JEE/NEET-ready summary of acidic, basic, neutral, and amphoteric oxides with examples + trends:


1. Acidic Oxides

Definition: Oxides that react with water to form acids or react with bases to give salt + water.

Examples:

  • CO₂, SO₂, SO₃, P₂O₅, N₂O₅, Cl₂O₇
  • Non-metal oxides

Reactions:

  • CO₂ + H₂O → H₂CO₃
  • SO₃ + 2NaOH → Na₂SO₄ + H₂O

Trend:

  • Non-metallic character ↑ ⇒ acidity ↑
  • Across a period: acidity increases
  • Down a group: acidity decreases

2. Basic Oxides

Definition: Oxides that react with water to form bases or react with acids to give salt + water.

Examples:

  • Na₂O, CaO, MgO, K₂O
  • Mostly metal oxides

Reactions:

  • Na₂O + H₂O → 2NaOH
  • CaO + 2HCl → CaCl₂ + H₂O

Trend:

  • Metallic character ↑ ⇒ basicity ↑
  • Down a group: basicity increases
  • Across a period: basicity decreases

3. Amphoteric Oxides

Definition: Oxides that react with both acids and bases.

Examples (VERY IMPORTANT for exams):

  • Al₂O₃, ZnO, BeO, SnO, PbO, Cr₂O₃

Reactions:

  • ZnO + 2HCl → ZnCl₂ + H₂O (acts basic)
  • ZnO + 2NaOH → Na₂ZnO₂ + H₂O (acts acidic)

Trend:

  • Found near metal–nonmetal boundary
  • Small size + high charge density → amphoteric behavior

4. Neutral Oxides

Definition: Oxides that do not react with acids or bases.

Examples:

  • CO, NO, N₂O

Trend:

  • Mostly simple molecules
  • No acidic or basic nature

IMPORTANT PERIODIC TREND (MOST ASKED)

Across a period (Left → Right):

Basic → Amphoteric → Acidic

Example (Period 3):
Na₂O → MgO → Al₂O₃ → SiO₂ → P₂O₅ → SO₃ → Cl₂O₇

Key shift:

  • Metals → Non-metals
  • Basic → Acidic

QUICK REVISION TABLE

TypeNatureExamples
BasicMetal oxidesNa₂O, CaO
AmphotericBothAl₂O₃, ZnO
AcidicNon-metal oxidesSO₃, CO₂
NeutralNo reactionCO, NO

EXAM TIPS

  • Al₂O₃, ZnO → always remember (amphoteric)
  • CO ≠ acidic (neutral!)
  • Higher oxidation state ⇒ more acidic oxide
    • Example: SO₂ < SO₃ (acidity increases)

MCQs

Q1. Arrange in increasing acidic nature:

A. CO₂ < SiO₂ < SO₃
B. SiO₂ < CO₂ < SO₃
C. SO₃ < CO₂ < SiO₂
D. CO₂ < SO₃ < SiO₂


Q2. Arrange in decreasing basic strength:

A. Na₂O > MgO > Al₂O₃
B. Al₂O₃ > MgO > Na₂O
C. MgO > Na₂O > Al₂O₃
D. Na₂O > Al₂O₃ > MgO


Q3. Arrange in increasing amphoteric character:

A. Na₂O < MgO < Al₂O₃
B. Al₂O₃ < MgO < Na₂O
C. MgO < Na₂O < Al₂O₃
D. Na₂O < Al₂O₃ < MgO


Q4. Arrange in increasing acidic strength:

A. SO₂ < SO₃ < Cl₂O₇
B. Cl₂O₇ < SO₃ < SO₂
C. SO₃ < SO₂ < Cl₂O₇
D. SO₂ < Cl₂O₇ < SO₃


Q5. Arrange in increasing basic character:

A. CaO < MgO < Na₂O
B. Na₂O < MgO < CaO
C. MgO < CaO < Na₂O
D. CaO < Na₂O < MgO


Q6. Arrange in increasing acidic nature:

A. P₂O₅ < SO₃ < Cl₂O₇
B. Cl₂O₇ < SO₃ < P₂O₅
C. SO₃ < P₂O₅ < Cl₂O₇
D. P₂O₅ < Cl₂O₇ < SO₃


Q7. Arrange in increasing basic strength:

A. BeO < MgO < CaO
B. CaO < MgO < BeO
C. MgO < BeO < CaO
D. BeO < CaO < MgO


Q8. Arrange in increasing acidic character:

A. N₂O₃ < NO₂ < N₂O₅
B. N₂O₅ < NO₂ < N₂O₃
C. NO₂ < N₂O₃ < N₂O₅
D. N₂O₃ < N₂O₅ < NO₂


ANSWERS

  1. B → SiO₂ < CO₂ < SO₃
  2. A → Na₂O > MgO > Al₂O₃
  3. A → Na₂O < MgO < Al₂O₃
  4. A → SO₂ < SO₃ < Cl₂O₇
  5. C → MgO < CaO < Na₂O
  6. A → P₂O₅ < SO₃ < Cl₂O₇
  7. A → BeO < MgO < CaO
  8. A → N₂O₃ < NO₂ < N₂O₅

SHORT TRICKS (VERY IMPORTANT)

  • Across period: Basic ↓, Acidic ↑
  • Down group: Basic ↑
  • Higher oxidation state ⇒ more acidic
  • Metal oxides → basic
  • Non-metal oxides → acidic
  • Border elements (Be, Al, Zn) → amphoteric

Electronegativity — JEE/NEET Complete Concepts

🔹 Definition

Electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself in a chemical bond.


Important Points

  • Applies only to bonded atoms
  • It is a relative (dimensionless) value
  • Most commonly used scale → Pauling scale

Trends in Periodic Table

Across a Period (→)

Increases

  • Reason: Increase in nuclear charge → stronger pull on electrons

Down a Group (↓)

Decreases

  • Reason: Increase in size → weaker attraction

🔹 Important Order

Fluorine (F) is the most electronegative elementF > O > N > Cl > Br > I\textbf{F > O > N > Cl > Br > I}


Factors Affecting Electronegativity

  1. Atomic Size
    • Smaller atom → higher electronegativity
  2. Nuclear Charge
    • More protons → stronger attraction
  3. Shielding Effect
    • More shielding → lower electronegativity
  4. Hybridisation
    • sp > sp² > sp³

More s-character → electrons closer to nucleus


Important Exceptions


❗ Noble Gases

  • Usually no electronegativity
  • (Because they rarely form bonds)

Pauling Scale (Electronegativity) — JEE/NEET Concept

The Pauling scale is the most commonly used scale to measure electronegativity, proposed by Linus Pauling.


Basic Idea

Electronegativity is calculated based on:

Bond energies (bond dissociation enthalpies)

  • If a bond A–B is stronger than expected, it means:
    A and B have different electronegativities

Formula (Conceptual)

Pauling related electronegativity difference to bond energy:χAχBDABDAA+DBB2\chi_A – \chi_B \propto \sqrt{D_{AB} – \frac{D_{AA} + D_{BB}}{2}}

Where:

  • DABD_{AB} = bond energy of A–B
  • DAA,DBBD_{AA}, D_{BB}​ = bond energies of A–A and B–B

Key Points

  1. Fluorine is highest
    • Value = 4.0 (maximum on scale)
  2. Values are relative
    • Not absolute, just comparison
  3. Dimensionless
    • No unit

🔹 Example Values

ElementElectronegativity
F4.0
O3.5
N3.0
Cl3.0
H2.1

Important Applications

1. Predict Bond Type

  • ΔEN ≈ 0 → Non-polar covalent
  • ΔEN small → Polar covalent
  • ΔEN large → Ionic

2. Predict Polarity

  • Larger difference → more polar bond

Mulliken Scale — Electronegativity (JEE/NEET)

The Mulliken scale was proposed by Robert S. Mulliken.


Basic Idea

Electronegativity depends on:

Ionisation Enthalpy (IE)
Electron Gain Enthalpy (EGE / Electron Affinity)

So, it considers both:

  • Tendency to lose electron (IE)
  • Tendency to gain electron (EGE)

🔹 Formula

χ=IE+EGE2\chi = \frac{\text{IE} + \text{EGE}}{2}(In some books, electron affinity is used instead of EGE)


🔹 Meaning

  • Higher IE → atom doesn’t lose electrons easily
  • More negative EGE → atom gains electrons easily

✔ So, higher value = higher electronegativity


🔹 Key Points

  1. Absolute scale
    • Based on measurable energies (unlike Pauling)
  2. Applies to isolated atoms
    • Not bond-based
  3. Units initially in energy
    • Often converted to dimensionless values

🔹 Comparison with Pauling Scale

FeatureMullikenPauling
Based onIE + EGEBond energy
TypeAbsoluteRelative
ConceptAtomic propertyBond property

Basic Relation

The two scales are related approximately by:χPaulingχMulliken2.8\chi_{\text{Pauling}} \approx \frac{\chi_{\text{Mulliken}}}{2.8}

Electron Gain Enthalpy (EGE) — JEE/NEET Complete Concepts

🔹 Definition

Electron Gain Enthalpy is the enthalpy change when an electron is added to an isolated gaseous atom.

  • Usually negative (energy released)
  • More negative = greater tendency to gain electron

🔹 Representation

X(g)+eX(g)X(g) + e^- \rightarrow X^-(g)


🔹 General Trends

Across a Period (Left → Right)

EGE becomes more negative

Reason:

  • Increase in nuclear charge → stronger attraction for incoming electron

Down a Group (Top → Bottom)

EGE becomes less negative

Reason:

  • Increase in size → electron added far from nucleus → less attraction

IMPORTANT EXCEPTIONS (Very Important for JEE/NEET)


❗ Exception 1: Be, Mg (Group 2)

EGE ≈ zero or slightly positive

Reason:

  • Stable ns² configuration
  • Incoming electron enters higher energy p-orbital

❗ Exception 2: N (Group 15)

EGE is less negative than expected

Reason:

  • Stable half-filled configuration (np³)
  • Adding electron causes electron-electron repulsion

❗ Exception 3: Noble Gases (Group 18)

EGE is positive

Reason:

  • Completely filled orbitals → very stable
  • Electron must enter new shell → requires energy

❗ Exception 4: Fluorine vs Chlorine

👉 Cl has more negative EGE than F

❌ Expected: F > Cl
✅ Actual: Cl > F

Reason:

  • F is very small → strong electron-electron repulsion in 2p orbital
  • Cl has larger size → less repulsion → easier electron addition

🔹 Important Order Example

Among halogens:
Cl > F > Br > I


Successive Electron Gain Enthalpy

👉 Adding 2nd electron:X+eX2X^- + e^- \rightarrow X^{2-}

❗ Always positive
Reason:

  • Electron added to already negative ion → strong repulsion

🔹 Elements in Group 16

  • Oxygen (O)
  • Sulfur (S)
  • Selenium (Se)
  • Tellurium (Te)
  • Polonium (Po)

Increases down the group
O < S < Se < Te


Why is the 2nd Electron Gain Enthalpy of Oxygen Positive?

Consider the process:O(g)+eO2(g)\text{O}^- (g) + e^- \rightarrow \text{O}^{2-} (g)


🔹 Key Reason: Electron–Electron Repulsion

After gaining one electron:

  • Oxygen becomes O⁻ (negatively charged)

Now, adding another electron:

  • Incoming electron is repelled by the already negative ion

✔ So, energy must be supplied → EGE becomes positive

Isoelectronic Species (JEE/NEET Concept)

Isoelectronic species are:

Atoms, ions, or molecules that have the same number of electrons.


🔹 Examples

  • N³⁻, O²⁻, F⁻, Ne, Na⁺, Mg²⁺

All of these have 10 electrons → so they are isoelectronic


How to identify?

Just count total electrons

Example:

  • O²⁻ → 8 + 2 = 10 electrons
  • F⁻ → 9 + 1 = 10 electrons
  • Na⁺ → 11 − 1 = 10 electrons

✔ Same electrons → Isoelectronic


🔹 Important Trend (VERY IMPORTANT for JEE/NEET)

In an isoelectronic series:

Size decreases as nuclear charge (Z) increases


Example Order of Size

For species with 10 electrons:

N³⁻ > O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺

Reason:

  • All have same electrons
  • But increasing protons (Z) pull electrons closer → size decreases

🔹 Key Exam Line

“Isoelectronic species have same number of electrons but different nuclear charge.”


Van der Waals radius (JEE/NEET Concept)

The Van der Waals radius is defined as:

Half of the distance between the nuclei of two non-bonded identical atoms when they are just touching each other.


Simple Understanding

  • When atoms are not chemically bonded (no covalent/ionic bond), they still can come close due to weak intermolecular forces.
  • The distance at this closest approach = Van der Waals distance
  • So,
    Van der Waals radius = (Van der Waals distance) / 2

Example

If two neon atoms (not bonded) are 3.2 Å apart:

  • Van der Waals radius of neon = 3.2 / 2 = 1.6 Å

Correct Order of Atomic Radii:

Van der Waals radius > Metallic radius > Covalent radius


🔹 Why this order?

  1. Covalent radius (smallest)
    • Atoms are strongly bonded
    • Nuclei pull shared electrons → atoms come closer
  2. Metallic radius (middle)
    • Metal atoms are packed in a lattice
    • Bonding is weaker than covalent → atoms are slightly farther apart
  3. Van der Waals radius (largest)
    • No bonding, only weak attraction
    • Atoms stay far apart

🔹 Comparison Table

Radius TypeConditionSmaller/Larger
Covalent radiusBonded atomsSmaller
Metallic radiusMetal latticeMedium
Van der WaalsNon-bonded atomsLargest

Mendeleev’s Periodic Table – Core Concept

  • Law: Properties of elements are a periodic function of their atomic masses
  • Elements arranged in increasing atomic mass
  • Similar properties appear at regular intervals → periodicity

🧪 Key Features

1. Groups & Periods

  • Vertical columns → Groups (I to VIII)
  • Horizontal rows → Periods
  • Elements in same group → similar valency & chemical properties

2. Subgroups (A & B)

  • Each group divided into:
    • A (main group) → normal elements
    • B (transition elements) → less regular properties

3. Gaps for Undiscovered Elements ⭐ (VERY IMPORTANT)

  • Mendeleev left blank spaces for unknown elements
  • Predicted their properties accurately
    Example:
    • Eka-aluminium → Gallium
    • Eka-silicon → Germanium

4. Correction of Atomic Mass

  • If element didn’t fit → atomic mass was corrected
    • Example: Tellurium–Iodine anomaly handled by placement based on properties

5. Periodicity Trend

  • Properties repeat due to similar valence shell configuration (not known then)

⚠️ Limitations (High-weightage for exams)

1. Hydrogen Position ❓

  • Resembles Group 1 and Group 17 → no fixed position

2. Isotopes Problem

  • Isotopes have same properties but different masses
  • Could not be placed properly

3. Increasing Atomic Mass Order Violations

  • Example:
    • Argon (39.9) before Potassium (39.1)
  • Order violated to maintain chemical similarity

4. No Explanation of Periodicity

  • Reason (electronic configuration) was unknown

5. Lanthanides & Actinides

  • No proper place in table

Hydrocarbon JEE/NEET Test Paper SET-1

Hydrocarbons form the fundamental backbone of Organic Chemistry, consisting only of carbon and hydrogen atoms.
They are broadly classified into alkanes, alkenes, alkynes, and aromatic hydrocarbons based on bonding and structure.
Understanding their properties and reactions is crucial for mastering concepts in competitive exams like JEE and NEET.
This test paper is designed to evaluate conceptual clarity, problem-solving ability, and application skills.
It covers important topics such as nomenclature, isomerism, and reaction mechanisms of hydrocarbons.

  1. Arrange the following alkenes in order of their stability

2. But-2-yne and hydrogen (one mole each) are separately treated with (i) P d / C and (ii) N a / liq. N H 3 to give the products X and Y respectively

Identify the incorrect statements.

A. X and Y are stereoisomers.

B. Dipole moment of X is zero

C. Boiling point of X is higher than Y .

D. X and Y react with O 3 / Z n + H 2 O to give different products.

Choose the correct answer from the options given below :

(1) B and C only

(2) B and D only

(3) A and B only

(4) A and C only

3.

The product Y formed is :

(1) 2-methylhex-2-yne

(2) 5-methylhex-2-yne

(3) 2-methylhex-3-yne

(4) Isopropylbut-1-yne

4

5. Which compound would give 3-methyl-6- oxoheptanal upon ozonolysis ?

6. Given below are two statements :

Statement (I) : Neopentane forms only one monosubstituted derivative.

Statement (II) : Melting point of neopentane is higher than n-pentane.

In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Statement I is correct but Statement II is incorrect

(2) Both Statement I and Statement II are correct

(3) Both Statement I and Statement II are incorrect

7. Total number of sigma (σ) and pi(π) bonds respectively present in hex-1-en-4-yne are:

(1) 13, 3

(2) 14, 3

(3) 3, 14

(4) 14, 13

8. Identify product [A], [B] and [C] in the following reaction sequence.

9.Statement I : Nitration of benzene involves the following step –

Statement II: Use of Lewis base promotes the electrophilic substitution of benzene. In the light of the above statements, choose the most appropriate answer from the options given below :

(1) Both Statement I and Statement II are incorrect

(2) Statement I is correct but Statement II is incorrect (3) Both Statement I and Statement II are correct (4) Statement I is incorrect but Statement II is correct

10.Compound A formed in the following reaction reacts with B gives the product C. Find out A and B

11. Product ‘A’ of following sequence of reactions is

12

A) Cyclohexanol
B) Hexane-1,6-dioic acid (Adipic acid)
C) Cyclohexanone
D) 1,2-diol (Glycol)

13.Consider the following reaction :

Which of these reaction(s) will not produce Saytzeff product ?

Which of these reaction(s) will not produce Saytzeff product ? (

1) (c) only

(2) (a) (c) and (d)

(3) (d) only

(4) (b) and (d)

15. In the following skew conformation of ethane, H’–C–C–H” dihedral angle is :

In the following skew conformation of ethane, H’–C–C–H” dihedral angle is :

(1) 120º

(2) 58º

(3) 151º

(4) 149º

Basic Idea of Wedge–Dash Notation

Writing wedge–dash (3D) configuration is a key skill in stereochemistry (very important for JEE/NEET). Let’s make it simple and practical.


It shows how atoms are arranged in 3D space around a carbon atom.

3 types of bonds:

  1. Straight line (—)
    → Bond lies in the plane of the paper
  2. Solid wedge (▲)
    → Bond coming out toward you
  3. Dashed wedge (▱ / —)
    → Bond going away from you

How to Draw Wedge–Dash Structure (Step-by-step)

Step 1: Identify the central atom

Usually a chiral carbon (sp³ hybridised) with 4 different groups.


Step 2: Draw two bonds in plane

Use simple straight lines for 2 groups


Step 3: Add wedge and dash

  • One group → solid wedge (toward you)
  • One group → dashed wedge (away from you)


Important Rules

✔ Carbon must be tetrahedral (109.5°)
✔ Always show 3D orientation clearly
✔ You can rotate the molecule, but relative positions must stay same


Interconversion Tip (Fischer → Wedge-Dash)

  • Horizontal bonds in Fischer → come out (wedge)
  • Vertical bonds → go back (dash)

Common Mistakes

❌ Putting all bonds in plane
❌ Using wedge/dash randomly
❌ Changing configuration while rotating


Exam Trick (JEE/NEET)

If lowest priority group (4th) is on:

  • Dash (away) → read order normally
  • Wedge (toward)reverse R/S result