Chemistry booster series
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#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
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#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

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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.


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#STOICHIOMETRY Topic 2

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

Steps in Stoichiometric Calculations
✔️ Write balanced chemical equation
✔️ Convert given data into moles
✔️ Use mole ratio from equation
✔️ Calculate required moles
✔️ Convert final answer into required unit
🧪 Mole Ratio
Ratio of coefficients of substances in a balanced equation.
📌 Example:
2H₂ + O₂ → 2H₂O
Mole ratio: H₂ : O₂ : H₂O = 2 : 1 : 2
🔹 Types of Stoichiometric Problems

1️⃣ Mass–Mass Relationship
✔️ Given mass of one reactant → find mass of product
✔️ Use moles as bridge
📌 Steps:
mass → moles → mole ratio → moles → mass

2️⃣ Mass–Volume Relationship (Gas)
✔️ Use 22.4 L at STP
📌 Formula:
Number of moles = Volume / 22.4

3️⃣ Volume–Volume Relationship (Gases)
✔️ Volumes are in same ratio as moles
✔️ Valid only for gases at same T & P

4️⃣ Limiting Reagent
✔️ Reactant consumed first
✔️ Decides amount of product formed
📌 Method:
Given moles / stoichiometric coefficient
→ Smaller value = limiting reagent
Percent Yield
✔️ Theoretical yield → from calculation
✔️ Actual yield → given in question
📌 Formula:
Percent yield = (Actual yield / Theoretical yield) × 100

Percentage Composition
✔️ Used to find composition of compound
📌 Formula:
Percentage of element = (Mass of element / Molar mass of compound) × 100

#NEETHOTPOINTS
✔️ Stoichiometry always uses balanced equation
✔️ All calculations done in moles first
✔️ Limiting reagent decides product
✔️ Volume ratio applies only to gases

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Question :1 Identify the correct set of statements:
Reaction: 2Al+ fe2o3 =Al2O3+ 2Fe
A. 54 g Al reacts completely with 160 g Fe₂O₃
B. 27 g Al produces 56 g Fe
C. If Al is excess, Fe₂O₃ is the limiting reagent
D. Mole ratio Al : Fe = 2 : 3
👉 Correct option(s):
1️⃣ A, B and C
2️⃣ A and B only
3️⃣ B and C only
4️⃣ A, C and D


Question 2 (Limiting Reagent )
A mixture of 4 g H₂ and 16 g O₂ reacts to form water.
A. H₂ is the limiting reagent
B. O₂ is the limiting reagent
C. 18 g H₂O is formed
D. Some H₂ remains unreacted
👉 Correct option(s):
1️⃣ A and C
2️⃣ B and D
3️⃣ A and D
4️⃣ B and C
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Extra P block grp number 13 afternoon
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P-BLOCK ELEMENTS
Group 13 – Boron Family
Elements of Group 13
✔️ Boron (B)
✔️ Aluminium (Al)
✔️ Gallium (Ga)
✔️ Indium (In)
✔️ Thallium (Tl)
⚛️ Electronic Configuration
📌 General configuration: ns² np¹
✔️ B → 1s² 2s² 2p¹
✔️ Al → [Ne] 3s² 3p¹
✔️ Ga → [Ar] 3d¹⁰ 4s² 4p¹
✔️ In → [Kr] 4d¹⁰ 5s² 5p¹
✔️ Tl → [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹
🔢 Oxidation States
✔️ Common oxidation state → +3
✔️ +1 oxidation state increases down the group
📌 Reason → Inert pair effect
Stability of +1 state:
Tl > In > Ga > Al > B

#PhysicalProperties (Trends)
✔️ Atomic size ↑ down the group Al, GA interchange
✔️ Boron → Non-metal
✔️ Al, Ga, In, Tl → Metals
📌 Melting point:
✔️ Boron → very high
✔️ Gallium → melts near room temperature

#ChemicalProperties
Reaction with Oxygen
✔️ Boron → B₂O₃
✔️ Aluminium → Al₂O₃ (protective oxide layer)
📌 Nature of oxides:
✔️ B₂O₃ → Acidic
✔️ Al₂O₃ → Amphoteric
✔️ Ga₂O₃, In₂O₃ → Amphoteric
✔️ Tl₂O₃ → Basic

ReactionwithAcids
✔️ Boron → does NOT react directly
✔️ Aluminium → reacts but shows passivation due to oxide layer

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#SpecialPointsaboutBoron
✔️ Forms only covalent compounds
✔️ Does NOT follow octet rule
✔️ Shows diagonal relationship with silicon
✔️ Does NOT form B³⁺ ion easily
✔️ Boric acid is weak monobasic acid
✔️ Acts as Lewis acid
Aluminium – NCERT Highlights
✔️ Most abundant metal in earth’s crust
✔️ Extracted by Hall–Héroult process
✔️ Al₂O₃ → amphoteric
✔️ Used in thermite reaction
📌 Thermite reaction:
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe + heat

#ImportantCompounds
Borax (Na₂B₄O₇·10H₂O)
✔️ Used in glass industry
✔️ Borax bead test
✔️ On heating → glassy bead

Boric Acid (H₃BO₃)
✔️ Weak acid
✔️ Lewis acid
✔️ Used as antiseptic

Aluminium Chloride (AlCl₃)
✔️ Covalent compound
✔️ Lewis acid
✔️ Exists as Al₂Cl₆ (dimer)

#NEETHOTPOINTS
✔️ Boron → non-metal, others metals
✔️ Inert pair effect → +1 oxidation state
✔️ AlCl₃ → Lewis acid
✔️ B₂O₃ acidic, Al₂O₃ amphoteric
✔️ Gallium melts in hand

@Ayano1me @Neetugpoll @Neetugquiz
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Question 1
Identify the correct set of statements related to Group 13 elements:
A. Boron shows anomalous behaviour in its group
B. Aluminium forms Al³⁺ ion easily in aqueous solution
C. Gallium has lower melting point than aluminium
D. Inert pair effect increases from B to Tl
👉 Correct option(s):
1️⃣ A, C and D
2️⃣ A and B
3️⃣ B and C
4️⃣ A, B, C and D


Question:2
Identify the correctly matched pairs:
A. Boron trichloride – strong Lewis acid
B. Aluminium chloride – exists as AlCl₃ in vapour phase
C. Diborane – electron deficient compound
D. Boric acid – monobasic acid
👉 Correct option(s):
1️⃣ A and C
2️⃣ A, C and D
3️⃣ B and D
4️⃣ 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. Tl³⁺ is a strong oxidising agent
D. Boron forms stable B³⁺ ion
👉 Correct option(s):
1️⃣ A, B and C
2️⃣ A and D
3️⃣ B and C
4️⃣ A, C and D
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A. Diborane has two 3-centre-2-electron bonds
B. All B–H bonds in diborane are equivalent
C. Boranes are electron-deficient compounds
D. Terminal B–H bonds are longer than bridge B–H bonds
👉 Correct option(s):
1️⃣ A and C
2️⃣ A, B and C
3️⃣ B and D
4️⃣ A, C and D


Identify the correct set regarding trihalides of Group 13:
A. Lewis acidity of BX₃ decreases from BF₃ to BI₃
B. BF₃ is a weaker Lewis acid than BCl₃
C. AlCl₃ acts as a Lewis acid
D. Boron trihalides undergo hydrolysis except BF₃
👉 Correct option(s):
1️⃣ A and C
2️⃣ B, C and D
3️⃣ A, B and D
4️⃣ B and C only

Identify the correct statements:
A. B₂O₃ is acidic in nature
B. Al₂O₃ is amphoteric
C. Tl₂O₃ is basic in nature
D. Basic character of oxides decreases down the group
👉 Correct option(s):
1️⃣ A and B
2️⃣ A, B and C
3️⃣ B and D
4️⃣ A, C and D
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thermodynamics.aac
1.6 MB
Topic cover thermodynamics system surrounding boundaries
Intensive extensive property & sign uses
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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 Al³⁺ ion easily in aqueous solution C. Gallium has lower melting point than aluminium D. Inert pair effect increases…
Answer: Option 1 (A, C and D)
Boron anomalous
Al³⁺ strongly hydrated → aqueous me stable nahi
Ga ka m.p. Al se kam
Inert pair effect down the group increase

Answer: Option 2 (A, C and D)
BCl₃ = strong Lewis acid
AlCl₃ vapour phase me dimer (Al₂Cl₆)
Diborane = electron deficient
Boric acid monobasic


Answer: Option 1 (A, B and C)
+3 down the group unstable
+1 stable due to inert pair
Tl³⁺ easily reduce hota ⇒ strong oxidiser
Boron B³⁺ ion nahi banata

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Chemistry booster series
A. Diborane has two 3-centre-2-electron bonds B. All B–H bonds in diborane are equivalent C. Boranes are electron-deficient compounds D. Terminal B–H bonds are longer than bridge B–H bonds 👉 Correct option(s): 1️⃣ A and C 2️⃣ A, B and C 3️⃣ B and D 4️⃣ A,…
Answer: Option 1 (A and C)
Diborane me 2 banana bonds
Terminal & bridge bonds same nahi
Boranes electron-deficient
Bridge bonds longer hote hain

Answer: Option 2 (B, C and D)
Lewis acidity: BF₃ < BCl₃ < BBr₃ < BI₃
AlCl₃ = Lewis acid
BF₃ hydrolysis nahi karta (others karte)
A statement ulta likha

Answer: Option 2 (A, B and C)
B₂O₃ acidic
Al₂O₃ amphoteric
Tl₂O₃ basic
Basic character increase hota hai down the group
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#ELECTROMAGNETICRADIATION
Electromagnetic radiation is a form of energy propagated as waves with electric and magnetic fields oscillating perpendicular to each other.

#️⃣ 1️⃣ Characteristics of EM Radiation

Wave Nature
✔️ Travels in space as waves
✔️ Characterized by wavelength (λ), frequency (ν), speed (c)
✔️ Relationship: c = λν

Particle Nature
✔️ Light also behaves as particles (photons)
✔️ Energy of photon: E = hν
✔️ Momentum of photon: p = h/λ
📌 Constants
✔️ h = Planck’s constant = 6.626 × 10⁻³⁴ J·s
✔️ c = speed of light = 3 × 10⁸ m/s

📌 Constants
✔️ h = Planck’s constant = 6.626 × 10⁻³⁴ J·s
✔️ c = speed of light = 3 × 10⁸ m/s

2#TypesofElectromagneticRadiation (NCERT)
📌 Order of increasing wavelength / decreasing frequency:
γ-rays < X-rays < UV < Visible < IR < Microwaves < Radio waves
✔️ UV, Visible, IR → Important in Atomic Spectra

3️⃣ #ImportantRelations (Atom Chapter)
✔️ Energy of photon: E = hν = hc/λ
✔️ Frequency & wavelength inversely proportional: ν = c/λ

4️⃣ #NEETIMPORTANTPOINTS
✔️ EM radiation has dual nature → wave + particle
✔️ Wave nature → explains interference, diffraction, refraction
✔️ Particle nature → explains photoelectric effect, Compton effect
✔️ Photon energy proportional to frequency (E ∝ ν)

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#PBLOCK ELEMENTSGROUP 14 (CARBON FAMILY)
Elements: C, Si, Ge, Sn, Pb
✔️ General electronic configuration: ns² np²
✔️ Valency: 4
✔️ Oxidation states: +4, +2
📌 Stability of +2 state increases down the group (inert pair effect)


1️⃣ #PHYSICALPROPERTIES
Atomic & Ionic Radii
✔️ Increases down the group (C < Si < Ge < Sn < Pb)
Ionisation Enthalpy
✔️ Decreases down the group
📌 Carbon has very high IE → strong covalent bonding
Electronegativity
✔️ Decreases down the group
📌 C highest, Pb lowest
Catenation (NEET HOT 🔥)
✔️ Ability to form long chains
📌 Order: C >> Si > Ge > Sn > Pb
📌 Strong C–C bond + small size


2️⃣ #ALLOTROPY
Carbon Allotropes
✔️ Crystalline → Diamond, Graphite, Fullerene
✔️ Amorphous → Coal, Coke, Charcoal
📌 Diamond
✔️ sp³ hybridised
✔️ Hardest substance
✔️ Electrical insulator
📌 Graphite
✔️ sp² hybridised
✔️ Good conductor of electricity
✔️ Layered structure


3️⃣ #CHEMICALPROPERTIES
Oxidation States
✔️ C, Si → mainly +4
✔️ Ge, Sn, Pb → +2 & +4
📌 +2 becomes more stable down the group (inert pair effect)
Inert Pair Effect
✔️ ns² electrons do not participate in bonding
📌 Maximum in Pb → Pb²⁺ more stable than Pb⁴⁺


4️⃣ #HYDRIDES (EH₄)
✔️ CH₄, SiH₄, GeH₄, SnH₄
📌 Trend:
✔️ Thermal stability ↓ down the group
✔️ Reducing character ↑ down the group

5️⃣ #HALIDES (MX₄ / MX₂)
Tetrahalides (MX₄)
✔️ CCl₄, SiCl₄ → covalent
✔️ Stability ↓ down the group
Dihalides (MX₂)
✔️ SnCl₂, PbCl₂ → more stable down the group
📌 Due to inert pair effect
📌 PbCl₄ unstable, PbCl₂ stable


6️⃣ #OXIDES (IMPORTANT 🔥)
✔️ CO₂, SiO₂ → Acidic
✔️ GeO₂ → Weakly acidic
✔️ SnO₂ → Amphoteric
✔️ PbO → Amphoteric / basic

📌 Acidity decreases down the group

7️⃣ #CARBONSPECIALCASE
Shows maximum catenation
Forms multiple bonds (C=C, C≡C)
Large number of organic compounds
No d-orbitals → strong π bonding

8️⃣ #COMPARISON ( NEET MCQ)
Property :Carbon :Lead
Catenation :Maximum :Negligible
Oxidation state :+4 :+2 more stable
Nature of oxide :Acidic :Amphoteric
Inert pair effect :Absent :Maximum

⚠️ #NCERTLINES
Pb⁴⁺ more stable than Pb²⁺ → WRONG
✔️ CCl₄ does NOT hydrolyse
✔️ SiCl₄ hydrolyses easily
✔️ Graphite conducts electricity, diamond does not

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Chemistry booster series pinned «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…»