A chemical bond is formed to lower the potential energy of the system and gain stability. Atoms combine to acquire the stable electronic configuration of the nearest noble gas (octet/duplet).
This occurs via:

• Transfer of electrons: One atom loses and another gains electrons (Ionic Bond).
• Sharing of electrons: Atoms share electron pairs (Covalent Bond).

Lewis symbols represent valence electrons as dots around the atomic symbol.

• Mg (Group 2): $\cdot Mg \cdot$ (2 valence electrons)
• Na (Group 1): $Na \cdot$ (1 valence electron)
• B (Group 13): $\cdot \dot{B} \cdot$ (3 valence electrons)
• O (Group 16): $: \ddot{O} \cdot$ (6 valence electrons)
• N (Group 15): $: \dot{N} \cdot$ (5 valence electrons)
• Br (Group 17): $: \ddot{Br} \cdot$ (7 valence electrons)

• S and $S^{2-}$:
  S (6 val e⁻): $: \ddot{S} \cdot$
  $S^{2-}$ (Gain 2e⁻, Octet complete): $[: \ddot{S} :]^{2-}$
• Al and $Al^{3+}$:
  Al (3 val e⁻): $\cdot \dot{Al} \cdot$
  $Al^{3+}$ (Lose 3e⁻): $[Al]^{3+}$
• H and $H^-$:
  H (1 val e⁻): $H \cdot$
  $H^-$ (Gain 1e⁻, Duplet): $[H:]^{-}$

Definition: Atoms tend to bond in such a way that they have eight electrons in their valence shell, achieving the stable electronic configuration of a noble gas.

Significance: It explains the chemical inertness of noble gases and the valency/bonding of many main group elements.

Limitations:

1. Incomplete Octet: Central atom has less than 8 e⁻ (e.g., $LiCl, BeH_2, BCl_3$).
2. Expanded Octet: Central atom has more than 8 e⁻ (e.g., $PF_5, SF_6$).
3. Odd-electron molecules: Molecules with odd number of total valence electrons (e.g., $NO, NO_2$).
4. It does not account for the shape or energy of molecules.

1. Low Ionization Enthalpy: Of the metal atom (cation), making it easy to lose electrons.
2. High Negative Electron Gain Enthalpy: Of the non-metal atom (anion), making it energetically favorable to gain electrons.
3. High Lattice Enthalpy: A large amount of energy released when the cation and anion combine to form the crystal lattice, stabilizing the compound.

• $BeCl_2$: $AB_2$ type. 2 bond pairs, 0 lone pairs. Linear.
• $BCl_3$: $AB_3$ type. 3 bp, 0 lp. Trigonal Planar.
• $SiCl_4$: $AB_4$ type. 4 bp, 0 lp. Tetrahedral.
• $AsF_5$: $AB_5$ type. 5 bp, 0 lp. Trigonal Bipyramidal.
• $H_2S$: $AB_2E_2$ type. 2 bp, 2 lp (on S). Bent / V-shaped.
• $PH_3$: $AB_3E$ type. 3 bp, 1 lp (on P). Trigonal Pyramidal.

Both have tetrahedral electron geometry ($sp^3$).

• Ammonia ($NH_3$): 3 bond pairs, 1 lone pair. Repulsion: $lp-bp > bp-bp$. Angle reduces from $109.5^\circ$ to $107^\circ$.
• Water ($H_2O$): 2 bond pairs, 2 lone pairs. Repulsion: $lp-lp > lp-bp$. The strong repulsion between the two lone pairs squeezes the bond angle further to $104.5^\circ$.

Bond Order is the number of bonds between two atoms.
Relationship: Bond Strength $\propto$ Bond Order.

As bond order increases (Single $\to$ Double $\to$ Triple), the bond enthalpy increases, meaning the bond becomes stronger and harder to break.

Bond Length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

In $CO_3^{2-}$, experimental data shows all three C-O bonds are of equal length (intermediate between single and double bond).
A single Lewis structure cannot explain this. Resonance assumes the ion is a hybrid of three equivalent structures where the double bond character is delocalized over all three oxygen atoms. This stabilizes the molecule and equalizes bond lengths.

No.
Resonance structures must involve only the movement of electrons (pi bonds/lone pairs). The position of nuclei (atoms) must remain fixed.
In the two structures of $H_3PO_3$ (usually represented as P-H vs P-OH forms), the position of the Hydrogen atom changes. This is an example of Tautomerism or structural isomerism, not resonance.

(a) K and S:
$K \cdot + \cdot \ddot{S} \cdot + \cdot K \rightarrow 2[K]^+ + [: \ddot{S} :]^{2-}$ (Formation of $K_2S$).

(b) Ca and O:
$\cdot Ca \cdot + \cdot \ddot{O} \cdot \rightarrow [Ca]^{2+} + [: \ddot{O} :]^{2-}$ (Formation of $CaO$).

(c) Al and N:
$\cdot \dot{Al} \cdot + : \dot{N} \cdot \rightarrow [Al]^{3+} + [: \ddot{N} :]^{3-}$ (Formation of $AlN$).

$CO_2$: Linear structure ($O=C=O$). The C-O bond dipoles are equal and opposite ($180^\circ$). They cancel each other out perfectly. Net Dipole $\mu = 0$.

$H_2O$: Bent structure (V-shape). The O-H bond dipoles and lone pair moments do not cancel; they add up vectorially to give a resultant dipole moment. Thus $\mu \neq 0$.

1. Predicting Molecular Geometry: (e.g., $\mu=0$ implies symmetry like linear/tetrahedral).
2. Determining Polarity: Higher dipole moment means higher polarity.
3. Calculating % Ionic Character: $\% \text{ Ionic} = \frac{\mu_{observed}}{\mu_{ionic}} \times 100$.
4. Distinguishing Isomers: Cis-isomers usually have higher dipole moments than Trans-isomers.

Electronegativity: Tendency of an atom in a bonded state (molecule) to attract shared electrons. It is a relative value, unitless.

Electron Gain Enthalpy: Energy released when an isolated gaseous atom accepts an electron. It is an absolute energy measurement, units kJ/mol.

A Polar Covalent Bond forms when electrons are shared between two atoms with different electronegativities. The electron pair shifts towards the more electronegative atom.

Example: HCl
Chlorine is more electronegative than Hydrogen. The electron cloud is pulled towards Cl.
Result: H gets partial positive charge ($\delta+$) and Cl gets partial negative charge ($\delta-$).
$H^{\delta+} – Cl^{\delta-}$.

Ionic character depends on the electronegativity difference ($\Delta EN$) between bonded atoms. Higher $\Delta EN$ = More Ionic.

Comparing molecules: $LiF, K_2O, N_2, SO_2, ClF_3$.

• $N_2$: Pure covalent ($\Delta EN = 0$).
• $SO_2, ClF_3$: Polar Covalent.
• $K_2O$: Ionic (Metal + Non-metal).
• $LiF$: Most Ionic (Most electropositive Li + Most electronegative F).

Order: $N_2 < SO_2 < ClF_3 < K_2O < LiF$.

Skeletal Structure issue: Usually, the acidic H is attached to Oxygen, not Carbon. (Though in acetic acid, methyl H are on C, acidic H is on O).
Formula: $CH_3COOH$.

According to VSEPR Theory, electron pairs around a central atom arrange themselves to be as far apart as possible to minimize repulsion.

• Square Planar: The H-C-H bond angle would be $90^\circ$.
• Tetrahedral: The H-C-H bond angle is $109.5^\circ$.

Since $109.5^\circ > 90^\circ$, the tetrahedral arrangement reduces electron-pair repulsion more effectively, making it the stable geometry.

Although the Be–H bond is polar (electronegativity difference exists), the molecule $BeH_2$ is Linear ($H-Be-H$, sp hybridisation).
The two individual bond dipole moments are equal in magnitude but opposite in direction ($180^\circ$). They cancel each other out vectorially, resulting in a net dipole moment of zero ($\mu = 0$).

Answer: $NH_3$ has a higher dipole moment.

• $NH_3$: Nitrogen is more electronegative than Hydrogen. Bond moments point towards N. The lone pair moment also points upward (away from N). All vectors reinforce each other ($\uparrow \uparrow \uparrow + \uparrow_{lp}$).
• $NF_3$: Fluorine is more electronegative than Nitrogen. Bond moments point towards F (downward). The lone pair moment points upward. The vectors oppose each other ($\downarrow \downarrow \downarrow + \uparrow_{lp}$), reducing the net dipole moment.

Hybridisation: The process of intermixing of atomic orbitals of slightly different energies to redistribute their energy and form a new set of orbitals of equivalent energy and shape.

• $sp$: Mixing 1s + 1p. Linear shape ($180^\circ$). Example: $BeCl_2$.
• $sp^2$: Mixing 1s + 2p. Trigonal Planar shape ($120^\circ$). Example: $BCl_3$.
• $sp^3$: Mixing 1s + 3p. Tetrahedral shape ($109.5^\circ$). Example: $CH_4$.

Reaction: $AlCl_3 + Cl^- \rightarrow [AlCl_4]^-$

• $AlCl_3$: Al has 3 bond pairs, 0 lone pairs. Hybridisation: $sp^2$ (Trigonal Planar).
• $[AlCl_4]^-$: Al forms 4 bonds (receiving a pair from $Cl^-$). 4 bond pairs, 0 lone pairs. Hybridisation: $sp^3$ (Tetrahedral).

Change: $sp^2 \rightarrow sp^3$.

Reaction: $BF_3 + NH_3 \rightarrow F_3B \leftarrow NH_3$

• Boron ($B$): In $BF_3$, it is $sp^2$ (3 bonds). In adduct, it forms 4 bonds. Change: $sp^2 \rightarrow sp^3$.
• Nitrogen ($N$): In $NH_3$, it is $sp^3$ (3 bonds + 1 lone pair). In adduct, lone pair becomes a bond pair. Still 4 domains. Change: Remains $sp^3$.

$C_2H_4$ (Ethene – Double Bond):
Carbon is $sp^2$ hybridised. One $sp^2-sp^2$ overlap forms a $\sigma$ bond. The unhybridised $p$-orbitals overlap sideways to form one $\pi$ bond. Total: 1 $\sigma$ + 1 $\pi$.

$C_2H_2$ (Ethyne – Triple Bond):
Carbon is $sp$ hybridised. One $sp-sp$ overlap forms a $\sigma$ bond. Two unhybridised $p$-orbitals on each carbon overlap sideways to form two $\pi$ bonds. Total: 1 $\sigma$ + 2 $\pi$.

(a) $C_2H_2$ ($H-C \equiv C-H$):
Single bonds (C-H): 2 ($\sigma$).
Triple bond (C-C): 1 ($\sigma$) + 2 ($\pi$).
Total: 3 $\sigma$, 2 $\pi$.

(b) $C_2H_4$ ($H_2C=CH_2$):
Single bonds (C-H): 4 ($\sigma$).
Double bond (C-C): 1 ($\sigma$) + 1 ($\pi$).
Total: 5 $\sigma$, 1 $\pi$.

Internuclear axis = x-axis.

• (a) $1s + 1s$: Spherical, forms $\sigma$.
• (b) $1s + 2p_x$: S + P along axis forms $\sigma$.
• (c) $2p_y + 2p_y$: These are perpendicular to the x-axis. They overlap sideways. Forms $\pi$ bond, not $\sigma$.
• (d) $1s + 2s$: Spherical, forms $\sigma$.

Answer: (c) because sideways overlap leads to pi bond formation.

Rule: 4 sigma = $sp^3$, 3 sigma = $sp^2$, 2 sigma = $sp$.

• (a) $CH_3-CH_3$: Both $sp^3$.
• (b) $CH_3-CH=CH_2$: $sp^3 – sp^2 – sp^2$.
• (c) $CH_3-CH_2-OH$: Both C are $sp^3$.
• (d) $CH_3-CHO$: Methyl C is $sp^3$, Carbonyl C ($C=O$) is $sp^2$.
• (e) $CH_3COOH$: Methyl C is $sp^3$, Carboxyl C is $sp^2$.

Bond Pair: A pair of valence electrons shared between two atoms to form a covalent bond. Example: The electrons in the C-H bond of Methane.

Lone Pair: A pair of valence electrons that is not shared with another atom. Example: The electron pair on Nitrogen in Ammonia ($NH_3$).

When two Hydrogen atoms approach each other, two forces operate:

1. Attractive forces: Between nucleus of one atom and electron of the other.
2. Repulsive forces: Between the two nuclei and the two electrons.

Initially, attraction dominates, lowering the system’s potential energy. At a critical distance (bond length, 74 pm), the energy is minimum (-435.8 kJ/mol), and a stable bond is formed. If brought closer, repulsion dominates, increasing energy sharply.

Linear Combination of Atomic Orbitals (LCAO) requires:

1. Same Energy: Combining atomic orbitals must have comparable energies (e.g., 1s with 1s).
2. Same Symmetry: Orbitals must have the same symmetry about the molecular axis (e.g., $2p_z$ with $2p_z$).
3. Maximum Overlap: The extent of overlap should be large for effective bond formation.

Be ($Z=4$). Electronic config: $1s^2 2s^2$. Total electrons in $Be_2 = 8$.
MO Configuration: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$.

Since Bond Order is zero, the molecule is unstable and does not exist.

Base $O_2$ (16e): BO = 2.0 (Paramagnetic, 2 unpaired).

• $O_2^-$ (Superoxide, 17e): Adds e to $\pi^*$. BO = 1.5. Paramagnetic (1 unpaired).
• $O_2^{2-}$ (Peroxide, 18e): Adds 2e to $\pi^*$. BO = 1.0. Diamagnetic (0 unpaired).

Stability Order (proportional to BO):
$O_2 (2.0) > O_2^- (1.5) > O_2^{2-} (1.0)$.

The plus (+) and minus (-) signs represent the phase of the wave function ($\Psi$) of the orbital. They do not represent electric charge.
– Constructive Interference (Bonding): Overlap of same signs (+ with +).
– Destructive Interference (Antibonding): Overlap of opposite signs (+ with -).

Hybridisation: P (Group 15) forms 5 sigma bonds. Requires 5 orbitals. $s + p + p + p + d = \mathbf{sp^3d}$. Shape: Trigonal Bipyramidal.

Axial vs Equatorial:
– 3 Equatorial bonds are at $120^\circ$.
– 2 Axial bonds are at $90^\circ$ to the equatorial plane.
The axial bond pairs suffer greater repulsion from the three equatorial bond pairs (at $90^\circ$) compared to the repulsion between equatorial bonds. To minimize this repulsion, axial bonds elongate. Thus, Axial > Equatorial.

Definition: The attractive force which binds the hydrogen atom of one molecule with the electronegative atom (F, O, N) of another molecule (intermolecular) or within the same molecule (intramolecular).

Comparison: Hydrogen bond is stronger than van der Waals forces (dipole-dipole or dispersion) because it involves specific interaction with hydrogen attached to high electronegative atoms, but it is weaker than covalent or ionic bonds.

Bond Order: Half the difference between the number of electrons in bonding and antibonding molecular orbitals.

• $N_2$ (14e): $KK (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p)^4 (\sigma 2p)^2$.
  $N_b=10, N_a=4$. BO = $(10-4)/2 = \mathbf{3}$.
• $O_2$ (16e): Adds 2e to $\pi^*$.
  $N_b=10, N_a=6$. BO = $(10-6)/2 = \mathbf{2}$.
• $O_2^+$ (15e): Loses 1e from $\pi^*$.
  $N_b=10, N_a=5$. BO = $(10-5)/2 = \mathbf{2.5}$.
• $O_2^-$ (17e): Gains 1e to $\pi^*$.
  $N_b=10, N_a=7$. BO = $(10-7)/2 = \mathbf{1.5}$.