JEE Main Chemical Bonding Previous Year Questions with Solutions

The formation of a chemical bond between two or more atoms, molecules, or ions to give rise to a chemical compound is known as chemical bonding. These chemical bonds keep the atoms together in the resulting compound. The attractive force which holds various constituents together and stabilizes them by the overall loss of energy is known as chemical bonding. Therefore, it can be understood that chemical compounds are reliant on the strength of the chemical bonds between their constituents. The stronger the bonding between the constituents, the more stable the resulting compound would be.

IIT JEE aspirants can find Chemical bonding JEE Main Chemistry previous year questions with solutions here. This set of questions helps aspirants to understand the type of questions asked in the JEE exams. Download JEE Main previous year Chemical bonding questions PDFs for free and crack JEE exam.

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JEE Main Previous Year Solved Questions on Chemical Bonding

1. The bond dissociation energy of B–F in BF₃ is 646 kJ mol^-1 whereas that of C–F in CF₄ is 515 kJ mol^–1. The correct reason for higher B–F bond dissociation energy as compared to that of C–F is

(1) Significant pπ – pπ interaction between B and F in BF₃ whereas there is no possibility of such interaction between C and F in CF₄.

(2) Lower degree of pπ – pπ interaction between B and F in BF₃ than that

between C and F in CF₄

(3) Smaller size of B-atom as compared to that of C-atom

(4) Stronger bond between B and F in BF₃ as compared to that between C and F in CF₄.

Solution:

Because of pπ – pπ back bonding in BF₃ molecule, all B-F bonds have partial double bond character.

Hence option (1) is the answer.

2. Among the following species which two have a trigonal bipyramidal shape?

(1) NI₃ (2) I₃^– (3) SO₃^2- (4) NO₃^–

(1) II and III

(2) III and IV

(3) I and IV

(4) I and III

Solution:

Let us find the hybridization (H) and shape of given species.

(1) For NI₃, H = ½ (5+3) = 8/2 = 4 → sp³ hybridized state. It is trigonal pyramidal in shape.

(2) For I₃^–, H = ½ (7+2+1) = 10/2 = 5 → sp³d hybridized state. It is linear in shape.

(3) For SO₃^2-, H = ½ (6+2) = 8/2 = 4 → sp³ hybridized state. It is trigonal pyramidal in shape.

(4) For NO₃^–, H = ½ (5+1) = 6/2 = 3 → sp² hybridized state. It is trigonal planar in shape.

Hence option (4) is the answer.

3. Using MO theory, predict which of the following species has the shortest bond length?

(1) O₂^–

(2) O₂^2-

(3) O₂²⁺

(4) O₂⁺

Solution:

Therefore bond length order O₂^2- >O₂^– >O₂⁺ >O₂²⁺

Hence option (3) is the answer.

4. Among the following, the species having the smallest bond is :

(1) NO

(2) NO⁺

(3) O₂

(4) NO^–

Solution:

The bond order of given molecules are:

NO = 2.5, NO⁺ = 3, O₂ = 2, NO^– = 2

Larger the bond order, the smaller the bond length.

NO⁺ has the largest bond order 3.

Therefore, it will have the smallest bond.

Hence option (2) is the answer.

5. The hybridisation of orbitals of N atom in NO₃^–, NO₂⁺, NH₄⁺ are respectively:

(1) sp², sp³, sp

(2) sp, sp³, sp²

(3) sp, sp², sp³

(4) sp², sp, sp³

Solution:

In NO₃, the central N atom has 3 bonding domains and zero lone pairs of electrons.

In NO₂, the central N atom has 2 bonding domains and zero lone pairs of electrons.

In NH₄, the central N atom has 4 bonding domains and zero lone pairs of electrons.

The Hybridization of N atom in NO₃^–, NO₂⁺, NH₄⁺ are sp², sp, sp³ respectively.

Hence option (4) is the answer.

6. Based on lattice energy and other considerations, which one of the following alkali metal chlorides is expected to have the highest melting point?

(1) RbCl

(2) LiCl

(3) KCl

(4) NaCl

Solution:

NaCl has the highest melting point.

Hence option (4) is the answer.

7. The structure of IF₇ is :

(1) octahedral

(2) pentagonal bipyramid

(3) square pyramid

(4) trigonal bipyramid

Solution:

For IF₇, hybridisation – sp³d³. The shape is pentagonal bipyramidal.

Hence option (2) is the answer.

8. Which of the following has the square planar structure :

(1) NH₄⁺

(2) CCl₄

(3) XeF₄

(4) BF₄^–

Solution:

Hybridization of XeF₄ sp³d²

It has a square planar shape.

Hence option (3) is the answer.

9. Among the following the maximum covalent character is shown by the compound :

(1) AlCl₃

(2) MgCl₂

(3) FeCl₂

(4) SnCl₂

Solution:

Al⁺³ is having the highest polarizing power than other compounds having greater covalent character.

Hence option (1) is the answer.

10. The compound of Xenon with zero dipole moment is :

(1) XeO₃

(2) XeO₂

(3) XeF₄

(4) XeOF₄

Solution:

XeF₄ has dipole moment zero.

Hence option (3) is the answer.

11. Which of the following has a maximum number of lone pairs associated with Xe?

(1) XeO₃

(2) XeF₄

(3) XeF₆

(4) XeF₂

Solution:

XeO₃ has 1 lone pair of electrons. XeF₄ has 2 lone pairs of electrons. XeF₆ has 1 lone pair of electrons. XeF₂ has 3 lone pairs of electrons. XeF₂ has a maximum number of lone pairs of electrons.

Hence option (4) is the answer.

12. Among the following the molecule with the lowest dipole moment is :

(1) CHCl₃

(2) CH₂Cl₂

(3) CCl₄

(4) CH₃Cl

Solution:

The order of the dipole moment is CCl₄ < CHCl₃< CH₂Cl₂ < CH₃Cl. So CCl₄ has the lowest dipole moment.

Hence option (3) is the answer.

13. The number of types of bonds between two carbon atoms in calcium carbide is

(1) One sigma, two pi

(2) One sigma, one pi

(3) Two sigma, one pi

(4) Two sigma, two pi

Solution:

CaC₂ → Ca⁺² +C₂^2-

^–C ≡ C^–

Number of sigma bond is 1 and number of pi bond is 2.

Hence option (1) is the answer.

14. The formation of molecular complex BF₃ – NH₃ results in a change in the hybridisation of boron

(1) From sp³ to sp³d

(2) From sp² to dsp²

(3) From sp³ to sp²

(4) From sp² to sp³

Solution:

In BF₃, Boron atom has 3 bond pairs of electrons and 0 lone pairs of electrons. It is sp² hybridized. In F₃B ← NH₃, Boron atom has 4 bond pairs of electrons and 0 lone pairs of electrons. It is sp³ hybridized. So the formation of molecular complex results in a change in the hybridization of boron from sp² to sp³.

Hence option (4) is the answer.

15. The molecule having the smallest bond angle is :

(1) PCl₃

(2) NCl₃

(3) AsCl₃

(4) SbCl₃

Solution:

Bond angle order NCl₃ > PCl₃ > AsCl₃ > SbCl₃.

Hence option (4) is the answer.

16. In which of the following pairs the two species are not isostructural?

(1) AlF₆^3- and SF₆

(2) CO₃^2- and NO₃^–

(3) PCl₄⁺ and SiCl₄

(4) PF₅ and BrF₅

Solution:

PF₅ has a trigonal bipyramidal shape. BrF₅ has a square pyramidal shape.

Hence option (4) is the answer.

17. Which one of the following molecules is expected to exhibit diamagnetic behaviour?

(1) C₂

(2) N₂

(3) O₂

(4) S₂

Solution:

C₂ and N₂ have no unpaired electrons. So they exhibit diamagnetic behaviour.

18. Which of the following is the wrong statement?

(1) ONCl and ONO^– are not isoelectronic

(2) O₃ molecule is bent

(3) Ozone is violet-black in solid-state

(4) Ozone is diamagnetic gas

Solution:

In the given options all are correct statements.

19. Stability of the species Li₂, Li₂^– and Li₂⁺ increases in the order of :

(1) Li₂ < Li₂⁺ < Li₂^–

(2) Li₂^– < Li₂⁺ < Li₂

(3) Li₂ < Li₂ < Li₂⁺

(4) Li₂^– < Li₂ < Li₂⁺

Solution:

The bond order of Li₂ is 1. The bond order of Li₂⁺ is 0.5. The bond order of Li₂^– is 0.5. Stability will depend on the bond order. Li₂⁺ is more stable than Li₂^– because the higher interelectronic repulsion in Li₂^– makes it the least stable. So the order is Li₂ > Li₂⁺ >Li₂^–.

Hence option (2) is the answer.

20. In which of the following pairs of molecules/ions, both species are not likely to exist?

(1) H₂⁺, He₂^2-

(2) H₂^–, He₂^2-

(3) H₂²⁺, He₂

(4) H₂^–, He₂²⁺

Solution:

The bond order of H₂²⁺ and He₂ is zero. So these molecules do not exist.

Hence option (3) is the answer.

21. Bond distance in HF is 9.17 × 10^-11m. Dipole moment of HF is 6.104 × 10^–30 Cm. The per cent ionic character in HF will be : (electron charge = 1.60 × 10^-19 C)

(1) 61.0%

(2) 38.0%

(3) 35.5%

(4) 41.5%

Solution:

Given Bond distance = 9.17 × 10^-11m.

Dipole moment = 6.104 × 10^–30 Cm

% iconic character = 6.104 × 10^–30× 100 / (1.60 × 10^-19×9.17 × 10^-11)

= 41.5%

Hence option (4) is the answer.

22. In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic?

(1) NO → NO⁺

(2) O₂ → O₂⁺

(3) N₂ → N₂⁺

(4) C₂ → C₂⁺

Solution:

During the ionisation of NO → NO⁺, the bond order changes from 2.5 to 3. Also magnetic character changes from paramagnetic to diamagnetic.

During the ionisation of O₂ → O₂⁺, the bond order increases from 2 to 2.5 and the magnetic character changes from paramagnetic to diamagnetic.

During the ionisation of N₂ → N₂⁺, the bond order decreases from 3 to 2.5 and the magnetic behaviour changes from diamagnetic to paramagnetic.

During the ionisation of C₂→ C₂⁺, the bond order decreases from 2 to 1.5 and the magnetic behaviour changes from diamagnetic to paramagnetic.

Hence option (1) is the answer.

23. Which one of the following molecules is paramagnetic?

(1) NO

( 2) O₃

(3) N₂

(4) CO

Solution:

NO has an unpaired electron. So it is paramagnetic in nature.

Hence option (1) is the answer.

24. The catenation tendency of C, Si and Ge is in the order Ge < Si < C. The bond energies (in kJ $mol^-1 of C — C, Si —Si and Ge—Ge bonds are respectively :

(1) 348, 260, 297

(2) 348, 297, 260

(3) 297, 348, 260

(4) 260, 297, 348

Solution:

Bond energy order is C – C > Si – Si > Ge – Ge.

Hence option (2) is the answer.

25. Oxidation state of sulphur in anions SO₃^2-, S₂O₄^2- and S₂O₆^2- increases in the orders

(1) S₂O₆^2- < S₂O₄^2- < SO₃^2-

(2) SO₃^2- < S₂O₄^2- < S₂O₆^2-

(3) S₂O₄^2- < SO₃^2- < S₂O₆^2-

(4) S₂O₄^2- < S₂O₆^2- < SO₃^2-

Solution:

The oxidation state of sulphur in SO₃^2- is +4. The Oxidation state of sulphur in S₂O₄^2- is +3 and in S₂O₆^2- is +5. So the order is S₂O₄^2- < SO₃^2- < S₂O₆^2-

Hence option (3) is the answer.

26. In which of the following species is the underlined carbon having sp³ hybridisation?

(1) CH₃COOH

(2) CH₃CH₂OH

(3) CH₃COCH₃

(4) CH₂=CH–CH₃

Solution:

Only in CH₃CH₂OH, carbon has sp³ hybridisation.

In other molecules, the carbon atom has multiple bonds,

Hence option (2) is the answer.

27. In which of the following sets, all the given species are isostructural?

(1) BF₃, NF₃, PF₃, AlF₃

(2) PCl₃, AlCl₃, BCl₃, SbCl₃

(3) BF₄^–, CCl₄, NH₄⁺,PCl₄⁺

(4) CO₂, NO₂, ClO₂, SiO₂

Solution:

BF₄^–, CCl₄, NH₄⁺,PCl₄⁺ are tetrahedral.

Hence option (3) is the answer.

28. In XeF₂, XeF₄, XeF₆ the number of lone pairs of Xe are respectively

(1) 2, 3, 1

(2) 1, 2, 3

(3) 4, 1, 2

(4) 3, 2, 1

Solution:

XeF₂ has 3 lone pairs of electrons. XeF₄ has 2 lone pairs of electrons. XeF₆ has 1 lone pair of electrons.

Hence option (4) is the answer.

29. Which of the following statements is true?

(1) HF is less polar than HBr

(2) absolutely pure water does not contain any ions

(3) chemical bond formation take place when forces of attraction overcome the forces of repulsion

(4) in covalency transference of electron takes place

Solution:

Chemical bond formation takes place when forces of attraction overcome the forces of repulsion.

Hence option (3) is the answer.

30. Which one of the following pairs of molecules will have permanent dipole moments for both members?

(1) N0₂ and C0₂

(2) N0₂ and 0₃

(3) SiF₄ and C0₂

(4) SiF₄ and N0₂

Solution:

N0₂ and 0₃ have angular shapes. So they will have a net dipole moment.

Hence option (2) is the answer.

31. The states of hybridization of boron and oxygen atoms in boric acid (H₃B0₃) are respectively

(1) sp² and sp² (2) sp³ and sp³

(3) sp³ and sp² (4) sp² and sp³

Solution:

Hybridization of B is sp² and O is sp³

Hence option (4) is the answer.

32. The maximum number of 90° angles between bond pair of electrons is observed in

(1) dsp³ hybridization

(2) sp³d² hybridization

(3) dsp² hybridization

(4) sp³d hybridization

Solution:

sp³d² hybridisation has an octahedral configuration. All the bond angles are 90° in the structure.

Hence option (2) is the answer.

33. Which of the following are arranged in an increasing order of their bond strengths?

(1) O₂^– < O₂ < O₂⁺ < O₂^2-

(2) O₂^2- < O₂^– < O₂ < O₂⁺

(3) O₂^– < O₂^2- < O₂ < O₂⁺

(4) O₂⁺ < O₂ < O₂^– < O₂^2-

Solution:

Higher the bond order, stronger the bonds. The increasing order is O₂^2- < O₂^– < O₂ < O₂⁺.

Hence option (2) is the answer.

34. Bond order and magnetic nature of CN^– are respectively

(1) 3, diamagnetic

(2) 2.5, paramagnetic

(3) 3, paramagnetic

(4) 2.5, diamagnetic

Solution:

Bond order = ½ [n_b – n_a]

= ½ [ 10-4]

= ½ (6)

= 3

It does not have unpaired electrons. So, it is diamagnetic.

Hence option (1) is the answer.

35. The bond order in NO is 2.5 while that in NO⁺ is 3. Which of the following statements is true for these two species?

(1)Bond length in NO⁺ is greater than in NO

(2)Bond length is unpredictable

(3)Bond length in NO⁺ in equal to that in NO

(4)Bond length in NO is greater than in NO⁺

Solution:

When bond order increases, bond length decreases. So the bond length in NO is greater than in NO⁺.

Hence option (4) is the answer.

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