CBSE JEE Main & Advanced Chemistry: Chemical Bonding And Molecular Structure Important Questions & Complete Chapter Notes

Correct Answer: d

Explanation: O$_{2}$ < O$_{3}$ < H$_{2}$O$_{2}$ Bond lengths of O – O in O$_{2}$ is 1.21Å , H$_{2}$O$_{2}$ is 1.48Å and O$_{3}$ is 1.28Å. Therefore,correct order of the O – O bond length is H$_{2}$O$_{2}$ > O$_{3}$ > O$_{2}$.

Correct Answer: (c)

Explanation: Concept: VSEPR theory predicts molecular shapes based on lone pairs and bond pairs. Hybridisation corresponds to the steric number (bond pairs + lone pairs). Explanation: For $\mathrm{IF}_3$ : Iodine has 7 valence electrons. Three bonds and two lone pairs give steric number 5. Thus $sp^3d$ hybridisation and T‑shaped geometry. Correct. For $\mathrm{IF}_5$ : Iodine forms five bonds and has one lone pair. Steric number 6 leads to $sp^3d^2$ hybridisation and square pyramidal shape. Correct. For $\mathrm{IF}_7$ : Iodine forms seven bonds and has no lone pair. Steric number 7 gives $sp^3d^3$ hybridisation and pentagonal bipyramidal shape. Correct. For $\mathrm{ClO}_4^-$ : Chlorine has 7 valence electrons plus one negative charge. Four bonds and no lone pair give steric number 4. Hybridisation is $sp^3$ , geometry is tetrahedral, not square planar. Hence D is incorrect.  Answer: A, B and C Only (Option C).

Correct Answer: (c)

Explanation: Explanation of Bond Order Calculation **Bond Order (B.O.)** is calculated using the formula: $ \text{Bond Order} = \frac{1}{2} (\text{number of electrons in bonding orbitals} - \text{number of electrons in antibonding orbitals}) $ Let's analyze each set: Set (a): $\mathrm{B}_2, \mathrm{CN}^-, \mathrm{O}_2^{2-}$ **$\mathrm{B}_2$**: $ \text{Number of bonding electrons} = 4, \text{Number of antibonding electrons} = 2 $ $ \text{B.O.} = \frac{1}{2}(4-2) = 1 $ **$\mathrm{CN}^-$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $ $ \text{B.O.} = \frac{1}{2}(10-4) = 3 $ **$\mathrm{O}_2^{2-}$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $ $ \text{B.O.} = \frac{1}{2}(10-8) = 1 $ Total Bond Order for Set (a): $ 1 + 3 + 1 = 5 $ Set (b): $\mathrm{O}_2, \mathrm{F}_2, \mathrm{O}_2^{2+}$ **$\mathrm{O}_2$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $ $ \text{B.O.} = \frac{1}{2}(10-6) = 2 $ **$\mathrm{F}_2$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 8 $ $ \text{B.O.} = \frac{1}{2}(10-8) = 1 $ **$\mathrm{O}_2^{2+}$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $ $ \text{B.O.} = \frac{1}{2}(10-4) = 3 $ Total Bond Order for Set (b): $ 2 + 1 + 3 = 6 $ Set (c): $\mathrm{O}_2^-, \mathrm{N}_2, \mathrm{O}_2^+$ **$\mathrm{O}_2^-$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 7 $ $ \text{B.O.} = \frac{1}{2}(10-7) = 1.5 $ **$\mathrm{N}_2$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 4 $ $ \text{B.O.} = \frac{1}{2}(10-4) = 3 $ **$\mathrm{O}_2^+$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 5 $ $ \text{B.O.} = \frac{1}{2}(10-5) = 2.5 $ Total Bond Order for Set (c): $ 1.5 + 3 + 2.5 = 7 $ Set (d): $\mathrm{C}_2, \mathrm{O}_2, \mathrm{He}_2^{2+}$ **$\mathrm{C}_2$**: $ \text{Number of bonding electrons} = 8, \text{Number of antibonding electrons} = 4 $ $ \text{B.O.} = \frac{1}{2}(8-4) = 2 $ **$\mathrm{O}_2$**: $ \text{Number of bonding electrons} = 10, \text{Number of antibonding electrons} = 6 $ $ \text{B.O.} = \frac{1}{2}(10-6) = 2 $ **$\mathrm{He}_2^{2+}$**: $ \text{Number of bonding electrons} = 2, \text{Number of antibonding electrons} = 0 $ $ \text{B.O.} = \frac{1}{2}(2-0) = 1 $ Total Bond Order for Set (d): $ 2 + 2 + 1 = 5 $ Conclusion Set (c) has the maximum total bond order sum of $7$.

Correct Answer: d

Explanation: All are correct Hydrogen bonding is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.  At high pressure or in the vapor phase, formic acid consists of hydrogen bonded dimers rather than individual molecules.

Correct Answer: (c)

Explanation: As the size of the cation decrease, lattice enthalpy increases. Hence, $\mathrm{BeF}_2$ exhibits the highest lattice enthalpy among the given.