100k Followers100k500k Followers500k+1 (510) 706-9331+1 (510) 706-9331
Schedule Your Free Exam Readiness Analysis Session!
Testprepkart Logo
Sign InEnroll NowEnroll
Select an exam to view its content.
  • Blog
  • Download
  • Course
  • Result
  • Video Library
  • Pages
  • Notifications

Loading...

Preparing content

Testprepkart Logo

Enabling students prepare and crack toughest examinations worldwide for over a decade with problem solving aptitude!

Contact Us

Useful Links

  • Connect With Counselor
  • University Admissions
  • Prime Videos
  • Enrollment Form
  • Online Fee Payment
  • Testprepkart Operations
  • Faculty Registration

Our Company

  • Contact Us
  • Work With Us
  • Blogs
  • Facultie
  • Partner

Contact Details

  • Phone: +91 0120 4525484
  • Whatsapp: +1 (510) 706-9331
  • Admission: +91 8800123492
  • E-mail: info@testprepkart.com
  • Head Office: F 377, Sector 63, Noida, Uttar Pradesh, India

Copyright Ā© 2024 CounselKart Educational Services Pvt. Ltd.. All Rights Reserved

Terms of service|Privacy policy|Refund Policy|Login & Register

Chemical Bonding

NEET > Chemistry > Chemical Bonding And Molecular Structure

Unit Progress

0%

Overview content

Chapter Snapshot - Chemical Bonding

A conceptually dense and high-weightage chapter covering the forces that hold atoms together in molecules and crystals. Ionic bond, covalent bond (sigma and pi), coordinate bond, metallic bond, hybridisation (sp, sp2, sp3, sp3d, sp3d2), VSEPR theory for molecular geometry, Molecular Orbital Theory (bond order and magnetic behaviour), Fajan's rule, dipole moment, and hydrogen bonding form the twelve pillars of NEET questions. The chapter demands both conceptual clarity on bonding models and quick recall of molecular shapes, bond angles, and MO configurations. Most NEET marks are lost from confusing lone pair effects on geometry, misapplying the N/2 hybridisation shortcut, or forgetting the MO energy order switch at O2.

āœ“ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
3-4
NEET consistently draws 3 to 4 questions from Chemical Bonding. One on hybridisation and molecular shape (VSEPR prediction), one on Molecular Orbital Theory (bond order or magnetic character), one on hydrogen bonding or dipole moment, and occasionally one on Fajan's rule or coordinate bond formation.
Time Required (Practical)
ā±
12-15 hrs
Lewis structures and octet rule 1 hr; ionic, covalent, coordinate bonds 2 hrs; dipole moment 1 hr; Fajan's rule 1 hr; hybridisation types with examples 2 hrs; VSEPR theory with all molecular shapes 2 hrs; MOT with bond order and configurations 2.5 hrs; hydrogen bonding 1.5 hrs; MCQ practice 2 hrs.
Difficulty Level
⚔
Moderate-Hard
This chapter bridges quantum concepts with structural chemistry. Hybridisation and VSEPR require spatial reasoning. MOT demands careful electron counting and knowledge of which energy order applies. The conceptual density is high but the number of formulas is low, making it a chapter where understanding beats rote learning.
Most Asked Style: Conceptual MCQ: predict molecular shape from hybridisation or VSEPR; determine bond order and magnetic nature from MO configuration; identify correct hybridisation of central atom in a given molecule or ion; rank compounds by dipole moment or ionic/covalent character using Fajan's rule.Biggest Trap: Using the wrong MO energy order. For molecules up to N2, the pi2p orbitals are lower in energy than sigma2px. For O2 and beyond, sigma2px is lower than pi2p. Applying the wrong order gives incorrect bond order and wrong magnetic character prediction. NEET distractors exploit this switch point.Fast Win: Memorise the N/2 shortcut for hybridisation: N = (valence electrons of central atom + number of bonded atoms plus or minus charge) / 2. N/2 = 2 gives sp (linear), 3 gives sp2 (trigonal planar), 4 gives sp3 (tetrahedral), 5 gives sp3d (trigonal bipyramidal), 6 gives sp3d2 (octahedral). Then adjust shape based on lone pairs. This single shortcut solves 80% of VSEPR/hybridisation questions.Revision-Friendly: Yes. Build one reference card with: hybridisation table (N/2 vs shape), MO energy orders (two sequences: up to N2 and O2 onwards), Fajan's rule four factors, hydrogen bond strength order F-H-F > O-H-O > N-H-N. A 30-minute sweep of this card plus two worked MOT problems covers the full scoring range.

Subtopics - Chemical Bonding (NEET)

Nine interconnected topic blocks: Lewis theory and octet rule, ionic/covalent/coordinate/metallic bond types, dipole moment as a measure of bond polarity, physical properties comparison of ionic and covalent compounds, Fajan's rule for covalent character prediction, five types of hybridisation with geometry, VSEPR theory for molecular shape prediction, Molecular Orbital Theory with bond order and magnetism, and hydrogen bonding with intermolecular and intramolecular types.

Revision tip: Before answering any molecular geometry question: (1) count total electron pairs around central atom using N/2 shortcut, (2) identify hybridisation from the table, (3) subtract lone pairs from total to get bonding pairs, (4) name the shape from the VSEPR chart. This four-step protocol eliminates the most common NEET errors in this chapter.
NCERT LinesMCQsQuick Test

1) Lewis Theory

Explains covalent bonding through shared electron pairs. The octet rule states that atoms form bonds until surrounded by eight electrons (hydrogen needs only two). Lewis dot structures use dashes for bonds and dots for lone pairs. A systematic method determines bonding and non-bonding electrons from total valence electron count.

Octet ruleLewis dot structuresFormal charge calculationLone pairs as dots
›
Octet Rule and Lewis Dot StructuresAtoms continue to form bonds until they achieve an octet of eight electrons, made up from electrons totally owned and shared. Lewis structures are drawn using a five-step method: count total valence electrons (n1), calculate bonding electrons (n3 = n2 minus n1), find non-bonding electrons (n4 = n1 minus n3), arrange bonds around the central atom, and assign formal charge as (valence electrons) minus (bonds) minus (unshared electrons). Hydrogen is the primary exception, stable with only two electrons.

2) Types of Chemical Bonds

Covers four fundamental bond types: ionic bond (electron transfer between electropositive and electronegative atoms), covalent bond (electron sharing with sigma and pi overlap), coordinate bond (both shared electrons from one donor atom), and metallic bond (delocalised electron sea model). Each bond type is linked to element electronegativity character.

Ionic: electron transferCovalent: sigma + pi overlapCoordinate: donor-acceptorMetallic: electron sea
›
Ionic BondFormed by complete transfer of electrons from an electropositive to an electronegative atom. Both ions attain stable noble gas configuration. The electrostatic attraction between cation and anion holds the crystal lattice together. Example: Na loses one electron to Cl forming Na-plus and Cl-minus in NaCl. Ionic bond strength depends on charge magnitude and ionic radii.
›
Covalent BondFormed when two electronegative atoms share electron pairs to achieve noble gas configuration. Sigma bonds form by head-on overlap (s-s in H2, s-p in HF, p-p in F2). Pi bonds form by lateral overlap of p-orbitals perpendicular to the internuclear axis. Double bonds consist of one sigma plus one pi bond (O2). Triple bonds consist of one sigma plus two pi bonds (N2). Polar covalent bonds arise when bonded atoms differ in electronegativity, creating partial charges delta-plus and delta-minus.
›
Coordinate Bond (Dative Bond)A covalent bond where both shared electrons are donated by a single atom (the donor) to another atom (the acceptor). Once formed, the coordinate bond is identical to a normal covalent bond. Represented by an arrow from donor to acceptor. Examples: NH3 donates a lone pair to H-plus forming NH4-plus; NH3 donates to BF3 enabling boron to achieve an octet.
›
Metallic BondThe bond holding metal atoms together in a lattice. Cannot be ionic (no electronegativity difference between identical atoms), covalent (Li has only one valence electron, cannot form eight bonds), or van der Waals (metals are too strong). Involves delocalised valence electrons mobile throughout the lattice, forming an electron sea. Over 80 elements are metals; all are solids at room temperature except Hg and Ga.

3) Dipole Moment

Measures the degree of polarity in a molecule. Dipole moment mu = charge (e) times distance (d), expressed in Debye (D) where 1D = 10 to the minus 18 esu-cm. It is a vector quantity and the net dipole moment is the vector sum of individual bond moments. Zero net dipole indicates molecular symmetry.

mu = e times dUnit: Debye (D)Vector quantityZero mu means symmetric
›
Dipole Moment Definition and MeasurementDipole moment (mu) is defined as the product of net positive or negative charge and the distance between the two charged ends (bond length). Measured in Debye: 1D = 10 to the minus 18 esu-cm = 3.33 times 10 to the minus 30 C-m. In H-X bonds hydrogen is the positive end. In C-X bonds (X is not H or C) carbon is the positive end. Dipole moment is a vector: net dipole is the vector sum of all bond dipoles. A molecule with zero dipole moment is symmetric (e.g., CO2, BF3, CCl4).

4) Physical Properties of Ionic and Covalent Compounds

Compares the physical behaviour of ionic and covalent compounds across three properties: melting point (ionic compounds have high MP due to strong lattice forces, covalent compounds have low MP due to weak van der Waals forces), conductivity (ionic compounds conduct when molten or dissolved, covalent compounds do not), and solubility (like dissolves like principle with dielectric constant).

Ionic: high MP, conduct when moltenCovalent: low MP, non-conductorIonic: polar solvent solubleCovalent: non-polar solvent soluble
›
Melting PointIonic compounds are solids with high melting and boiling points because breaking the lattice of non-directional electrostatic attractions requires large energy. Covalent compounds are gases, liquids, or low-melting solids because they consist of discrete molecules held by weak van der Waals forces that need little energy to overcome.
›
ConductivityIonic compounds conduct electricity when melted or dissolved in water because ions migrate toward electrodes under electric potential. In solid state ions are fixed in the lattice and cannot conduct. Covalent compounds contain no ions or mobile electrons, so they do not conduct in any state.
›
SolubilityIonic compounds dissolve in polar solvents with high dielectric constants such as water. The high dielectric constant reduces electrostatic attraction between ions, allowing solvation. Covalent compounds dissolve in non-polar solvents with low dielectric constants such as benzene or CCl4, following the like-dissolves-like principle.

5) Fajan's Rule

Explains why ionic bonds develop partial covalent character. When a cation polarises the electron cloud of an anion, the bond gains covalent character. Four factors favour covalent character: small cation (high polarising power), large anion (high polarisability), high charge on either ion, and cation with pseudo noble gas configuration (18 electrons in outermost shell).

Small cation = more covalentLarge anion = more covalentHigh charge = more covalentPseudo noble gas config > noble gas config
›
Factors Favouring Covalent Character in Ionic CompoundsFajan's rule: covalent character increases with (1) smaller cation size (higher charge density, greater polarising power; LiCl more covalent than KCl), (2) larger anion size (electrons loosely held, more easily distorted; iodides are most covalent among halides), (3) higher charge on either ion (NaCl < MgCl2 < AlCl3), (4) pseudo noble gas configuration of cation with 18 outer electrons is more polarising than noble gas configuration with 8 electrons at same size and charge (CuCl more covalent than NaCl despite similar ionic radii). Greater polarisation lowers melting point and increases solubility in non-polar solvents.

6) Hybridisation

The intermixing of atomic orbitals of the same atom having similar energies, followed by redistribution to form new hybrid orbitals of identical energy and shape. Five types: sp3 (tetrahedral, 109.5 degrees), sp2 (trigonal planar, 120 degrees), sp (linear, 180 degrees), sp3d (trigonal bipyramidal), sp3d2 (octahedral). The number of hybrid orbitals equals the number of atomic orbitals mixed.

sp3: tetrahedral 109.5 degreessp2: trigonal planar 120 degreessp: linear 180 degreessp3d: trigonal bipyramidalsp3d2: octahedral
›
sp3 HybridizationOne s and three p orbitals of the same principal quantum number mix to form four sp3 hybrid orbitals directed toward corners of a regular tetrahedron with bond angle 109 degrees 28 minutes. Example: CH4 where carbon forms four equivalent C-H bonds. Lone pairs on the central atom reduce the bond angle: NH3 is trigonal pyramidal (107 degrees) with one lone pair; H2O is bent (104.5 degrees) with two lone pairs.
›
sp2 HybridizationOne s and two p orbitals mix to form three sp2 hybrid orbitals oriented toward corners of an equilateral triangle with bond angle 120 degrees. Example: BF3 has trigonal planar geometry. In ethene (C2H4) each carbon is sp2 hybridised and the unhybridised p orbital overlaps sideways with the adjacent carbon to form a pi bond, creating a double bond.
›
sp HybridizationOne s and one p orbital mix to form two sp hybrid orbitals arranged linearly at 180 degrees. Example: BeCl2 has linear geometry. In ethyne (C2H2) each carbon is sp hybridised with two unhybridised p orbitals that overlap sideways to form two pi bonds, creating a triple bond.
›
sp3d HybridizationOne s, three p, and one d orbital mix to form five sp3d hybrid orbitals directed toward corners of a trigonal bipyramid. The three equatorial positions are in a plane at 120 degrees, while two axial positions are at 90 degrees to the equatorial plane. Example: PCl5 where phosphorus bonds to five chlorine atoms.
›
sp3d2 HybridizationOne s, three p, and two d orbitals mix to form six sp3d2 hybrid orbitals directed toward corners of a regular octahedron with all bond angles at 90 degrees. Example: SF6 where sulfur bonds to six fluorine atoms in a perfectly octahedral arrangement.

7) VSEPR Theory

Predicts molecular geometry from electron pair repulsions in the valence shell. Lone pair repulsion is stronger than bond pair repulsion (lp-lp > lp-bp > bp-bp), causing bond angle reduction from ideal values. The N/2 shortcut determines hybridisation and the number of bonded atoms versus lone pairs fixes the molecular shape from a comprehensive chart covering linear through pentagonal bipyramidal geometries.

lp-lp > lp-bp > bp-bpN/2 shortcut for hybridisationShape from bond pairs onlyLone pairs reduce bond angle
›
VSEPR Principles and Molecular ShapesVSEPR theory: the geometric arrangement of atoms is determined by repulsions between all electron pairs in the valence shell. Lone pairs occupy more space than bond pairs, so lp-lp repulsion > lp-bp repulsion > bp-bp repulsion. Lone pairs on the central atom distort bond angles below ideal values. Hybridisation shortcut: N = (valence electrons of central atom + number of bonded atoms plus or minus charge); N/2 = 2 sp linear, 3 sp2 trigonal planar, 4 sp3 tetrahedral, 5 sp3d trigonal bipyramidal, 6 sp3d2 octahedral. Molecular shape is named from atom positions only: 4 electron pairs with 1 lone pair gives trigonal pyramidal (NH3), 4 electron pairs with 2 lone pairs gives bent (H2O), 5 with 1 lone pair gives seesaw (SF4), 5 with 2 lone pairs gives T-shaped (ClF3), 6 with 1 lone pair gives square pyramidal (IF5), 6 with 2 lone pairs gives square planar (XeF4).

8) Molecular Orbital Theory

Treats electrons as waves that undergo constructive interference (bonding MO, lower energy) and destructive interference (anti-bonding MO, higher energy, marked with asterisk). Bond order = half the difference between bonding and anti-bonding electrons. Two energy level sequences exist: one for molecules up to N2 (pi before sigma-2p) and one for O2 onwards (sigma-2p before pi). Unpaired electrons in the MO configuration indicate paramagnetism.

Bond order = (Nb minus Na)/2Two MO energy ordersUnpaired e = paramagneticBond order 0 = molecule unstable
›
Bond OrderBonding MOs result from constructive interference of atomic orbitals and are lower in energy. Anti-bonding MOs result from destructive interference and are higher in energy, denoted with an asterisk. Bond order = half times (number of bonding electrons minus number of anti-bonding electrons). Integral bond order equals the number of bonds; fractional bond order means the molecule exists with intermediate stability. Higher bond order means greater stability and shorter bond length. Zero bond order means the molecule does not exist. Energy order for molecules up to N2: sigma-1s, sigma-star-1s, sigma-2s, sigma-star-2s, pi-2py = pi-2pz, sigma-2px, pi-star-2py = pi-star-2pz, sigma-star-2px. For O2 and beyond: sigma-2px comes before pi-2py = pi-2pz.
›
Application of MOT to Homonuclear Diatomic MoleculesH2 (2 electrons): configuration sigma-1s-squared, bond order 1, diamagnetic. H2-plus ion (1 electron): sigma-1s-1, bond order 0.5, paramagnetic (single electron bond, weakest known bond). Li2 (6 electrons): KK sigma-2s-squared, bond order 1, diamagnetic. C2 (12 electrons): bond order 2, diamagnetic. N2 (14 electrons): bond order 3, diamagnetic, strongest homonuclear bond. O2 (16 electrons): 2 unpaired electrons in pi-star-2p orbitals, bond order 2, paramagnetic. The paramagnetism of O2 is a key triumph of MOT that VBT cannot explain.

9) Hydrogen Bonding

A weak attachment (2 to 10 kcal/mol) formed when hydrogen bonded to a strongly electronegative atom (F, O, or N) interacts with another electronegative atom. Requires high electronegativity and small size of the electronegative element. Strength order: F-H-F > O-H-O > N-H-N. Two types: intermolecular (between molecules, raises boiling point) and intramolecular (within same molecule, causes chelation). Critical for biological structures like protein alpha-helices.

F, O, N form H-bondsF-H-F strongestIntermolecular raises BPIntramolecular causes chelation
›
Conditions for Hydrogen BondingTwo conditions: (a) hydrogen must be covalently linked to a highly electronegative element, (b) the electronegative element must be small in size. Only F (EN 4.0), O (EN 3.5), and N (EN 3.0) satisfy both conditions. Although N and Cl have equal electronegativity (3.0), Cl does not form effective hydrogen bonds because its larger atomic radius reduces the charge concentration needed to attract hydrogen. Strength order: F-H-F > O-H-O > N-H-N, following electronegativity. Hydrogen bond strength is 2 to 10 kcal/mol, far weaker than covalent bonds at 50 to 100 kcal/mol.
›
Types of Hydrogen BondingIntermolecular hydrogen bonding occurs between two molecules of the same or different types. Water forms extensive hydrogen-bonded networks; even at 90 degrees C most water molecules remain hydrogen bonded, explaining its anomalously high boiling point. HF forms zig-zag polymer chains (HF)n. Intramolecular hydrogen bonding occurs within the same molecule between atoms at different sites, forming a closed ring structure called chelation. Examples: o-nitrophenol and salicylaldehyde form intramolecular hydrogen bonds that prevent intermolecular association, lowering their boiling points compared to their para-isomers.
›
Importance of Hydrogen Bonding in Biological SystemsProteins consist of amino acid chains arranged in a spiral helix (alpha-helix structure). The N-H group of each amino acid unit forms an N-H---O hydrogen bond with the C=O group of the fourth amino acid along the chain. These hydrogen bonds are partly responsible for the stability of the helical protein structure. DNA double helix is also stabilised by hydrogen bonds between complementary base pairs.

Chemical Bonding Download Notes & Weightage Plan

For each topic in the Chemical Bonding chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Lewis Theory

Foundation for understanding covalent bonding: shared electron pairs, octet rule, Lewis dot structures, and formal charge calculation.

Conceptual foundationFormal charge calculationExceptions to octet ruleLow standalone Q frequency

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Octet rule: atoms form bonds until surrounded by 8 electrons (H needs 2). Lewis structure method: n1 = total valence electrons, n2 = 2(H atoms) + 8(other atoms), bonding electrons n3 = n2 minus n1, non-bonding n4 = n1 minus n3. Formal charge = valence electrons minus bonds minus unshared electrons. Exceptions: H (duet), B/Be (incomplete octet), PCl5/SF6 (expanded octet).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Practise Lewis structures of 5 molecules: CO2, SO2, NO2, PCl5, XeF4. For each, calculate formal charges and identify exceptions to the octet rule. The systematic five-step method is faster than trial-and-error for inorganic species.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Rarely tested as a standalone question. Lewis structures appear as part of hybridisation or molecular shape prediction questions. Formal charge may appear in assertion-reason format.
Time Required1 hr30 min theory on octet rule and systematic method; 30 min practising Lewis structures of inorganic molecules.
DifficultyEasyStraightforward counting of electrons. The systematic method eliminates guesswork. The only pitfall is forgetting to include ionic charge in total valence electron count.
  • Scoring Focus: Formal charge calculation and knowing when octet rule is violated. Elements in Period 3 and beyond can expand their octet using d-orbitals.
  • High-risk Area: Forgetting to adjust total valence electrons for ionic charge: add electrons for negative charge, subtract for positive charge. Missing this step gives wrong Lewis structure.
  • Best Practice Style: Always count total valence electrons first, then apply the five-step method. Check formal charges on every atom to validate the structure.
Priority rule: Low priority. Quick prerequisite topic. Master Lewis structures in 1 hour then move to hybridisation and VSEPR which carry direct NEET marks.

Types of Chemical Bonds

Four fundamental bond types: ionic (electron transfer), covalent (sigma and pi overlap), coordinate (donor-acceptor), and metallic (delocalised electrons). Foundation for understanding molecular properties.

Sigma: head-on overlapPi: lateral overlapCoordinate: both e from donorMetallic: electron sea model

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Ionic bond: complete electron transfer, electropositive + electronegative, crystal lattice, high MP. Covalent bond: electron sharing, sigma (s-s, s-p, p-p head-on overlap) and pi (lateral p-p overlap). Double bond = 1 sigma + 1 pi. Triple bond = 1 sigma + 2 pi. Polar covalent: unequal sharing from EN difference, partial charges delta-plus and delta-minus. Coordinate: donor provides both electrons, acceptor needs electrons. NH3 donates to H-plus or BF3. Metallic bond: electron sea in metal lattice, explains conductivity and malleability.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw orbital overlap diagrams for each sigma type (s-s in H2, s-p in HF, p-p in F2) and pi bond formation in O2 and N2. List 3 examples of coordinate bond compounds. Summarise ionic vs covalent vs metallic properties in a table.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question on identifying bond type in a given molecule, counting sigma and pi bonds, or recognising coordinate bond formation in species like NH4-plus, H3O-plus, or BF3-NH3 adduct.
Time Required2 hrs45 min ionic and covalent bond theory with overlap diagrams; 30 min coordinate bond with examples; 15 min metallic bond concept; 30 min MCQ practice.
DifficultyEasy-ModerateBond types are conceptually clear but counting sigma and pi bonds in complex molecules like benzene or acetic acid requires practice. Coordinate bond identification in ions trips students.
  • Scoring Focus: Counting sigma and pi bonds in a molecule: every single bond is one sigma, every double bond is one sigma plus one pi, every triple bond is one sigma plus two pi. NEET asks total sigma and pi bonds in molecules like CH3-CH=CH-C(triple bond)CH.
  • High-risk Area: Forgetting that a coordinate bond, once formed, is identical to a covalent bond. All four N-H bonds in NH4-plus are equivalent despite one originating as a coordinate bond.
  • Best Practice Style: For sigma/pi counting: draw the structural formula, count single bonds as sigma, double bonds as sigma+pi, triple bonds as sigma+2pi. Total up.
Priority rule: Medium priority. Bond type identification and sigma/pi counting are regularly tested. Overlaps with hybridisation topic.

Dipole Moment

Quantifies bond polarity. mu = e times d in Debye units. Vector quantity requiring vector addition of individual bond moments to find net molecular dipole.

mu = charge times distanceVector addition requiredZero mu = symmetric moleculeH is positive end in H-X

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Dipole moment mu = e times d. Unit: 1 Debye = 10 to the minus 18 esu-cm = 3.33 times 10 to the minus 30 C-m. Vector quantity: net mu is vector sum of bond moments. Zero mu indicates symmetric molecule (BF3, CCl4, CO2, BeF2). In H-X bonds, H is the positive end. In C-X bonds (X not H or C), C is positive end. Direction of C-H dipole depends on hybridisation of carbon.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: List molecules with zero dipole moment and explain the symmetry: CO2 (linear, equal C=O), BF3 (trigonal planar, equal B-F), CCl4 (tetrahedral, equal C-Cl). Contrast with molecules having nonzero dipole: H2O (bent), NH3 (pyramidal), CHCl3 (distorted tetrahedron). Practise vector addition for 2-3 molecules.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasionally tested as which molecule has zero dipole moment or which has highest dipole moment. Often combined with molecular shape questions.
Time Required1 hr30 min concept and vector addition principles; 30 min practising dipole moment comparison MCQs.
DifficultyEasyConceptual topic. Knowing molecular geometry automatically tells you whether dipole moment is zero (symmetric) or nonzero (asymmetric).
  • Scoring Focus: Zero dipole moment identification. If all bond dipoles cancel by symmetry, mu = 0. This links directly to molecular shape from VSEPR.
  • High-risk Area: Assuming all tetrahedral molecules have zero dipole. Only symmetric tetrahedrals like CCl4 have zero mu. CHCl3 is tetrahedral but asymmetric, so mu is nonzero.
  • Best Practice Style: First determine molecular geometry, then check if all surrounding atoms are identical. If yes, mu = 0. If any atom differs, mu is nonzero.
Priority rule: Medium priority. Quick conceptual topic that overlaps with VSEPR. Master molecular geometry first, then dipole moment follows logically.

Physical Properties of Ionic and Covalent Compounds

Contrasts physical properties across three dimensions: melting point, electrical conductivity, and solubility, linking each to the nature of bonding and intermolecular forces.

Ionic: high MP, conduct in melt/solutionCovalent: low MP, non-conductorLike dissolves likeDielectric constant governs solubility

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Ionic: high MP and BP (strong non-directional lattice forces), conduct electricity when molten or in solution (ion migration), soluble in polar solvents with high dielectric constant (water). Covalent: gases/liquids/low-MP solids (weak van der Waals forces between discrete molecules), non-conductors in any state (no ions or mobile electrons), soluble in non-polar solvents with low dielectric constant (benzene, CCl4).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw a three-row comparison table: Property | Ionic | Covalent. Fill in MP, conductivity, solubility with reasoning. Memorise the key distinction: ionic compounds are lattice-based (bulk property), covalent compounds are molecule-based (intermolecular forces govern).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Rarely a standalone NEET question but appears as assertion-reason or in combination with Fajan's rule. The conceptual knowledge is essential background for multiple topics.
Time Required30 minQuick read and table construction. No calculations needed.
DifficultyEasyPure conceptual recall. No numerical component.
  • Scoring Focus: The key distinction: ionic compounds conduct when molten or dissolved (not in solid state). NEET distractors include solid ionic compounds conducting, which is false.
  • High-risk Area: Stating that ionic compounds conduct in solid state. They do not; ions are locked in the lattice. Only when molten or dissolved do ions become mobile.
  • Best Practice Style: Always specify the state when discussing conductivity of ionic compounds: molten state or aqueous solution allows conduction, solid state does not.
Priority rule: Low priority. Quick topic with no direct calculations. Spend 30 minutes then move to Fajan's rule.

Fajan's Rule

Predicts covalent character in ionic compounds based on ion polarisation. Four factors: cation size, anion size, ionic charge, and electronic configuration of cation.

Small cation: more covalentLarge anion: more covalentHigh charge: more covalent18-electron config > 8-electron config

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Fajan's rule: covalent character of ionic bonds increases with polarisation of anion by cation. Four factors: (1) Small cation = high polarising power (LiCl more covalent than KCl). (2) Large anion = high polarisability (iodides most covalent among halides). (3) High charge on either ion increases covalency (NaCl < MgCl2 < AlCl3). (4) Pseudo noble gas config (18 electrons, e.g. Cu-plus) polarises more than noble gas config (8 electrons, e.g. Na-plus) at same size/charge. Greater polarisation lowers MP and increases non-polar solvent solubility. Li halide MP order: LiF 870 C > LiCl 613 C > LiBr 547 C > LiI 446 C.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Memorise the four factors as a mnemonic: Small Cat, Big An, High Charge, Pseudo Config. Practise ordering compounds by covalent character: LiCl vs NaCl vs KCl (effect of cation size). NaCl vs NaBr vs NaI (effect of anion size). NaCl vs MgCl2 vs AlCl3 (effect of charge).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question per paper on ordering compounds by ionic/covalent character or identifying which factor explains a given trend (e.g., why AgCl is more covalent than NaCl).
Time Required1 hr30 min theory on four Fajan's rule factors with examples; 30 min MCQ practice on ordering compounds.
DifficultyEasy-ModerateThe four rules are straightforward to memorise but applying them to rank compounds requires combining multiple factors simultaneously.
  • Scoring Focus: Ranking compounds by covalent character using Fajan's rule. Common NEET series: LiF < LiCl < LiBr < LiI (increasing covalent character with increasing anion size).
  • High-risk Area: Confusing polarising power (cation property) with polarisability (anion property). Small cation has high polarising power. Large anion has high polarisability. Both increase covalent character.
  • Best Practice Style: For any comparison: identify which Fajan factor differs between the two compounds, then apply that factor. If multiple factors differ, the dominant factor usually determines the answer.
Priority rule: Medium priority. Regularly tested in NEET as a conceptual MCQ. Quick to master with the four-factor framework.

Hybridisation

Five hybridisation types: sp3, sp2, sp, sp3d, sp3d2 producing tetrahedral, trigonal planar, linear, trigonal bipyramidal, and octahedral geometries. The N/2 shortcut enables rapid determination.

sp3: 109.5 degrees tetrahedralsp2: 120 degrees trigonalsp: 180 degrees linearN/2 shortcutLone pairs modify ideal shape

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Hybridisation: intermixing of atomic orbitals of same atom with similar energies to form equivalent hybrid orbitals. sp3: 1s + 3p = 4 orbitals, tetrahedral, 109.5 degrees (CH4). sp2: 1s + 2p = 3 orbitals, trigonal planar, 120 degrees (BF3); in C2H4 unhybridised p forms pi bond. sp: 1s + 1p = 2 orbitals, linear, 180 degrees (BeCl2); in C2H2 two unhybridised p orbitals form two pi bonds. sp3d: 1s + 3p + 1d = 5, trigonal bipyramidal (PCl5). sp3d2: 1s + 3p + 2d = 6, octahedral (SF6). Number of hybrid orbitals always equals number of orbitals hybridised.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Build a five-row table: hybridisation | orbitals mixed | geometry | bond angle | example. Use the N/2 shortcut on 10 molecules: NH3, H2O, BF3, PCl5, SF6, XeF2, XeF4, IF5, ClF3, SF4. For each, predict hybridisation, number of lone pairs, and actual molecular shape.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2One to two questions asking to identify the hybridisation of the central atom in a molecule or ion, or to predict bond angles. Often combined with VSEPR shape prediction.
Time Required2 hrs1 hr on five hybridisation types with orbital diagrams; 30 min N/2 shortcut practice; 30 min MCQ on identifying hybridisation.
DifficultyModerateRequires remembering the correspondence between N/2 value and hybridisation type, and knowing how lone pairs modify ideal geometry.
  • Scoring Focus: N/2 shortcut is the fastest way to identify hybridisation in NEET. Apply it to central atom: count valence electrons plus bonded atoms plus or minus charge, divide by 2, match to table.
  • High-risk Area: Forgetting to include the ionic charge when using the N/2 shortcut. For NH4-plus: N = 5 + 4 minus 1 = 8, N/2 = 4, sp3 (correct). Omitting the charge gives N/2 = 4.5 which is meaningless.
  • Best Practice Style: For every hybridisation question: (1) identify central atom, (2) count N using shortcut, (3) divide by 2, (4) match to hybridisation table, (5) subtract lone pairs for molecular shape.
Priority rule: Highest priority. Hybridisation identification is one of the most frequently asked NEET topics in Chemical Bonding. Master the N/2 shortcut thoroughly.

VSEPR Theory

Predicts molecular geometry based on electron pair repulsions. Lone pair repulsion exceeds bond pair repulsion. Combined with hybridisation to give exact molecular shapes and bond angles for any molecule or ion.

lp-lp > lp-bp > bp-bpLone pairs reduce angleShape from atom positions only14 geometries in the chart

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)VSEPR: molecular shape determined by repulsion between all electron pairs in valence shell. Repulsion order: lp-lp > lp-bp > bp-bp. Lone pairs cause bond angle reduction from ideal. Shapes by total pairs and lone pairs: 2 total/0 lp = linear (BeCl2), 3/0 = trigonal planar (BF3), 3/1 = bent (SnCl2), 4/0 = tetrahedral (CH4), 4/1 = trigonal pyramidal (NH3), 4/2 = bent (H2O), 5/0 = trigonal bipyramidal (PCl5), 5/1 = seesaw (SF4), 5/2 = T-shaped (ClF3), 5/3 = linear (XeF2), 6/0 = octahedral (SF6), 6/1 = square pyramidal (IF5), 6/2 = square planar (XeF4).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Memorise the full VSEPR chart with 14 entries. For each entry know: total pairs, lone pairs, hybridisation, molecular shape, one example molecule. Quick mnemonics: TBP with 1 lp = seesaw, TBP with 2 lp = T-shape, Octahedral with 2 lp = square planar. Practise predicting shapes for XeF2, XeF4, IF5, ClF3.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question asking to predict molecular shape of a given species (e.g., What is the shape of XeF4? Answer: square planar). May also ask for bond angle comparison.
Time Required2 hrs1 hr learning the VSEPR chart with all 14 shapes and example molecules; 30 min on repulsion order and bond angle effects; 30 min MCQ practice.
DifficultyModerateThe chart has many entries to memorise. The conceptual principle is simple (lone pairs take more space) but applying it to expanded octet molecules (SP3d, sp3d2) requires systematic thinking.
  • Scoring Focus: Shape names for molecules with lone pairs: NH3 is trigonal pyramidal (NOT tetrahedral), H2O is bent (NOT tetrahedral), SF4 is seesaw (NOT tetrahedral), XeF4 is square planar (NOT octahedral). Name the shape from atom positions, not from total electron pair geometry.
  • High-risk Area: Naming the electron pair geometry instead of the molecular geometry. VSEPR names molecular shapes from atom positions only. NH3 has tetrahedral ELECTRON geometry but trigonal pyramidal MOLECULAR geometry.
  • Best Practice Style: For every molecule: (1) find total electron pairs from N/2, (2) determine lone pairs = total pairs minus bonded atoms, (3) look up molecular shape from VSEPR chart. Answer with the molecular geometry name, not the electron pair arrangement.
Priority rule: Highest priority alongside hybridisation. These two topics together yield 1 to 2 guaranteed NEET questions. Master the full VSEPR chart with examples.

Molecular Orbital Theory

Models bonding through constructive (bonding MO) and destructive (antibonding MO) interference of atomic orbital waves. Bond order, magnetic character, and molecular stability are determined from the MO electronic configuration.

Two MO energy ordersBond order = (Nb minus Na)/2Unpaired e = paramagneticO2 paramagnetism explained

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)MO Theory: atomic orbitals combine to form molecular orbitals. Bonding MO (constructive interference, lower energy) and antibonding MO (destructive, higher energy, asterisk). Bond order = (Nb minus Na)/2. Energy order up to N2: sigma-1s < sigma-star-1s < sigma-2s < sigma-star-2s < pi-2py = pi-2pz < sigma-2px < pi-star-2py = pi-star-2pz < sigma-star-2px. For O2 and above: sigma-2px comes before pi-2py = pi-2pz. Key molecules: H2 (BO=1, diamagnetic), H2-plus (BO=0.5, paramagnetic), Li2 (BO=1, diamagnetic), N2 (BO=3, diamagnetic), O2 (BO=2, paramagnetic with 2 unpaired e). Unpaired electrons in configuration indicate paramagnetism.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the two MO energy orders side by side and label the switch point. Fill electrons for H2, N2, O2, F2 using both orders. Calculate bond order for each and determine magnetic nature. The O2 paramagnetism explanation is the most commonly tested MOT concept.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question on bond order calculation, magnetic character determination, or identifying which molecule has the highest bond strength among a series. O2 paramagnetic nature is a perennial NEET favourite.
Time Required2.5 hrs1 hr on MO formation and energy level orders; 45 min filling electron configurations for H2 through Ne2; 45 min MCQ practice on bond order and magnetism.
DifficultyHardThe most conceptually challenging topic in Chemical Bonding. Two different energy orders must be memorised. Filling electrons correctly into degenerate pi orbitals (Hund's rule) is essential for magnetic character prediction.
  • Scoring Focus: Bond order of O2 = 2 with 2 unpaired electrons (paramagnetic). Bond order of N2 = 3, diamagnetic. These two molecules are tested repeatedly. Also know: if bond order = 0, molecule does not exist (He2, Ne2).
  • High-risk Area: Using the wrong MO energy order. For molecules up to N2, pi-2p orbitals fill BEFORE sigma-2px. For O2 onwards, sigma-2px fills BEFORE pi-2p. Applying the wrong order gives wrong bond order and wrong magnetic prediction.
  • Best Practice Style: First check: is the molecule N2 or lighter? Use order 1 (pi before sigma). Is it O2 or heavier? Use order 2 (sigma before pi). Then fill electrons following aufbau principle and Hund's rule.
Priority rule: High priority. MOT bond order and magnetic character questions appear frequently in NEET. The O2 paramagnetism explanation is a classic exam question.

Hydrogen Bonding

Weak electrostatic attraction between H bonded to F, O, or N and another electronegative atom. Two types: intermolecular (raises BP, causes association) and intramolecular (chelation, lowers BP relative to intermolecular variant). Essential for biological structure stability.

F, O, N only2-10 kcal/mol strengthIntermolecular: high BPIntramolecular: chelation

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Hydrogen bond: X-H---Y where X, Y are F, O, or N. Conditions: (1) H bonded to highly electronegative atom, (2) electronegative atom must be small. Strength: F-H-F > O-H-O > N-H-N, matching electronegativity order. Cl has same EN as N (3.0) but larger size prevents effective H-bonding. Strength 2-10 kcal/mol vs covalent 50-100 kcal/mol. Intermolecular: between molecules, causes association, raises BP (water, HF polymer). Intramolecular: within same molecule, chelation, closed ring (o-nitrophenol, salicylaldehyde). Biological role: N-H---O bonds stabilise protein alpha-helix structure.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: List substances showing hydrogen bonding: H2O, HF, NH3, alcohols, carboxylic acids, amines. Know why H2O has anomalously high BP. Distinguish ortho-nitrophenol (intramolecular H-bond, lower BP, steam-volatile) from para-nitrophenol (intermolecular H-bond, higher BP). This ortho vs para comparison is a NEET favourite.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question on hydrogen bonding: identifying which substance shows hydrogen bonding, ordering boiling points based on hydrogen bond strength, or distinguishing intermolecular from intramolecular hydrogen bonding.
Time Required1.5 hrs30 min conditions and strength order; 30 min types with examples; 30 min MCQ practice on BP ordering and ortho/para comparison.
DifficultyEasy-ModerateThe conditions are simple (F, O, N) but applying them to compare boiling points of related compounds requires careful reasoning about intermolecular vs intramolecular types.
  • Scoring Focus: Boiling point ordering based on hydrogen bond strength and type. Ortho-nitrophenol vs para-nitrophenol comparison is a classic: ortho has intramolecular H-bond (lower BP, volatile in steam), para has intermolecular H-bond (higher BP, non-volatile in steam).
  • High-risk Area: Assuming Cl forms hydrogen bonds because its EN equals N (3.0). Cl does NOT form effective H-bonds due to its larger size. Only F, O, and N form hydrogen bonds in the periodic table.
  • Best Practice Style: For every H-bonding question: (1) check if H is bonded to F, O, or N, (2) check if the interacting atom is also F, O, or N, (3) determine type (inter or intra), (4) predict property accordingly.
Priority rule: Medium priority. Hydrogen bonding questions are straightforward scoring opportunities. The ortho/para nitrophenol comparison and HF/H2O boiling point anomaly are tested most often.

Chemical Bonding Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Chemical Bonding chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
MO Energy Order and O2 Paramagnetism
NEETMolecular Orbital TheoryBond orderParamagnetism

Mistake Snapshot (What Students Do Wrong)

  • Using the wrong MO energy level order for O2: Two MO energy orders exist. For molecules up to N2, pi-2p fills before sigma-2px. For O2 and beyond, sigma-2px fills before pi-2p. Using the wrong order places electrons in incorrect orbitals, giving wrong bond order and magnetic character.
  • Predicting O2 as diamagnetic based on Lewis structure: The Lewis structure of O2 shows a double bond with all paired electrons, which wrongly suggests diamagnetic character. MOT correctly places 2 electrons in degenerate pi-star orbitals with parallel spins (Hund's rule), predicting paramagnetic behaviour confirmed experimentally.
2–3 Line Example (Typical Error)

O2 has 16 electrons. Using the correct MO order for O2: KK sigma-2s2 sigma-star-2s2 sigma-2px2 pi-2py2 pi-2pz2 pi-star-2py1 pi-star-2pz1. Bond order = (10 minus 6)/2 = 2. Two unpaired electrons in pi-star orbitals make O2 paramagnetic. If you mistakenly use the N2 order (pi before sigma), you get the same bond order but might misplace electrons and miss the unpaired electron count.

How NEET Frames The Trap

NEET asks the magnetic nature of O2 or bond order of a diatomic molecule. Students who use one universal MO order get wrong results for molecules near the N2/O2 boundary.

NEET-Style Trap Question Format

Q. Which diatomic species is paramagnetic with bond order 2?
A. N2   B. O2   C. F2   D. C2  
Trick: N2 has bond order 3, diamagnetic. O2 has bond order 2, paramagnetic (2 unpaired electrons in pi-star orbitals). F2 has bond order 1, diamagnetic. C2 has bond order 2 but is diamagnetic (no unpaired electrons). Students confuse C2 and O2 because both have bond order 2, but only O2 is paramagnetic.

Quick rule: For magnetic nature: fill MO configuration, check for unpaired electrons in pi-star orbitals. Use the correct energy order: up to N2 use pi-before-sigma, for O2 onwards use sigma-before-pi.
VSEPR Molecular Shape vs Electron Geometry
NEETVSEPRMolecular geometryHybridisation

Mistake Snapshot (What Students Do Wrong)

  • Naming electron pair geometry instead of molecular shape: VSEPR names molecular shapes from atom positions only, excluding lone pairs. NH3 is trigonal pyramidal (not tetrahedral). H2O is bent (not tetrahedral). Students who report the electron pair arrangement instead of the molecular geometry lose the mark.
  • Forgetting to include ionic charge in the N/2 hybridisation shortcut: For ions like NH4-plus, the charge must be subtracted from N: N = 5 + 4 minus 1 = 8, N/2 = 4 (sp3 tetrahedral). For SO4-2-minus, add 2 for the negative charge. Omitting the charge gives wrong N/2 and wrong hybridisation.
2–3 Line Example (Typical Error)

What is the shape of SF4? S has 6 valence electrons, bonds to 4 F atoms. N = 6 + 4 = 10, N/2 = 5, so sp3d hybridisation with 5 electron pairs. 5 pairs minus 4 bonded = 1 lone pair. Shape is seesaw (not trigonal bipyramidal). Naming it trigonal bipyramidal gives the electron pair geometry, not the molecular shape.

How NEET Frames The Trap

NEET asks for the shape of molecules with lone pairs on the central atom. The electron pair arrangement and the molecular shape have different names. Distractors use the electron pair geometry name.

NEET-Style Trap Question Format

Q. The shape of XeF4 molecule is
A. Tetrahedral   B. Octahedral   C. Square planar   D. Square pyramidal  
Trick: Xe has 8 valence electrons, bonds to 4 F atoms. N = 8 + 4 = 12, N/2 = 6, sp3d2 hybridisation. 6 pairs minus 4 bonded = 2 lone pairs. Octahedral electron geometry with 2 lone pairs gives square planar molecular shape. Option B (octahedral) is the electron pair geometry, not the molecular shape. Option D (square pyramidal) has only 1 lone pair.

Quick rule: Molecular shape = atom arrangement only, ignoring lone pairs. Electron geometry = all electron pairs. NEET always asks for molecular shape. After finding total pairs and lone pairs, use the VSEPR chart to name the molecular shape.
Hybridisation and Bond Angle Reduction by Lone Pairs
NEETHybridisationVSEPRBond angle

Mistake Snapshot (What Students Do Wrong)

  • Assuming ideal bond angles when lone pairs are present: Lone pairs occupy more space than bond pairs, compressing bond angles below the ideal value. sp3 ideal angle is 109.5 degrees, but NH3 is 107 degrees (1 lone pair) and H2O is 104.5 degrees (2 lone pairs). NEET distractors offer 109.5 for both.
  • Not recognising that multiple bonds cause greater repulsion than single bonds: A double bond has greater electron density than a single bond and repels more strongly. In molecules like SO2 (two S=O double bonds plus one lone pair), bond angle is close to 120 degrees but not exactly, due to double bond repulsion exceeding single bond repulsion.
2–3 Line Example (Typical Error)

Bond angle in NH3 vs CH4 vs H2O. CH4 has 4 bond pairs, ideal tetrahedral angle 109.5 degrees. NH3 has 3 bond pairs and 1 lone pair; lone pair compresses bond angle to 107 degrees. H2O has 2 bond pairs and 2 lone pairs; two lone pairs compress angle further to 104.5 degrees. All three are sp3 hybridised but bond angles differ.

How NEET Frames The Trap

NEET asks to arrange NH3, H2O, CH4 in order of increasing or decreasing bond angle. Students who give the same angle (109.5) for all three because they are all sp3 lose the mark.

NEET-Style Trap Question Format

Q. Arrange in decreasing order of bond angle: CH4, NH3, H2O
A. CH4 > NH3 > H2O   B. H2O > NH3 > CH4   C. NH3 > CH4 > H2O   D. All have equal bond angles  
Trick: All are sp3. CH4 has 0 lone pairs (109.5 degrees). NH3 has 1 lone pair (107 degrees). H2O has 2 lone pairs (104.5 degrees). Decreasing order: CH4 > NH3 > H2O. Option D (equal angles) ignores lone pair compression. Option B reverses the order.

Quick rule: More lone pairs on the central atom means smaller bond angle. For sp3: 0 lp = 109.5 degrees, 1 lp = approximately 107 degrees, 2 lp = approximately 104.5 degrees. Lone pair repulsion is always greater than bond pair repulsion.
Fajan's Rule and Covalent Character Ordering
NEETFajan's RuleIonic characterPolarisation

Mistake Snapshot (What Students Do Wrong)

  • Confusing polarising power with polarisability: Polarising power is a cation property (ability to distort anion electron cloud). Polarisability is an anion property (susceptibility to distortion). Mixing them up leads to wrong ordering. Small cation = high polarising power. Large anion = high polarisability.
  • Ignoring electronic configuration effect in Fajan's rule: Cations with pseudo noble gas configuration (18 electrons, e.g., Cu-plus, Ag-plus) are more polarising than noble gas configuration cations (8 electrons, e.g., Na-plus, K-plus) of the same size and charge. Cu-plus (0.96 A) and Na-plus (0.95 A) have nearly identical radii and same charge, but CuCl is more covalent than NaCl.
2–3 Line Example (Typical Error)

Order LiCl, NaCl, KCl by covalent character. All have Cl-minus as anion (same polarisability). Cation size: Li-plus < Na-plus < K-plus. Smaller cation = higher polarising power = more covalent character. Order: LiCl > NaCl > KCl. Students who think larger cation polarises more get the reverse order.

How NEET Frames The Trap

NEET gives a series of compounds and asks to arrange by covalent or ionic character. Distractors reverse the effect of cation or anion size.

NEET-Style Trap Question Format

Q. Among NaCl, MgCl2, and AlCl3, which has the most covalent character?
A. NaCl   B. MgCl2   C. AlCl3   D. All are equally ionic  
Trick: Cation charge increases: Na-plus (1+), Mg-2-plus (2+), Al-3-plus (3+). Higher charge plus smaller cation radius gives greater polarising power. AlCl3 has the highest covalent character. NaCl has the least. Students who focus only on size and forget charge effect may pick MgCl2.

Quick rule: Four checks in order: (1) smaller cation = more covalent, (2) larger anion = more covalent, (3) higher charge = more covalent, (4) 18-electron config > 8-electron config at same size/charge. Apply the factor that differs between the compounds being compared.
Hydrogen Bonding: Chlorine vs Nitrogen
NEETHydrogen bondElectronegativityAtomic size

Mistake Snapshot (What Students Do Wrong)

  • Assuming Cl forms hydrogen bonds because it has the same electronegativity as N: Both N and Cl have electronegativity 3.0 on the Pauling scale. However, hydrogen bond formation requires BOTH high electronegativity AND small atomic size. Cl is much larger than N, so the charge concentration around Cl is insufficient to form effective hydrogen bonds. Only F, O, and N form hydrogen bonds.
  • Confusing intermolecular and intramolecular hydrogen bonding effects on boiling point: Intermolecular hydrogen bonding raises boiling point by causing molecular association. Intramolecular hydrogen bonding (chelation) forms a closed ring within the same molecule and PREVENTS intermolecular association, actually LOWERING boiling point. Ortho-nitrophenol (intramolecular H-bond) has a lower boiling point than para-nitrophenol (intermolecular H-bond).
2–3 Line Example (Typical Error)

Compare boiling points of o-nitrophenol and p-nitrophenol. In o-nitrophenol, the OH group forms an intramolecular hydrogen bond with the adjacent NO2 group (chelation). This internal bond prevents intermolecular hydrogen bonding. p-nitrophenol has no intramolecular H-bond, so it forms strong intermolecular H-bonds, raising its BP above the ortho isomer. Students who assume both isomers have similar BP miss this distinction.

How NEET Frames The Trap

NEET asks which isomer has a higher boiling point or which is steam-volatile. Students who do not distinguish between inter and intramolecular hydrogen bonding answer incorrectly.

NEET-Style Trap Question Format

Q. Which compound is steam-volatile: o-nitrophenol or p-nitrophenol?
A. o-nitrophenol   B. p-nitrophenol   C. Both are steam-volatile   D. Neither is steam-volatile  
Trick: o-Nitrophenol forms intramolecular hydrogen bonds (chelation), preventing intermolecular association. It behaves as a discrete molecule with low effective molecular mass. o-Nitrophenol is steam-volatile. p-Nitrophenol forms intermolecular H-bonds, creating molecular association and higher effective BP, making it non-volatile in steam.

Quick rule: Intermolecular H-bonding increases BP and molecular association. Intramolecular H-bonding forms a closed ring and DECREASES BP by preventing molecular association. Ortho-substituted phenols with nearby H-bond donors and acceptors favour intramolecular H-bonding.
Previous
Atomic Structure > Nuclear Chemistry
Next
Chemical Thermodynamics > Thermodynamics and Thermochemistry

Loading tests...

NEET > Chemistry > Chemical Bonding And Molecular Structure Chapters

Review your status and progress for each chapter in this unit. Use the slider to set progress or click "Mark as Done" to complete.

ChapterStatusProgress

Chemical Bonding

Weightage: 02.2K
0%

Comments

Leave a comment

0/2000Comments are moderated

You can comment without logging in. We'll ask for your name and email before submitting.

Comments (0)

No comments yet. Be the first to comment!