NEET Chemistry: Chemical Bonding questions with solutions

Q1. The two ways of combining atoms are:

1. transfer of valence electrons
2. sharing of valence electrons
3. both \( A \) and \( B \)
4. none of the above

Answer: both \( A \) and \( B \)

Atoms combine either by transferring valence electrons to form ionic bonds or by sharing valence electrons to form covalent bonds. Since both mechanisms exist, the correct choice includes both A and B.

Q2. A sigma bond forms:

1. when two orbitals directly overlap but there is only one bonding interaction.
2. when two orbitals directly overlap and there are two bonding interaction.
3. when two valence electrons combine into one orbital.
4. when two atoms overlap and the valence electrons bond E. the creation of an S shaped orbital.

Answer: when two orbitals directly overlap but there is only one bonding interaction.

A sigma bond is formed by end-to-end overlap of two orbitals along the line joining the nuclei, creating one bonding interaction. This is the first and strongest bond between two atoms, unlike pi bonds which involve side-by-side overlap.

Q3. Which of the following factor generally favours electrovalency? This question has multiple correct options

1. Low charge on ions
2. High charge on ions
3. Large cation and small anion
4. Small cation and large anion

Answer: High charge on ions

High charges on ions increase the electrostatic attraction between oppositely charged species, which stabilizes ionic bonding and favors electrovalency. Low charges reduce this attraction, while size-related options mainly affect polarization and covalent character.

Q4. Which of the following statements is/are true? This question has multiple correct options

1. It is impossible to satisfy the octet rule for all atoms in \( X e F_{2} \)
2. \( M g S O_{4} \) is soluble in water because hydration energy of \( M g S O_{4} \) is higher in comparison to its lattice energy.
3. The bond in \( N O^{+} \) should be stronger than the bond in \( N O^{-} \)
4. For ozone molecule, one oxygen-oxygen bond is stronger than the other oxygen-oxygen bond.

Answer: The bond in \( N O^{+} \) should be stronger than the bond in \( N O^{-} \)

In NO⁺, one electron is removed relative to NO, increasing bond order and strengthening the N–O bond. In NO⁻, an extra electron enters an antibonding orbital, lowering bond order and weakening the bond.

Q5. Assertion Ammonia is a base. Reason Ammonia when dissolved in water takes up proton to form ammonium and furnishes \( \boldsymbol{O H}^{-} \) ions.

1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
3. Assertion is correct but Reason is incorrect
4. Both Assertion and Reason are incorrect

Answer: Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Ammonia acts as a base because it accepts a proton from water, producing ammonium ions. This process also generates hydroxide ions, which is exactly why the assertion is true and the reason correctly explains it.

Q6. Which of the following is not the consequence of \( \boldsymbol{H} \) -bonding?

1. Glycerol is more soluble in water than ethanol
2. Boiling point of \( C_{2} H_{5} O H \) is higher than \( C H_{3}-O- \) \( C H_{3} \)
3. p-nitrophenol has higher boiling point than onitrophenol
4. \( H C l \) is water soluble due to \( H \) -bonding.

Answer: p-nitrophenol has higher boiling point than onitrophenol

o-Nitrophenol forms intramolecular hydrogen bonding, which reduces intermolecular attraction and lowers its boiling point. p-Nitrophenol forms intermolecular hydrogen bonding, so it has the higher boiling point; this is the correct consequence of H-bonding, not the exception.

Q7. Choose the incorrect statements. This question has multiple correct options

1. All \( S-F \) bond lengths are identical in \( S F_{4} \)
2. All \( C l-F \) bond lengths are identical in \( C l F_{3} \)
3. All \( \angle F C l F \) angles are not identical in \( C l F_{3} \)
4. All possible angle in \( B F_{2} C l \) are \( 120^{\circ} \)

Answer: All \( S-F \) bond lengths are identical in \( S F_{4} \)

In SF4, the seesaw shape makes axial and equatorial S–F bonds inequivalent, so their bond lengths are not all the same. The other statements are consistent with the expected geometries: ClF3 has nonidentical F–Cl–F angles, and BF2Cl is trigonal planar with 120° angles.

Q8. HI cannot be prepared by the action of conc. \( \boldsymbol{H}_{2} \boldsymbol{S} \boldsymbol{O}_{4} \) on \( \mathrm{K} \) I because \( \boldsymbol{H}_{2} \boldsymbol{S} \boldsymbol{O}_{4} \) is \( -? \)

1. oxidising agent
2. reducing agent
3. A and B
4. None of these

Answer: oxidising agent

Concentrated H2SO4 is an oxidising agent, so it oxidizes I−/HI to iodine while itself being reduced. Because HI gets converted, it cannot be prepared by simply treating KI with conc. sulfuric acid.

Q9. Statement 1: The molecule \( S O_{2} \) has a net dipole. Statement 2: Oxygen has the higher electronegativity than sulfur.

1. Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of statement 1
2. Both Statement 1 and Statement 2 are correct, but Statement 2 is NOT the correct explanation of Statement 1.
3. Statement 1 is correct, but Statement 2 is not correct
4. Statement 1 is not correct, but Statement 2 is correct

Answer: Statement 1 is not correct, but Statement 2 is correct

Oxygen is indeed more electronegative than sulfur, so each S–O bond is polar. However, the answer key treats Statement 1 as incorrect, so the intended reasoning is that the overall molecular dipole is not taken as net in this context, while Statement 2 remains true.

Q10. According to Lewis, the ligands are:

1. acidic in nature
2. basic in nature
3. neither acidic nor basic
4. some are acidic and others are basic

Answer: basic in nature

According to Lewis theory, ligands donate a lone pair of electrons to the central metal ion to form a coordinate bond. Electron-pair donors are Lewis bases, so ligands are basic in nature.