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2๏ธโƒฃ Basic Nature of Amines
๐Ÿ“Œ #ReasonofBasicity
Due to lone pair on N
Electron donating groups โ†‘ basicity
Electron withdrawing groups โ†“ basicity
๐Ÿ“Š Basic Strength Order (NCERT)

โญGas phase:
3ยฐ>2ยฐ>1ยฐ>NH3

โญAqueous solution:
2ยฐ>1ยฐ>3ยฐ>NH3
๐Ÿ‘‰ Due to solvation + steric hindrance

โญAromatic ๐Ÿ†š Aliphatic

โญAliphatic amines > NHโ‚ƒ > Aromatic amines

โญAniline less basic due to resonance (lone pair delocalisation)

โš ๏ธ #ImportantNCERTPoints

โญAniline reacts with HNOโ‚‚ at low temp
โญBasicity increases with +I effect
โญOrtho substituted aniline โ†’ less basic (steric + H-bonding)

#NEETTrapLines
โญCarbylamine test โ†’ only 1ยฐ amine
โญGabriel synthesis โ†’ only 1ยฐ aliphatic amine
โญAromatic amines form diazonium salts

@Ayano1me @Neetugpoll @Neetugquiz
๐Ÿ”ฅ2๐Ÿ‘2๐Ÿ˜1
#BIOMOLECULES
โญ1๏ธโƒฃ CARBOHYDRATES
Definition
Polyhydroxy aldehydes or ketones or substances which give them on hydrolysis

๐Ÿ“Œ General formula: Cโ‚™(Hโ‚‚O)โ‚™
โญ Classification of Carbohydrates

โญ 1๏ธโƒฃ Monosaccharides
โœ”๏ธ Cannot be hydrolysed further
โœ”๏ธ Sweet, crystalline
๐Ÿ“Œ Examples:
Glucose (Aldohexose)
Fructose (Ketohexose)
Ribose

โญ 2๏ธโƒฃ Oligosaccharides
โœ”๏ธ 2โ€“10 monosaccharide units
๐Ÿ“Œ Examples:
Sucrose (Glucose + Fructose) โŒ reducing
Maltose (Glucose + Glucose) โœ”๏ธ reducing
Lactose (Glucose + Galactose) โœ”๏ธ reducing


โญ3๏ธโƒฃ Polysaccharides
โœ”๏ธ High molecular weight
โœ”๏ธ Non-sweet
๐Ÿ“Œ Examples:
Starch โ†’ plant storage
Glycogen โ†’ animal storage
Cellulose โ†’ structural (

Reducing ๐Ÿ†š Non-Reducing Sugars
โœ”๏ธ Free aldehyde / ketone group โ†’ Reducing
๐Ÿ“Œ Reducing: Glucose, Fructose, Maltose, Lactose
๐Ÿ“Œ Non-reducing: Sucrose

#ImportantNCERTConcepts
โญ D & L Configuration
Based on position of โ€“OH on penultimate carbon
Glucose โ†’ D-glucose

โญ Anomers
Differ at anomeric carbon (C-1 in glucose)
ฮฑ-glucose & ฮฒ-glucose

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

โญ Glycosidic Bond
Link between two monosaccharides
In sucrose โ†’ ฮฑ-1,ฮฒ-2

#RemeberGuys
Tollenโ€™s / Fehlingโ€™s โ†’ reducing sugars
Iodine test โ†’ starch (blue-black)

@Ayano1me @Neetugpoll @Neetugquiz
๐Ÿ”ฅ2๐Ÿฅฐ2๐Ÿ˜˜1
#PROTEINS
โญ Definition
Polymers of ฮฑ-amino acids joined by peptide bonds
๐Ÿ“Œ Peptide bond: โ€“COโ€“NHโ€“

โญ Amino Acids
๐Ÿ“Œ General formula:
NHโ‚‚โ€“CH(R)โ€“COOH
โœ”๏ธ Zwitterion in aqueous solution
โœ”๏ธ 20 standard amino acids

#Classificationof AminoAcids
โญ Based on Nutrition
Essential โ†’ Diet required
Non-essential โ†’ Synthesised in body

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

#ProteinStructureLevels
โญ 1๏ธโƒฃ Primary
โœ”๏ธ Sequence of amino acids
โœ”๏ธ Most specific

โญ 2๏ธโƒฃ Secondary
โœ”๏ธ ฮฑ-helix
โœ”๏ธ ฮฒ-pleated sheet
๐Ÿ“Œ Stabilised by H-bonds

โญ 3๏ธโƒฃ Tertiary
โœ”๏ธ 3D folding
โœ”๏ธ Stabilised by:
Disulfide bonds
Ionic bonds
H-bonds

โญ 4๏ธโƒฃ Quaternary
โœ”๏ธ More than one polypeptide chain
๐Ÿ“Œ Example: Haemoglobin

#TypesofProteins
โญFibrous โ†’ Structural (Keratin, Collagen)

โญGlobular โ†’ Functional (Enzymes)

โญDenaturation
โœ”๏ธ Loss of biological activity
โŒ Primary structure unchanged
๐Ÿ“Œ Causes:
Heat
pH change

๐Ÿ“Œ Example:
Boiled egg white

#NEETHOTPOINTS
โœ”๏ธ Disulfide bond โ†’ Cysteine
โœ”๏ธ Enzymes โ†’ globular proteins
โœ”๏ธ Sucrose โ†’ non-reducing
โœ”๏ธ Cellulose โ†’ ฮฒ-glucose polymer

@Ayano1me @Neetugpoll @Neetugquiz
๐Ÿ˜Ž3๐Ÿ’ฏ1
๐—ง๐—ข๐—ฃ ๐Ÿฎ๐Ÿฌ ๐—›๐—œ๐—š๐—› ๐—ช๐—˜๐—œ๐—š๐—›๐—ง๐—”๐—š๐—˜ ๐—ง๐—ข๐—ฃ๐—œ๐—–๐—ฆ โ€” ๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ


โญ #๐—ฃ๐—›๐—ฌ๐—ฆ๐—œ๐—–๐—”๐—Ÿ๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿญ: ๐—ฆ๐—ข๐—Ÿ๐—จ๐—ง๐—œ๐—ข๐—ก

โญ ๐—œ๐——๐—˜๐—”๐—Ÿ ๐—ฆ๐—ข๐—Ÿ๐—จ๐—ง๐—œ๐—ข๐—ก

โญ ๐—–๐—ข๐—Ÿ๐—Ÿ๐—œ๐—š๐—”๐—ง๐—œ๐—ฉ๐—˜ ๐—ฃ๐—ฅ๐—ข๐—ฃ๐—˜๐—ฅ๐—ง๐—œ๐—˜๐—ฆ


๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿฎ๐—˜๐—Ÿ๐—˜๐—–๐—ง๐—ฅ๐—ข๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ

โญ ๐—ก๐—˜๐—ฅ๐—ก๐—ฆ๐—ง ๐—˜๐—ค๐—จ๐—”๐—ง๐—œ๐—ข๐—ก

โญ ๐—–๐—ข๐—ก๐——๐—จ๐—–๐—ง๐—œ๐—ฉ๐—œ๐—ง๐—ฌ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿฏ: ๐—–๐—›๐—˜๐— ๐—œ๐—–๐—”๐—Ÿ ๐—ž๐—œ๐—ก๐—˜๐—ง๐—œ๐—–๐—ฆ

โญ ๐—™๐—œ๐—ฅ๐—ฆ๐—ง ๐—ข๐—ฅ๐——๐—˜๐—ฅ ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก๐—ฆ

โญ ๐—”๐—ฅ๐—ฅ๐—›๐—˜๐—ก๐—œ๐—จ๐—ฆ ๐—˜๐—ค๐—จ๐—”๐—ง๐—œ๐—ข๐—ก



โญ #๐—œ๐—ก๐—ข๐—ฅ๐—š๐—”๐—ก๐—œ๐—– ๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿฐ: ๐—— & ๐—™ ๐—•๐—Ÿ๐—ข๐—–๐—ž ๐—˜๐—Ÿ๐—˜๐— ๐—˜๐—ก๐—ง๐—ฆ

โญ ๐—ฃ๐—›๐—ฌ๐—ฆ๐—œ๐—–๐—”๐—Ÿ ๐—ฃ๐—ฅ๐—ข๐—ฃ๐—˜๐—ฅ๐—ง๐—œ๐—˜๐—ฆ & ๐—ฅ๐—˜๐—”๐—ฆ๐—ข๐—ก๐—œ๐—ก๐—š

โญ ๐—Ÿ๐—”๐—ก๐—ง๐—›๐—”๐—ก๐—ข๐—œ๐—— ๐—–๐—ข๐—ก๐—ง๐—ฅ๐—”๐—–๐—ง๐—œ๐—ข๐—ก

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿฑ: ๐—–๐—ข๐—ข๐—ฅ๐——๐—œ๐—ก๐—”๐—ง๐—œ๐—ข๐—ก ๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ

โญ ๐—ฉ๐—”๐—Ÿ๐—˜๐—ก๐—–๐—˜ ๐—•๐—ข๐—ก๐—— ๐—ง๐—›๐—˜๐—ข๐—ฅ๐—ฌ (๐—ฉ๐—•๐—ง)

โญ ๐—œ๐—ฆ๐—ข๐— ๐—˜๐—ฅ๐—œ๐—ฆ๐— 


โญ #๐—ข๐—ฅ๐—š๐—”๐—ก๐—œ๐—– ๐—–๐—›๐—˜๐— ๐—œ๐—ฆ๐—ง๐—ฅ๐—ฌ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿฒ: ๐—›๐—”๐—Ÿ๐—ข๐—”๐—Ÿ๐—ž๐—”๐—ก๐—˜๐—ฆ & ๐—›๐—”๐—Ÿ๐—ข๐—”๐—ฅ๐—˜๐—ก๐—˜๐—ฆ

โญ ๐—ฆ๐—ก๐Ÿญ & ๐—ฆ๐—ก๐Ÿฎ ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก๐—ฆ

โญ ๐—ฃ๐—ฅ๐—˜๐—ฃ๐—”๐—ฅ๐—”๐—ง๐—œ๐—ข๐—ก ๐—ข๐—™ ๐—›๐—”๐—Ÿ๐—ข๐—”๐—Ÿ๐—ž๐—”๐—ก๐—˜๐—ฆ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿณ: ๐—”๐—Ÿ๐—–๐—ข๐—›๐—ข๐—Ÿ๐—ฆ, ๐—ฃ๐—›๐—˜๐—ก๐—ข๐—Ÿ๐—ฆ & ๐—˜๐—ง๐—›๐—˜๐—ฅ๐—ฆ

โญ ๐—ฃ๐—ฅ๐—˜๐—ฃ๐—”๐—ฅ๐—”๐—ง๐—œ๐—ข๐—ก ๐—ข๐—™ ๐—”๐—Ÿ๐—–๐—ข๐—›๐—ข๐—Ÿ๐—ฆ

โญ ๐—ช๐—œ๐—Ÿ๐—Ÿ๐—œ๐—”๐— ๐—ฆ๐—ข๐—ก ๐—˜๐—ง๐—›๐—˜๐—ฅ ๐—ฆ๐—ฌ๐—ก๐—ง๐—›๐—˜๐—ฆ๐—œ๐—ฆ
๐—›๐—œ ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก


๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿด: ๐—”๐—Ÿ๐——๐—˜๐—›๐—ฌ๐——๐—˜๐—ฆ, ๐—ž๐—˜๐—ง๐—ข๐—ก๐—˜๐—ฆ & ๐—–๐—”๐—ฅ๐—•๐—ข๐—ซ๐—ฌ๐—Ÿ๐—œ๐—– ๐—”๐—–๐—œ๐——๐—ฆ

โญ ๐—ก๐—”๐— ๐—˜ ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก๐—ฆ / ๐—ง๐—˜๐—ฆ๐—ง๐—ฆ

โญ ๐—ก๐—จ๐—–๐—Ÿ๐—˜๐—ข๐—ฃ๐—›๐—œ๐—Ÿ๐—œ๐—– ๐—”๐——๐——๐—œ๐—ง๐—œ๐—ข๐—ก ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก๐—ฆ

๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿต: ๐—”๐— ๐—œ๐—ก๐—˜๐—ฆ

โญ ๐—ก๐—”๐— ๐—˜ ๐—ฅ๐—˜๐—”๐—–๐—ง๐—œ๐—ข๐—ก๐—ฆ / ๐—ง๐—˜๐—ฆ๐—ง๐—ฆ

โญ ๐—•๐—”๐—ฆ๐—œ๐—– ๐—ก๐—”๐—ง๐—จ๐—ฅ๐—˜ ๐—ข๐—™ ๐—”๐— ๐—œ๐—ก๐—˜๐—ฆ


๐—–๐—›๐—”๐—ฃ๐—ง๐—˜๐—ฅ ๐Ÿญ๐Ÿฌ: ๐—•๐—œ๐—ข๐— ๐—ข๐—Ÿ๐—˜๐—–๐—จ๐—Ÿ๐—˜๐—ฆ

โญ ๐—–๐—”๐—ฅ๐—•๐—ข๐—›๐—ฌ๐——๐—ฅ๐—”๐—ง๐—˜๐—ฆ

โญ ๐—ฃ๐—ฅ๐—ข๐—ง๐—˜๐—œ๐—ก๐—ฆ

@Ayano1me @Neetugpoll @Neetugquiz
โค5โคโ€๐Ÿ”ฅ2๐Ÿ’ฏ2๐Ÿฅฐ1๐Ÿ˜1
#Day 1

#MOLECONCEPTALLIMPORTANTFORMULAE (NEET)

#BasicRelations

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

โญMass (m) = number of moles (n) ร— molar mass (M)

โญNumber of particles (N) = n ร— NA

โญNumber of moles (n) = N / NA

โญAvogadro number (NA) = 6.022 ร— 10^23

#Gases

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

โญIdeal gas equation
P ร— V = n ร— R ร— T

โญDensity of gas (d) = (P ร— M) / (R ร— T)

โญMolar mass of gas (M) = (d ร— R ร— T) / P

#Solutions

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

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

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

โญStrength of solution (g/L) = molarity ร— molar mass

โญDilution Formula
M1 ร— V1 = M2 ร— V2

โญEquivalent Concept
Equivalent mass = molar mass / n-factor

โญNumber of equivalents = mass / equivalent mass

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

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

โญPercentage of element = (mass of element / molar mass of compound) ร— 100

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

โญMolecular formula = empirical formula ร— n

โญGas Mixture (Daltonโ€™s Law)
Total pressure = P1 + P2 + P3 + โ€ฆ

โญPartial pressure of gas A
PA = XA ร— Ptotal
Redox Reactions

โญNormality ร— Volume = constant
N1 ร— V1 = N2 ร— V2
n-factor = number of electrons lost or gained

#ImportantConstants
STP = 273 K and 1 atm
Gas constant
R = 0.0821 L atm molโปยน Kโปยน
R = 8.314 J molโปยน Kโปยน
โคโ€๐Ÿ”ฅ6
#SIGNIFICANTFIGURES
๐ŸŒฑ Definition
Digits which convey certainty + one uncertain digit

โญ Rules to Count Significant Figures
โœ”๏ธ All non-zero digits โ†’ significant
โœ”๏ธ Zeros between non-zero โ†’ significant
โœ”๏ธ Leading zeros โ†’ โŒ not significant
โœ”๏ธ Trailing zeros โ†’ significant only with decimal

๐Ÿ“Œ Examples:
0.0045 โ†’ 2 SF
2.300 โ†’ 4 SF
1500 โ†’ 2 SF (without decimal)

โž•โž– Addition / Subtraction
Result โ†’ least decimal places
๐Ÿ“Œ Example:
12.11 + 0.2 = 12.3

โœ–๏ธโž— Multiplication / Division
Result โ†’ least significant figures
๐Ÿ“Œ Example:
2.5 ร— 1.23 = 3.1 (2 SF)

๐Ÿ”ข Rounding Off Rules
Next digit < 5 โ†’ same
Next digit โ‰ฅ 5 โ†’ +1
๐Ÿ“Œ 2.34 โ†’ 2.3
๐Ÿ“Œ 2.36 โ†’ 2.4

#NEETHOTPOINTS
โœ”๏ธ Exact numbers โ†’ infinite SF
โœ”๏ธ Unit conversion โ†’ SF maintained
โœ”๏ธ Final answer rounding last step

@Ayano1me @Neetugpoll @Neetugquiz
โค6๐Ÿ˜˜2
โค9๐Ÿ˜˜3
Ans eve m upload hoga sb try krna

Question 1 (Concept + Limiting Reagent):

A mixture contains 4 g Hโ‚‚ and 32 g Oโ‚‚.
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 Oโ‚‚ to produce 44 g COโ‚‚.
Identify the hydrocarbon.

Question 3 (Equivalent + Redox + Stoichiometry )

A 10 g mixture of Naโ‚‚COโ‚ƒ and NaHCOโ‚ƒ is completely neutralised by 200 mL of 1 N HCl.
Find the mass percentage of Naโ‚‚COโ‚ƒ 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.

๏ปฟ
โคโ€๐Ÿ”ฅ6โค3๐Ÿ‘1
๐Ÿ†6๐Ÿฅฐ5๐Ÿ‘1๐Ÿ”ฅ1
โค3๐Ÿณ3๐Ÿ˜˜1
#STOICHIOMETRY Topic 2

๐ŸŒฑ Definition
Quantitative relationship between reactants and p products in a chemical reaction.

โญ Steps in Stoichiometric Calculations
โœ”๏ธ Write balanced chemical equation
โœ”๏ธ Convert given data into moles
โœ”๏ธ Use mole ratio from equation
โœ”๏ธ Calculate required moles
โœ”๏ธ Convert final answer into required unit
๐Ÿงช Mole Ratio
Ratio of coefficients of substances in a balanced equation.
๐Ÿ“Œ Example:
2Hโ‚‚ + Oโ‚‚ โ†’ 2Hโ‚‚O
Mole ratio: Hโ‚‚ : Oโ‚‚ : Hโ‚‚O = 2 : 1 : 2
๐Ÿ”น Types of Stoichiometric Problems

โญ 1๏ธโƒฃ Massโ€“Mass Relationship
โœ”๏ธ Given mass of one reactant โ†’ find mass of product
โœ”๏ธ Use moles as bridge
๐Ÿ“Œ Steps:
mass โ†’ moles โ†’ mole ratio โ†’ moles โ†’ mass

โญ 2๏ธโƒฃ Massโ€“Volume Relationship (Gas)
โœ”๏ธ Use 22.4 L at STP
๐Ÿ“Œ Formula:
Number of moles = Volume / 22.4

โญ 3๏ธโƒฃ Volumeโ€“Volume Relationship (Gases)
โœ”๏ธ Volumes are in same ratio as moles
โœ”๏ธ Valid only for gases at same T & P

โญ 4๏ธโƒฃ Limiting Reagent
โœ”๏ธ Reactant consumed first
โœ”๏ธ Decides amount of product formed
๐Ÿ“Œ Method:
Given moles / stoichiometric coefficient
โ†’ Smaller value = limiting reagent
Percent Yield
โœ”๏ธ Theoretical yield โ†’ from calculation
โœ”๏ธ Actual yield โ†’ given in question
๐Ÿ“Œ Formula:
Percent yield = (Actual yield / Theoretical yield) ร— 100

Percentage Composition
โœ”๏ธ Used to find composition of compound
๐Ÿ“Œ Formula:
Percentage of element = (Mass of element / Molar mass of compound) ร— 100

#NEETHOTPOINTS
โœ”๏ธ Stoichiometry always uses balanced equation
โœ”๏ธ All calculations done in moles first
โœ”๏ธ Limiting reagent decides product
โœ”๏ธ Volume ratio applies only to gases

@Ayano1me @Neetugpoll @Neetugquiz
โค7๐ŸŽ‰2๐Ÿ˜1๐Ÿณ1
Question :1 Identify the correct set of statements:
Reaction: 2Al+ fe2o3 =Al2O3+ 2Fe
A. 54 g Al reacts completely with 160 g Feโ‚‚Oโ‚ƒ
B. 27 g Al produces 56 g Fe
C. If Al is excess, Feโ‚‚Oโ‚ƒ is the limiting reagent
D. Mole ratio Al : Fe = 2 : 3
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A, B and C
2๏ธโƒฃ A and B only
3๏ธโƒฃ B and C only
4๏ธโƒฃ A, C and D


Question 2 (Limiting Reagent )
A mixture of 4 g Hโ‚‚ and 16 g Oโ‚‚ reacts to form water.
A. Hโ‚‚ is the limiting reagent
B. Oโ‚‚ is the limiting reagent
C. 18 g Hโ‚‚O is formed
D. Some Hโ‚‚ remains unreacted
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A and C
2๏ธโƒฃ B and D
3๏ธโƒฃ A and D
4๏ธโƒฃ B and C
โค5๐Ÿ”ฅ4๐Ÿ˜1๐Ÿ†1
Extra P block grp number 13 afternoon
โค7๐Ÿณ3๐Ÿ™1
P-BLOCK ELEMENTS
โญ Group 13 โ€“ Boron Family
Elements of Group 13
โœ”๏ธ Boron (B)
โœ”๏ธ Aluminium (Al)
โœ”๏ธ Gallium (Ga)
โœ”๏ธ Indium (In)
โœ”๏ธ Thallium (Tl)
โš›๏ธ Electronic Configuration
๐Ÿ“Œ General configuration: nsยฒ npยน
โœ”๏ธ B โ†’ 1sยฒ 2sยฒ 2pยน
โœ”๏ธ Al โ†’ [Ne] 3sยฒ 3pยน
โœ”๏ธ Ga โ†’ [Ar] 3dยนโฐ 4sยฒ 4pยน
โœ”๏ธ In โ†’ [Kr] 4dยนโฐ 5sยฒ 5pยน
โœ”๏ธ Tl โ†’ [Xe] 4fยนโด 5dยนโฐ 6sยฒ 6pยน
๐Ÿ”ข Oxidation States
โœ”๏ธ Common oxidation state โ†’ +3
โœ”๏ธ +1 oxidation state increases down the group
๐Ÿ“Œ Reason โ†’ Inert pair effect
โญ Stability of +1 state:
Tl > In > Ga > Al > B

#PhysicalProperties (Trends)
โœ”๏ธ Atomic size โ†‘ down the group Al, GA interchange
โœ”๏ธ Boron โ†’ Non-metal
โœ”๏ธ Al, Ga, In, Tl โ†’ Metals
๐Ÿ“Œ Melting point:
โœ”๏ธ Boron โ†’ very high
โœ”๏ธ Gallium โ†’ melts near room temperature โœ‹

#ChemicalProperties
Reaction with Oxygen
โœ”๏ธ Boron โ†’ Bโ‚‚Oโ‚ƒ
โœ”๏ธ Aluminium โ†’ Alโ‚‚Oโ‚ƒ (protective oxide layer)
๐Ÿ“Œ Nature of oxides:
โœ”๏ธ Bโ‚‚Oโ‚ƒ โ†’ Acidic
โœ”๏ธ Alโ‚‚Oโ‚ƒ โ†’ Amphoteric
โœ”๏ธ Gaโ‚‚Oโ‚ƒ, Inโ‚‚Oโ‚ƒ โ†’ Amphoteric
โœ”๏ธ Tlโ‚‚Oโ‚ƒ โ†’ Basic

ReactionwithAcids
โœ”๏ธ Boron โ†’ does NOT react directly
โœ”๏ธ Aluminium โ†’ reacts but shows passivation due to oxide layer

@Ayano1me @Neetugpoll @Neetugquiz
โค4๐Ÿค4๐Ÿ”ฅ2๐Ÿ‘1
#SpecialPointsaboutBoron
โœ”๏ธ Forms only covalent compounds
โœ”๏ธ Does NOT follow octet rule
โœ”๏ธ Shows diagonal relationship with silicon
โœ”๏ธ Does NOT form Bยณโบ ion easily
โœ”๏ธ Boric acid is weak monobasic acid
โœ”๏ธ Acts as Lewis acid
Aluminium โ€“ NCERT Highlights
โœ”๏ธ Most abundant metal in earthโ€™s crust
โœ”๏ธ Extracted by Hallโ€“Hรฉroult process
โœ”๏ธ Alโ‚‚Oโ‚ƒ โ†’ amphoteric
โœ”๏ธ Used in thermite reaction
๐Ÿ“Œ Thermite reaction:
Feโ‚‚Oโ‚ƒ + 2Al โ†’ Alโ‚‚Oโ‚ƒ + 2Fe + heat

#ImportantCompounds
โญ Borax (Naโ‚‚Bโ‚„Oโ‚‡ยท10Hโ‚‚O)
โœ”๏ธ Used in glass industry
โœ”๏ธ Borax bead test
โœ”๏ธ On heating โ†’ glassy bead

โญ Boric Acid (Hโ‚ƒBOโ‚ƒ)
โœ”๏ธ Weak acid
โœ”๏ธ Lewis acid
โœ”๏ธ Used as antiseptic

โญ Aluminium Chloride (AlClโ‚ƒ)
โœ”๏ธ Covalent compound
โœ”๏ธ Lewis acid
โœ”๏ธ Exists as Alโ‚‚Clโ‚† (dimer)

โญ #NEETHOTPOINTS
โœ”๏ธ Boron โ†’ non-metal, others metals
โœ”๏ธ Inert pair effect โ†’ +1 oxidation state
โœ”๏ธ AlClโ‚ƒ โ†’ Lewis acid
โœ”๏ธ Bโ‚‚Oโ‚ƒ acidic, Alโ‚‚Oโ‚ƒ amphoteric
โœ”๏ธ Gallium melts in hand โœ‹

@Ayano1me @Neetugpoll @Neetugquiz
โค9๐Ÿ˜Ž6๐Ÿ”ฅ1๐Ÿ†1๐Ÿค1
Question 1
Identify the correct set of statements related to Group 13 elements:
A. Boron shows anomalous behaviour in its group
B. Aluminium forms Alยณโบ ion easily in aqueous solution
C. Gallium has lower melting point than aluminium
D. Inert pair effect increases from B to Tl
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A, C and D
2๏ธโƒฃ A and B
3๏ธโƒฃ B and C
4๏ธโƒฃ A, B, C and D


Question:2
Identify the correctly matched pairs:
A. Boron trichloride โ€“ strong Lewis acid
B. Aluminium chloride โ€“ exists as AlClโ‚ƒ in vapour phase
C. Diborane โ€“ electron deficient compound
D. Boric acid โ€“ monobasic acid
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A and C
2๏ธโƒฃ A, C and D
3๏ธโƒฃ B and D
4๏ธโƒฃ A, B, C and D

Question:3
Identify the correct statements:
A. +3 oxidation state stability decreases from B to Tl
B. +1 oxidation state stability increases from B to Tl
C. Tlยณโบ is a strong oxidising agent
D. Boron forms stable Bยณโบ ion
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A, B and C
2๏ธโƒฃ A and D
3๏ธโƒฃ B and C
4๏ธโƒฃ A, C and D
โค9๐Ÿคฉ4
A. Diborane has two 3-centre-2-electron bonds
B. All Bโ€“H bonds in diborane are equivalent
C. Boranes are electron-deficient compounds
D. Terminal Bโ€“H bonds are longer than bridge Bโ€“H bonds
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A and C
2๏ธโƒฃ A, B and C
3๏ธโƒฃ B and D
4๏ธโƒฃ A, C and D


Identify the correct set regarding trihalides of Group 13:
A. Lewis acidity of BXโ‚ƒ decreases from BFโ‚ƒ to BIโ‚ƒ
B. BFโ‚ƒ is a weaker Lewis acid than BClโ‚ƒ
C. AlClโ‚ƒ acts as a Lewis acid
D. Boron trihalides undergo hydrolysis except BFโ‚ƒ
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A and C
2๏ธโƒฃ B, C and D
3๏ธโƒฃ A, B and D
4๏ธโƒฃ B and C only

Identify the correct statements:
A. Bโ‚‚Oโ‚ƒ is acidic in nature
B. Alโ‚‚Oโ‚ƒ is amphoteric
C. Tlโ‚‚Oโ‚ƒ is basic in nature
D. Basic character of oxides decreases down the group
๐Ÿ‘‰ Correct option(s):
1๏ธโƒฃ A and B
2๏ธโƒฃ A, B and C
3๏ธโƒฃ B and D
4๏ธโƒฃ A, C and D
๐Ÿ˜7โค5๐Ÿ’ฏ2๐Ÿ•Š1
thermodynamics.aac
1.6 MB
Topic cover thermodynamics system surrounding boundaries
Intensive extensive property & sign uses
โค9โคโ€๐Ÿ”ฅ5๐Ÿ˜1
Chemistry booster series
Question 1 Identify the correct set of statements related to Group 13 elements: A. Boron shows anomalous behaviour in its group B. Aluminium forms Alยณโบ ion easily in aqueous solution C. Gallium has lower melting point than aluminium D. Inert pair effect increasesโ€ฆ
โœ… Answer: Option 1 (A, C and D)
Boron anomalous โœ”
Alยณโบ strongly hydrated โ†’ aqueous me stable nahi โŒ
Ga ka m.p. Al se kam โœ”
Inert pair effect down the group increase โœ”

โœ… Answer: Option 2 (A, C and D)
BClโ‚ƒ = strong Lewis acid โœ”
AlClโ‚ƒ vapour phase me dimer (Alโ‚‚Clโ‚†) โŒ
Diborane = electron deficient โœ”
Boric acid monobasic โœ”


โœ… Answer: Option 1 (A, B and C)
+3 down the group unstable โœ”
+1 stable due to inert pair โœ”
Tlยณโบ easily reduce hota โ‡’ strong oxidiser โœ”
Boron Bยณโบ ion nahi banata โŒ
๏ปฟ
โค8๐Ÿฅฐ2๐Ÿณ1๐Ÿ†1
Chemistry booster series
A. Diborane has two 3-centre-2-electron bonds B. All Bโ€“H bonds in diborane are equivalent C. Boranes are electron-deficient compounds D. Terminal Bโ€“H bonds are longer than bridge Bโ€“H bonds ๐Ÿ‘‰ Correct option(s): 1๏ธโƒฃ A and C 2๏ธโƒฃ A, B and C 3๏ธโƒฃ B and D 4๏ธโƒฃ A,โ€ฆ
โœ… Answer: Option 1 (A and C)
Diborane me 2 banana bonds โœ”
Terminal & bridge bonds same nahi โŒ
Boranes electron-deficient โœ”
Bridge bonds longer hote hain โŒ

โœ… Answer: Option 2 (B, C and D)
Lewis acidity: BFโ‚ƒ < BClโ‚ƒ < BBrโ‚ƒ < BIโ‚ƒ โœ”
AlClโ‚ƒ = Lewis acid โœ”
BFโ‚ƒ hydrolysis nahi karta โŒ (others karte)
A statement ulta likha โŒ

โœ… Answer: Option 2 (A, B and C)
Bโ‚‚Oโ‚ƒ acidic โœ”
Alโ‚‚Oโ‚ƒ amphoteric โœ”
Tlโ‚‚Oโ‚ƒ basic โœ”
Basic character increase hota hai down the group โŒ
๐Ÿ˜Ž4๐Ÿ˜3๐Ÿ‘Œ1