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

45 NEET-level MCQs covering Solutions, Classification, d-block elements, and Coordination Compounds

SECTION 1: SOLUTIONS (10 MCQs)

Which of the following is an ideal solution?
A) Acetone + Chloroform
B) Benzene + Toluene
C) Ethanol + Water
D) Water + HCl
Raoult’s law is applicable to:
A) Non-volatile solutes
B) Electrolytes
C) Ideal solutions
D) Strong acids
Mass of glucose $(C_{6} H_{12} O_{6})$ required to be dissolved to prepare one litre of its solution which is isotonic with 15 g $L^{- 1}$ solution of urea $({NH}_{2} {CONH}_{2})$ is(Given: Molar mass in g ${mol}^{- 1}$ C : 12, H : 1, O : 16, N : 14)
A) 55g
B) 30g
C) 15g
D) 45g
The Henry’s law constant (KH) values of three gases (A, B, C) in water are 145, 2 × 10^-5 and 35 kbar, respectively. The solubility of these gases in water follow the order:
A) B > A > C (B) B > C > A (C) A > C > B (D) A > B > C
The plot of osmotic pressure ( ∏ ) vs concentration ( mol^{− 1} ) for a solution gives a straight line with slope 25.73 L bar mol⁻¹ . The temperature at which the osmotic pressure measurement is done is (Use R = 0.083 L bar mol⁻¹ K ⁻¹ )
A) 1
B) 2
C) 3
D) 0
Osmotic pressure is given by:

$\pi = iCRT$

A) Boyle’s law
B) Ideal gas equation
C) van’t Hoff equation
D) Dalton’s law

Maximum boiling azeotrope shows:
A) Positive deviation
B) Negative deviation
C) No deviation
D) Infinite deviation
Which increases boiling point?
A) Adding volatile solute
B) Adding non-volatile solute
C) Decreasing pressure
D) Increasing solvent
Henry’s law relates:
A) Pressure and solubility
B) Volume and temperature
C) Mass and density
D) Temperature and pressure
Which amongst the following aqueous solution of electrolytes will have minimum elevation in boiling point? Choose the correct option :-
A) 0.05 M NaCl
B) 0.1 M KCl
C) 0.1MgSO₄
D) 1M NaCl

SECTION 2: CLASSIFICATION OF ELEMENTS (10 MCQs)

Which of the following statements are true?
A) Unlike Ga that has a very high melting point, Cs has a very low melting point.
B) On Pauling scale, the electronegativity values of N and Cl are not the same
C) A r , K⁺ , C l⁻ , C a^{2 +}, and S ^{2 −}are all isoelectronic species.
D) The correct order of the first ionization enthalpies of N a , M g , A l and Si is S i > A l > M g > Na

12. Which among the following electronic configurations belong to main group elements?

Choose the correct answer from the option given below :

(1) B and E only

(2) A and C only

(3) D and E only

(4) A, C and D

13.The correct decreasing order of atomic radii (pm) of Li, Be, B and C is

(1) Be > Li > B > C

(2) Li > Be > B > C

(3) C > B > Be > Li

(4) Li > C > Be > B

14.Match List-I with List-II :

Identify the correct answer from the options given below :

(1) A-III, B-I, C-IV, D-II

(2) A-I, B-IV, C-III, D-II

(3) A-II, B-IV, C-I, D-III

(4) A-III, B-IV, C-I, D-II

15.Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N

Choose the correct answer from the options given below:

(1) Li < Be < B < C < N

(2) Li < B < Be < C < N

(3) Li < Be < C < B < N
(4) Li < Be < N < B < CR

16. Arrange the following elements in increasing order of electronegativity:

N, O, F, C, Si

Choose the correct answer from the options given below :

Choose the correct answer from the options given below:

(1) Si < C < N < O < F

(2) Si < C < O < N < F

(3) O < F < N < C < Si

(4) F < O < N < C < Si

17 Which of the following is correctly matched?

18. The correct sequence given below containing neutral, acidic, basic and amphoteric oxide each, respectively, is

Options: A. NO, ZnO, CO₂, CaO

B. ZnO, NO, CaO, CO₂

C. NO, CO₂, ZnO, CaO

D. NO, CO₂, CaO, ZnO

19. For elements B, C, N, Li, Be, O and F the correct order of first ionization enthalpy is

(1) Li < Be < B < C < N < O < F

(2) B > Li > Be > C > N > O > F

(3) Li < B < Be < C < O < N < F

(4) Li < Be < B < C < O < N < F

20. Decrease in size from left to right in actinoid series is greater and gradual than that in lanthanoid series due to

1.4f orbitals are second to last

2. 4f orbitals have greater shielding effect

3. 5f orbitals have poor shielding effect

4.5f orbitals have greater shielding effect

SECTION 3: d-BLOCK ELEMENTS (10 MCQs)

Transition elements have:
A) Completely filled d-orbitals
B) Partially filled d-orbitals
C) No d-orbitals
D) Only s-orbitals
General configuration:
A) (n-1)d¹–¹⁰ ns¹–²
B) ns² np⁶
C) ns¹
D) np⁵
Highest oxidation state of Mn:
A) +2
B) +4
C) +7
D) +6
Color in transition metals due to:
A) s-electrons
B) d-d transition
C) p-electrons
D) nuclear charge
Which is not transition element?
A) Zn
B) Fe
C) Cu
D) Cr
Catalytic property due to:
A) Low density
B) Variable oxidation states
C) High melting point
D) Atomic size
Magnetic behavior depends on:
A) paired electrons
B) unpaired electrons
C) protons
D) neutrons
Which is diamagnetic?
A) Fe²⁺
B) Mn²⁺
C) Zn²⁺
D) Cu²⁺
Alloy of Cu and Zn:
A) Bronze
B) Brass
C) Steel
D) Solder
Highest melting point:
A) Hg
B) Fe
C) W
D) Cu

SECTION 4: COORDINATION COMPOUNDS (15 MCQs)

Coordination number of central atom is:
A) number of ligands
B) number of bonds
C) number of donor atoms
D) oxidation state
Ligand donating one pair:
A) Bidentate
B) Monodentate
C) Tridentate
D) Polydentate
Example of bidentate ligand:
A) NH₃
B) H₂O
C) en
D) Cl⁻
IUPAC name of [Cu(NH₃)₄]²⁺:
A) Copper tetraammine
B) Tetraammine copper(II)
C) Ammine copper
D) Copper ammonia
Oxidation state of Fe in [Fe(CN)₆]³⁻:
A) +2
B) +3
C) +6
D) −3
Coordination compounds show isomerism:
A) Yes
B) No
C) Rarely
D) Only optical
Geometrical isomerism in:
A) tetrahedral
B) square planar
C) linear
D) trigonal
Strong field ligand:
A) Cl⁻
B) H₂O
C) CN⁻
D) Br⁻
Crystal field splitting occurs in:
A) s-orbitals
B) p-orbitals
C) d-orbitals
D) f-orbitals
High spin complex has:
A) pairing
B) maximum unpaired electrons
C) no electrons
D) stable structure
Hybridization of [Ni(CN)₄]²⁻:
A) sp³
B) dsp²
C) sp²
D) d²sp³
Werner theory explains:
A) valency
B) bonding
C) structure of complexes
D) reactivity
Chelate effect increases:
A) instability
B) stability
C) reactivity
D) acidity
[Co(NH₃)₆]³⁺ is:
A) paramagnetic
B) diamagnetic
C) ferromagnetic
D) antiferromagnetic
Which shows optical isomerism?
A) [Co(en)₃]³⁺
B) [NiCl₄]²⁻
C) [PtCl₄]²⁻
D) [Zn(NH₃)₄]²⁺

Organic chemistry – Some Basic Principles and techniques JEE/NEET Concepts

Methods of Purification of Organic Compounds

Basic Idea of Wedge–Dash Notation

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

Conformations of Propane (JEE/NEET Concepts)

Stereoisomerism organic chemistry JEE/NEET Concepts

Key Concepts to Arrange Nucleophile Strength (NEET/JEE)

Key Concepts to Arrange Basic Strength (NEET/JEE)

Degree of Unsaturation (DU) formula for organic compound

Methods of Purification of Organic Compounds

Sublimation

What is Sublimation?

Sublimation is the process in which a solid directly changes into vapour without becoming liquid.

Example:

Ammonium chloride
Iodine
Camphor
Naphthalene

Why does it happen?

Some solids have:

Weak intermolecular forces
High vapour pressure even at lower temperatures

So, when heated, they directly escape into vapour form.

Sublimation as a Separation Technique

It is used to purify substances:

✔ Sublimable substance → turns into vapour
✔ Non-sublimable impurities → remain as solid

Then vapour is cooled → forms pure solid again.

Simple Setup

Heat the mixture
Sublimable substance vaporizes
Vapour condenses on a cold surface
Pure solid is collected

Example Use

Separating:

Ammonium chloride + sand

Ammonium chloride sublimates, sand stays behind.

Crystallisation

What is Crystallisation?

A purification technique used to purify solid organic compounds based on difference in solubility.

Principle

Substance is less soluble at low temperature
Substance is more soluble at high temperature

So:

Dissolve at high temp
Cool → pure crystals form

Steps

Dissolve impure solid in hot solvent
Make a saturated solution
Filter (if needed, remove insoluble impurities)
Cool the solution
Pure crystals separate out
Filter and dry crystals

Important Terms

Mother liquor → remaining liquid after crystallisation
Activated charcoal → removes colored impurities

🔹 Examples

Example 1: Benzoic Acid purification

Impure benzoic acid dissolved in hot water
On cooling → pure crystals of benzoic acid form

When repeated crystallisation is needed?

When

Simple Distillation (large boiling point difference)

When used:

Difference in b.p. > 25–30°C
Or liquid + non-volatile impurity

Examples:

Example 1:

Water + salt
Water distils, salt remains

Example 2:

Chloroform (334 K) + Aniline (457 K)
Easily separated

Example 3:

Alcohol + sugar solution
Alcohol vaporises, sugar stays

Example 4:

Acetone + water
Acetone (low b.p.) comes first

Fractional Distillation (close boiling points)

When used:

Difference in b.p. < 25°C

Examples:

Example 1:

Ethanol (78°C) + Water (100°C)
Cannot separate by simple distillation

Example 2:

Benzene (80°C) + Toluene (110°C)
Need fractionating column

Example 3:

Hexane + Heptane
Very close boiling points

Example 4 (Industrial):

Crude oil refining
Petrol, diesel, kerosene separated

3. Steam Distillation (heat-sensitive substances)

When used:

Substance:
- Immiscible with water
- Volatile with steam
- Decomposes at high temperature

Examples:

Example 1:

Extraction of essential oils
From plants (e.g., clove oil, eucalyptus oil)

Example 2:

Aniline purification
High b.p. but steam volatile

Example 3:

Nitrobenzene
Separated using steam

Example 4:

Rose oil extraction (attar)
Used in perfumes

Quick Comparison (Very Important)

Method	Condition	Example
Simple Distillation	Large b.p. difference	Water + salt
Fractional Distillation	Small b.p. difference	Ethanol + water
Steam Distillation	Heat-sensitive, steam volatile	Essential oils

Distillation Under Reduced Pressure (Vacuum Distillation)

Statement:

Distillation under reduced pressure → Glycerol + spent-lye

Why this method is used?

Some liquids:

Have very high boiling points
Decompose before boiling

Example: Glycerol

Principle

When pressure is reduced:
➡ Boiling point decreases

So, substance can distil at lower temperature without decomposition

Application in given case

Mixture:

Glycerol (high b.p., heat sensitive)
Spent-lye (impurities)

What happens?

Pressure ↓
Glycerol boils at lower temp
Vapours collected → pure glycerol
Impurities remain

Why not simple distillation?

❌ Glycerol decomposes at high temperature
✔ So vacuum distillation is required

🔹 More Examples

Example 1:

Glycerol purification (most common)

Example 2:

High boiling oils
Example 3:
Fatty acids

Example 4:

Petroleum residues

Quick Summary

Method	Used for
Simple distillation	Low b.p. liquids
Fractional distillation	Close b.p. liquids
Steam distillation	Heat-sensitive, steam volatile
Reduced pressure distillation	High b.p., decomposing liquids

Differential Extraction

When an organic compound is present in an aqueous medium, it is separated by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and aqueous solution should be immiscible so that they form two distinct layers which can be separated using a separating funnel. The organic solvent is later removed by distillation or evaporation to obtain the compound.

Concept: Differential (Liquid–Liquid) Extraction

Principle

Based on difference in solubility of a compound in two immiscible liquids

Usually:

Water (aqueous layer)
Organic solvent (like ether, benzene)

Key Conditions

✔ Two liquids must be immiscible
✔ Compound should be more soluble in organic solvent

Process (Separating Funnel)

Take mixture in separating funnel
Add organic solvent
Shake well
Allow layers to separate

Two layers form:

Upper layer (usually organic solvent)
Lower layer (water)

Separate layers
Evaporate solvent → get pure compound

Example

Example 1:

Iodine in water + carbon tetrachloride (CCl₄)
→ Iodine moves to organic layer

Example 2:

Benzoic acid from water using ether

Continuous Extraction

When used?

When compound is slightly soluble in solvent

Why?

Single extraction inefficient
Repeated extraction increases yield

Same solvent reused again and again

Advantages

Efficient separation
Better recovery of compound
Widely used in organic chemistry labs

Chromatography

What is Chromatography?

A separation technique where components of a mixture are separated based on their different affinities between two phases.

Principle

Based on distribution of components between two phases:

Stationary phase → fixed (solid or liquid)
Mobile phase → moving (liquid or gas)

Different substances move at different speeds → separation occurs

Types of Chromatography

1. Paper Chromatography

Stationary phase → paper
Mobile phase → solvent

Used for:

Ink separation
Plant pigments

2. Thin Layer Chromatography (TLC)

Stationary phase → silica gel layer
Faster and more accurate than paper chromatography

3. Column Chromatography

Column filled with adsorbent (silica/alumina)
Used for larger scale separation

4. Gas Chromatography (GC)

Mobile phase → gas
Used for volatile compounds

How Separation Happens?

Component with more attraction to stationary phase → moves slowly
Component with more attraction to mobile phase → moves faster

Example

Ink Separation

Black ink → separates into different colors on paper

Plant Pigments

Chlorophyll, carotene separated using chromatography

Important Term: Rf Value

$Rf = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent}}$

Uses

✔ Separation of mixtures
✔ Purification
✔ Checking purity
✔ Drug testing
✔ Food analysis

Stephen Reaction (Stephen Aldehyde Synthesis)

tephen Reaction (Stephen Aldehyde Synthesis) – Mechanism (JEE/NEET Concept)

Definition:
Conversion of nitriles (R–C≡N) into aldehydes (R–CHO) using SnCl₂/HCl, followed by hydrolysis.

Overall Reaction

$R-C \equiv N \xrightarrow[\text{HCl}]{\text{SnCl}_2} R-CH=NH \cdot HCl \xrightarrow{\text{H}_2O} R-CHO$

Stepwise Mechanism

Step 1: Formation of Iminium Salt (Reduction Step)

SnCl₂ (mild reducing agent) donates electrons.
Nitrile gets partially reduced.
Formation of iminium chloride salt:

$R-C \equiv N \xrightarrow{\text{SnCl}_2/HCl} R-CH=NH \cdot HCl$

Key point:

Triple bond (C≡N) → double bond (C=NH)
Controlled reduction (not all the way to amine)

Step 2: Hydrolysis of Iminium Salt

Iminium salt reacts with water.
Produces aldehyde + ammonium salt

$R-CH=NH \cdot HCl + H_2O \rightarrow R-CHO + NH_4Cl$ R−CH=NH⋅HCl+H2O→R−CHO+NH4Cl

Important Points (Exam-Oriented)

Works best for alkyl nitriles (aryl nitriles give poor yield).
Stops at aldehyde stage (unlike strong reduction → amine).
SnCl₂/HCl = selective mild reducing system
Intermediate = iminium salt (not imine directly)

Shortcut Trick (Memory Tip)

“Nitrile + Mild Reduction = Aldehyde”
Think: Stephen = Stops Early (Aldehyde, not amine)

Example

$CH_3CN \xrightarrow{\text{SnCl}_2/HCl} CH_3CHO$

Gattermann–Koch Reaction (JEE/NEET Must-Know)

What is it?

Introduction of a formyl group (–CHO) into an aromatic ring using:

CO + HCl in presence of Aluminium chloride and Cuprous chloride

Mechanism (Stepwise)

Formation of Electrophile

CO + HCl react in presence of AlCl₃/CuCl
Generate formyl cation

$CO + HCl \rightarrow HCO^+ \ (\text{electrophile})$

Electrophilic Attack (EAS Step)

Benzene ring attacks HCO⁺
Forms sigma complex (arenium ion)

Deprotonation

Loss of H⁺ restores aromaticity
Forms benzaldehyde complex

Hydrolysis

Final hydrolysis gives aldehyde (–CHO)

Key Points for Exams

✔ Type of Reaction

Electrophilic Aromatic Substitution (EAS)

✔ Important Conditions

CO must be dry
Presence of both AlCl₃ and CuCl is essential

❌ Limitations

Does NOT work with strongly deactivated rings
(e.g., nitrobenzene ❌)

🔁 Comparison Trick

Reaction	Reagent	Product
Gattermann–Koch	CO + HCl	–CHO
Friedel–Crafts	R–Cl / RCOCl	–R / –COR

🔹 One-Line Revision

Gattermann–Koch introduces –CHO on benzene via formyl cation (HCO⁺) using CO + HCl / AlCl₃–CuCl

Etard reaction mechanism

Oxidation of a benzylic methyl group (–CH₃ attached to benzene) to an aldehyde (–CHO) using
Chromyl chloride in a non-aqueous solvent (like CCl₄).

Mechanism (Stepwise)

Formation of Etard Complex

Chromyl chloride attacks benzylic hydrogen
Forms a brown complex (Etard complex)

$Ar-CH_3 \rightarrow Ar-CH(OCrOCl_2)_2$

Hydrolysis of Complex

On hydrolysis (H₂O), the complex breaks
Gives aldehyde

$Ar-CH(OCrOCl_2)_2 \xrightarrow{H_2O} Ar-CHO$

The unsaturated ether on acidic hydrolysis produces carbonyl compounds

Key Idea

Unsaturated ethers (enol ethers) on acidic hydrolysis give carbonyl compounds (aldehydes/ketones).

Why does this happen?

Unsaturated ethers are basically enol forms of carbonyl compounds.

General structure: $R – CH = CH – OR’$

Under acidic conditions:

Protonation of ether oxygen
Formation of unstable enol
Tautomerism (enol → carbonyl)

🔹 Important Reaction

Example: $CH_2 = CH – OCH_3 \xrightarrow{H^+, H_2O} CH_3CHO + CH_3OH$

Product:

Aldehyde/Ketone (carbonyl compound)
Alcohol

🔹 Mechanism Shortcut (Exam Trick ⚡)

Break C–O bond
Form enol
Enol → carbonyl (keto form)

One-line Concept (for revision)

Unsaturated ether + acidic hydrolysis → enol → carbonyl compound

Important for Exams

Only vinyl / allyl ethers show this behavior
Due to keto-enol tautomerism
Very common in mechanism-based questions

Some Questions from Aldehydes, Ketone and carboxylic acids.

The major product formed in the following conversion is

2.The product formed from the following reaction sequence is

3.The correct order of strengths of the carboxylic acids

Q 4.Given below are two statements :

Statement I : Cross aldol condensation between two different aldehydes will always produce four different products.

Statement II : When semicarbazide reacts with a mixture of benzaldehyde and acetophenone under optimum pH , it forms a condensation product with acetophenone only.

Q.5

In the light of the above statements, choose the correct answer from the options given below

A student is given one compound among the following compounds that gives positive test with Tollen’s reagent

Cannizzaro Reaction (JEE/NEET Concepts)

Cannizzaro Reaction (JEE/NEET Concepts)

Definition:
Cannizzaro reaction is a disproportionation reaction in which an aldehyde without α-hydrogen undergoes self oxidation–reduction in the presence of strong base (NaOH/KOH).

General Reaction

$2RCHO + OH^- \rightarrow RCH_2OH + RCOO^-$

(One molecule is reduced → alcohol, other oxidized → carboxylate salt)

Key Conditions (VERY IMPORTANT for JEE)

Aldehyde must have NO α-H (alpha hydrogen)
Strong base: conc. NaOH or KOH
Usually occurs with:
- Formaldehyde (HCHO)
- Benzaldehyde (C₆H₅CHO)

Example

$2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa$

(Benzaldehyde → benzyl alcohol + sodium benzoate)

Mechanism (Conceptual Steps)

Nucleophilic attack of OH⁻ on aldehyde → alkoxide intermediate
Hydride transfer (H⁻ shift) from one molecule to another
Formation of:
- Alcohol (reduction)
- Carboxylate ion (oxidation)

Types of Cannizzaro Reaction

Self Cannizzaro
Same aldehyde reacts
Example: benzaldehyde
Cross Cannizzaro (Important!)
Two different aldehydes
- One should be formaldehyde (HCHO) (best reducing agent)

$HCHO + C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + HCOONa$

(Formaldehyde gets oxidized, other gets reduced)

Epoxide + Grignard Reagent (Complete JEE/NEET Concept)

This is a high-weightage concept from alcohols + organometallics.

General Reaction:

$\text{Epoxide} + RMgX \xrightarrow{\text{dry ether}} \text{alkoxide} \xrightarrow{H_3O^+} \text{alcohol}$

Core Mechanism:

Grignard reagent (RMgX) behaves like R⁻ (strong nucleophile)
Attacks epoxide → ring opening (SN2 type)
Final step: acidic hydrolysis → alcohol

Golden Rules (VERY IMPORTANT):

Attack Position:

✔ Always attacks LESS substituted carbon (SN2 mechanism)

Product Type:

✔ Alcohol is formed at the carbon where O⁻ was present
✔ Nature depends on epoxide structure:

Ethylene oxide → Primary alcohol
Substituted epoxide → Secondary / tertiary alcohol

Chain Extension Trick:

✔ Adds +2 carbons (only with ethylene oxide clearly)
✔ With other epoxides → chain increases but depends on structure

Important Cases:

1. Ethylene Oxide:

$RMgX \rightarrow R-CH_2-CH_2-OH$

Always primary alcohol
+2 carbon extension (fixed)

2. Unsymmetrical Epoxide (e.g., Propylene oxide):

$RMgX \rightarrow R-CH_2-CH(R’)-OH$

Attack at less hindered carbon
Usually secondary alcohol

JEE/NEET Quick Summary:

✔ RMgX = nucleophile (R⁻)
✔ Epoxide opening = SN2 reaction
✔ Attack = less substituted carbon
✔ Product = alcohol after H₃O⁺