Lewis Octet Rule: Atoms combine either by transfer of valence electrons (gaining or losing) or by sharing valence electrons in order to attain a stable octet ($8\mathrm{e^-}$) in their outermost shell (duplet $2\mathrm{e^-}$ for Hydrogen).
The formal charge ($FC$) of an atom in a polyatomic ion or molecule is defined as:
Where: $V = \text{Valence } \mathrm{e^-}$ of isolated atom, $L = \text{Lone pair (non-bonding) } \mathrm{e^-}$, $S = \text{Shared (bonding) } \mathrm{e^-}$.
Q. Calculate formal charge on each oxygen atom in Ozone ($\mathrm{O_3}$): $\mathrm{O_{(1)} = O_{(2)} - O_{(3)}}$.
| Limitation | Explanation | Key Standard Examples |
|---|---|---|
| 1. Incomplete Octet | Central atom has fewer than 8 valence electrons. | $\mathrm{LiCl}$ ($2\mathrm{e^-}$), $\mathrm{BeH_2}$ ($4\mathrm{e^-}$), $\mathrm{BF_3, BCl_3}$ ($6\mathrm{e^-}$) |
| 2. Odd-Electron Molecules | Total valence electrons is odd, so octet cannot be satisfied for all atoms. | Nitric oxide ($\mathrm{NO}$), Nitrogen dioxide ($\mathrm{NO_2}$) |
| 3. Expanded Octet | Central atom (Period 3+) uses vacant $d$-orbitals, accommodating $> 8$ electrons. | $\mathrm{PF_5}$ ($10\mathrm{e^-}$), $\mathrm{SF_6}$ ($12\mathrm{e^-}$), $\mathrm{H_2SO_4}$ ($12\mathrm{e^-}$) |
Lattice Enthalpy ($\Delta_{\text{lattice}}H^\circ$): Energy required to completely separate 1 mole of a solid ionic compound into gaseous constituent ions (or energy released when 1 mole ionic solid forms from gaseous ions).
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No bond is 100% ionic. Cation distorts electron cloud of anion — a phenomenon called Polarization.
Q. Why is $\mathrm{LiI}$ more covalent than $\mathrm{LiF}$? Why is $\mathrm{CuCl}$ more covalent than $\mathrm{NaCl}$?
Dipole Moment ($\mu$): $\mu = Q \times r$ (Unit: Debye, $1\text{ D} = 3.3356 \times 10^{-30}\text{ C m}$). It is a vector quantity pointing from $\delta+$ to $\delta-$.
| Molecule | Geometry | Dipole Moment ($\mu$) | Exam Reason / Explanation |
|---|---|---|---|
| $\mathrm{CO_2}$ | Linear | $\mu = 0$ | Two equal bond dipoles ($\mathrm{C=O}$) point in opposite directions ($180^\circ$) and cancel. |
| $\mathrm{BF_3}$ | Trigonal Planar | $\mu = 0$ | Resultant of two $\mathrm{B-F}$ dipoles equals and opposes the third $\mathrm{B-F}$ dipole ($120^\circ$). |
| $\mathrm{NH_3}$ vs $\mathrm{NF_3}$ | Pyramidal | $\mathrm{NH_3} (1.47\text{D}) > \mathrm{NF_3} (0.23\text{D})$ | In $\mathrm{NH_3}$, orbital dipole of lone pair & $\mathrm{N-H}$ bond dipoles reinforce; in $\mathrm{NF_3}$, $\mathrm{N-F}$ dipoles oppose lone pair dipole. |
Q. Why is dipole moment ($\mu$) of $\mathrm{CCl_4}$ zero, whereas that of $\mathrm{CHCl_3}$ is non-zero ($1.04\text{ D}$)? How is % ionic character calculated?
Resonance: When a single Lewis structure cannot accurately describe a molecule, two or more canonical structures with identical atomic positions but different electron distributions are drawn. The actual molecule is a resonance hybrid of all canonical forms.
Resonance Bond Order Formula: $\text{Bond Order} = \frac{\text{Total number of bonds between two atoms in all canonical forms}}{\text{Total number of canonical forms}}$.
Example ($\mathrm{CO_3^{2-}}$): $\mathrm{C-O}$ Bond Order = $\frac{4}{3} = \mathbf{1.33}$.
| Type | bp | lp | Electron Geometry | Molecular Shape | Bond Angle | Standard Examples |
|---|---|---|---|---|---|---|
| $\mathrm{AB_2}$ | 2 | 0 | Linear | Linear | $180^\circ$ | $\mathrm{BeCl_2, CO_2, HgCl_2}$ |
| $\mathrm{AB_3}$ | 3 | 0 | Trigonal Planar | Trigonal Planar | $120^\circ$ | $\mathrm{BF_3, BCl_3}$ |
| $\mathrm{AB_2E}$ | 2 | 1 | Trigonal Planar | Bent / V-shaped | $< 120^\circ$ ($119.5^\circ$) | $\mathrm{SO_2, O_3}$ |
| $\mathrm{AB_4}$ | 4 | 0 | Tetrahedral | Tetrahedral | $109.5^\circ$ | $\mathrm{CH_4, NH_4^+}$ |
| $\mathrm{AB_3E}$ | 3 | 1 | Tetrahedral | Trigonal Pyramidal | $107^\circ$ | $\mathrm{NH_3, H_3O^+}$ |
| $\mathrm{AB_2E_2}$ | 2 | 2 | Tetrahedral | Bent / V-shaped | $104.5^\circ$ | $\mathrm{H_2O, H_2S}$ |
| $\mathrm{AB_5}$ | 5 | 0 | Trigonal Bipyramidal | Trigonal Bipyramidal | $120^\circ, 90^\circ$ | $\mathrm{PCl_5}$ |
| $\mathrm{AB_4E}$ | 4 | 1 | Trigonal Bipyramidal | See-Saw | $< 120^\circ, < 90^\circ$ | $\mathrm{SF_4}$ |
| $\mathrm{AB_3E_2}$ | 3 | 2 | Trigonal Bipyramidal | T-Shaped | $< 90^\circ$ ($87.5^\circ$) | $\mathrm{ClF_3, BrF_3}$ |
| $\mathrm{AB_2E_3}$ | 2 | 3 | Trigonal Bipyramidal | Linear | $180^\circ$ | $\mathrm{XeF_2, I_3^-}$ |
| $\mathrm{AB_6}$ | 6 | 0 | Octahedral | Octahedral | $90^\circ$ | $\mathrm{SF_6}$ |
| $\mathrm{AB_5E}$ | 5 | 1 | Octahedral | Square Pyramidal | $< 90^\circ$ | $\mathrm{BrF_5, IF_5}$ |
| $\mathrm{AB_4E_2}$ | 4 | 2 | Octahedral | Square Planar | $90^\circ$ | $\mathrm{XeF_4}$ |
Q. Why are axial $\mathrm{P-Cl}$ bonds in $\mathrm{PCl_5}$ longer and weaker than equatorial bonds?
Q. Explain why $\mathrm{CH_4, NH_3}$, and $\mathrm{H_2O}$ all have $sp^3$ central atoms, but their bond angles decrease: $\mathrm{CH_4} (109.5^\circ) > \mathrm{NH_3} (107^\circ) > \mathrm{H_2O} (104.5^\circ)$.
| Feature | $\sigma$ (Sigma) Bond | $\pi$ (Pi) Bond |
|---|---|---|
| Type of Overlap | Head-on / End-to-end (axial) overlap along internuclear axis ($s-s, s-p_z, p_z-p_z$). | Sideways / Lateral (parallel) overlap perpendicular to axis ($p_x-p_x, p_y-p_y$). |
| Extent of Overlap | Greater overlap $\implies$ Strong bond. | Smaller overlap $\implies$ Weaker bond. |
| Electron Density | Cylindrically symmetrical around internuclear axis. | Two saucer-like clouds above and below the nodal plane. |
| Rotation | Free rotation about $\sigma$-axis is possible. | Free rotation is restricted. |
| Existence | Can exist independently as a single bond. | Always formed alongside a $\sigma$ bond in multiple bonds ($\mathrm{C=C}$ has $1\sigma + 1\pi$; $\mathrm{C\equiv C}$ has $1\sigma + 2\pi$). |
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Hybridisation: The process of intermixing of atomic orbitals of slightly different energies of an atom to form a new set of equivalent orbitals of identical shape and energy called hybrid orbitals.
| Hybridisation | Orbitals Mixed | Geometry | Bond Angle | Examples |
|---|---|---|---|---|
| $sp$ | $1s + 1p_z$ | Linear | $180^\circ$ | $\mathrm{BeCl_2, C_2H_2, CO_2}$ |
| $sp^2$ | $1s + 2p$ ($p_x, p_y$) | Trigonal Planar | $120^\circ$ | $\mathrm{BCl_3, C_2H_4, BF_3}$ |
| $sp^3$ | $1s + 3p$ | Tetrahedral | $109.5^\circ$ | $\mathrm{CH_4, C_2H_6, NH_3, H_2O}$ |
| $sp^3d$ | $1s + 3p + 1d_{z^2}$ | Trigonal Bipyramidal | $120^\circ, 90^\circ$ | $\mathrm{PCl_5, SF_4, ClF_3}$ |
| $sp^3d^2$ | $1s + 3p + 2d$ ($d_{z^2}, d_{x^2-y^2}$) | Octahedral | $90^\circ$ | $\mathrm{SF_6, XeF_4}$ |
Q. Find the total number of $\sigma$ and $\pi$ bonds in: (a) Ethene ($\mathrm{C_2H_4}$), (b) Ethyne ($\mathrm{C_2H_2}$), (c) Benzene ($\mathrm{C_6H_6}$), and (d) Propadiene ($\mathrm{CH_2=C=CH_2}$).
| Feature | Bonding Molecular Orbital (BMO) | Antibonding Molecular Orbital (ABMO) |
|---|---|---|
| Wave Function | Formed by constructive interference ($\psi = \psi_A + \psi_B$). | Formed by destructive interference ($\psi^* = \psi_A - \psi_B$). |
| Energy & Stability | Lower energy than parent AOs $\implies$ High stability. | Higher energy than parent AOs $\implies$ Lower stability. |
| Electron Density | High electron density concentrated between the nuclei. | Zero electron density between nuclei (possesses a Nodal Plane). |
In lighter homonuclear diatomics ($Z \le 7$: $\mathrm{Li_2, Be_2, B_2, C_2, N_2}$), the energy gap between $2s$ and $2p$ atomic orbitals is small, leading to significant $2s-2p$ orbital mixing. This pushes the $\sigma 2p_z$ MO higher in energy than the degenerate $\pi 2p_x = \pi 2p_y$ pair. In heavier diatomics ($Z > 7$: $\mathrm{O_2, F_2}$), the $2s-2p$ energy gap is large $\implies$ negligible mixing $\implies \sigma 2p_z$ remains lower in energy than $\pi 2p$.
1. For $Z \le 7$ ($14\mathrm{e^-}$ or fewer: $\mathrm{Li_2, Be_2, B_2, C_2, N_2}$):
$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$2. For $Z > 7$ (More than $14\mathrm{e^-}$: $\mathrm{O_2, F_2, Ne_2}$):
$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$| Species | Total $\mathrm{e^-}$ | MO Electronic Configuration | $N_b$ | $N_a$ | Bond Order | Magnetic Character & Notes |
|---|---|---|---|---|---|---|
| $\mathrm{H_2}$ | 2 | $\sigma 1s^2$ | 2 | 0 | 1 | Diamagnetic |
| $\mathrm{He_2}$ | 4 | $\sigma 1s^2 \, \sigma^* 1s^2$ | 2 | 2 | 0 | Does not exist ($BO=0$) |
| $\mathrm{Li_2}$ | 6 | $\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2$ | 4 | 2 | 1 | Diamagnetic (Stable in gas phase) |
| $\mathrm{Be_2}$ | 8 | $\sigma 1s^2 \, \sigma^* 1s^2 \, \sigma 2s^2 \, \sigma^* 2s^2$ | 4 | 4 | 0 | Does not exist ($BO=0$) |
| $\mathrm{B_2}$ | 10 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ $(\pi 2p_x^1 = \pi 2p_y^1)$ | 6 | 4 | 1 | Paramagnetic (2 unpaired $\mathrm{e^-}$) |
| $\mathrm{C_2}$ | 12 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ $(\pi 2p_x^2 = \pi 2p_y^2)$ | 8 | 4 | 2 | Diamagnetic (Both bonds are $\pi$ bonds!) |
| $\mathrm{N_2}$ | 14 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ $(\pi 2p_x^2 = \pi 2p_y^2) \, \sigma 2p_z^2$ | 10 | 4 | 3 | Diamagnetic ($1\sigma + 2\pi$, highest $BO$) |
| $\mathrm{O_2}$ | 16 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2$ $(\pi 2p_x^2 = \pi 2p_y^2)$ $(\pi^* 2p_x^1 = \pi^* 2p_y^1)$ | 10 | 6 | 2 | Paramagnetic (2 unpaired $\mathrm{e^-}$) |
| $\mathrm{F_2}$ | 18 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2$ $(\pi 2p_x^2 = \pi 2p_y^2)$ $(\pi^* 2p_x^2 = \pi^* 2p_y^2)$ | 10 | 8 | 1 | Diamagnetic |
| $\mathrm{Ne_2}$ | 20 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2$ $(\pi 2p_x^2 = \pi 2p_y^2)$ $(\pi^* 2p_x^2 = \pi^* 2p_y^2) \, \sigma^* 2p_z^2$ | 10 | 10 | 0 | Does not exist ($BO=0$) |
| Species | Total $\mathrm{e^-}$ | MO Electronic Configuration | $N_b$ | $N_a$ | Bond Order | Magnetic Character & Notes |
|---|---|---|---|---|---|---|
| $\mathrm{CO}$ | 14 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ $(\pi 2p_x^2 = \pi 2p_y^2) \, \sigma 2p_z^2$ | 10 | 4 | 3 | Diamagnetic (Isoelectronic with $\mathrm{N_2}$) |
| $\mathrm{NO}$ | 15 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2$ $(\pi 2p_x^2 = \pi 2p_y^2) \, (\pi^* 2p_x^1)$ | 10 | 5 | 2.5 | Paramagnetic (1 unpaired $\mathrm{e^-}$ in $\pi^* 2p$) |
| $\mathrm{NO^+}$ | 14 | $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ $(\pi 2p_x^2 = \pi 2p_y^2) \, \sigma 2p_z^2$ | 10 | 4 | 3 | Diamagnetic (More stable than $\mathrm{NO}$) |
Q. Why is $\mathrm{O_2}$ paramagnetic according to MOT while VBT predicts it to be diamagnetic?
Q. Arrange $\mathrm{O_2, O_2^+, O_2^-, O_2^{2-}}$ in increasing order of bond length and stability.
Q. Calculate bond order of $\mathrm{N_2, N_2^+}$, and $\mathrm{N_2^-}$. Compare their stabilities and magnetic behaviors.
Hydrogen Bond: Electrostatic attraction between hydrogen atom covalently bonded to a highly electronegative atom ($\mathrm{F, O, N}$) and another electronegative atom with a lone pair. Energy $\approx 10 - 40\text{ kJ/mol}$.
| Type | Description & Mechanism | Examples & Physical Effect |
|---|---|---|
| Intermolecular H-Bond | Occurs between different molecules of same or different compounds. | $\mathrm{H_2O, HF, NH_3}$, alcohols. Effect: Raises b.p. ($\mathrm{H_2O}$ liquid while $\mathrm{H_2S}$ gas). |
| Intramolecular H-Bond | Occurs within the same molecule forming a 5/6-membered chelate ring. | $o$-nitrophenol, salicylaldehyde. Effect: Lowers b.p., makes $o$-nitrophenol steam volatile. |
Q. Why does $o$-nitrophenol have a lower boiling point and higher steam volatility than $p$-nitrophenol?
Q1. Why is $\mathrm{H_2O}$ a liquid while $\mathrm{H_2S}$ is a gas at room temperature?
Q2. Why does ice float on water and why is water's density maximum at $4^\circ\text{C}$?
| Concept | High-Yield Exam Formula / Key Result | Exam Tip |
|---|---|---|
| Formal Charge | $FC = V - L - \frac{1}{2}S$ | Sum of FCs equals total charge on ion. |
| Dipole Moment | $\mu = q \times r$ (Debye, D) | $\mu = 0$ for symmetrical molecules ($\mathrm{CO_2, BF_3, CCCl_4}$). |
| VSEPR Repulsion | $\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$ | Lone pairs squeeze bond angles ($\mathrm{CH_4} 109.5^\circ \to \mathrm{NH_3} 107^\circ \to \mathrm{H_2O} 104.5^\circ$). |
| Bond Order (MOT) | $BO = \frac{N_b - N_a}{2}$ | $BO \propto \text{Bond Enthalpy} \propto \frac{1}{\text{Bond Length}}$. |
| Steric Number | $SN = \frac{1}{2}[V + M - C + A]$ | $SN = 2(sp), 3(sp^2), 4(sp^3), 5(sp^3d), 6(sp^3d^2)$. |
VARDAAN COMET