Assertion A: o-Nitrophenol is more volatile than p-nitrophenol. Reason R: Molecules of o-nitrophenol associate through intermolecular hydrogen bonds. Choose the correct answer.

A is correct but R is not correct

A is not correct but R is correct

Both A and R are correct but R is not the correct explanation of A

Both A and R are correct and R is the correct explanation of A

Chemical Bonding & Molecular Structure

13 NEET-UG practice questions on Chemical Bonding & Molecular Structure, part of Chemistry. The first 13 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
Which species is linear although its central atom carries three lone pairs?
1. ACO₂
2. BXeF₂
3. CBeCl₂
4. DHCN
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Answer: B · XeF₂
Xe in XeF₂ has 2 bond pairs and 3 lone pairs, which is five electron pairs. The lone pairs fill the three equatorial positions of a trigonal bipyramid, leaving the two fluorine atoms opposite each other.
- A: CO₂ is linear, but its carbon has no lone pairs.
- C: BeCl₂ is linear, but its beryllium has no lone pairs.
- D: HCN is linear, but its carbon has no lone pairs.
Remember this
Five electron pairs with three lone pairs gives a linear shape, as in XeF₂.
Question 2 · Medium
A Lewis structure of the nitrate ion, NO₃⁻, shows one N=O bond and two N-O single bonds. Find the formal charges on the nitrogen atom and on a singly bonded oxygen atom, in that order.
1. A0 and −1
2. B−1 and 0
3. C+1 and −1
4. D+2 and −1
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Answer: C · +1 and −1
Formal charge = valence electrons − lone pair electrons − half the bonding electrons. For N: 5 − 0 − 4 = +1. For a singly bonded O: 6 − 6 − 1 = −1. The total, +1 − 1 − 1 + 0, equals the ion's charge of −1.
- A: This treats nitrogen as neutral, so the total would not equal −1.
- B: This reverses the signs and treats oxygen as neutral.
- D: This counts only three bonds on nitrogen rather than four.
Remember this
The sum of formal charges in a species must equal its overall charge.
Question 3 · Medium
Assertion A: o-Nitrophenol is more volatile than p-nitrophenol. Reason R: Molecules of o-nitrophenol associate through intermolecular hydrogen bonds. Choose the correct answer.
1. AA is not correct but R is correct
2. BBoth A and R are correct but R is not the correct explanation of A
3. CBoth A and R are correct and R is the correct explanation of A
4. DA is correct but R is not correct
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Answer: D · A is correct but R is not correct
The ortho isomer forms an intramolecular hydrogen bond between OH and NO₂, so its molecules do not associate and it is more volatile. It is the para isomer that associates through intermolecular hydrogen bonds.
- A: A is correct, because the ortho isomer boils lower.
- B: R is false, because the ortho hydrogen bond is intramolecular.
- C: R is false, so it cannot explain A.
Remember this
Intramolecular hydrogen bonding lowers boiling point; intermolecular hydrogen bonding raises it.
Question 4 · Easy
Which molecule is NOT linear?
1. ABeCl₂
2. BC₂H₂
3. CCO₂
4. DSO₂
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Answer: D · SO₂
Sulphur in SO₂ has two bonding regions and one lone pair, so its shape is bent. The central atoms of the other three have no lone pairs and only two electron regions.
- A: Gaseous BeCl₂ has sp beryllium with no lone pair, so it is linear.
- B: Each carbon of C₂H₂ is sp hybridised, so the molecule is linear.
- C: CO₂ has an sp carbon with no lone pair, so it is linear.
Remember this
A lone pair on the central atom bends a molecule with two bonded regions.
Question 5 · Hard
Six electron pairs surround the central atom of a molecule, and the molecule has a net non-zero dipole moment. Which molecule fits both conditions?
1. ABrF₅
2. BXeF₄
3. CSF₆
4. DClF₃
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Answer: A · BrF₅
BrF₅ has 5 bond pairs and 1 lone pair, so its shape is square pyramidal and the bond moments do not cancel. XeF₄ and SF₆ also have six pairs, but their symmetrical shapes cancel the bond moments.
- B: XeF₄ has six pairs, but its square planar shape cancels the bond moments.
- C: SF₆ has six pairs, but its octahedral shape cancels the bond moments.
- D: ClF₃ is polar, but it has only five electron pairs.
Remember this
One lone pair on an octahedral arrangement gives a polar square pyramid; two opposite lone pairs give a non-polar square plane.
Question 6 · Hard
Which of the following statements are correct according to molecular orbital theory? A. O₂ is paramagnetic with two unpaired electrons. B. Li₂ is paramagnetic. C. The bond order of N₂⁺ is greater than that of N₂. D. C₂ has a bond order of 2 and is diamagnetic.
1. AA, B and D only
2. BB and C only
3. CA, C and D only
4. DA and D only
Answer and explanation in LearnFree account
Question 7 · Easy
According to molecular orbital theory, the bond order of N₂ is
1. A3
2. B1
3. C2
4. D2.5
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Question 8 · Medium
Match List-I with List-II. List-I (molecule) A. XeF₄ B. SF₄ C. ClF₃ D. PCl₅ List-II (shape) I. T-shaped II. Trigonal bipyramidal III. Square planar IV. See-saw
1. AA-III, B-IV, C-I, D-II
2. BA-III, B-I, C-IV, D-II
3. CA-II, B-IV, C-I, D-III
4. DA-IV, B-III, C-I, D-II
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Question 9 · Medium
Arrange the following species in increasing order of bond length. A. O₂⁻ B. O₂²⁻ C. O₂⁺ D. O₂
1. AD, C, A, B
2. BB, A, D, C
3. CC, A, D, B
4. DC, D, A, B
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Question 10 · Medium
Given below are two statements: Statement I: The dipole moment of NH₃ is greater than that of NF₃. Statement II: In NH₃, the orbital dipole of the lone pair opposes the resultant of the N-H bond moments. Choose the correct answer.
1. AStatement I is incorrect but Statement II is correct
2. BStatement I is correct but Statement II is incorrect
3. CBoth Statement I and Statement II are incorrect
4. DBoth Statement I and Statement II are correct
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Question 11 · Hard
Treat each C–Cl bond as a vector of the same magnitude. Which of the molecules drawn have zero dipole moment?
Figure: MedExams
1. AB and D only
2. BA, B and D only
3. CA, B, C and D
4. DA and B only
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Question 12 · Easy
Intramolecular hydrogen bonding is present in which of the structures drawn?
Figure: MedExams
1. AStructure B
2. BStructure A
3. CStructure C
4. DStructure D
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Question 13 · Medium
The figure gives the filling of the valence molecular orbitals of a homonuclear diatomic oxygen species (σ1s and σ*1s are full and not drawn). State its bond order and magnetic behaviour.
Figure: MedExams
1. A2, paramagnetic
2. B1.5, paramagnetic
3. C1.5, diamagnetic
4. D2.5, paramagnetic
Answer and explanation in LearnFree account

Chemical Bonding & Molecular Structure

Which species is linear although its central atom carries three lone pairs?

A Lewis structure of the nitrate ion, NO₃⁻, shows one N=O bond and two N-O single bonds. Find the formal charges on the nitrogen atom and on a singly bonded oxygen atom, in that order.

Assertion A: o-Nitrophenol is more volatile than p-nitrophenol. Reason R: Molecules of o-nitrophenol associate through intermolecular hydrogen bonds. Choose the correct answer.

Which molecule is NOT linear?

Six electron pairs surround the central atom of a molecule, and the molecule has a net non-zero dipole moment. Which molecule fits both conditions?

Which of the following statements are correct according to molecular orbital theory? A. O₂ is paramagnetic with two unpaired electrons. B. Li₂ is paramagnetic. C. The bond order of N₂⁺ is greater than that of N₂. D. C₂ has a bond order of 2 and is diamagnetic.

According to molecular orbital theory, the bond order of N₂ is

Match List-I with List-II. List-I (molecule) A. XeF₄ B. SF₄ C. ClF₃ D. PCl₅ List-II (shape) I. T-shaped II. Trigonal bipyramidal III. Square planar IV. See-saw

Arrange the following species in increasing order of bond length. A. O₂⁻ B. O₂²⁻ C. O₂⁺ D. O₂

Given below are two statements: Statement I: The dipole moment of NH₃ is greater than that of NF₃. Statement II: In NH₃, the orbital dipole of the lone pair opposes the resultant of the N-H bond moments. Choose the correct answer.

Treat each C–Cl bond as a vector of the same magnitude. Which of the molecules drawn have zero dipole moment?

Intramolecular hydrogen bonding is present in which of the structures drawn?

The figure gives the filling of the valence molecular orbitals of a homonuclear diatomic oxygen species (σ1s and σ*1s are full and not drawn). State its bond order and magnetic behaviour.