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 its 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 bond having partial double bond character.

Hence option (1) is the answer.

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

(1) NI3 (2) I3 (3) SO32- (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 SO32-, 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) O22-

(3) O22+

(4) O2+

Solution:

Chemical species O2 O22- O22+ O2+
Bond order 1.5 1 3 2.5

Therefore bond length order O22- >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:

Larger the bond order, smaller the bond length. NO+ has bond order 3.

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 chloride 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. 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 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 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 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 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 +C22-

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 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 hybridization of boron from sp2 to sp3.

Hence option (4) is the answer.

15. The molecule having 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) AlF63- and SF6

(2) CO32- and NO3

(3) PCl4+ and SiCl4

(4) PF5 and BrF5

Solution:

PF5 has trigonal bipyramidal shape. BrF5 has 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 electron. 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:

Bond order of Li2 is 1. Bond order of Li2+ is 0.5. Bond order of Li2is 0.5. Stability will depend on the bond order. Li2+ is more stable than Li2 because the higher interelectronic repulsion in Li2 makes it 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 the species are not likely to exist ?

(1) H2+, He22-

(2) H2, He22-

(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 percent 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 SO32-, S2O42- and S2O62- increases in the orders

(1) S2O62- < S2O42- < SO32-

(2) SO32- < S2O42- < S2O62-

(3) S2O42- < SO32- < S2O62-

(4) S2O42- < S2O62- < SO32-

Solution:

The oxidation state of sulphur in SO32- is +4. The Oxidation state of sulphur in S2O42- is +3 and in S2O62- is +5. So the order is S2O42- < SO32- < S2O62-

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 a multiple bond,

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 shape. 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 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+ < O22-

(2) O2 < O2 < O2 < O2+

(3) O2 < O22- < O2 < O2+

(4) O2+ < O2 < O2 < O22-

Solution:

Higher the bond order, stronger the bonds. The increasing order is O2 < 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 [nb – na]

= 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.