Q4. Explain the important aspects of resonance with reference to the CO32− ion (NCERT Exercise 4.11, page 134).
No single Lewis structure (one C=O, two C–O) fits, because all three C–O bonds are found to be equal. The ion is a hybrid of three canonical forms that differ only in where the double bond and the negative charges are; the positions of the nuclei are the same in all. The hybrid is more stable than any one form, each C–O bond has bond order 1⅓, and the mean formal-charge label on each O is −⅔; this is not its measured partial charge. The canonical forms have no separate existence.
Q5. H3PO3 can be drawn as structure (1), with one H bonded to P, or structure (2), with all three H on oxygen atoms. Can these be canonical forms of one resonance hybrid (NCERT Exercise 4.12, page 134)?
No. Canonical forms must have the same positions of all nuclei and differ only in the arrangement of electrons. In (1) one hydrogen is bonded to P and in (2) to O, so the atoms are in different places: these are different molecules (isomers), not resonance forms.
Q6. Write the resonance structures of SO3, NO2 and NO3− (NCERT Exercise 4.13, page 134).
SO3: in the octet-based representation, three equivalent forms, each with one S=O and two S–O bonds, the double bond on a different O each time; S is +2 and each single-bonded O −1. NO2: two forms, O=N–O and O–N=O, the single-bonded O carrying −1 and N +1 (N also carries an odd electron). NO3−: three forms, each with one N=O and two N–O bonds, the double bond on a different O each time; N is +1 and each single-bonded O −1.
Q7. Explain why BeH2 has zero dipole moment although its Be–H bonds are polar (NCERT Exercise 4.22, page 134).
An isolated BeH2 molecule is linear (H–Be–H at 180°); solid BeH2 is polymeric. In the isolated molecule so the two equal bond dipoles point in opposite directions and cancel, giving a resultant of zero.
Q8. Arrange the bonds in LiF, K2O, N2, SO2 and ClF3 in order of increasing ionic character (NCERT Exercise 4.19, page 134).
Use the electronegativity difference (NCERT Table 3.8): N–N 0; S–O 1.0; Cl–F 1.0; K–O 2.7; Li–F 3.0. So N2 < SO2 ≈ ClF3 < K2O < LiF. On NCERT’s values the S–O and Cl–F differences are equal (1.0) and cannot separate the middle two; revised Pauling values (S 2.58, O 3.44, Cl 3.16, F 3.98) give 0.86 and 0.82, putting SO2 just above ClF3.
Q9. Write the significance or applications of dipole moment (NCERT Exercise 4.16, page 134).
It shows whether a molecule is polar; it helps decide shapes (CO2 with μ = 0 must be linear, H2O with μ ≠ 0 bent); it gives the percentage ionic character of a bond; and it helps tell isomers apart, such as the dichlorobenzenes (para has μ = 0; ortho > meta) and cis/trans-1,2-dichloroethene (trans has zero).
Q10. Define bond length and give one way it is measured (based on NCERT Exercise 4.10, page 134).
The equilibrium distance between the nuclei of two bonded atoms in a molecule. It is measured by spectroscopic, X-ray diffraction or electron diffraction methods.