Q.Explain with the help of a suitable example a polar covalent bond.

Concept and Intuition

Intermolecular force laws depend on whether molecules can freely rotate or are held fixed:

• London dispersion forces arise from instantaneous induced dipoles and their interaction energy always falls off as $r^{-6}$, regardless of phase.
• Dipole-dipole interactions: if the dipoles are free to rotate/tumble (as in gases and liquids), thermal averaging over all orientations gives a net energy $\propto r^{-6}$ (the Keesom formula). But if the dipoles are held in a fixed, favourable orientation — as happens for molecules locked into a lattice in a solid — there's no rotational averaging, and the direct dipole-dipole interaction energy varies as $r^{-3}$.

Step-by-Step Solution

1. Recall the dispersion (London) energy formula: $E_{London} \propto -\dfrac{1}{r^6}$ — this is true in all phases. Statement I ✓.
2. Recall the dipole-dipole interaction: instantaneous energy between two fixed dipoles at a given orientation is $E \propto \dfrac{1}{r^3}$; only when free rotational averaging is invoked (gas/liquid, Boltzmann-weighted) does this become $r^{-6}$. …

Concept and Intuition

Dipole moment is a vector sum of individual bond dipoles. Even if a molecule has polar bonds, the molecule can be net non-polar if its geometry is symmetric enough that the bond dipole vectors cancel exactly. Common zero-dipole geometries: linear symmetric (like $CO_2$, $O=C=O$), trigonal planar symmetric (like $BF_3$), and tetrahedral with identical substituents (like $CH_4$). In contrast, molecules with a lone pair distorting the geometry (pyramidal $NH_3$, $PF_3$; bent $H_2O$) or with an asymmetric substituent (like $NF_3$, where the lone pair and F atoms don't cancel) are polar.

Step-by-Step Solution

1. Option A ($CO_2$, $NH_3$, $H_2O$): $CO_2$ is zero, but $NH_3$ (pyramidal) and $H_2O$ (bent) are both polar → rejected.
2. Option B ($NH_3$, $NF_3$, $BF_3$): $BF_3$ is zero, but $NH_3$ and $NF_3$ (both pyramidal with a lone pair) are polar → rejected.
3. Option C ($PF_3$, $NH_3$, $CH_4$): $CH_4$ is zero, but $PF_3$ and $NH_3$ (pyramidal) are polar → rejected. …

Concept and Intuition

A polar diatomic molecule can be modelled as two point charges $+\delta$ and $-\delta$ separated by the bond length $d$. The dipole moment is $\mu = \delta \times d$. If the bond were perfectly ionic, $\delta$ would equal a full electronic charge $e$; the ratio $\delta/e$ tells us the actual fractional (partial) ionic character of the bond.

Step-by-Step Solution

1. Convert bond length: $d = 1.0\,\text{Å} = 1.0\times10^{-10}\,\mathrm{m}$.
2. Given $\mu = 4\times10^{-30}\,\mathrm{C\,m}$. …

Concept and Intuition

Dipole moment depends on both bond polarity and lone-pair contribution, all added vectorially. In NH₃, the lone pair on N points away from the three N–H bonds (same general direction as the resultant bond dipole), so they reinforce, giving a sizeable dipole (1.47 D). In NF₃, F is more electronegative than N, so the N–F bond dipoles point toward F (away from N), while the lone pair on N points the opposite way — these two contributions largely cancel, leaving NF₃ with an unusually small dipole moment (0.23 D) despite the highly polar N–F bonds.

Step-by-Step Solution

1. Recall standard dipole moments (Debye): $\mathrm{HCl} \approx 1.07\,D$, $\mathrm{NH_3} \approx 1.47\,D$, $\mathrm{H_2O} \approx 1.85\,D$, $\mathrm{NF_3} \approx 0.23\,D$. …

Concept and Intuition

Net molecular dipole moment depends both on individual bond dipoles and molecular geometry. A perfectly symmetric molecule (like trigonal planar $\text{BF}_3$) has bond dipoles that cancel, giving zero net dipole moment. Pyramidal molecules with a lone pair (NH$_3$, NF$_3$, H$_2$O) have a net dipole, but its magnitude depends on whether the lone-pair dipole reinforces or opposes the bond dipoles.

Step-by-Step Solution

1. $\text{BF}_3$: trigonal planar, all B–F bond dipoles cancel by symmetry → $\mu = 0$ D → matches I.
2. $\text{NF}_3$: pyramidal, but F is more electronegative than N, so the bond dipoles point away from N while the lone pair also points away from the F atoms — the lone-pair dipole partly opposes/cancels the bond dipoles, giving a small net moment, $\mu = 0.23$ D → matches II.
3. $\text{NH}_3$: pyramidal, H is less electronegative than N, so bond dipoles point toward N, and the lone pair adds in the same general direction, giving a moderate net moment, $\mu = 1.47$ D → matches III. …

Concept and Intuition

To vaporise a liquid, energy must be supplied to overcome the intermolecular attractions holding molecules together in the liquid state. The stronger these attractions (here assumed to be dominated by dipole-dipole forces), the more energy (higher $\Delta_{vap}H^\circ$) is required. So a higher enthalpy of vaporisation directly implies stronger dipole-dipole attraction.

Step-by-Step Solution

1. List the given values: $\Delta_{vap}H^\circ(A) = 23.3$, $\Delta_{vap}H^\circ(B) = 41$, $\Delta_{vap}H^\circ(C) = 29$ kJ/mol. …

Concept and Intuition

A molecule has zero net dipole moment only when its bond dipoles cancel by symmetry. Benzene's perfectly symmetric hexagonal, non-polar C–H/C–C framework gives zero net dipole. $CO$, despite looking "balanced," actually has a small residual dipole due to electronegativity difference and lone-pair/back-bonding effects. $SO_2$ is bent (not linear), so its two S=O dipoles do not cancel. $NH_3$ is pyramidal with an asymmetric lone pair, giving a clear net dipole.

Step-by-Step Solution

1. $C_6H_6$: symmetric planar ring, all C–H bond dipoles cancel by the hexagonal symmetry $\Rightarrow \mu = 0$.
2. $CO$: two atoms, different electronegativities, no symmetry to cancel $\Rightarrow \mu \ne 0$ (small, ~0.11 D). …

Concept and Intuition

Net dipole moment is the vector sum of individual bond dipoles. Perfectly symmetric linear molecules ($CO_2$: O=C=O) cancel completely. Molecules with a bent or pyramidal shape (a lone pair breaking symmetry) retain a significant net dipole.

Step-by-Step Solution

1. $CO_2$: linear, O=C=O, the two C=O bond dipoles point in opposite directions and cancel exactly ⇒ $\mu = 0$.
2. $N_2O$: linear (N=N=O), but the two ends are chemically different, so the bond dipoles don't fully cancel — still, the net dipole is quite small ($\approx 0.17$ D) since the molecule is nearly balanced.
3. $NH_3$: trigonal pyramidal with a lone pair on N; bond dipoles and the lone pair's contribution add up to a moderate net dipole ($\approx 1.47$ D). …

Concept and Intuition

For a 1,4-disubstituted benzene with two IDENTICAL substituents, the molecule has a centre of symmetry if the substituent's own dipole lies exactly along the C–X bond axis (like Cl, or the linear C≡N group) — the two bond dipoles then point in exactly opposite directions and cancel completely, giving $\mu=0$ (this is the well-known case of p-dichlorobenzene). However, groups like –OH and –SH are bent (the O–H or S–H bond makes an angle with the C–O/C–S axis), so the group's own dipole has a component off the ring axis. Even in the para arrangement, this off-axis component does not exactly cancel between the two substituents, so the molecule retains a small net dipole moment.

Step-by-Step Solution

1. (i) 1,4-dichlorobenzene: Cl is axially symmetric on the C–Cl bond; the two bond dipoles are collinear and opposite → cancel → $\mu=0$.
2. (ii) 1,4-dicyanobenzene: C≡N is a linear group along the C–C(N) axis; same cancellation as Cl → $\mu=0$. …

Concept and Intuition

If a bond were purely ionic, the two atoms would carry full charges $+e$ and $-e$, giving a theoretical (maximum possible) dipole moment $\mu_{ionic}=e\times d$, where $d$ is the bond length. In a real (partially covalent) bond, the measured dipole moment $\mu_{obs}$ is smaller than this theoretical maximum, because the charge separation is only partial. The ratio $\mu_{obs}/\mu_{ionic}$, expressed as a percentage, is defined as the percent ionic character of the bond.

Step-by-Step Solution

1. Theoretical 100%-ionic dipole moment: $\mu_{ionic}=e\times d = (1.6\times10^{-19}\ \text{C})(150\times10^{-12}\ \text{m}) = 2.4\times10^{-29}$ C·m. …

Concept and Intuition

Molecular geometry is determined by VSEPR theory: sulphur in $H_2S$ has 6 valence electrons, forms 2 bond pairs with the two hydrogens, and retains 2 lone pairs. Four electron domains around a central atom arrange themselves tetrahedrally, but with two lone pairs pushed into two of the tetrahedral positions, the observed molecular shape (ignoring the lone pairs, which are not "seen" in molecular geometry) is bent/angular, not linear. Whenever a molecule is bent (asymmetric charge distribution) rather than linear/symmetric, the individual bond dipole moments do not fully cancel, so the molecule has a net dipole moment.

Step-by-Step Solution

1. Determine the electron geometry: S has 6 valence electrons; 2 are used to bond with 2 H atoms, leaving 2 lone pairs — total 4 electron domains around S.
2. Four electron domains adopt (approximately) a tetrahedral arrangement, but the molecular shape is described only by the positions of the atoms (H atoms), not the lone pairs.
3. With 2 bonded atoms and 2 lone pairs on a central atom with 4 electron domains, the resulting molecular geometry is bent/angular (analogous to water, though $H_2S$'s H–S–H bond angle, ~92°, is smaller than water's ~104.5° because sulphur's lone pairs occupy more nearly pure p-orbitals with less s-character mixing). …