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 BF _{3} is 646 kJ mol^{-1} whereas that of C–F in CF_{4} 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 BF_{3 }whereas there is no possibility of such interaction between C and F in CF_{4}.

(2) Lower degree of pπ – pπ interaction between B and F in BF_{3} than that

between C and F in CF_{4}

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

(4) Stronger bond between B and F in BF_{3} as compared to that between C and F in CF_{4}.

**Solution:**

Because of pπ – pπ back bonding in BF_{3} 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) NI _{3} (2) I_{3}^{–} (3) SO_{3}^{2-} (4) NO_{3}^{–}**

(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 NI_{3}, H = ½ (5+3) = 8/2 = 4 → sp^{3} hybridized state. It is trigonal pyramidal in shape.

(2) For I_{3}^{–}, H = ½ (7+2+1) = 10/2 = 5 → sp^{3}d hybridized state. It is linear in shape.

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

(4) For NO_{3}^{–}, H = ½ (5+1) = 6/2 = 3 → sp^{2} 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) O_{2}^{–}

(2) O_{2}^{2-}

(3) O_{2}^{2+}

(4) O_{2}^{+}

**Solution:**

Chemical species | O_{2}^{–} |
O_{2}^{2-} |
O_{2}^{2+} |
O_{2}^{+} |

Bond order | 1.5 | 1 | 3 | 2.5 |

Therefore bond length order O_{2}^{2-} >O_{2}^{–} >O_{2}^{+} >O_{2}^{2+}

Hence option (3) is the answer.

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

(1) NO

(2) NO^{+}

(3) O_{2}

(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 NO _{3}^{–}, NO_{2}^{+}, NH_{4}^{+} are respectively:**

(1) sp^{2}, sp^{3}, sp

(2) sp, sp^{3}, sp^{2}

(3) sp, sp^{2}, sp^{3}

(4) sp^{2}, sp, sp^{3}

**Solution:**

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

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

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

The Hybridization of N atom in NO_{3}^{–}, NO_{2}^{+}, NH_{4}^{+} are sp^{2}, sp, sp^{3} 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 IF _{7} is :**

(1) octahedral

(2) pentagonal bipyramid

(3) square pyramid

(4) trigonal bipyramid

**Solution:**

For IF_{7}, hybridisation – sp^{3}d^{3}. Shape is pentagonal bipyramidal.

Hence option (2) is the answer.

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

(1) NH_{4}^{+}

(2) CCl_{4}

(3) XeF_{4}

(4) BF_{4}^{–}

**Solution:**

Hybridization of XeF_{4 }sp^{3}d^{2}

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) AlCl_{3}

(2) MgCl_{2}

(3) FeCl_{2}

(4) SnCl_{2}

**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) XeO_{3}

(2) XeO_{2}

(3) XeF_{4}

(4) XeOF_{4}

**Solution:**

XeF_{4} 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) XeO_{3}

(2) XeF_{4}

(3) XeF_{6}

(4) XeF_{2}

**Solution:**

XeO_{3} has 1 lone pair of electrons. XeF_{4} has 2 lone pairs of electrons. XeF_{6} has 1 lone pair of electrons. XeF_{2} has 3 lone pairs of electrons. XeF_{2} 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) CHCl_{3}

(2) CH_{2}Cl_{2}

(3) CCl_{4}

(4) CH_{3}Cl

**Solution:**

The order of the dipole moment is CCl_{4} < CHCl_{3}< CH_{2}Cl_{2} < CH_{3}Cl. So CCl_{4} 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:**

CaC_{2} → Ca^{+2} +C_{2}^{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 BF _{3} – NH_{3} results in a change in hybridisation of boron**

(1) From sp^{3} to sp^{3}d

(2) From sp^{2} to dsp^{2}

(3) From sp^{3} to sp^{2}

(4) From sp^{2} to sp^{3}

**Solution:**

In BF_{3}, Boron atom has 3 bond pairs of electrons and 0 lone pairs of electrons. It is sp^{2} hybridized. In F_{3}B ← NH_{3}, Boron atom has 4 bond pairs of electrons and 0 lone pairs of electrons. It is sp^{3} hybridized. So the formation of molecular complex results in a change in hybridization of boron from sp^{2} to sp^{3}.

Hence option (4) is the answer.

**15. The molecule having smallest bond angle is :**

(1) PCl_{3}

(2) NCl_{3}

(3) AsCl_{3}

(4) SbCl_{3}

**Solution:**

Bond angle order NCl_{3} > PCl_{3} > AsCl_{3} > SbCl_{3}.

Hence option (4) is the answer.

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

(1) AlF_{6}^{3-} and SF_{6}

(2) CO_{3}^{2-} and NO_{3}^{–}

(3) PCl_{4}^{+} and SiCl_{4}

(4) PF_{5} and BrF_{5}

**Solution:**

PF_{5} has trigonal bipyramidal shape. BrF_{5} has square pyramidal shape.

Hence option (4) is the answer.

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

(1) C_{2}

(2) N_{2}

(3) O_{2}

(4) S_{2}

**Solution:**

C_{2} and N_{2} 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) O_{3} 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 Li _{2}, Li_{2}^{– } and Li_{2}^{+} increases in the order of :**

(1) Li_{2} < Li_{2}^{+} < Li_{2}^{–}

(2) Li_{2}^{–} < Li_{2}^{+} < Li_{2}

(3) Li_{2} < Li_{2} < Li_{2}^{+}

(4) Li_{2}^{–} < Li_{2} < Li_{2}^{+}

**Solution:**

Bond order of Li_{2} is 1. Bond order of Li_{2}^{+ } is 0.5. Bond order of Li_{2}^{– }is 0.5. Stability will depend on the bond order. Li_{2}^{+} is more stable than Li_{2}^{–} because the higher interelectronic repulsion in Li_{2}^{–} makes it least stable. So the order is Li_{2} > Li_{2}^{+} >Li_{2}^{–}.

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) H_{2}^{+}, He_{2}^{2-}

(2) H_{2}^{–}, He_{2}^{2-}

(3) H_{2}^{2+}, He_{2}

(4) H_{2}^{–}, He_{2}^{2+}

**Solution:**

The bond order of H_{2}^{2+} and He_{2 }is zero. So these molecules do not exist.

Hence option (3) is the answer.

**21. Bond distance in HF is 9.17 × 10 ^{-11}m. 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^{-11}m.

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) O_{2} → O_{2}^{+}

(3) N_{2} → N_{2}^{+}

(4) C_{2} → C_{2}^{+}

**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 O_{2} → O_{2}^{+}, the bond order increases from 2 to 2.5 and the magnetic character changes from paramagnetic to diamagnetic.

During the ionisation of N_{2} → N_{2}^{+}, the bond order decreases from 3 to 2.5 and the magnetic behaviour changes from diamagnetic to paramagnetic.

During the ionisation of C_{2}→ C_{2}^{+}, 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) O_{3}

(3) N_{2}

(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 SO _{3}^{2-}, S_{2}O_{4}^{2-} and S_{2}O_{6}^{2-} increases in the orders **

(1) S_{2}O_{6}^{2-} < S_{2}O_{4}^{2-} < SO_{3}^{2-}

(2) SO_{3}^{2-} < S_{2}O_{4}^{2-} < S_{2}O_{6}^{2-}

(3) S_{2}O_{4}^{2-} < SO_{3}^{2-} < S_{2}O_{6}^{2-}

(4) S_{2}O_{4}^{2- }< S_{2}O_{6}^{2-} < SO_{3}^{2-}

**Solution:**

The oxidation state of sulphur in SO_{3}^{2-} is +4. The Oxidation state of sulphur in S_{2}O_{4}^{2-} is +3 and in S_{2}O_{6}^{2-} is +5. So the order is S_{2}O_{4}^{2-} < SO_{3}^{2-} < S_{2}O_{6}^{2-}

Hence option (3) is the answer.

**26. In which of the following species is the underlined carbon having sp ^{3} hybridisation?**

(1) CH_{3}COOH

(2) CH_{3}CH_{2}OH

(3) CH_{3}COCH_{3}

(4) CH_{2}=CH–CH_{3}

**Solution:**

Only in CH_{3}CH_{2}OH, carbon has sp^{3} 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) BF_{3}, NF_{3}, PF_{3}, AlF_{3}

(2) PCl_{3}, AlCl_{3}, BCl_{3}, SbCl_{3}

(3) BF_{4}^{–}, CCl_{4}, NH_{4}^{+},PCl_{4}^{+}

(4) CO_{2}, NO_{2}, ClO_{2}, SiO_{2}

**Solution:**

BF_{4}^{–}, CCl_{4}, NH_{4}^{+},PCl_{4}^{+} are tetrahedral.

Hence option (3) is the answer.

**28. In XeF _{2}, XeF_{4}, XeF_{6} 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:**

XeF_{2} has 3 lone pairs of electrons. XeF_{4} has 2 lone pairs of electrons. XeF_{6} 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) N0_{2} and C0_{2 }

(2) N0_{2} and 0_{3}

(3) SiF_{4} and C0_{2}

(4) SiF_{4} and N0_{2}

**Solution:**

N0_{2} and 0_{3} 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 (H _{3}B0_{3}) are respectively**

(1) sp^{2} and sp^{2 } (2) sp^{3} and sp^{3}

(3) sp^{3 }and sp^{2} (4) sp^{2} and sp^{3}

**Solution:**

Hybridization of B is sp^{2} and O is sp^{3}

Hence option (4) is the answer.

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

(1) dsp^{3} hybridization

(2) sp^{3}d^{2} hybridization

(3) dsp^{2} hybridization

(4) sp^{3}d hybridization

**Solution:**

sp^{3}d^{2} 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) O_{2}^{–} < O_{2} < O_{2}^{+} < O_{2}^{2-}

(2) O_{2}^{–} < O_{2}^{–} < O_{2} < O_{2}^{+}

(3) O_{2}^{–} < O_{2}^{2-} < O_{2} < O_{2}^{+}

(4) O_{2}^{+} < O_{2} < O_{2}^{–} < O_{2}^{2-}

**Solution:**

Higher the bond order, stronger the bonds. The increasing order is O_{2}^{–} < O_{2}^{–} < O_{2} < O_{2}^{+}.

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.