NEET Chemistry: Chemical Bonding & Molecular Structure – Complete Revision Guide, Formulas & Solved PYQs

Introduction to Chemical Bonding and Molecular Structure for NEET UG

Chemical Bonding and Molecular Structure is one of the most fundamental chapters in the Class 11 Chemistry syllabus (NCERT) and acts as the structural backbone for both Organic and Inorganic Chemistry. Every year, a significant portion of the NTA NEET UG exam tests students' conceptual clarity on how atoms interact, hybridize, form bonds, and adopt specific 3D geometries.

Understanding this chapter is essential not only for scoring full marks in physical/inorganic chemistry but also for mastering complex topics like Coordination Chemistry, Reaction Mechanisms, P-Block, and Biomolecules. Atoms combine to attain stability, lower their potential energy, and achieve a noble gas electronic configuration. Whether through ionic, covalent, or metallic bonding, the driving force remains electrostatic attraction and quantum mechanical stability. In this detailed guide, we break down core principles, mathematical formulas, molecular orbital diagrams, previous years' questions (PYQs), and practice MCQs to ensure you secure top marks in NEET UG.

NEET Weightage & Expected Questions

Chemical Bonding consistently holds high weightage in the NEET UG exam paper. It is considered a 'high-yield' topic because concepts learned here directly apply to multiple other chapters.

• Expected Number of Questions: 4 to 5 direct questions (yielding 16 to 20 marks).
• Indirect Application: 3 to 4 additional questions across Organic Reaction Mechanisms, Coordination Compounds, and Block Elements.
• Difficulty Level: Moderate to Conceptual. Questions are predominantly based on NCERT statements, structure determination, magnetic behavior, and bond order calculations.
• AIR Rank Impact: Scoring 100% accuracy in this chapter provides a massive boost to your All India Rank (AIR), as these questions are straightforward if key rules (VSEPR, MOT, Hybridization, Fajan's Rules) are thoroughly understood.

Core Concepts & Key Mechanisms Explained

1. Octet Rule and Formal Charge

Kossel and Lewis proposed that atoms combine to complete eight electrons in their valence shell (octet rule). However, exceptions exist: incomplete octet ($ \text{LiCl}, \text{BeH}_2, \text{BF}_3$), expanded octet ($ \text{PF}_5, \text{SF}_6, \text{H}_2 \text{SO}_4$), and odd-electron molecules ($ \text{NO}, \text{NO}_2$).

To evaluate the stability of Lewis structures, we calculate the Formal Charge (FC) of an atom in a polyatomic ion or molecule using the equation:

$$ \text{FC} = V - L - \frac{1}{2}B$$

Where $V$ = total valence electrons in free atom, $L$ = number of non-bonding (lone pair) electrons, and $B$ = number of bonding (shared) electrons.

2. VSEPR Theory (Valence Shell Electron Pair Repulsion)

Proposed by Gillespie and Nyholm, VSEPR theory predicts the shape of covalent molecules based on electron pair repulsions in the valence shell. The order of repulsive interactions is:

$$ \text{Lone Pair - Lone Pair (lp-lp)} > \text{Lone Pair - Bond Pair (lp-bp)} > \text{Bond Pair - Bond Pair (bp-bp)}$$

Key Molecular Shapes under VSEPR:

• $ \text{AB}_2 \text{E}_0$: Linear ($ \text{BeCl}_2, \text{CO}_2$), $180^ \text{o}$
• $ \text{AB}_3 \text{E}_0$: Trigonal Planar ($ \text{BF}_3$), $120^ \text{o}$
• $ \text{AB}_2 \text{E}_1$: Bent/V-shaped ($ \text{SO}_2, \text{O}_3$), $<120^ \text{o}$
• $ \text{AB}_4 \text{E}_0$: Tetrahedral ($ \text{CH}_4$), $109.5^ \text{o}$
• $ \text{AB}_3 \text{E}_1$: Trigonal Pyramidal ($ \text{NH}_3$), $107^ \text{o}$
• $ \text{AB}_2 \text{E}_2$: Bent/V-shaped ($ \text{H}_2 \text{O}$), $104.5^ \text{o}$
• $ \text{AB}_5 \text{E}_0$: Trigonal Bipyramidal ($ \text{PCl}_5$), $90^ \text{o} \text{ and } 120^ \text{o}$
• $ \text{AB}_4 \text{E}_1$: See-Saw ($ \text{SF}_4$)
• $ \text{AB}_3 \text{E}_2$: T-shaped ($ \text{ClF}_3$)
• $ \text{AB}_2 \text{E}_3$: Linear ($ \text{XeF}_2$)
• $ \text{AB}_6 \text{E}_0$: Octahedral ($ \text{SF}_6$), $90^ \text{o}$
• $ \text{AB}_5 \text{E}_1$: Square Pyramidal ($ \text{BrF}_5$)
• $ \text{AB}_4 \text{E}_2$: Square Planar ($ \text{XeF}_4$)

3. Valence Bond Theory (VBT) and Hybridization

VBT states that covalent bonds form via the overlap of atomic orbitals containing unpaired electrons. Overlap along the internuclear axis produces strong **sigma ($\boldsymbol{ \text{σ}}$) bonds**, while lateral/sideways overlap forms weaker **pi ($\boldsymbol{ \text{π}}$) bonds**.

Hybridization is the intermixing of atomic orbitals of slightly different energies to form an equal number of new hybridized orbitals of equivalent energy and shape.

Formula to calculate steric number ($Z$) for central atom hybridization:

$$Z = \frac{1}{2} \times \big[ V + M - C + A \big]$$

Where $V$ = valence electrons of central atom, $M$ = number of monovalent surrounding atoms ($ \text{H, F, Cl, Br, I}$), $C$ = cationic charge, and $A$ = anionic charge.

• $Z = 2 ightarrow sp$ (Linear)
• $Z = 3 ightarrow sp^2$ (Trigonal Planar)
• $Z = 4 ightarrow sp^3$ (Tetrahedral)
• $Z = 5 ightarrow sp^3d$ (Trigonal Bipyramidal)
• $Z = 6 ightarrow sp^3d^2$ (Octahedral)
• $Z = 7 ightarrow sp^3d^3$ (Pentagonal Bipyramidal)

4. Molecular Orbital Theory (MOT)

Developed by Hund and Mulliken, MOT explains paramagnetism and bond order accurately. Atomic orbitals combine linearly (LCAO) to form Bonding Molecular Orbitals (BMO - lower energy) and Antibonding Molecular Orbitals (ABMO - higher energy).

Electronic Configuration Sequence:

• For molecules with $\boldsymbol{ \text{total electrons } \text{N} \times 2 \textbf{ ≤ 14}}$ (e.g., $ \text{B}_2, \text{C}_2, \text{N}_2$):
  $ \text{σ}1s < \text{σ}^*1s < \text{σ}2s < \text{σ}^*2s < ( \text{π}2p_x = \text{π}2p_y) < \text{σ}2p_z < ( \text{π}^*2p_x = \text{π}^*2p_y) < \text{σ}^*2p_z$
• For molecules with $\boldsymbol{ \text{total electrons } > 14}$ (e.g., $ \text{O}_2, \text{F}_2$):
  $ \text{σ}1s < \text{σ}^*1s < \text{σ}2s < \text{σ}^*2s < \text{σ}2p_z < ( \text{π}2p_x = \text{π}2p_y) < ( \text{π}^*2p_x = \text{π}^*2p_y) < \text{σ}^*2p_z$

Key Formulae in MOT:

$$ \text{Bond Order (B.O.)} = \frac{N_b - N_a}{2}$$

Where $N_b$ = number of bonding electrons, and $N_a$ = number of antibonding electrons.

• Higher Bond Order $ ightarrow$ Higher Bond Dissociation Energy $ ightarrow$ Shorter Bond Length.
• If unpaired electrons exist $ ightarrow$ Paramagnetic (e.g., $ \text{O}_2, \text{B}_2$).
• If all electrons are paired $ ightarrow$ Diamagnetic (e.g., $ \text{N}_2, \text{C}_2$).

5. Dipole Moment and Hydrogen Bonding

Dipole moment ($\boldsymbol{ \text{μ}}$) measures bond polarity: $\boldsymbol{ \text{μ}} = q \times d$ (in Debye, D). Non-polar molecules have $\boldsymbol{ \text{μ}} = 0$ (e.g., $ \text{CO}_2, \text{BF}_3, \text{CCl}_4$), while polar molecules have $\boldsymbol{ \text{μ}} > 0$ (e.g., $ \text{H}_2 \text{O}, \text{NH}_3$). Note that $\boldsymbol{ \text{μ}}( \text{NH}_3) > \boldsymbol{ \text{μ}}( \text{NF}_3)$ due to directional addition of lone pair moments in $ \text{NH}_3$.

Hydrogen Bonding: Special attractive dipole force occurring when Hydrogen is covalently bonded to highly electronegative elements ($ \text{F}, \text{O}, \text{N}$).

• Intermolecular H-bonding: Between different molecules (e.g., $ \text{H}_2 \text{O}$, $ \text{HF}$, alcohol). Increases boiling point and solubility.
• Intramolecular H-bonding: Within the same molecule (e.g., o-nitrophenol). Decreases boiling point relative to inter-counterparts.

Important Formulas & Key Terms Table

Solved Step-by-Step Previous Years Questions (PYQs)

PYQ 1 (NEET 2023): Which of the following species is paramagnetic in nature?

Options: (A) $ \text{N}_2$    (B) $ \text{O}_2$    (C) $ \text{CN}^-$    (D) $ \text{CO}$

Solution:
Let's count the total electrons for each species:

• $ \text{N}_2$: $7 + 7 = 14$ electrons (All paired in MO diagram $ ightarrow$ Diamagnetic)
• $ \text{O}_2$: $8 + 8 = 16$ electrons. According to MOT configuration for $>14 e^-$:
  $ \text{σ}1s^2 \text{σ}^*1s^2 \text{σ}2s^2 \text{σ}^*2s^2 \text{σ}2p_z^2 ( \text{π}2p_x^2 = \text{π}2p_y^2) ( \text{π}^*2p_x^1 = \text{π}^*2p_y^1)$
  Since it contains two unpaired electrons in antibonding $ \text{π}^*2p$ orbitals, $ \text{O}_2$ is paramagnetic.
• $ \text{CN}^-$: $6 + 7 + 1 = 14$ electrons (Diamagnetic)
• $ \text{CO}$: $6 + 8 = 14$ electrons (Diamagnetic)

Correct Answer: (B) $ \text{O}_2$

PYQ 2 (NEET 2022): Match List-I with List-II for hybridization and shape of molecules:

• (a) $ \text{PCl}_5$      (i) $sp^3d^2$, Square planar
• (b) $ \text{SF}_6$       (ii) $sp^3d$, Trigonal bipyramidal
• (c) $ \text{BrF}_5$      (iii) $sp^3d^2$, Octahedral
• (d) $ \text{XeF}_4$      (iv) $sp^3d^2$, Square pyramidal

Solution:
Using steric number formula $Z = \frac{1}{2}[V + M]$:

• (a) $ \text{PCl}_5$: $Z = \frac{1}{2}[5 + 5] = 5 ightarrow sp^3d$, 0 lone pairs $ ightarrow$ Trigonal bipyramidal $ ightarrow$ (ii)
• (b) $ \text{SF}_6$: $Z = \frac{1}{2}[6 + 6] = 6 ightarrow sp^3d^2$, 0 lone pairs $ ightarrow$ Octahedral $ ightarrow$ (iii)
• (c) $ \text{BrF}_5$: $Z = \frac{1}{2}[7 + 5] = 6 ightarrow sp^3d^2$, 1 lone pair $ ightarrow$ Square pyramidal $ ightarrow$ (iv)
• (d) $ \text{XeF}_4$: $Z = \frac{1}{2}[8 + 4] = 6 ightarrow sp^3d^2$, 2 lone pairs $ ightarrow$ Square planar $ ightarrow$ (i)

Correct Matching: (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

PYQ 3 (NEET 2020): Identify the molecule which does NOT exist based on Molecular Orbital Theory:

Options: (A) $ \text{He}_2$    (B) $ \text{Li}_2$    (C) $ \text{C}_2$    (D) $ \text{O}_2$

Solution:
A diatomic molecule cannot exist if its Bond Order ($ \text{B.O.}$) is zero ($0$).
For $ \text{He}_2$: Total electrons = $2 + 2 = 4$.
Electronic configuration: $ \text{σ}1s^2 \text{σ}^*1s^2$
Number of bonding electrons ($N_b$) = 2, Antibonding electrons ($N_a$) = 2.
$$ \text{B.O.} = \frac{N_b - N_a}{2} = \frac{2 - 2}{2} = 0$$
Since the bond order is 0, $ \text{He}_2$ has no net binding energy and does not exist in nature.

Correct Answer: (A) $ \text{He}_2$

Common NEET Traps & Mistakes to Avoid

• Equatorial vs Axial Bonds in $ \text{PCl}_5$: Remember that axial bonds ($2$ bonds) experience more repulsive forces from equatorial bonds ($3$ bonds) and are longer and weaker than equatorial bonds. Therefore, $ \text{PCl}_5$ readily dissociates into $ \text{PCl}_3 + \text{Cl}_2$.
• Dipole Moment Comparison of $ \text{NH}_3$ and $ \text{NF}_3$: Even though Fluorine is more electronegative than Nitrogen, $ \text{NH}_3$ has a higher dipole moment ($1.47 \text{ D}$) than $ \text{NF}_3$ ($0.23 \text{ D}$). In $ \text{NH}_3$, the orbital dipole due to the lone pair reinforces the resultant dipole of $ \text{N-H}$ bonds. In $ \text{NF}_3$, the lone pair dipole opposes the resultant dipole of $ \text{N-F}$ bonds.
• $ \text{C}_2$ Molecule Bonding: In $ \text{C}_2$, according to MOT, both of the bonds formed are **pi ($\boldsymbol{ \text{π}}$) bonds** because all four valence electrons occupy the degenerate $ \text{π}2p_x$ and $ \text{π}2p_y$ molecular orbitals. This is a favorite trick question in NEET.
• Formal Charge vs Oxidation Number: Do not confuse formal charge with oxidation state. Formal charge assumes equal sharing of covalent electrons, whereas oxidation state assigns electrons to the more electronegative atom.
• O-nitrophenol vs P-nitrophenol: Intramolecular H-bonding makes o-nitrophenol steam-volatile with a lower boiling point, whereas intermolecular H-bonding in p-nitrophenol increases its boiling point.

High-Yield NEET Practice MCQs with Answer Keys

Q1. Which of the following species possesses maximum bond order?

(A) $ \text{O}_2$
(B) $ \text{O}_2^+$
(C) $ \text{O}_2^-$
(D) $ \text{O}_2^{2-}$

Detailed Solution:
Let's calculate the bond order for each oxygen species using MOT rules:

• $ \text{O}_2$ ($16 e^-$): $N_b = 10, N_a = 6 ightarrow \text{B.O.} = \frac{10 - 6}{2} = 2.0$
• $ \text{O}_2^+$ ($15 e^-$): $N_b = 10, N_a = 5 ightarrow \text{B.O.} = \frac{10 - 5}{2} = 2.5$
• $ \text{O}_2^-$ ($17 e^-$): $N_b = 10, N_a = 7 ightarrow \text{B.O.} = \frac{10 - 7}{2} = 1.5$
• $ \text{O}_2^{2-}$ ($18 e^-$): $N_b = 10, N_a = 8 ightarrow \text{B.O.} = \frac{10 - 8}{2} = 1.0$

Hence, $ \text{O}_2^+$ has the maximum bond order ($2.5$) and the strongest, shortest bond.

Answer: (B)

Q2. Among $ \text{SF}_4, \text{BF}_3, \text{ClF}_3,$ and $ \text{XeF}_4$, the molecules with T-shape and See-saw shape respectively are:

(A) $ \text{ClF}_3$ and $ \text{SF}_4$
(B) $ \text{SF}_4$ and $ \text{ClF}_3$
(C) $ \text{BF}_3$ and $ \text{XeF}_4$
(D) $ \text{ClF}_3$ and $ \text{XeF}_4$

Detailed Solution:
Using VSEPR theory:

• $ \text{ClF}_3$: Steric number = $\frac{1}{2}[7 + 3] = 5$ ($sp^3d$). Contains $3$ bond pairs and $2$ lone pairs $ ightarrow$ T-shaped.
• $ \text{SF}_4$: Steric number = $\frac{1}{2}[6 + 4] = 5$ ($sp^3d$). Contains $4$ bond pairs and $1$ lone pair $ ightarrow$ See-saw shaped.

Therefore, T-shape is $ \text{ClF}_3$ and See-saw is $ \text{SF}_4$.

Answer: (A)

Q3. Which of the following compounds displays the highest covalent character according to Fajan's Rules?

(A) $ \text{LiF}$
(B) $ \text{LiCl}$
(C) $ \text{LiBr}$
(D) $ \text{LiI}$

Detailed Solution:
According to Fajan's rules, covalent character increases with:

• Smaller size of cation (Here, cation $ \text{Li}^+$ is constant).
• Larger size of anion ($ \text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$).
• Higher charge on either ion.

Since Iodide ($ \text{I}^-$) is the largest anion, it undergoes maximum polarization by $ \text{Li}^+$, imparting the highest covalent character to $ \text{LiI}$.

Answer: (D)

Q4. Hybridization state of central Xenon atom in $ \text{XeF}_2$ and its geometry are:

(A) $sp^3d$, Linear
(B) $sp^3$, Bent
(C) $sp^3d^2$, Linear
(D) $sp^3$, Linear

Detailed Solution:
Steric number $Z = \frac{1}{2}[V + M] = \frac{1}{2}[8 + 2] = 5$.
$Z = 5$ corresponds to $sp^3d$ hybridization.
Xenon forms $2$ bond pairs with Fluorine and retains $3$ lone pairs ($5 - 2 = 3$).
To minimize lone pair repulsions, the 3 lone pairs occupy equatorial positions at $120^ \text{o}$, making the overall molecular geometry Linear.

Answer: (A)

Q5. The molecular orbital with two nodal planes passing through the internuclear axis is:

(A) $ \text{σ}2p_z$
(B) $ \text{π}2p_x$
(C) $ \text{π}^*2p_x$
(D) $ \text{σ}^*2s$

Detailed Solution:
An antibonding $ \text{π}^*$ orbital formed by side-ways overlap of $p_x$ or $p_y$ atomic orbitals contains two nodal planes: one containing the internuclear axis and another perpendicular to the internuclear axis between the nuclei. Thus, $ \text{π}^*2p_x$ possesses two nodal planes.

Answer: (C)

Summary & Final NEET Revision Tips

• Master the Steric Number Formula: Save time during the exam by using $Z = \frac{1}{2}[V + M - C + A]$ directly instead of drawing complete Lewis structures.
• Remember MOT Electron Tricks: For $14$ electrons, Bond Order is $3.0$. For every electron added or removed from $14$, reduce the bond order by $0.5$ (e.g., $13 e^- ightarrow 2.5$, $15 e^- ightarrow 2.5$, $16 e^- ightarrow 2.0$).
• Paramagnetic Exception Shortcuts: Molecules/ions with $10$ electrons ($ \text{B}_2$) and $16$ electrons ($ \text{O}_2, \text{S}_2$) are **paramagnetic** despite having an even number of total electrons.
• NCERT Line-by-Line: Re-read NCERT tables on molecular geometry and hydrogen bonding carefully. Direct factual questions are frequently framed from these sections.
• Focus on Hybridization of Interhalogens & Noble Gas Compounds: Compounds like $ \text{XeF}_2, \text{XeF}_4, \text{XeOF}_4, \text{ClF}_3, \text{IF}_7$ appear almost every alternate year in NEET UG papers!