General Physical Properties (d-Block)
⭐ Atomic & Ionic Size
✔️ Size decreases from left → right (↑ nuclear charge)
✔️ After middle → almost constant (poor shielding by d-electrons)

⭐ Density
✔️ Increases from 3d → 4d → 5d
📌 Reason:
Increase in atomic mass
Lanthanide contraction (5d elements smaller → higher density)

⭐ Melting & Boiling Points
✔️ Very high
📌 Reason:
Strong metallic bonding
Involvement of d-electrons
📌 Exceptions:
Zn, Cd, Hg → low m.p.
✔️ Hg → liquid (weak metallic bonding)

⭐ Variable Oxidation States
✔️ Due to similar energy of (n−1)d & ns electrons
📌 Example:
Fe → +2, +3
Mn → +2 to +7

⭐ Magnetic Properties
✔️ Paramagnetic → unpaired electrons
✔️ Diamagnetic → all paired
📌 Magnetic moment:
U=√n(n+2)

⭐ Colour
✔️ Due to d–d transitions
✔️ Zn²⁺, Cd²⁺ → colourless (d¹⁰)
@Ayano1me @Neetugpoll @neetugquiz

🌟 Lanthanide Contraction (VERY HIGH YIELD 🔥)
🌱 Definition
Gradual decrease in atomic & ionic radii of lanthanides from La → Lu
#Reason
✔️ Poor shielding effect of 4f-electrons
✔️ Effective nuclear charge increases
🔥 Consequences of Lanthanide Contraction
1️⃣ Similar size of 4d & 5d elements
📌 Zr ≈ Hf
📌 Chemical properties similar
2️⃣ High density of 5d elements
📌 Reason:
Smaller size
Higher mass
3️⃣ Difficulty in separation of lanthanides
📌 Very similar radii & properties
4️⃣ Basic strength of hydroxides decreases
📌 La(OH)₃ > Lu(OH)₃
5️⃣ Decrease in ionic radii of Ln³⁺ ions
📌 Affects coordination number & complex formation
#NEETHOTPOINTS
✔️ Poor shielding → 4f electrons
✔️ Cause of Zr–Hf similarity → Lanthanide contraction
✔️ Colour in d-block → d–d transition
✔️ Colour in f-block → f–f transition
✔️ Most stable oxidation state of lanthanides → +3
@Ayano1me @Neetugpoll @neetugquiz

#CoordinationCompounds
Valence Bond Theory (VBT) & Isomerism (NCERT • NEET)

🌱 Coordination Compound
A compound in which: ✔️ Central metal atom/ion
✔️ Surrounded by ligands
✔️ Linked by coordinate bonds
📌 Example: [Co(NH₃)₆]Cl₃

⭐ #ValenceBondTheory (VBT)
🌱 Explains: ✔️ Geometry
✔️ Magnetic behaviour
✔️ Hybridisation

🔹 Postulates of VBT
✔️ Metal ion provides empty orbitals
✔️ Ligands donate lone pair
✔️ Overlap → coordinate bond
✔️ Paired/unpaired electrons decide magnetic nature

⭐ #Hybridisation&Geometry
Hybridisation :Geometry
d²sp³ / sp³d² :Octahedral
sp³ : Tetrahedral
dsp² : Square planar

#InnerOrbitalvsOuterOrbitalComplex
⭐ Inner Orbital Complex (Low spin)
✔️ Uses (n−1)d orbitals
✔️ Pairing of electrons occurs
✔️ Strong field ligands
📌 Example:
[Co(NH₃)₆]³⁺ → d²sp³ (octahedral)

⭐ Outer Orbital Complex (High spin)
✔️ Uses nd orbitals
✔️ No pairing
✔️ Weak field ligands
📌 Example:
[FeF₆]³⁻ → sp³d² (octahedral)

#MagneticNature (VBT)
✔️ Unpaired e⁻ → Paramagnetic
✔️ Paired e⁻ → Diamagnetic
📌 Example:
[Ni(CN)₄]²⁻ → Diamagnetic (dsp²)
[NiCl₄]²⁻ → Paramagnetic (sp³)
⚠️ Limitations of VBT
❌ Cannot explain colour
❌ Cannot explain strong vs weak ligands clearly
❌ No quantitative explanation of spectra

@Ayano1me @Neetugpoll @Neetugquiz

Isomerism in Coordination Compounds
🌱 Compounds with same formula but different arrangement

⭐ #Structural Isomerism
1️⃣ Ionisation Isomerism
✔️ Exchange between ligand & counter ion
📌 Example:
[Co(NH₃)₅SO₄]Br
[Co(NH₃)₅Br]SO₄

2️⃣ Solvate (Hydrate) Isomerism
✔️ Water inside or outside coordination sphere
📌 Example:
[Cr(H₂O)₆]Cl₃
[Cr(H₂O)₅Cl]Cl₂·H₂O

3️⃣ Linkage Isomerism
✔️ Ambidentate ligands
📌 Example:
NO₂⁻ → nitro / nitrito
SCN⁻ → thiocyanato / isothiocyanato

4️⃣ Coordination Isomerism
✔️ Between cationic & anionic complexes
📌 Example:
[Co(NH₃)₆][Cr(CN)₆]
[Cr(NH₃)₆][Co(CN)₆]

⭐ #Stereoisomerism
1️⃣ Geometrical Isomerism
✔️ cis–trans
📌 Examples:
[Pt(NH₃)₂Cl₂]
[Co(NH₃)₄Cl₂]⁺

2️⃣ Optical Isomerism
✔️ Non-superimposable mirror images
✔️ d & l forms
📌 Example:
[Co(en)₃]³⁺

#NEETHOTPOINTS
✔️ Square planar → dsp²
✔️ Strong ligand → pairing → inner orbital
✔️ Ambidentate ligand → linkage isomerism
✔️ cis–trans in square planar & octahedral
✔️ Optical isomerism → no plane of symmetry

@Ayano1me @Neetugpoll @Neetugquiz

#Haloalkanes
Alkanes in which H is replaced by halogen (F, Cl, Br, I)
📌 General formula: R–X
🔄 Nucleophilic Substitution Reactions
⭐ SN1 Reaction (Unimolecular)
🌱 Rate depends on only one species
📌 Rate = k[R–X]
🔹 Mechanism (2 Step)
1️⃣ R–X → R⁺ + X⁻ (Slow, RDS)
2️⃣ R⁺ + Nu⁻ → R–Nu (Fast)
⭐ Key Features
✔️ Formation of carbocation
✔️ Rearrangement possible
✔️ Racemisation occurs
✔️ Favoured by polar protic solvents
⭐ Order of Reactivity
3° > 2° > 1° > CH₃
📌 Reason: Carbocation stability
⭐ SN2 Reaction (Bimolecular)
🌱 Single step reaction
📌 Rate = k[R–X][Nu⁻]
🔹 Mechanism
✔️ Backside attack
✔️ Transition state formation
✔️ Simultaneous bond making & breaking
⭐ Key Features
✔️ No carbocation
✔️ No rearrangement
✔️ Inversion of configuration (Walden inversion)
✔️ Favoured by polar aprotic solvents
⭐ Order of Reactivity
CH₃ > 1° > 2° >> 3°
📌 Reason: Steric hindrance

SN1 vs SN2 (Direct MCQ)
Feature :SN1 :SN2
Steps :2 :1
Rate depends on :R–X :R–X & Nu⁻
Intermediate :Carbocation :None
Rearrangement :✔️ :❌
Stereochemistry :Racemisation
: Inversion
Favoured by ;3° haloalkane
: 1° haloalkane

@Ayano1me @Neetugpoll @Neetugquiz

#PreparationofHaloalkanes
⭐ 1️⃣ From Alcohols
✔️ Using HX
📌 Reactivity:
3° > 2° > 1°
✔️ Using PCl₃ / PCl₅ / SOCl₂
📌 SOCl₂ best → gaseous by-products
⭐ 2️⃣ From Alkenes
✔️ Addition of HX (Markovnikov rule)
✔️ Anti-Markovnikov (HBr + peroxide)
⭐ 3️⃣ From Alkanes
✔️ Free radical halogenation
✔️ Less selective
⭐ 4️⃣ Finkelstein Reaction
✔️ R–Cl / R–Br + NaI (acetone) → R–I
⭐ 5️⃣ Swarts Reaction
✔️ R–Cl / R–Br → R–F
✔️ Using AgF / Hg₂F₂
#NEETHOTPOINTS
✔️ SN1 → Carbocation intermediate
✔️ SN2 → Walden inversion
✔️ 3° haloalkane → SN1 favoured
✔️ SOCl₂ → best for R–Cl
✔️ Anti-Markovnikov → HBr + peroxide

@Ayano1me @Neetugpoll @Neetugquiz

#Alcohols
Organic compounds containing –OH group attached to sp³ carbon
📌 General formula: R–OH

⭐ #PreparationofAlcohols
1️⃣ From Alkenes
✔️ Acid-catalysed hydration
RCH=CH2+H+/H20 = RCH(OH) -CH3
📌 Follows Markovnikov’s rule

Hydroboration–Oxidation
RCH=CH2+BH3/THF, H2O2/OH^- = RCH2-CH2OH
📌 Anti-Markovnikov, no rearrangement

2️⃣ From Haloalkanes
R-X+Aq KOH= ROH

3️⃣ From Aldehydes & Ketones
✔️ Reduction using:
NaBH₄
LiAlH₄
📌 Aldehyde → 1° alcohol
📌 Ketone → 2° alcohol

4️⃣ From Grignard Reagent
RMGX+ HCHO=1° alcohol
RMGX+ RCHO=2°alcohol
RMGX+R2CO=3°alcohol


#NEETHOTPOINTS (Alcohol)
✔️ Hydroboration → Anti-Markovnikov
✔️ NaBH₄ milder than LiAlH₄
✔️ Grignard reagent + HCHO → 1° alcohol

@Ayano1me @Neetugpoll @Neetugquiz

Ethers
Organic compounds with –O– linkage
📌 General formula: R–O–R′

Williamson Ether Synthesis (Very High Yield 🔥)
📌 Best method to prepare ethers
RONA+R'X = ROR' + Nax
✔️ Involves SN2 mechanism
✔️ Best with 1° haloalkane
❌ 3° haloalkane → Elimination occurs
📌 For unsymmetrical ether:
Use bulky group as alkoxide
Smaller group as haloalkane

⭐ Reaction of Ether with HI
📌 Ether cleavage by HI / HBr
🔹 Mechanism
✔️ Protonation of ether oxygen
✔️ I⁻ attacks alkyl group

#CaseWiseReaction

1️⃣ Symmetrical Ether
R-O-R+ 2HI = 2RI + H20

2️⃣ Unsymmetrical Ether
✔️ If one group is 3° → Cleavage at 3° carbon (SN1)
✔️ If both are 1° → I⁻ attacks less hindered carbon (SN2)
📌 Example:

CH3-O-C2H5 + HI = Ch3I + C2H5OH

NEETHOTPOINTS (Ether)
✔️ Williamson → SN2 reaction
✔️ Best haloalkane → 1°
✔️ Ether cleavage → HI > HBr
✔️ 3° ether → SN1 cleavage
✔️ Unsymmetrical ether → less hindered side breaks

@Ayano1me @Neetugpoll @Neetugquiz

Aldehyde Ketone & Carbohydrates

1️⃣ Important Name Reactions & Tests
✅ #Aldehydes
⭐Tollens’ Test
Aldehyde + [Ag(NH3) 2}^+→ Silver mirror
👉 Ketone ❌ (except α-hydroxy ketone

⭐Fehling’s Test 🔵➡️🔴
Aliphatic aldehyde → Red ppt (Cu₂O)
👉 Aromatic aldehyde ❌

⭐Benedict’s Test
Similar to Fehling’s (for reducing sugars)

⭐Schiff’s Test
Pink colour → Aldehyde present


#Ketones
⭐Iodoform Test 🟡
Group present: –COCH₃
Positive for:
Acetone
Ethanol
2° alcohol with –CHOH–CH₃

⭐ 2,4-DNP Test
Orange / yellow ppt → Carbonyl present (Ald/Ket)

#ImportantNameReactions

⭐Aldol Condensation
Aldehyde/Ketone with α-H → β-hydroxy aldehyde → α,β-unsat compound

⭐ Cannizzaro Reaction
Aldehyde without α-H → Alcohol + Acid (base medium)

⭐Clemmensen Reduction
Zn-Hg /HCL - C=O → –CH₂–

⭐Wolff–Kishner Reduction
NH2NH2/KOH→ C=O → –CH₂–

#CarbohydratesImportantTests
Molisch Test → Violet ring (general test)
Fehling / Benedict → Red ppt (reducing sugar)
Barfoed Test → Monosaccharide (+ fast)
Iodine Test → Blue colour (starch)
Osazone Test → Needle shaped crystals (glucose = fructose)

@Ayano1me @Neetugpoll @Neetugquiz

2: #NucleophilicAdditionReactions
📌 General Reaction
RCHO/R-CO-R'+NU^- = Additional product

📌 Reactivity Order
Aldehyde > Ketone
(H- > CH₃- > bulky alkyl)

#ImportantNucleophiles
HCN → Cyanohydrin
NaHSO₃ → Bisulphite compound
RMgX (Grignard)
HCHO → 1° alcohol
Aldehyde → 2° alcohol
Ketone → 3° alcohol
NH₂OH → Oxime
NH₂NH₂ → Hydrazone
2,4-DNP → Orange ppt

#NCERTDirectPoints
Aldehydes oxidise easily ✔️
Ketones resist oxidation ❌
Formaldehyde is most reactive
Benzaldehyde ❌ Fehling test

@Ayano1me @Neetugpoll @Neetugquiz

#AMINES Topic : 9

1️⃣ Important Name Reactions & Tests
#NameReactions

Hoffmann Bromamide Reaction

1⭐ Amide → 1° amine (one C less)
RCONH2+ BR2+KOH = RNH2

2⭐Gabriel Phthalimide Synthesis
Alkyl halide → 1° aliphatic amine only
👉 Aromatic ❌

3⭐ Carbylamine Reaction (Confirmatory for 1° amine)
(bad smell)
R-NH2+CHCl3+KOH=R-NC
Hinsberg Test
1° amine → soluble sulphonamide
2° amine → insoluble
3° amine → no reaction
Diazotisation Reaction
Aromatic 1° amine + Nano2/HCL(0°-5°)→ Diazonium salt

#Tests

⭐Nitrous Acid Test
1° aliphatic → N₂ gas
1° aromatic → diazonium salt
2° → nitrosoamine
3° → salt formation

⭐Acetylation Test
1° & 2° amines react; 3° ❌

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2️⃣ Basic Nature of Amines
📌 #ReasonofBasicity
Due to lone pair on N
Electron donating groups ↑ basicity
Electron withdrawing groups ↓ basicity
📊 Basic Strength Order (NCERT)

⭐Gas phase:
3°>2°>1°>NH3

⭐Aqueous solution:
2°>1°>3°>NH3
👉 Due to solvation + steric hindrance

⭐Aromatic 🆚 Aliphatic

⭐Aliphatic amines > NH₃ > Aromatic amines

⭐Aniline less basic due to resonance (lone pair delocalisation)

⚠️ #ImportantNCERTPoints

⭐Aniline reacts with HNO₂ at low temp
⭐Basicity increases with +I effect
⭐Ortho substituted aniline → less basic (steric + H-bonding)

#NEETTrapLines
⭐Carbylamine test → only 1° amine
⭐Gabriel synthesis → only 1° aliphatic amine
⭐Aromatic amines form diazonium salts

@Ayano1me @Neetugpoll @Neetugquiz

#BIOMOLECULES
⭐1️⃣ CARBOHYDRATES
Definition
Polyhydroxy aldehydes or ketones or substances which give them on hydrolysis

📌 General formula: Cₙ(H₂O)ₙ
⭐ Classification of Carbohydrates

⭐ 1️⃣ Monosaccharides
✔️ Cannot be hydrolysed further
✔️ Sweet, crystalline
📌 Examples:
Glucose (Aldohexose)
Fructose (Ketohexose)
Ribose

⭐ 2️⃣ Oligosaccharides
✔️ 2–10 monosaccharide units
📌 Examples:
Sucrose (Glucose + Fructose) ❌ reducing
Maltose (Glucose + Glucose) ✔️ reducing
Lactose (Glucose + Galactose) ✔️ reducing


⭐3️⃣ Polysaccharides
✔️ High molecular weight
✔️ Non-sweet
📌 Examples:
Starch → plant storage
Glycogen → animal storage
Cellulose → structural (

Reducing 🆚 Non-Reducing Sugars
✔️ Free aldehyde / ketone group → Reducing
📌 Reducing: Glucose, Fructose, Maltose, Lactose
📌 Non-reducing: Sucrose

#ImportantNCERTConcepts
⭐ D & L Configuration
Based on position of –OH on penultimate carbon
Glucose → D-glucose

⭐ Anomers
Differ at anomeric carbon (C-1 in glucose)
α-glucose & β-glucose

⭐ Epimers
Differ at one carbon (except anomeric)
Glucose & Galactose (C-4)

⭐ Glycosidic Bond
Link between two monosaccharides
In sucrose → α-1,β-2

#RemeberGuys
Tollen’s / Fehling’s → reducing sugars
Iodine test → starch (blue-black)

@Ayano1me @Neetugpoll @Neetugquiz

#PROTEINS
⭐ Definition
Polymers of α-amino acids joined by peptide bonds
📌 Peptide bond: –CO–NH–

⭐ Amino Acids
📌 General formula:
NH₂–CH(R)–COOH
✔️ Zwitterion in aqueous solution
✔️ 20 standard amino acids

#Classificationof AminoAcids
⭐ Based on Nutrition
Essential → Diet required
Non-essential → Synthesised in body

⭐Based on Charge
Acidic: Aspartic, Glutamic
Basic: Lysine, Arginine
Neutral: Glycine, Alanine

#ProteinStructureLevels
⭐ 1️⃣ Primary
✔️ Sequence of amino acids
✔️ Most specific

⭐ 2️⃣ Secondary
✔️ α-helix
✔️ β-pleated sheet
📌 Stabilised by H-bonds

⭐ 3️⃣ Tertiary
✔️ 3D folding
✔️ Stabilised by:
Disulfide bonds
Ionic bonds
H-bonds

⭐ 4️⃣ Quaternary
✔️ More than one polypeptide chain
📌 Example: Haemoglobin

#TypesofProteins
⭐Fibrous → Structural (Keratin, Collagen)

⭐Globular → Functional (Enzymes)

⭐Denaturation
✔️ Loss of biological activity
❌ Primary structure unchanged
📌 Causes:
Heat
pH change

📌 Example:
Boiled egg white

#NEETHOTPOINTS
✔️ Disulfide bond → Cysteine
✔️ Enzymes → globular proteins
✔️ Sucrose → non-reducing
✔️ Cellulose → β-glucose polymer

@Ayano1me @Neetugpoll @Neetugquiz

𝗧𝗢𝗣 𝟮𝟬 𝗛𝗜𝗚𝗛 𝗪𝗘𝗜𝗚𝗛𝗧𝗔𝗚𝗘 𝗧𝗢𝗣𝗜𝗖𝗦 — 𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬


⭐ #𝗣𝗛𝗬𝗦𝗜𝗖𝗔𝗟𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟭: 𝗦𝗢𝗟𝗨𝗧𝗜𝗢𝗡

⭐ 𝗜𝗗𝗘𝗔𝗟 𝗦𝗢𝗟𝗨𝗧𝗜𝗢𝗡

⭐ 𝗖𝗢𝗟𝗟𝗜𝗚𝗔𝗧𝗜𝗩𝗘 𝗣𝗥𝗢𝗣𝗘𝗥𝗧𝗜𝗘𝗦


𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟮𝗘𝗟𝗘𝗖𝗧𝗥𝗢𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬

⭐ 𝗡𝗘𝗥𝗡𝗦𝗧 𝗘𝗤𝗨𝗔𝗧𝗜𝗢𝗡

⭐ 𝗖𝗢𝗡𝗗𝗨𝗖𝗧𝗜𝗩𝗜𝗧𝗬

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟯: 𝗖𝗛𝗘𝗠𝗜𝗖𝗔𝗟 𝗞𝗜𝗡𝗘𝗧𝗜𝗖𝗦

⭐ 𝗙𝗜𝗥𝗦𝗧 𝗢𝗥𝗗𝗘𝗥 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡𝗦

⭐ 𝗔𝗥𝗥𝗛𝗘𝗡𝗜𝗨𝗦 𝗘𝗤𝗨𝗔𝗧𝗜𝗢𝗡


⭐ #𝗜𝗡𝗢𝗥𝗚𝗔𝗡𝗜𝗖 𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟰: 𝗗 & 𝗙 𝗕𝗟𝗢𝗖𝗞 𝗘𝗟𝗘𝗠𝗘𝗡𝗧𝗦

⭐ 𝗣𝗛𝗬𝗦𝗜𝗖𝗔𝗟 𝗣𝗥𝗢𝗣𝗘𝗥𝗧𝗜𝗘𝗦 & 𝗥𝗘𝗔𝗦𝗢𝗡𝗜𝗡𝗚

⭐ 𝗟𝗔𝗡𝗧𝗛𝗔𝗡𝗢𝗜𝗗 𝗖𝗢𝗡𝗧𝗥𝗔𝗖𝗧𝗜𝗢𝗡

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟱: 𝗖𝗢𝗢𝗥𝗗𝗜𝗡𝗔𝗧𝗜𝗢𝗡 𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬

⭐ 𝗩𝗔𝗟𝗘𝗡𝗖𝗘 𝗕𝗢𝗡𝗗 𝗧𝗛𝗘𝗢𝗥𝗬 (𝗩𝗕𝗧)

⭐ 𝗜𝗦𝗢𝗠𝗘𝗥𝗜𝗦𝗠


⭐ #𝗢𝗥𝗚𝗔𝗡𝗜𝗖 𝗖𝗛𝗘𝗠𝗜𝗦𝗧𝗥𝗬

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟲: 𝗛𝗔𝗟𝗢𝗔𝗟𝗞𝗔𝗡𝗘𝗦 & 𝗛𝗔𝗟𝗢𝗔𝗥𝗘𝗡𝗘𝗦

⭐ 𝗦𝗡𝟭 & 𝗦𝗡𝟮 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡𝗦

⭐ 𝗣𝗥𝗘𝗣𝗔𝗥𝗔𝗧𝗜𝗢𝗡 𝗢𝗙 𝗛𝗔𝗟𝗢𝗔𝗟𝗞𝗔𝗡𝗘𝗦

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟳: 𝗔𝗟𝗖𝗢𝗛𝗢𝗟𝗦, 𝗣𝗛𝗘𝗡𝗢𝗟𝗦 & 𝗘𝗧𝗛𝗘𝗥𝗦

⭐ 𝗣𝗥𝗘𝗣𝗔𝗥𝗔𝗧𝗜𝗢𝗡 𝗢𝗙 𝗔𝗟𝗖𝗢𝗛𝗢𝗟𝗦

⭐ 𝗪𝗜𝗟𝗟𝗜𝗔𝗠𝗦𝗢𝗡 𝗘𝗧𝗛𝗘𝗥 𝗦𝗬𝗡𝗧𝗛𝗘𝗦𝗜𝗦
𝗛𝗜 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟴: 𝗔𝗟𝗗𝗘𝗛𝗬𝗗𝗘𝗦, 𝗞𝗘𝗧𝗢𝗡𝗘𝗦 & 𝗖𝗔𝗥𝗕𝗢𝗫𝗬𝗟𝗜𝗖 𝗔𝗖𝗜𝗗𝗦

⭐ 𝗡𝗔𝗠𝗘 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡𝗦 / 𝗧𝗘𝗦𝗧𝗦

⭐ 𝗡𝗨𝗖𝗟𝗘𝗢𝗣𝗛𝗜𝗟𝗜𝗖 𝗔𝗗𝗗𝗜𝗧𝗜𝗢𝗡 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡𝗦

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟵: 𝗔𝗠𝗜𝗡𝗘𝗦

⭐ 𝗡𝗔𝗠𝗘 𝗥𝗘𝗔𝗖𝗧𝗜𝗢𝗡𝗦 / 𝗧𝗘𝗦𝗧𝗦

⭐ 𝗕𝗔𝗦𝗜𝗖 𝗡𝗔𝗧𝗨𝗥𝗘 𝗢𝗙 𝗔𝗠𝗜𝗡𝗘𝗦

𝗖𝗛𝗔𝗣𝗧𝗘𝗥 𝟭𝟬: 𝗕𝗜𝗢𝗠𝗢𝗟𝗘𝗖𝗨𝗟𝗘𝗦

⭐ 𝗖𝗔𝗥𝗕𝗢𝗛𝗬𝗗𝗥𝗔𝗧𝗘𝗦

⭐ 𝗣𝗥𝗢𝗧𝗘𝗜𝗡𝗦

@Ayano1me @Neetugpoll @Neetugquiz

#Day 1

#MOLECONCEPTALLIMPORTANTFORMULAE (NEET)

#BasicRelations

⭐Number of moles (n) = mass (m) / molar mass (M)

⭐Mass (m) = number of moles (n) × molar mass (M)

⭐Number of particles (N) = n × NA

⭐Number of moles (n) = N / NA

⭐Avogadro number (NA) = 6.022 × 10^23

#Gases

⭐Number of moles at STP (n) = volume (V in litres) / 22.4

⭐Ideal gas equation
P × V = n × R × T

⭐Density of gas (d) = (P × M) / (R × T)

⭐Molar mass of gas (M) = (d × R × T) / P

#Solutions

⭐Molarity (M) = number of moles of solute / volume of solution (in litre)

⭐Molality (m) = number of moles of solute / mass of solvent (in kg)

⭐Normality (N) = number of equivalents / volume of solution (in litre)

⭐Strength of solution (g/L) = molarity × molar mass

⭐Dilution Formula
M1 × V1 = M2 × V2

⭐Equivalent Concept
Equivalent mass = molar mass / n-factor

⭐Number of equivalents = mass / equivalent mass

⭐Mole Fraction
Mole fraction of A (XA) = moles of A / (moles of A + moles of B)
XA + XB = 1

⭐LimitingReagent
Required moles = given moles / stoichiometric coefficient
The reactant with least required moles is the limiting reagent
Percentage Composition

⭐Percentage of element = (mass of element / molar mass of compound) × 100

⭐Empirical and Molecular Formula
Empirical formula mass = sum of atomic masses in empirical formula
n = molecular mass / empirical formula mass

⭐Molecular formula = empirical formula × n

⭐Gas Mixture (Dalton’s Law)
Total pressure = P1 + P2 + P3 + …

⭐Partial pressure of gas A
PA = XA × Ptotal
Redox Reactions

⭐Normality × Volume = constant
N1 × V1 = N2 × V2
n-factor = number of electrons lost or gained

#ImportantConstants
STP = 273 K and 1 atm
Gas constant
R = 0.0821 L atm mol⁻¹ K⁻¹
R = 8.314 J mol⁻¹ K⁻¹

#SIGNIFICANTFIGURES
🌱 Definition
Digits which convey certainty + one uncertain digit

⭐ Rules to Count Significant Figures
✔️ All non-zero digits → significant
✔️ Zeros between non-zero → significant
✔️ Leading zeros → ❌ not significant
✔️ Trailing zeros → significant only with decimal

📌 Examples:
0.0045 → 2 SF
2.300 → 4 SF
1500 → 2 SF (without decimal)

➕➖ Addition / Subtraction
Result → least decimal places
📌 Example:
12.11 + 0.2 = 12.3

✖️➗ Multiplication / Division
Result → least significant figures
📌 Example:
2.5 × 1.23 = 3.1 (2 SF)

🔢 Rounding Off Rules
Next digit < 5 → same
Next digit ≥ 5 → +1
📌 2.34 → 2.3
📌 2.36 → 2.4

#NEETHOTPOINTS
✔️ Exact numbers → infinite SF
✔️ Unit conversion → SF maintained
✔️ Final answer rounding last step

@Ayano1me @Neetugpoll @Neetugquiz

Ans eve m upload hoga sb try krna

Question 1 (Concept + Limiting Reagent):

A mixture contains 4 g H₂ and 32 g O₂.
They react according to:
2H2+O2= 2H2O
Find:
(i) Limiting reagent
(ii) Mass of water formed
(iii) Mass of excess reactant left

Question 2 (Gas + Stoichiometry + Trick)

At STP, 11.2 L of a gaseous hydrocarbon reacts completely with excess O₂ to produce 44 g CO₂.
Identify the hydrocarbon.

Question 3 (Equivalent + Redox + Stoichiometry )

A 10 g mixture of Na₂CO₃ and NaHCO₃ is completely neutralised by 200 mL of 1 N HCl.
Find the mass percentage of Na₂CO₃ in the mixture.

#SIGNIFICANTFIGURES

Question :1 Evaluate the result with correct significant figures:
(2.36+0.040) +1.2

Question 2
The mass of a cube is measured as 2.50 g and each edge is measured as 1.20 cm.
Calculate the density of the cube with correct significant figures.



⭐ Day 1 Biology
⭐Day 1Chemistry
⭐Day1 physics

Notes + question🙋