Solved Examples (Dedicated Problem Set)

How to Use This Problem Set

This section is a single, dense workout covering every problem-type in Chemical Bonding. The problems progress from easy warm-ups (direct concept checks) through medium application problems to hard, multi-step challenges of JEE/NEET difficulty.

Work each one with paper and pencil before reading the solution. For every problem we state the relevant formula or rule first, then solve step-by-step, and finish with a one-line takeaway. The handful of formulas you'll keep reusing:

• Formal charge: $FC = V - L - B/2$
• Hybridization steric number: $H = \tfrac{1}{2}(V + M - C + A)$
• Bond order (MOT): $\tfrac{1}{2}(N_b - N_a)$
• Reaction enthalpy: $\Delta_r H = \sum BE_{broken} - \sum BE_{formed}$
• % ionic character: $\dfrac{\mu_{obs}}{\mu_{ionic}} \times 100$

Key Point: Speed in this chapter comes from the shortcut formulas. Drill them here until they are automatic.

Solved Examples

Example 1: Electrons in 1 coulomb (easy)

How many electrons make up a charge of 1 C? ($e = 1.6 \times 10^{-19}$ C)

Solution:

1. Formula: $n = Q/e$.
2. Substitute: $n = 1 / (1.6 \times 10^{-19})$.
3. Compute: $n = 6.25 \times 10^{18}$ electrons.

Takeaway: $Q = ne$ links macroscopic charge to electron count.

Example 2: Predict the ionic formula (easy)

Predict the formula of the compound between aluminium and oxygen.

Solution:

1. Ions: $Al^{3+}$ and $O^{2-}$.
2. Cross-over charges: 2 Al and 3 O.
3. Formula: $Al_2O_3$.

Takeaway: Balance total positive and negative charge to get the neutral formula.

Example 3: Lewis structure electron count (easy)

How many valence electrons must the Lewis structure of $SO_4^{2-}$ account for?

Solution:

1. S(6) + 4×O(24) = 30.
2. Add 2 for the 2− charge: 32.
3. Pairs: 16.

Takeaway: Adjust the total for ionic charge before drawing.

Example 4: Bond order from Lewis structure (easy)

State the bond order of the C–O bonds in $CO_2$.

Solution:

1. Structure: $O=C=O$, each C–O is a double bond.
2. Bond order: 2.

Takeaway: In $CO_2$ each C–O bond has order 2 (a double bond).

Example 5: Identify hybridization (easy)

What is the hybridization of carbon in $CH_4$?

Solution:

1. Steric number: $\tfrac{1}{2}(4+4) = 4$.
2. Hybridization: $sp^3$.

Takeaway: Four σ bonds, no lone pairs → $sp^3$.

Example 6: Polar or non-polar (easy)

Is $BF_3$ polar or non-polar?

Solution:

1. Shape: trigonal planar, symmetric.
2. Bond dipoles cancel by symmetry.
3. Conclusion: non-polar ($\mu = 0$).

Takeaway: Symmetric $AX_3$ planar molecules are non-polar.

Example 7: Formal charge on central O of ozone (medium)

Find the formal charge on the central oxygen in $O_3$.

Solution:

1. Central O: 1 lone pair ($L=2$), 3 bonds ($B=6$), $V=6$.
2. $FC = 6 - 2 - 3 = +1$.

Takeaway: Central O of ozone carries a $+1$ formal charge.

Example 8: VSEPR shape of SF₄ (medium)

Predict the shape of $SF_4$.

Solution:

1. Steric number: 4 bonds + 1 lone pair = 5 ($sp^3d$).
2. Lone pair goes equatorial.
3. Shape: see-saw.

Takeaway: $AX_4E$ → see-saw.

Example 9: Bond order of O₂ by MOT (medium)

Calculate the bond order of $O_2$.

Solution:

1. $N_b = 8, N_a = 4$.
2. Bond order $= \tfrac{1}{2}(8-4) = 2$.
3. Two unpaired electrons in $\pi^*$: paramagnetic.

Takeaway: $O_2$: bond order 2, paramagnetic.

Example 10: Reaction enthalpy from bond energies (medium)

Find $\Delta_r H$ for $H_2 + Cl_2 \to 2HCl$. ($H-H=435, Cl-Cl=242, H-Cl=431$ kJ/mol)

Solution:

1. Broken: $435 + 242 = 677$.
2. Formed: $2(431) = 862$.
3. $\Delta_r H = 677 - 862 = -185$ kJ/mol.

Takeaway: 'Broken − formed'; negative ⇒ exothermic.

Example 11: % ionic character (medium)

$\mu_{obs}(HCl) = 1.03$ D; 100%-ionic value $= 6.09$ D. Find % ionic character.

Solution:

1. $\% = (1.03/6.09)\times100$.
2. Compute: $\approx 16.9\%$.

Takeaway: $HCl$ is ~17% ionic.

Example 12: Hybridization of Xe in XeF₄ (medium)

Find the hybridization and shape of $XeF_4$.

Solution:

1. Steric number: 4 bonds + 2 lone pairs = 6.
2. Hybridization: $sp^3d^2$.
3. Shape: square planar.

Takeaway: $AX_4E_2$ → square planar, $sp^3d^2$.

Example 13: Born-Haber lattice energy (medium)

For $NaCl$: sublimation 108, ionisation 496, ½ dissociation 121, electron gain −349, formation −411 (kJ/mol). Find lattice enthalpy.

Solution:

1. $-411 = 108 + 496 + 121 - 349 - \Delta_{lattice}H$.
2. $-411 = 376 - \Delta_{lattice}H$.
3. $\Delta_{lattice}H = 787$ kJ/mol.

Takeaway: Born-Haber = Hess's law; watch the signs.

Example 14: Bond order in carbonate (medium)

Find the C–O bond order in $CO_3^{2-}$.

Solution:

1. Resonance: 4 bonds over 3 positions.
2. Bond order $= 4/3 = 1.33$.

Takeaway: Three equivalent resonance forms → bond order 1.33.

Example 15: σ and π bonds in a molecule (medium)

How many σ and π bonds in propyne $CH_3-C \equiv CH$?

Solution:

1. σ bonds: 3 (C–H of CH₃) + 1 (C–C single) + 1 (C–H terminal) + 1 (C–C of triple) = 6 σ.
2. π bonds: 2 (from the triple bond).
3. Total: 6 σ + 2 π.

Takeaway: Count one σ per bonded atom-pair; extra bonds in the triple bond are π.

Example 16: Dipole moment comparison (medium)

Why is the dipole moment of $NH_3$ greater than that of $NF_3$?

Solution:

1. $NH_3$: bond dipoles point toward N, same direction as lone-pair dipole → add.
2. $NF_3$: bond dipoles point toward F, opposing the lone-pair dipole → partly cancel.
3. Result: $\mu(NH_3) > \mu(NF_3)$.

Takeaway: Direction of bond dipoles vs lone pair decides the net μ.

Example 17: Octet exception classification (medium)

Classify $BCl_3$, $NO$, and $SF_6$ by their octet-rule exception.

Solution:

1. $BCl_3$: incomplete octet (6 e⁻ on B).
2. $NO$: odd-electron (11 valence e⁻).
3. $SF_6$: expanded octet (12 e⁻ on S).

Takeaway: Three distinct ways to break the octet rule.

Example 18: Bond angle ordering (medium)

Arrange $H_2O$, $NH_3$, $CH_4$ by increasing bond angle.

Solution:

1. Lone pairs: $H_2O$ (2), $NH_3$ (1), $CH_4$ (0).
2. More lone pairs → smaller angle.
3. Increasing angle: $H_2O\ (104.5°) < NH_3\ (107°) < CH_4\ (109.5°)$.

Takeaway: Bond angle falls as lone-pair count rises.

Example 19: Hybridization of a polyatomic ion (medium)

Find the hybridization of S in $SO_4^{2-}$.

Solution:

1. Steric number: 4 σ bonds to O, no lone pair on S → 4.
2. Hybridization: $sp^3$.
3. Shape: tetrahedral.

Takeaway: $SO_4^{2-}$ is $sp^3$, tetrahedral (ignore π bonds).

Example 20: Compare lattice enthalpies (medium)

Which has the larger lattice enthalpy, $NaF$ or $KF$?

Solution:

1. Same charges and anion; cations differ: $Na^+$ smaller than $K^+$.
2. Smaller cation → shorter inter-ionic distance → larger lattice enthalpy.
3. Conclusion: $NaF > KF$.

Takeaway: Smaller ions → larger lattice enthalpy (for the same charges).

Example 21: Bond order of O₂⁺, O₂, O₂⁻ (hard)

Compute and rank the bond orders of $O_2^+$, $O_2$, $O_2^-$.

Solution:

1. $O_2^+$ (15 e⁻): $\tfrac{1}{2}(8-3) = 2.5$.
2. $O_2$ (16 e⁻): $\tfrac{1}{2}(8-4) = 2.0$.
3. $O_2^-$ (17 e⁻): $\tfrac{1}{2}(8-5) = 1.5$.
4. Rank: $O_2^+ > O_2 > O_2^-$.

Takeaway: Adding antibonding electrons lowers bond order and lengthens the bond.

Example 22: Multi-step shape + polarity (hard)

Predict the shape and polarity of $XeF_2$.

Solution:

1. Xe valence e⁻: 8; 2 in bonds, leaving 6 → 3 lone pairs.
2. Steric number: 2 + 3 = 5 ($sp^3d$); lone pairs equatorial.
3. Shape: linear ($AX_2E_3$).
4. Polarity: the two Xe–F dipoles are equal and opposite → cancel → non-polar.

Takeaway: $XeF_2$ is linear and non-polar.

Example 23: Bond enthalpy of combustion (hard)

Estimate $\Delta_r H$ for $CH_4 + 2O_2 \to CO_2 + 2H_2O$. ($C-H=414, O=O=499, C=O=799, O-H=463$ kJ/mol)

Solution:

1. Broken: 4(C–H) + 2(O=O) = $4(414) + 2(499) = 1656 + 998 = 2654$.
2. Formed: 2(C=O) + 4(O–H) = $2(799) + 4(463) = 1598 + 1852 = 3450$.
3. $\Delta_r H = 2654 - 3450 = -796$ kJ/mol.

Takeaway: Combustion is strongly exothermic; account for every bond on both sides.

Example 24: Fajans' rules application (hard)

Arrange $LiCl$, $LiBr$, $LiI$ in increasing order of covalent character.

Solution:

1. Fajans: larger anion → more polarisable → more covalent.
2. Anion size: $Cl^- < Br^- < I^-$.
3. Covalent character increases: $LiCl < LiBr < LiI$.

Takeaway: Bigger, more polarisable anions add covalent character.

Example 25: Paramagnetism prediction (hard)

Determine the number of unpaired electrons in $N_2$, $O_2$, and $B_2$.

Solution:

1. $N_2$: all paired → 0 unpaired (diamagnetic).
2. $O_2$: 2 in $\pi^*$ singly → 2 unpaired (paramagnetic).
3. $B_2$: 2 in $\pi$ singly → 2 unpaired (paramagnetic).

Takeaway: $N_2$ diamagnetic; $O_2$ and $B_2$ paramagnetic.

Example 26: Resonance bond order in benzene (hard)

What is the C–C bond order in benzene, and why are all bonds equal?

Solution:

1. Two Kekulé structures delocalise 3 π bonds over 6 C–C bonds.
2. Each bond: one σ + half a π on average → bond order 1.5.
3. All six equal ($139\,pm$).

Takeaway: Benzene's delocalisation gives every C–C bond order 1.5.

Example 27: Dipole vector addition (hard)

The two O–H bond dipoles in water are each $1.5$ D with an angle of $104.5°$ between them. Estimate the net dipole moment. (Use $\mu_{net} = 2\mu\cos(\theta/2)$.)

Solution:

1. Formula: $\mu_{net} = 2\mu\cos(\theta/2)$.
2. $\theta/2 = 52.25°$, $\cos 52.25° \approx 0.612$.
3. $\mu_{net} = 2(1.5)(0.612) \approx 1.84$ D.

Takeaway: The resultant of two equal dipoles is $2\mu\cos(\theta/2)$ — matches water's observed $1.85$ D.

Example 28: Hybridization with the formula on an oxoanion (hard)

Find the hybridization of P in $PO_4^{3-}$.

Solution:

1. Formula: $H = \tfrac{1}{2}(V + M - C + A) = \tfrac{1}{2}(5 + 0 - 0 + 3) = 4$.
2. Hybridization: $sp^3$.
3. Shape: tetrahedral.

Takeaway: $PO_4^{3-}$ is $sp^3$, tetrahedral; add the 3 for the 3− charge in the formula.

Example 29: Comparing bond strength via MOT (hard)

Which is more stable, $N_2$ or $N_2^+$, and what happens to bond length on ionisation?

Solution:

1. $N_2$: bond order 3.
2. $N_2^+$: removing a bonding (σ2pz) electron → bond order 2.5.
3. Conclusion: $N_2$ is more stable; ionisation increases bond length (weaker bond).

Takeaway: Removing a bonding electron weakens and lengthens the bond.

Example 30: Isoelectronic species (hard)

Identify the bond order and an isoelectronic partner for $CO$.

Solution:

1. CO has 10 valence electrons → bond order 3 (like $N_2$).
2. Isoelectronic species: $N_2$, $NO^+$, $CN^-$ (all 10 valence e⁻, bond order 3).

Takeaway: Isoelectronic species share bond order and similar bond parameters.

Example 31: Combined shape, hybridization, polarity (hard)

For $PCl_5$, state the hybridization, shape, and whether it is polar.

Solution:

1. Steric number: 5 → $sp^3d$.
2. Shape: trigonal bipyramidal.
3. Polarity: symmetric arrangement → bond dipoles cancel → non-polar.

Takeaway: $PCl_5$ is $sp^3d$, trigonal bipyramidal, and non-polar.

Example 32: H-bonding and boiling point (hard)

Arrange $H_2O$, $H_2S$, $H_2Se$ in order of boiling point and explain the anomaly.

Solution:

1. By molar mass alone: $H_2Se > H_2S > H_2O$ would be expected.
2. But $H_2O$ has strong hydrogen bonding, raising its boiling point dramatically.
3. Actual order: $H_2O\ (100°C) > H_2Se\ (-41°C) > H_2S\ (-60°C)$.

Takeaway: Hydrogen bonding makes water's boiling point anomalously high within its group.