\)
B.sp\(^{3}\)d\(^{2}\)
C.sp\(^{3}\)d
D.sp\(^{3}\)

Exam Year(s)

GATE Life Sciences - 2012,

Answer: C

Sulphur in SF₄ is sp\(^{3}\)d hybridised.

Sulphur has six valence electrons. Four form bonds to fluorine and the remaining two make one lone pair, giving five electron pairs around the central atom. Five pairs need five hybrid orbitals, sp³d, arranged as a trigonal bipyramid; the lone pair occupies an equatorial position, so the molecule has a see-saw shape.

sp³ would allow only four pairs (as in SCl₂ with two lone pairs, or a tetrahedral molecule), dsp² gives a square planar set of four, and sp³d² six pairs, as in SF₆.

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As predicted by MO theory, the bond order and magnetic nature of NO⁺ are

A.three and paramagnetic
B.two and diamagnetic
C.two and paramagnetic
D.three and diamagnetic

Exam Year(s)

GATE Life Sciences - 2012,

Answer: D

NO⁺ has bond order three and is diamagnetic.

NO⁺ has 10 valence electrons (5 from N, 6 from O, minus 1), the same as N₂ and CO. Filling the molecular orbitals:

\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2} \\ &\pi_{2p}^{4}\,\sigma_{2p}^{2} \end{aligned} \]

\[ \text{bond order} = \frac{8 - 2}{2} = 3 \]

All electrons are paired, so the ion is diamagnetic. Neutral NO has one more electron, in a \(\pi^{*}\) orbital, giving bond order 2.5 and paramagnetism; options with bond order two or paramagnetism describe neither species correctly.

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Among the following, the one having smallest bond angle is

A.\(\mathrm{PH_{3}}\)
B.\(\mathrm{PF_{3}}\)
C.\(\mathrm{NF_{3}}\)
D.\(\mathrm{NH_{3}}\)

Exam Year(s)

GATE Life Sciences - 2022,

Answer: A

The smallest bond angle is in PH₃, about 93°.

The measured angles are roughly NH₃ 107°, NF₃ 102°, PF₃ 98° and PH₃ 93°.

Going down the group from N to P, the central atom becomes larger and less electronegative. Its bonding pairs sit farther from the nucleus and repel each other less, and the bonds use orbitals of nearly pure p character, so the angle closes toward 90°. That puts both phosphorus compounds below both nitrogen compounds.

Within the nitrogen pair, fluorine pulls the bonding pairs away from N, reducing their mutual repulsion, so NF₃ has a smaller angle than NH₃. Between the phosphorus compounds, PH₃ is the smaller; PF₃ is opened slightly by the bulk of the fluorine atoms and by their lone pairs interacting with phosphorus.

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The bond order in \(\mathrm{N_{2}^{2-}}\) species is

A.2
B.2.5
C.3
D.3.5

Exam Year(s)

GATE Life Sciences - 2025,

Answer: A

The bond order of \(\mathrm{N_{2}^{2-}}\) is 2.

N₂ has 10 valence electrons; the dianion has 12. Filling the molecular orbitals in the order used for N₂:

\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2p}^{4} \\ &\sigma_{2p}^{2}\,\pi_{2p}^{*2} \end{aligned} \]

That is 8 bonding and 4 antibonding electrons, so

\[ \text{bond order} = \frac{8 - 4}{2} = 2 \]

3 is the bond order of neutral N₂, and 2.5 is that of N₂⁻, where only one electron has entered \(\pi^{*}\). Each extra electron goes into an antibonding orbital and lowers the bond order by half, so 3.5 is impossible here.

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10.

A few species are given in Column I. Column II contains the hybrid orbitals used by the central atom of the species for bonding.

The CORRECT match for the species to their central atom hybridization is

(Given: Atomic numbers of B: 5; C: 6; O: 8; F: 9; P: 15; Cl: 17; I: 53)

Column I Species	Column II Hybrid orbitals used by the central atom for bonding
i. I₃⁻	a. sp
ii. PCl₃	b. sp²
iii. BF₃	c. sp³
iv. CO₂	d. sp³d

A.i-d, ii-c, iii-b, iv-a
B.i-a, ii-d, iii-b, iv-c
C.i-d, ii-c, iii-a, iv-b
D.i-d, ii-b, iii-c, iv-a

Exam Year(s)

GATE Life Sciences - 2025,

Answer: A

The correct match is i-d, ii-c, iii-b, iv-a.

Count the electron domains (bond pairs plus lone pairs) on the central atom; that number fixes the hybridization.

I₃⁻: the central I has 2 bonds and 3 lone pairs, 5 domains, so sp³d (linear ion, lone pairs equatorial).
PCl₃: 3 bonds and 1 lone pair, 4 domains, so sp³ (pyramidal).
BF₃: 3 bonds and no lone pair, so sp² (trigonal planar).
CO₂: two double bonds, 2 domains, so sp (linear).

The other options give I₃⁻ sp (counting only its two bonds), or swap PCl₃ with BF₃, or give CO₂ a trigonal hybrid; each fails one of these counts.

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