Top 20 Most Expected Questions from Chemical Bonding Class 11 Chemistry NEET | NEET 2023 Exam

NEET 2023 is approaching, and it is time to do thorough revisions and give your best for the entrance exam. This comprehensive session consists of the 20 most expected questions from Class 11 Chemistry Unit 4 Chemical Bonding for NEET 2023. Some of the topics discussed are chemical covalent bond, molecular orbitals, hybridisation, and bond order.

Question 1: The number and type of bond between two carbon atoms in CaC₂ are:

1. One given sigma and one pi bond
2. One sigma and two pi bonds
3. One sigma and one half pi bond
4. One sigma bond

Answer: b) One sigma and two pi bonds

Explanation: Calcium carbide (CaC₂) is an ionic compound (Ca²⁺C²⁻₂) having anion C²⁻₂. The calcium atom is bonded to the two carbon atoms through ionic bonding. It has one sigma and two pi bonds [:C ≡ C:]²⁻.

Question 2: The type of chemical covalent bond between carbon-carbon atom in C₂H₄ is:

1. sp² – sp² – σ bond, 2pπ – 2pπ bond
2. 2p – 2p – σ bond, 2pπ – 2π bond
3. sp³ – sp³ – σ bond, 2pπ – 2pπ bond
4. None of these

Answer: a) sp² – sp² – σ bond, 2pπ – 2pπ bond

Explanation: The carbon atoms in the ethene molecule are sp² hybridised. The p orbital still has one unpaired electron.

Each carbon joins three rather than four other atoms to form ethylene.

Between two carbon atoms, a sigma bond and a pi bond are formed.

Question 3: The element that shows greater ability to form pπ-pπ multiple bonds is:

1. Si
2. C
3. Sn
4. Ge

Answer: b) C

Explanation: Due to its size, the carbon atom’s 2p orbitals can form the strongest pπ−pπ bonds.

Question 4: Which of the following reactions is feasible?

1. KF(s) + HF(l) → KHF₂
2. KCl(s) + HCl(l) → KHCl₂
3. KBr(s) + HBr(l) → KHBr₂
4. Kl(s) + HI(l) → KHl₂

Answer: a) KF(s) + HF(l) → KHF₂

Explanation: KF will ionise to K⁺ and F⁻. A Hydrogen bond will be formed between F⁻ and H⁺ resulting in HF₂⁻.

Question 5: How many molecular orbitals are formed?

1. same as atomic orbitals
2. more than atomic orbitals
3. less than atomic orbitals
4. none of these

Answer: a) same as atomic orbitals

Explanation: The total number of atomic orbitals that make up a molecule determines how many molecular orbitals it has. Let’s use the hydrogen molecule as an example. Two hydrogen atoms make up a hydrogen molecule. Atoms of hydrogen have 1s orbitals. Because two atoms are fused together to form the hydrogen molecule, it has two molecular orbits.

Question 6: In which of the following species are the bonds non-directional?

1. NCl₃
2. RbCl
3. BeCl₂
4. BCl₃

Answer: b) RbCl

Explanation: The electrostatic force between cations (positively charged ions) and anions (negatively charged ions) is known as an ionic bond. The bonds are non-directional because there is a constant force of attraction acting in all directions. Covalent bonds are created when orbitals overlap, which can happen either axially or laterally. Covalent bonds are hence directional by nature.

a) NCl₃: Since both nitrogen and chlorine are nonmetals, NCl₃ is a covalent compound.

b) RbCl: Chlorine is a nonmetal, while rubidium is a metal; hence, RbCl is an ionic compound.

c) BeCl₂: Despite the fact that beryllium is a metal and chlorine is a nonmetal, they do not combine to create an ionic compound since there is a difference in electronegativity of less than 1.7. Be’s smaller size also contributes to its high polarising power, which accounts for its significant covalent character.

d) BCl₃: It is a covalent molecule because both boron and chlorine are nonmetals.

Question 7: Decreasing bond order of CO, CO₂ and CO₃^2– is:

1. CO, CO₂ and CO₃^2–
2. CO₃^2–, CO₂ and CO
3. CO, CO₃^2– and CO₂
4. CO₂, CO and CO₃^2–

Answer: a) CO, CO₂ and CO₃^2–

Explanation: CO has a triple bond. While carbonate ion is a resonance hybrid of three structures with partial double bonds, CO₂ has a double bond. Thus, the bond length decreases.

Question 8: Which of the following consist of equal no. of pπ–pπ & pπ–dπ bonds?

1. SO₃
2. SO₂
3. CO₂
4. SO₄^2–

Answer: b) SO₂

Explanation: In the compound SO₂, the sulphur atom contains one lone pair, six valence electrons (3s²3p⁴), and forms bonds with two oxygen atoms. It also forms one sigma bond and one pi bond with each oxygen atom. The 3s orbital has two paired electrons, while the 3p orbital contains four electrons. It requires four unpaired electrons in order to establish four bonds. As a result, one 3p electron moves to a vacant 3d orbital. In the current state, there are four unpaired electrons: three in three 3p orbitals and one in a 3d orbital.

Three equal sp² hybrid orbitals are formed by the hybridisation of one 3s and two 3p orbitals. The remaining 3p and 3d orbitals are still unhybridised. While the unpaired electron in the d orbital forms a dπ–pπ bond, the unpaired electron in the 3p orbital generates a pπ–pπ bond. There will only be one dπ–pπ bond as a result.

Question 9: Hybridisation involves:

1. Addition of an electron pair
2. Mixing up of atomic orbitals
3. Removal of an electron pair
4. Separation of orbitals

Answer: b) Mixing up of atomic orbitals

Explanation: Hybridisation is the process of combining (mixing) dissimilar orbitals with different energies to produce new orbitals. The newly created orbitals are referred to as hybrid orbitals and share the same energy.

Question 10: Carbon atoms in C₂(CN)₄ are:

1. sp hybridised
2. sp² hybridised
3. sp and sp² hybridised
4. sp, sp² and sp³ hybridised

Answer: c) sp and sp² hybridised

Explanation:

Question 11: In which of the following species can a maximum atom lie in the same plane?

1. XeF₂O₂
2. PCl₅
3. AsH₄⁺
4. XeF₄

Answer: d) XeF₄

Explanation: In XeF₄, Xenon is sp³d² hybridised and has square planar geometry. Its structure is given by,

The valence shell of the Xe atom has two electrons in the d-orbital and six electrons in the 5p-orbital. The fifth shell of d-orbitals and f-orbitals is free of electrons. Four F atoms bind with four half-filled orbitals, placing the fluorine atoms on either side of the central atom.

To fill the empty 5d-orbital space during the synthesis of XeF₄, 5p-orbital electrons went to an excited state. Four unpaired electrons are formed, two of which are filled with 5p-orbitals and two with 5d-orbitals, resulting in sp³d² hybridisation.

Question 12: The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF₄, respectively, are:

1. sp³d and 2
2. sp³d and 1
3. sp³d² and 2
4. sp³d² and 1

Answer: d) sp³d² and 1

Explanation: XeOF₄ → sp³d² hybridisation

Question 13: sp³d² hybridisation is not displayed by:

1. BrF₅
2. PF₅
3. SF₆
4. XeOF₄

Answer: b) PF₅

Explanation: The term “sp³d² hybridisation” refers to six sets of electron pairs that show an octahedral shape.

The centre P atom in PF₅ only carries 5 electron pairs, and this corresponds to sp³d hybridisation, which indicates a triangular bipyramidal shape. As a result, PF₅ does not exhibit sp³d² hybridisation.

Question 14: Species having a maximum number of 180° bond angles in the following is:

1. XeF₄
2. PCl₅
3. SiCl₄
4. IF₇

Answer: a) XeF₄

Explanation: There are two lone pair electrons in XeF₄. The overall electronic repulsions must be at a minimum if we are to follow the VSEPR theory. They will achieve a steady condition as a result of this. The lone pairs are arranged in an octahedral layout opposite (180 degrees) from one another in order to achieve this. The molecular geometry of XeF₄ is hence square planar.

Question 15: In PO₄³⁻, the formal charge on each oxygen atom and the P−O bond order, respectively, are:

1. –0.75; 0.6
2. –0.75; 1.25
3. –1.0; 1.0
4. –3.1; 1.25

Answer: b) –0.75; 1.25

Explanation: A formal charge is also referred to as a fake charge. The formal charge over an atom in a polyatomic molecule or ion is equal to the difference between that atom’s valence electron in the elemental state and the number of electrons allocated to that atom in the Lewis structure.

The formula to calculate formal charge is as follows:

F.C. = [Total number of valence electrons in the free state] – [total number of electrons assigned in Lewis structure]

F.C. = [Total number of valence electrons in the free state] – [total number of non-bonding pair electrons (lone pair)] – 1/2 [total number of bonding electrons]

The factor of ½ is attached to the no. of bonding e– because bonding e– is shared between two atoms.

Calculating the bond order of the phosphate ion using the formula,

Bond order = 5/4

Bond order = 1.25

For the atom of oxygen that makes a double bond with the atom of phosphorus,

F.C = [6] – [4 + 2]

F.C = [6] – [6]

F.C = 0

For a single bond between an oxygen atom and a phosphorus atom,

F.C = [6] – [6 + 1]

F.C = [6] – [7]

F.C = –1

A total of -3 charge is spread over four oxygen atoms in the resonance structure. As a result, each oxygen atom’s formal charge is,

F.C = –3/4

F.C = –0.75

Question 16: When the hybridisation state of a carbon atom changes from sp³ to sp² to sp, the angle between the hybridised orbitals:

1. Decreases gradually
2. Increases gradually
3. Decreases considerably
4. All of the above

Answer: b) Increases gradually

Explanation: The angle between the hybridised orbitals gradually increases when the hybridisation state of the carbon atom shifts from sp³ to sp² and then to sp. In sp³, sp², and sp-hybridisation, the bond angles are 109°, 120°, and 180°, respectively.

Question 17: Which of the following species show deviation from the octet’s rule?

1. AlCl₃
2. SF₆
3. NO
4. All of the above

Answer: d) All of the above

Explanation: All three molecules, AlCl₃, SF₆, and NO, deviate from the octet rule, as the central atom in them does not contain 8 electrons. Thus, this is an example of an expanded octet.

The central atom in AlCl₃ has 6 valence electrons.

The central atom in SF₆ has 12 valence electrons.

The central atom in NO has 7 valence electrons.

Question 18: Arrange the following in increasing order of stability:

O₂, N₂, Li₂, H₂

1. Li₂ < H₂ < O₂ < N₂
2. H₂ < Li₂ < O₂ < N₂
3. N₂ < O₂ < Li₂ < H₂
4. H₂ < Li₂ < N₂ < O₂

Answer: b) H₂ < Li₂ < O₂ < N₂

Explanation: Between the N atoms in N₂, there is a triple bond.

Between the O atoms in O₂, there is a double bond.

The atoms of Li₂ and H₂ are connected by a single bond.

The bond energy will increase as the number of bonds between atoms increases.

Between Li₂ and H₂, Li₂ (6e⁻) has more bonding electrons (1s² 1s² 2s²) than H₂ (2e) (1s₂). Both Li₂ and H₂ have similar bond order, but Li₂ has more B.E. than H₂. Thus, the order of stability in increasing order is H₂ < Li₂ < O₂ < N₂.

Question 19: If x-axis is the molecular axis, then π-molecular orbitals are formed by the overlap of:

1. s + p_x
2. p_x + p_y
3. p_z + p_z
4. p_x + p_x

Answer: c) p_z + p_z

Explanation: The orbital overlapping must be sideways for the creation of the π-bond. So p_z + p_z will give a π – bond.

Question 20: Among the following compounds, the one that is polar and has the central atoms with sp² hybridisation is:

1. H₂CO₃
2. SiF₄
3. C₂H₂
4. HClO₂

Answer: a) H₂CO₃

Explanation: Let’s first use VSEPR theory to determine the structure of each chemical in this situation. Consider H₂CO₃, where C is the main atom. Electron pairs (ep) result in trigonal planar geometry ⇒ sp² hybridisation.

We know that number of hybrid orbitals = Number of bonded atoms + number of lone pairs

⇒ Number of hybrid orbitals = 3 ⇒ sp²

(b) SiF₄ (Si is the central atom)

⇒ Number of hybrid orbitals = 4 ⇒ sp³

(c) C₂H₂ (C is the central atom)

The carbon atom requires additional electrons to establish four bonds with hydrogen and other carbon atoms in the synthesis of C₂H₂. The result is the transfer of one 2s² pair to the vacant 2pz orbital.

Each carbon has two sp hybrid orbitals after the 2s orbital in each atom combines with one of the 2p orbitals. The two carbon atoms in ethyne are connected by triple bonds.

Due to the double bond, the molecular geometry of C₂H₂ is linear and has a 180° bond angle. Thus, the number of hybrid orbitals is sp.

(d) HClO₂ (Cl is the central atom)

Number of hybrid orbitals = 2 + 2 = 4 ⇒ sp³

Now, we know that in H₂CO₃, the central atom (C) is sp² hybridised.

In addition, the dipole moments do not cancel one another in H₂CO₃.

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