Chemical Bonding and Molecular Structure JEE Main Questions 2025: Important Practice Questions with PYQs

1.

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

NaCl has the highest melting point.

Hence option (4) is the answer.

2.

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

(1) O2– < O2 < O2+ < O22-

(2) O22- < O2– < O2 < O2+

(3) O2– < O22- < O2 < O2+

(4) O2+ < O2 < O2– < O22-

Higher the bond order, stronger the bonds. The increasing order is O22- < O2– < O2 < O2+.

Hence option (2) is the answer.

3.

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%

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.

4.

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

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.

5.

Which of the following has the square planar structure :

(1) NH4+

(2) CCl4

(3) XeF4

(4) BF4–

Hybridization of XeF4 sp3d2

It has a square planar shape.

Hence option (3) is the answer.

6.

2. Among the following species which two have a 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

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.

7.

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

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

Hence option (2) is the answer.

8.

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

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

Hence option (2) is the answer.

9.

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

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.

10.

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+

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.

11.

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

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

Hence option (3) is the answer.

12.

Bond order and magnetic nature of CN– are respectively

(1) 3, diamagnetic

(2) 2.5, paramagnetic

(3) 3, paramagnetic

(4) 2.5, diamagnetic

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.

13.

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

(1) NO

(2) NO+

(3) O2

(4) NO–

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.

14.

The compound of Xenon with zero dipole moment is :

(1) XeO3

(2) XeO2

(3) XeF4

(4) XeOF4

XeF4 has dipole moment zero.

Hence option (3) is the answer.

15.

The molecule having the smallest bond angle is :

(1) PCl3

(2) NCl3

(3) AsCl3

(4) SbCl3

Bond angle order NCl3 > PCl3 > AsCl3 > SbCl3.

Hence option (4) is the answer.

16.

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+

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.

17.

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

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

Hence option (3) is the answer.

18.

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+

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

Hence option (4) is the answer.

19.

The structure of IF7 is :

(1) octahedral

(2) pentagonal bipyramid

(3) square pyramid

(4) trigonal bipyramid

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

Hence option (2) is the answer.

20.

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

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.

21.

Which one of the following molecules is paramagnetic?

(1) NO

( 2) O3

(3) N2

(4) CO

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

Hence option (1) is the answer.

22.

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

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

Hence option (2) is the answer.

23.

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

(1) AlCl3

(2) MgCl2

(3) FeCl2

(4) SnCl2

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

Hence option (1) is the answer.

24.

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

(1) C2

(2) N2

(3) O2

(4) S2

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

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-

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.

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

Hybridization of B is sp2 and O is sp3

Hence option (4) is the answer.

27.

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.

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

Hence option (1) is the answer.

28.

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

(1) CHCl3

(2) CH2Cl2

(3) CCl4

(4) CH3Cl

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

Hence option (3) is the answer.

29.

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

(1) CH3COOH

(2) CH3CH2OH

(3) CH3COCH3

(4) CH2=CH–CH3

Only in CH3CH2OH, carbon has sp3 hybridisation.

In other molecules, the carbon atom has multiple bonds,

Hence option (2) is the answer.

30.

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

(1) XeO3

(2) XeF4

(3) XeF6

(4) XeF2

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.