Chemical Bonding and Molecular Structure Class 11 Notes

Valence Shell Electron Pair Repulsion (VSEPR) & Hybridization

VSEPR Theory is essential to predict the geometrical shapes of molecules. It is built on the premise that valence shell electron pairs (both bonding and non-bonding) surround a central atom and orient themselves in space to minimize electrostatic repulsion. The relative strength of electron pair repulsions follows this order: Lone Pair - Lone Pair (lp-lp) > Lone Pair - Bond Pair (lp-bp) > Bond Pair - Bond Pair (bp-bp). Consequently, the presence of lone pairs causes compression in regular geometries, altering the expected bond angles.

To complement this, Hybridization explains the bonding mechanism mathematically. It involves the intermixing of atomic orbitals of slightly different energies to form equivalent hybrid orbitals with definite spatial orientations. By calculating the steric number, students can instantly deduce the hybridization state (e.g., $sp$, $sp^2$, $sp^3$, $sp^3d$, $sp^3d^2$) and the baseline molecular geometry.

Definitions Glossary

Octet Rule

The chemical principle stating that atoms tend to combine in such a way that they each have eight electrons in their valence shells, achieving a stable noble gas electronic configuration.

Lattice Enthalpy

The amount of energy released when one mole of a crystalline ionic solid is formed from its constituent gaseous ions under standard conditions.

Dipole Moment (\mu)

The mathematical product of the magnitude of the charge (q) and the distance (d) separating the centers of positive and negative charges. It is measured in Debye (D) units.

Hybridization

The phenomenon of mixing atomic orbitals of comparable energies in a central atom to generate a completely new set of degenerate hybrid orbitals of equivalent shapes and energies.

Resonance

The representation of a single molecule by multiple contributing Lewis structures (canonical structures) because no single structure can fully explain all its observed properties.

Bond Order

According to Molecular Orbital Theory, it is half the difference between the number of electrons in bonding molecular orbitals ($N_b$) and antibonding molecular orbitals ($N_a$).

Must Remember | Key Points

• Exceptions to the Octet Rule exist as: electron-deficient molecules (e.g., $BCl_3$, $BeCl_2$), odd-electron molecules (e.g., $NO$, $NO_2$), and expanded octet molecules (e.g., $PCl_5$, $SF_6$, $IF_7$).
• The dipole moment is a vector quantity pointing from the electropositive atom towards the electronegative atom (chemists draw it from positive to negative with a crossed arrow).
• Steric Number ($H$) Formula: $H = \frac{1}{2} [V + M - C + A]$, where $V$ is valence electrons of the central atom, $M$ is the number of monovalent atoms, $C$ is cationic charge, and $A$ is anionic charge.
• For diatomic molecules containing $\le 14$ electrons (like $B_2, C_2, N_2$), the $\sigma 2p_z$ molecular orbital is higher in energy than $\pi 2p_x$ and $\pi 2p_y$ due to intense $s$-$p$ orbital mixing.
• Bond Order calculation: $\text{BO} = \frac{1}{2}(N_b - N_a)$. A positive bond order indicates a stable molecule, while a zero or negative bond order implies the molecule cannot exist.
• Bond Order is directly proportional to Bond Strength/Dissociation Enthalpy and inversely proportional to Bond Length.
• Hydrogen bonding can be Intermolecular (between different molecules, like HF or $H_2O$) or Intramolecular (within the same molecule, like o-nitrophenol), drastically affecting melting/boiling points.

Comparison: Sigma (\sigma) vs Pi (\pi) Bonds

Process: Step-by-Step Guide to Predict Hybridization & Shape

1. Identify the Central Atom & Valence Electrons — Locate the central atom and determine its group number to find the count of valence electrons ($V$).
2. Calculate the Steric Number (H) — Apply the steric formula: $H = \frac{1}{2} [V + M - C + A]$. Count only monovalent surrounding atoms ($H, F, Cl, Br, I$); ignore divalent oxygen or sulfur.
3. Assign the Hybridization State — Based on $H$, assign: $H=2 \to sp$ (linear), $H=3 \to sp^2$ (trigonal planar), $H=4 \to sp^3$ (tetrahedral), $H=5 \to sp^3d$ (trigonal bipyramidal), $H=6 \to sp^3d^2$ (octahedral).
4. Determine Lone Pairs and Final Shape — Calculate Lone Pairs ($LP$) = $H - (\text{number of actual bonded atoms})$. Use VSEPR geometry tables to find the resulting molecular shape (e.g., if $H=4, LP=1$, the shape is Trigonal Pyramidal, like $NH_3$).

Solved Exam-Style Examples

• {"title":"Example 1: Compare Dipole Moments of Ammonia ($NH_3$) and Nitrogen Trifluoride ($NF_3$)","problem":"Explain why the net dipole moment of $NH_3$ (1.47 D) is significantly higher than that of $NF_3$ (0.23 D), despite Fluorine being highly electronegative.","solution":"In $NH_3$, Nitrogen is more electronegative than Hydrogen. The three $N-H$ bond dipole vectors point inward towards Nitrogen. The lone pair orbital dipole also points upward away from the nucleus. Since all vectors point in the same general direction, they reinforce each other, resulting in a large net dipole. \nIn $NF_3$, Fluorine is more electronegative than Nitrogen. The three $N-F$ bond dipoles point downward towards Fluorine, directly opposing the upward-pointing orbital dipole of the lone pair. These opposing vectors partially cancel out, resulting in a very low net dipole moment."}
• {"title":"Example 2: Determining Hybridization and Shape of Sulfur Tetrafluoride ($SF_4$)","problem":"Predict the hybridization state, number of lone pairs, and exact molecular shape of $SF_4$.","solution":"1. Central atom is Sulfur ($S$), which has $V = 6$ valence electrons.\n2. Number of monovalent Fluorine atoms attached is $M = 4$. No charge ($C=0, A=0$).\n3. Steric Number $H = \\frac{1}{2}[6 + 4] = 5$. Thus, hybridization is $sp^3d$.\n4. Number of Lone Pairs ($LP$) = $5 - 4$ (surrounding atoms) = $1$ lone pair.\n5. With 5 orbital directions and 1 lone pair, the lone pair occupies an equatorial position to minimize repulsions. The final molecular shape is See-saw (or distorted tetrahedral)."}
• {"title":"Example 3: Molecular Orbital Theory Analysis of Oxygen Species ($O_2$ vs $O_2^+$)","problem":"Calculate the bond order and determine the magnetic behavior of $O_2$ and $O_2^+$ using Molecular Orbital configurations.","solution":"Total electrons in $O_2$ = 16. MO Configuration:\n$\\sigma 1s^2 \\ \\sigma^ 1s^2 \\ \\sigma 2s^2 \\ \\sigma^ 2s^2 \\ \\sigma 2p_z^2 \\ (\\pi 2p_x^2 = \\pi 2p_y^2) \\ (\\pi^ 2p_x^1 = \\pi^ 2p_y^1)$\nBonding electrons ($N_b$) = 10, Antibonding electrons ($N_a$) = 6.\nBond Order of $O_2$ = $\\frac{1}{2}(10 - 6) = 2.0$. It contains 2 unpaired electrons, making it paramagnetic.\n\nFor $O_2^+$, total electrons = 15. The electron is removed from the antibonding $\\pi^*$ orbital.\n$N_b$ = 10, $N_a$ = 5.\nBond Order of $O_2^+$ = $\\frac{1}{2}(10 - 5) = 2.5$. It contains 1 unpaired electron, making it paramagnetic."}

Exam Tips: Avoid Common Board Traps

1. The 14-Electron Rule Trap: When writing molecular orbital energy levels, remember that for $N_2$ and elements before it ($Z \le 7$), the $\sigma 2p_z$ orbital is placed higher in energy than the $\pi 2p_x = \pi 2p_y$ orbitals due to $s$-$p$ mixing. For $O_2$ and $F_2$, the $\sigma 2p_z$ orbital is lower in energy than the $\pi 2p$ orbitals.
2. Formal Charge Calculations: CBSE regularly asks for the formal charge on individual atoms in molecules like Ozone ($O_3$) or Carbonate ion ($CO_3^{2-}$). Use the formula:

$\text{Formal Charge} = [\text{Valence } e^-] - [\text{Non-bonding lone } e^-] - \frac{1}{2}[\text{Bonding shared } e^-]$ to verify your Lewis dot structures and secure full marks.

Practice Questions with Solutions

• Why is Carbon Dioxide ($CO_2$) linear, whereas Sulfur Dioxide ($SO_2$) is bent/angular? In $CO_2$, Carbon has no lone pairs and forms two double bonds (Steric number = 2), resulting in linear $sp$ hybridization with a $180^\circ$ bond angle. In $SO_2$, Sulfur has one lone pair and two double bonds (Steric number = 3), resulting in $sp^2$ hybridization. The lp-bp repulsion pushes the bonds down, resulting in a bent/angular structure.
• How does intermolecular hydrogen bonding affect the physical state of Water ($H_2O$) compared to Hydrogen Sulfide ($H_2S$)? Oxygen has a very high electronegativity and small atomic size, enabling water molecules to form extensive network-like intermolecular hydrogen bonds, making $H_2O$ a liquid at room temperature. Sulfur is much less electronegative and cannot form hydrogen bonds, meaning $H_2S$ molecules are held only by weak van der Waals forces, making it a gas.
• Calculate the formal charge of the central oxygen atom in Ozone ($O_3$). The central Oxygen atom in Ozone forms one single bond and one double bond, and holds 1 lone pair (2 non-bonding electrons). Formal Charge = $6 - 2 - \frac{1}{2}(6) = +1$.
• Explain why $He_2$ molecule does not exist in nature using MOT. Helium ($He$) has 2 electrons, so $He_2$ would have 4 electrons. The MO electronic configuration is $\sigma 1s^2 \ \sigma^* 1s^2$. Here, $N_b = 2$ and $N_a = 2$. Bond Order = $\frac{1}{2}(2 - 2) = 0$. Since the bond order is zero, the molecule is unstable and cannot exist.

Frequently Asked Questions