Chemical Bonding – JEE Advanced Previous Year Questions with Solutions

Chemical Bonding JEE Advanced Previous Year Questions with Solutions are provided here to make studying an easy and interesting process. Students will find all the important previous year questions for the Chemical Bonding chapter in JEE Advanced syllabus. We have also provided the solutions which will further serve as excellent study material that will help JEE aspirants to develop the right approach to tackle different types of questions.

Chemical Bonding JEE Advanced Previous Year Questions with Solutions are aimed at helping students solve even the most complicated problems with speed and accuracy. Students can download the questions and solutions as free PDf downloads. More significantly, students will be greatly benefited and they will understand how easy it is to analyze the questions and score high marks in the entrance exam.

Download Chemical Bonding Previous Year Solved Questions PDF

JEE Advanced Previous Year Questions on Chemical Bonding

Question 1. Consider the following compounds in the liquid form:

O₂, HF, H₂O, NH₃, H₂O₂, CCl₄, CHCl₃, C₆H₆, C₆H₅Cl

When a charged comb is brought near their flowing stream, how many of them show deflection as per the following figure?

Solution: (6)

Only polar molecules are deflected by charged comb.

Polar molecules: HF, H₂O, NH₃, H₂O₂, CHCl₃, C₆H₅Cl

Non-polar molecules: O₂, CCl₄, Benzene

Therefore, when a charged comb is brought near their flowing stream, six compounds will show deflection.

Question 2. Among B₂H₆, B₃N₃H₆, N₂O, N₂O₄, H₂S₂O₃ and H₂S₂O₈, the total number of molecules containing covalent bond between two atoms of the same kind is_____.

Solution: (4)

∴ Number of compounds having bond between the same type of atoms = 4

Question 3. Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules possess permanent dipole moment at room temperature.

A. NO₂, NH₃, POCl₃, CH₃Cl

B. BF₃, O₃, SF₆, XeF₆

C. BeCl₂, CO₂, BCl₃, CHCl₃

D. SO₂, C₆H₅Cl, H₂Se, BrF₅

Solution: (A, D)

If two identical dipole vectors are oriented at an angle of 180^o to each other, they cancel each other resulting in zero net dipole moment. Example. CO₂ and BeCl₂

If three identical dipole vectors are oriented at an angle of 120^o, they cancel each other out resulting in zero net dipole moment. eg BF₃

SF₆ also has a perfectly symmetrical octahedral geometry. Hence it also does not possess a permanent dipole moment at room temperature.

Question 4. The sum of the number of lone pairs of electrons on each central atom in the following species is.

[TeBr₆]^2-, [BrF₂]⁺, SNF₃ and [XeF₃]^–

(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)

Solutions: (6)

From the diagrams we can say that;

Te atom in [TeBr₆]^2- has 1 lone pair

Br atom in [BrF₂]⁺ has 2 lone pairs

S atom in SNF₃ has 0 lone pair

Xe atom in [XeF₃]^– has 3 lone pairs

Therefore, the sum of the number of lone pairs of electrons on each central atom in the given species = 1 + 2 + 0 + 3 = 6.

Question 5. Among the triatomic molecules/ions, Be Cl₂, N^3-, N₂O, NO⁺², O₃, SCl₂, ICl₂, l₃^– and XeF₂, and XeF₂, the total number of linear molecules(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is [Atomic number:S = 16, Cl = 17, I = 53 and Xe = 54]

Solution: (4)

The total number of the linear molecule(s)/ion(s) where the hybridization of the central atom does not have a contribution from the d-orbital(s) is 4.

They are BeCl₂, N₃^–, N₂O, NO²⁺.

In the remaining molecules/ions, O₃ is bent, SCl₂ is V-shaped with sp³ hybridization, ICl₂^–, I₃^– and XeF₂, are linear with sp³d hybridization.

Question 6. A list of species having the formula XZ₄ is given below:

XeF₄, SF₄, SiF₄, BF₄^–, BrF₄^–, [Cu(NH₃)₄]²⁺, [FeCl₄]^2-, CoCl₄^2- and [PtCl₄]^2-

Defining shape on the basis of the location of X and Z atoms, the total number of species having a square planar shape is.

Solution: (4)

XeF₄ = Square planar

SF₄ = See-saw

SiF₄ = Tetrahedral

BF₄^– = Tetrahedral

BrF₄^– = Square planar

CoCl₄^2- = Tetrahedral

Question 7. Based on VSEPR theory, the number of 90 degrees F-Br-F angles in BrF₅, is.

A. 8

B. 4

C. 0

D. 6

Solution: (C)

In BrF₅ there are 5 bond pairs and 1 lone pair on Br atom and hence the hybridization is sp³d² geometry and shape are octahedral and square pyramidal respectively. Due to lp-bp repulsions, the bond angles are lower than 90^o (as in octahedral). Hence there are no bond angles with 90^o.

Question 8. The compound (s) with two lone pairs of electrons on the central atom is (are)

A. BrF₅

B. CIF₃

C. XeF₄

D. Sf₄

Solution: (B, C)

Only, CIF₃ and XeF₄ contain two pairs of electrons on the central atom.

Question 9. Assuming 2s – 2p mixing is not operative the paramagnetic species among the following is:

A. Be₂

B. B₂

C. C₂

D. N₂

Solution: (C)

If 2 s − 2 p 2s-2p mixing is not operative , then molecular orbitals may be arranged in order of energy as follows:

σ(1s), σ^*(1s), σ(2s), σ^*(2s), σ(2pz), π(2px) = (2py), π^*(2px) = π^*(2py), σ^*(2pz)

Applying this configuration, BE₂, B₂ and N₂ will be diamagnetic but C₂ will be paramagnetic.

Question 10. Among H₂, He₂⁺, Li₂, Be₂, B₂, C₂, N₂,O₂^– and F₂, the number of diamagnetic species is (Atomic numbers: H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6, N = 7, O = 8, F = 9)

Solution: (6)

Molecular orbitals configurations:

Diamagnetic species → All electrons paired

H₂, Li₂, Be₂, C₂, N₂ and F₂ are diamagnetic species.

However, the bond order of Be₂ is zero thus Be₂ does not exist. So, the number of diamagnetic species is 6.

Also Read:-