The geometry of a molecule of type AB3E2 with 3 bonding pairs and 2 lone pairs is

An AB3E2 molecule is T-shaped.Five electron pairs around the central atom adopt a trigonal bipyramidal arrangement. The two lone pairs occupy equatorial positions, where they have the most room, leaving the three bonding pairs in a T. ClF3 is the standard example.Trigonal bipyramidal is the arrangement of all five pairs, not the shape of the atoms. See-saw is AB4E, square pyramidal is AB5E, and trigonal pyramidal is AB3E with four pairs.

Molecular Structure and Bonding Theory Questions and Answers

Q: What is the shape of chlorine trifluoride?

Chlorine trifluoride is T-shaped.The chlorine atom in ClF3 has 7 valence electrons: three form bonds with fluorine and four remain as two lone pairs. That is five electron pairs, arranged in a trigonal bipyramid. The two lone pairs take equatorial positions to minimise repulsion, leaving the three fluorine atoms in a T shape.Trigonal planar would need three bond pairs and no lone pairs (as in BF3). Tetrahedral, see-saw and square pyramidal shapes come from four, five and six electron pairs with different numbers of lone pairs.

Q: The bond order in NO+ ion is

The bond order in NO+ is 3.NO has 15 electrons, so NO+ has 14, the same as N2. Filling molecular orbitals gives 10 electrons in bonding orbitals and 4 in antibonding orbitals.\[ \begin{aligned} \text{bond order} &= \frac{N_b - N_a}{2} \\ &= \frac{10 - 4}{2} = 3 \end{aligned} \]2.5 is the bond order of neutral NO, which has one more electron in an antibonding orbital; removing that electron raises the bond order to 3.

Q: Intramolecular hydrogen bond is present in

Intramolecular hydrogen bonding is present in o-nitrophenol.In the ortho isomer the -OH and -NO2 groups are next to each other, so the hydrogen of -OH can bond to an oxygen of the nitro group within the same molecule, forming a six-membered ring (chelation). This is why o-nitrophenol is steam volatile and has a lower boiling point than the para isomer.In p-nitrophenol the groups are too far apart, so it forms intermolecular hydrogen bonds instead. Phenol and the cresols have no acceptor group next to the -OH.

Q: Which of the following pair of elements have greater ability to form \(p\pi - p\pi\) multiple bonds?

The pair is Carbon and Oxygen.Strong \(p\pi - p\pi\) multiple bonds need small atoms with compact 2p orbitals that can overlap sideways effectively. Second-period elements such as C, N and O do this well, which is why C=O, C=C and N=N double bonds are common.Each other pair includes a heavier element (Al, P, Cl, Si) whose larger, more diffuse 3p orbitals overlap poorly sideways, so they rarely form \(p\pi - p\pi\) bonds. Boron and fluorine are also poor choices: boron forms few multiple bonds and fluorine forms single bonds.

Q: Which of the following set of compounds does not follow octet rule?

The set that does not follow the octet rule is PF5, H2SO4 and SF6.In each of these the central atom has more than eight electrons in its valence shell (an expanded octet): phosphorus has 10 in PF5, sulphur has 12 in SF6, and in the usual Lewis structure of H2SO4 sulphur also has 12. Elements of the third period can do this by using their d orbitals.SCl2 obeys the octet rule (two bonds and two lone pairs on S), and so does CH4, so every set containing either of them is wrong.

Q: Which of the following statement is NOT the postulate of VSEPR theory?

The statement that is not a postulate is Where two or more resonance structures can represent a molecule, the VSEPR model is not applicable to any such structure.The VSEPR postulates state the opposite: where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.The other four statements are genuine postulates: a multiple bond is treated as a single super pair, the shape depends on the number of valence shell electron pairs around the central atom, the pairs repel one another, and they localise on a sphere at maximum distance from one another.

Q: The correct increasing order of number of electrons around the central atom in the following molecules is

The increasing order is \(\mathrm{SCl_{2} \lt PF_{5} \lt SF_{6}}\).Count the valence electrons around each central atom. In SCl2, sulphur has two bond pairs and two lone pairs: 8 electrons. In PF5, phosphorus forms five bonds: 10 electrons. In SF6, sulphur forms six bonds: 12 electrons.So the order is \(8 \lt 10 \lt 12\). PF5 and SF6 are the standard examples of an expanded octet; SCl2 obeys the octet rule, so it must come first, which rules out every other order.

Q: Which of the following molecule has double bond with two π bonds?

The molecule is C2.In the molecular orbital picture, C2 has 12 electrons, and its bonding electrons occupy the two π2p orbitals (π2px2 π2py2) with no electrons in σ2p. Its bond order is 2, and both bonds are π bonds, an unusual double bond with no σ component.O2 also has bond order 2, but as one σ and one π bond. B2 and Li2 have bond order 1, and Ne2 has bond order 0 and does not exist as a stable molecule.

41.

What is the shape of chlorine trifluoride?

A.Tetrahedral
B.T-shape
C.Trigonal planar
D.Seasaw
E.Square pyramidal

Exam Year(s)

KEAM - 2026,

Answer: B

Chlorine trifluoride is T-shaped.

The chlorine atom in ClF₃ has 7 valence electrons: three form bonds with fluorine and four remain as two lone pairs. That is five electron pairs, arranged in a trigonal bipyramid. The two lone pairs take equatorial positions to minimise repulsion, leaving the three fluorine atoms in a T shape.

Trigonal planar would need three bond pairs and no lone pairs (as in BF₃). Tetrahedral, see-saw and square pyramidal shapes come from four, five and six electron pairs with different numbers of lone pairs.

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42.

The bond order in NO⁺ ion is

A.3
B.2
C.1
D.2.5
E.1.5

Exam Year(s)

KEAM - 2026,

Answer: A

The bond order in NO⁺ is 3.

NO has 15 electrons, so NO⁺ has 14, the same as N₂. Filling molecular orbitals gives 10 electrons in bonding orbitals and 4 in antibonding orbitals.

\[ \begin{aligned} \text{bond order} &= \frac{N_b - N_a}{2} \\ &= \frac{10 - 4}{2} = 3 \end{aligned} \]

2.5 is the bond order of neutral NO, which has one more electron in an antibonding orbital; removing that electron raises the bond order to 3.

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43.

Intramolecular hydrogen bond is present in

A.phenol
B.o-nitrophenol
C.p-nitrophenol
D.o-cresol
E.p-cresol

Exam Year(s)

KEAM - 2026,

Answer: B

Intramolecular hydrogen bonding is present in o-nitrophenol.

In the ortho isomer the -OH and -NO₂ groups are next to each other, so the hydrogen of -OH can bond to an oxygen of the nitro group within the same molecule, forming a six-membered ring (chelation). This is why o-nitrophenol is steam volatile and has a lower boiling point than the para isomer.

In p-nitrophenol the groups are too far apart, so it forms intermolecular hydrogen bonds instead. Phenol and the cresols have no acceptor group next to the -OH.

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44.

Which of the following pair of elements have greater ability to form \(p\pi - p\pi\) multiple bonds?

A.Carbon and Oxygen
B.Boron and Aluminium
C.Nitrogen and Phosphorus
D.Fluorine and Chlorine
E.Carbon and Silicon

Exam Year(s)

KEAM - 2026,

Answer: A

The pair is Carbon and Oxygen.

Strong \(p\pi - p\pi\) multiple bonds need small atoms with compact 2p orbitals that can overlap sideways effectively. Second-period elements such as C, N and O do this well, which is why C=O, C=C and N=N double bonds are common.

Each other pair includes a heavier element (Al, P, Cl, Si) whose larger, more diffuse 3p orbitals overlap poorly sideways, so they rarely form \(p\pi - p\pi\) bonds. Boron and fluorine are also poor choices: boron forms few multiple bonds and fluorine forms single bonds.

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45.

Which of the following statements are correct about the PCl₅ molecule?

(i) It has trigonal bipyramidal geometry.
(ii) It has three equatorial and two axial bonds.
(iii) The equatorial bond pairs suffer more repulsive interaction from the axial bond pair.
(iv) The equatorial bonds are slightly weaker than axial bonds.
(v) The hybridization involved in the molecule is sp³d.

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

Exam Year(s)

KEAM - 2026,

Answer: E

The correct statements are (i), (ii) and (v), so the answer is (i), (ii), (v).

In PCl₅ the phosphorus uses sp³d hybrid orbitals (v), giving a trigonal bipyramidal shape (i) with three equatorial and two axial P-Cl bonds (ii).

Statements (iii) and (iv) are reversed. It is the axial bond pairs that suffer more repulsion, since each has three bond pairs at 90°, while an equatorial pair has only two. As a result the axial bonds are slightly longer and weaker than the equatorial bonds, not the other way round.

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46.

The geometry of a molecule of type AB₃E₂ with 3 bonding pairs and 2 lone pairs is

A.T-shape
B.trigonal pyramidal
C.trigonal bi-pyramidal
D.square pyramidal
E.see saw

Exam Year(s)

KEAM - 2026,

Answer: A

An AB₃E₂ molecule is T-shaped.

Five electron pairs around the central atom adopt a trigonal bipyramidal arrangement. The two lone pairs occupy equatorial positions, where they have the most room, leaving the three bonding pairs in a T. ClF₃ is the standard example.

Trigonal bipyramidal is the arrangement of all five pairs, not the shape of the atoms. See-saw is AB₄E, square pyramidal is AB₅E, and trigonal pyramidal is AB₃E with four pairs.

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47.

Which of the following set of compounds does not follow octet rule?

A.\(\mathrm{SCl_{2}}\), \(\mathrm{PF_{5}}\) and \(\mathrm{SF_{6}}\)
B.\(\mathrm{CH_{4}}\), \(\mathrm{PF_{5}}\) and \(\mathrm{SF_{6}}\)
C.\(\mathrm{PF_{5}}\), \(\mathrm{H_{2}SO_{4}}\) and \(\mathrm{SF_{6}}\)
D.\(\mathrm{PF_{5}}\), \(\mathrm{H_{2}SO_{4}}\) and \(\mathrm{SCl_{2}}\)
E.\(\mathrm{SCl_{2}}\), \(\mathrm{H_{2}SO_{4}}\) and \(\mathrm{SF_{6}}\)

Exam Year(s)

KEAM - 2026,

Answer: C

The set that does not follow the octet rule is PF₅, H₂SO₄ and SF₆.

In each of these the central atom has more than eight electrons in its valence shell (an expanded octet): phosphorus has 10 in PF₅, sulphur has 12 in SF₆, and in the usual Lewis structure of H₂SO₄ sulphur also has 12. Elements of the third period can do this by using their d orbitals.

SCl₂ obeys the octet rule (two bonds and two lone pairs on S), and so does CH₄, so every set containing either of them is wrong.

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48.

Which of the following statement is NOT the postulate of VSEPR theory?

A.Where two or more resonance structures can represent a molecule, the VSEPR model is not applicable to any such structure.
B.A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.
C.The shape of a molecule depends upon the number of valence shell electron pairs around the central atom.
D.Pairs of electrons in the valence shell repel one another since their electron clouds are negatively charged.
E.The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at maximum distance from one another.

Exam Year(s)

KEAM - 2026,

Answer: A

The statement that is not a postulate is Where two or more resonance structures can represent a molecule, the VSEPR model is not applicable to any such structure.

The VSEPR postulates state the opposite: where two or more resonance structures can represent a molecule, the VSEPR model is applicable to any such structure.

The other four statements are genuine postulates: a multiple bond is treated as a single super pair, the shape depends on the number of valence shell electron pairs around the central atom, the pairs repel one another, and they localise on a sphere at maximum distance from one another.

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49.

The correct increasing order of number of electrons around the central atom in the following molecules is

A.\(\mathrm{SF_{6} \lt SCl_{2} \lt PF_{5}}\)
B.\(\mathrm{SCl_{2} \lt PF_{5} \lt SF_{6}}\)
C.\(\mathrm{PF_{5} \lt SCl_{2} \lt SF_{6}}\)
D.\(\mathrm{SCl_{2} \lt SF_{6} \lt PF_{5}}\)
E.\(\mathrm{PF_{5} \lt SF_{6} \lt SCl_{2}}\)

Exam Year(s)

KEAM - 2026,

Answer: B

The increasing order is \(\mathrm{SCl_{2} \lt PF_{5} \lt SF_{6}}\).

Count the valence electrons around each central atom. In SCl₂, sulphur has two bond pairs and two lone pairs: 8 electrons. In PF₅, phosphorus forms five bonds: 10 electrons. In SF₆, sulphur forms six bonds: 12 electrons.

So the order is \(8 \lt 10 \lt 12\). PF₅ and SF₆ are the standard examples of an expanded octet; SCl₂ obeys the octet rule, so it must come first, which rules out every other order.

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50.

Which of the following molecule has double bond with two π bonds?

A.\(\mathrm{Ne_{2}}\)
B.\(\mathrm{O_{2}}\)
C.\(\mathrm{B_{2}}\)
D.\(\mathrm{C_{2}}\)
E.\(\mathrm{Li_{2}}\)

Exam Year(s)

KEAM - 2026,

Answer: D

The molecule is C₂.

In the molecular orbital picture, C₂ has 12 electrons, and its bonding electrons occupy the two π2p orbitals (π2p_x² π2p_y²) with no electrons in σ2p. Its bond order is 2, and both bonds are π bonds, an unusual double bond with no σ component.

O₂ also has bond order 2, but as one σ and one π bond. B₂ and Li₂ have bond order 1, and Ne₂ has bond order 0 and does not exist as a stable molecule.

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