Author: Sarvan Kumar

Sarvan Kumar http://sarvankumar.wordpress.com My name is Sarvan Kumar, I am post graduate in chemistry (MSc). besides teaching chemistry at various schools and coaching centres I have been giving home tuitions to students for 10 years. I have helped students score good marks in chemistry not only in board examination but also JEE, NEET, SAT, IGCSE and IB examinations. Nearly 90% of my students have scored more than 95% in their CBSE board examination. Moreover they have also secured a seat in prestigious engineering and medical college. Since every student is different hence my unique style of teach

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

R and S Configuration (JEE/NEET Concept – Easy Explanation)

R/S configuration is used to describe the absolute configuration of a chiral carbon (asymmetric carbon with 4 different groups)

Step-by-Step Rules (Cahn–Ingold–Prelog Rules)

Step 1: Assign Priority (1 → 4)

  • Higher atomic number = higher priority
  • Example:
    I > Br > Cl > S > O > N > C > H

If same atom, move to next atoms (tie-breaker rule)

Step 2: Orient the Molecule

  • Keep the lowest priority group (4) away from you (behind the plane)

Step 3: Trace Path (1 → 2 → 3)

  • Clockwise direction → R (Rectus)
  • Anticlockwise direction → S (Sinister)

Important Shortcut

If the lowest priority (4) is towards you (front):

  • Reverse the result
    • Clockwise → S
    • Anticlockwise → R

Key Points for Exams

  • R/S is about 3D arrangement, not optical rotation (+/−)
  • Always check lowest priority position
  • Double/triple bonds are treated as multiple single bonds

Quick Trick (Exam Speed)

“4 behind → direct result“4 in front → reverse result”

The R and S configuration for each stereogenic centre in this from top to bottom is:

A.R, R, R
B.R, S, S
C.R, S, R
D.S, S, R

The correct option is (C)

The R and S Configuration Practice Problems

(i)

Groups: OH, CHO, CH₂OH, H
Priority: OH (1) > CHO (2) > CH₂OH (3) > H (4)

  • H is horizontal (front) → ❗reverse result
  • 1 → 2 → 3 = anticlockwise → reversed → R

✔️ Answer: (i) → R

(ii)

Groups: OH, COOH, CH₃, H
Priority: OH (1) > COOH (2) > CH₃ (3) > H (4)

  • H is horizontal (front) → ❗reverse
  • 1 → 2 → 3 = clockwise → reversed → S

✔️ Answer: (ii) → S

(iii)

Groups: NH₂, COOH, CH₃, H
Priority: NH₂ (1) > COOH (2) > CH₃ (3) > H (4)

  • H is horizontal (front) → ❗reverse
  • 1 → 2 → 3 = anticlockwise → reversed → R

✔️ Answer: (iii) → R

(iv) (Two chiral centers)

🔹 Upper carbon

Priority: NH₂ (1) > COOH (2) > lower C (3) > H (4)

  • H is horizontal → reverse
  • 1 → 2 → 3 = anticlockwise → reversed → R

🔹 Lower carbon

Priority: OH (1) > upper C (2) > CH₃ (3) > H (4)

  • H is horizontal → reverse
  • 1 → 2 → 3 = clockwise → reversed → S

✔️ Answer: (iv) → (R, S)

Final Answers:

  • (i) → R
  • (ii) → S
  • (iii) → R
  • (iv) → (R, S)

Position of Elements on Lothar Meyer Curve (JEE/NEET)

Lothar Meyer’s Contribution to Periodic Table (JEE/NEET Concepts)

Julius Lothar Meyer (1869) was a German chemist who independently worked on classification of elements, around the same time as Dmitri Mendeleev.

Key Idea of Lothar Meyer

He arranged elements based on:

  • Atomic mass
  • Valency (combining capacity)
  • Physical properties

Most Important Contribution

Atomic Volume vs Atomic Mass Graph

He plotted a graph between:

  • Atomic Volume (volume occupied by 1 mole of element)
  • Atomic Mass

This graph showed a periodic pattern, proving that properties repeat regularly.

On the atomic volume vs atomic mass curve given by
Julius Lothar Meyer, elements occupy characteristic positions depending on their properties.

1. Peaks (Maxima) → Highly Reactive Metals

Elements at the top of the curve

  • Alkali metals:
    • Li, Na, K, Rb, Cs

✔ Features:

  • Maximum atomic volume
  • Highly reactive
  • Valency = +1

2. Just After Peaks → Alkaline Earth Metals

Slightly lower than peaks

  • Be, Mg, Ca, Sr, Ba

✔ Features:

  • Less reactive than alkali metals
  • Valency = +2

3. Descending Region → Transition Elements

Moving downward from peak

  • Fe, Co, Ni, etc.

✔ Features:

  • Moderate atomic volume
  • Variable valency
  • Metallic character

4. Troughs (Minima) → Least Reactive Elements

Bottom of the curve

  • Elements like:
    • Cu, Zn (near minima region)

✔ Features:

  • Minimum atomic volume
  • Stable nature

5. Rising Region → Non-metals

Moving upward after minima

  • C, N, O, F

✔ Features:

  • Increasing electronegativity
  • Non-metallic character increases

Newlands’ Law of Octaves (JEE/NEET Concepts)

Newlands’ Law of Octaves (JEE/NEET Concepts)

Proposed by John Alexander Reina Newlands (1864), this was another early attempt to classify elements.

🔹 What is Newlands’ Law of Octaves?

Statement:
When elements are arranged in increasing order of atomic mass,
every 8th element has properties similar to the first, just like musical octaves 🎵

Why “Octaves”?

Inspired by music:

  • In music, the 8th note repeats
  • Similarly, in elements:
    • 1st ≈ 8th
    • 2nd ≈ 9th
    • and so on

🔹 Example (Important for Exams)

Element No.ElementSimilar To
1LiNa (8th)
2BeMg
3BAl
4CSi
5NP
6OS
7FCl

Example:

  • Li and Na → similar properties
  • Be and Mg → similar properties

Key Features

  • First to show periodicity in a systematic way
  • Applicable mainly to lighter elements (up to Ca)
  • Based on atomic mass

🔹 Limitations (VERY IMPORTANT )

  1. Worked only up to Calcium (Ca)
  2. No space for new elements
  3. Dissimilar elements grouped together:
    • Co and Ni placed with F, Cl
  4. Ignored transition elements
  5. Assumed only 56 elements existed

🔹 Why Important?

  • First clear attempt to show periodic repetition
  • Helped in development of:
    • Dmitri Mendeleev Periodic Table

Dobereiner’s Triads (JEE/NEET Concepts)

Johann Wolfgang Döbereiner (1829) proposed one of the earliest attempts to classify elements based on their properties.

What are Dobereiner’s Triads?

A triad is a group of three elements having:

  • Similar chemical properties
  • Gradual change in physical properties
  • A special relationship in atomic masses

Key Rule (Important for exams):
The atomic mass of the middle element ≈ average of the other two

Examples of Dobereiner’s Triads

1. Alkali Metal Triad

  • Lithium (Li) = 7
  • Sodium (Na) = 23
  • Potassium (K) = 39

✔ Check:7+392=23≈Na\frac{7 + 39}{2} = 23 \approx \text{Na}27+39​=23≈Na

2. Alkaline Earth Metal Triad

  • Calcium (Ca) = 40
  • Strontium (Sr) = 88
  • Barium (Ba) = 137

✔ Check:40+1372=88.5≈Sr\frac{40 + 137}{2} = 88.5 \approx \text{Sr}

3. Halogen Triad

  • Chlorine (Cl) = 35.5
  • Bromine (Br) = 80
  • Iodine (I) = 127

✔ Check:35.5+1272=81.25≈Br\frac{35.5 + 127}{2} = 81.25 \approx \text{Br}

🔹 Important Characteristics

  • Properties change gradually across the triad
  • Middle element shows intermediate behavior
  • First hint toward periodicity in elements

🔹 Limitations (VERY IMPORTANT for NEET/JEE)

  • Only a few elements could be grouped
  • Many known elements did not fit into triads
  • Did not explain why periodicity occurs