❖ 11 years of Experience
❖ Mentored top AIR 100 in JEE & INCHO
gold medalists
❖ 500+ IIT Selections
❖ Certified By American Chemical Society
Mohit Ryan Sir
BIT Sindri
MASTER CLASS
TEACHER TEACHER
JEE 2023
Benefits of Pro Pro Lite Pro Classic Pro Plus
Subscription ₹2,700/- ₹3,600/- ₹5,400/-
LIVE Interactive Online Classes
Test series and analysis
Assignments and notes
Doubt solving during class
Doubt Solving on mobile app
Personal mentor
COUPON CODE:
MRPRO ₹53,999/- ₹67,499/- ₹1,07,999/-
VERY IMPORTANT NEWS AT THE END OF THE SESSION
How to Avail Discount ?
Special Discount for this class
Apply Coupon Code:
MRPRO
Visit: https://vdnt.in/YTPRO
Link in Description
If you put my code MRCC then additional reward is -
you get special support from me
Join me PERSONALLY ON TELEGRAM
https://t.me/BEASTCHEMYST
Redox reaction
An oxidation-reduction (redox) reaction is one in which Oxidation and Reduction
reactions take place simultaneously.
Redox reaction
Redox reaction
Redox reaction
Redox reaction
Oxidation: Addition of Oxygen or any electronegative element
2 Mg + O2 2 MgO
S + O2 SO2
Reduction: Removal of Oxygen or any electronegative element
2 HgO 2 Hg + O2
C + H2O H2 + CO
Redox reaction
Electronic Concept:
OXIDATION – Loss of electrons (De-electronation)
REDUCTION – Gain of electrons (Electronation)
Example of Oxidation Reactions
Na Na+ + e−
Zn Zn2+ + 2 e−
Example of Reduction Reactions
Cl2 + 2 e− 2 Cl−
S + 2 e− S2−
Redox reaction
Redox reaction
Redox reaction
Oxidizing Agents (OA) and Reducing Agents (RA)
Oxidizing Agent:
👉 The species which oxidizes a substance and gets reduced is called as Oxidizing
Agent.
👉 It gains electrons.
Reducing Agent:
👉 The species which reduces a substance and gets oxidized is called as Reducing
Agent.
👉 It loses electrons.
Oxidation Number
It is the charge (real or imaginary) which an atom appears to have when it is
in combination. It may be a whole no. or fractional. An element may have
different values of oxidation number depending. It depends on nature of
compound in which it is present. These are some operational rules to
determine oxidation number
Rules for Assigning Oxidation Number
👉 Oxidation number of free elements or atoms is zero
👉 Oxidation number of allotropes is zero
👉 Oxidation number of atoms in homo-nuclear molecules is zero.
👉 Oxidation number of mono-atomic ions is equal to the algebraic charge on
them
👉 Oxidation number of F in compounds is -1
Rules for Assigning OXidation Number
👉 Oxidation number of H in its compounds is +1, except in metal hydrides
where it is -1
👉 Oxidation number of O is -2 in its compounds, but in F2O it is +2 and in
peroxides it is -1 and -0.5 in KO2
👉 Oxidation number of alkali metals in their compounds is +1.
👉 Oxidation number of alkaline earth metals in their compounds is +2.
👉 The sum of oxidation number of all the atoms in a molecule should be zero
and in an ion equal to its charge
Calculate Oxidation Number of Mn in KMnO4
A x=+3
B x=+7
C x=+4
D x=-3
Calculate Oxidation Number of Mn in KMnO4
A x=+3
B x=+7
C x=+4
D x=-3
Solution :
O.N. of Mn in KMnO4 Standard Approach
Algebraic Method
Let the O.N. of Mn be x
O.N. of K + O.N. of Mn + 4(O.N. of O) = 0
(+1) + x + 4 (–2) =0
x=+7
Calculate Oxidation Number of Cr in Cr2O72−
A x =+6
B x= +4
C x=+2
D x=+1
Calculate Oxidation Number of Cr in Cr2O72−
A x =+6
B x= +4
C x=+2
D x=+1
Solution :
O.N. of Cr in Cr2O72−
Let the O.N. of Cr be x
2(O.N. of Cr) + 7(O.N. of O) = - 2
2x + 7 (–2) =-2
x= +6
Calculate Oxidation Number of S in H2SO4
A x=+3
B x=+6
C x=+2
D x=+7
Calculate Oxidation Number of S in H2SO4
A x=+3
B x=+6
C x=+2
D x=+7
Solution :
O.N. of S in H2SO4
Let the O.N. of S be x
2(O.N. of H) + O.N. of S + 4(O.N. of O) = 0
2 (+1) + x + 4 (-2) =0
x=+6
Calculate Oxidation Number of C in C6H12O6
A x=+2
B x=+1
C x=+3
D x=0
Calculate Oxidation Number of C in C6H12O6
A x=+2
B x=+1
C x=+3
D x=0
Solution :
O.N. of C in C6H12O6
Let the O.N. of C be x
6(O.N. of C) + 12(O.N. of H) + 6(O.N. of O) = 0
6x + 12 (+1) + 6 (-2) =0
x=0
What makes learning at Vedantu Interesting and effective
● Quality of Teacher
● Wave Platform
● In Class Doubts
● Doubts application
● Testing Platform and Reports
● Replay and Notes
● Batch Changes Policies
Equivalent weight
Molecular Weight
Equivalent Weight =
Valency Factor
Unit of Equivalent Weight : g/eq
Equivalent weight
Equivalent weight
Equivalent weight
n- factor
n factor here we mean a conversion factor by which we divide molar mass
of substance to get equivalent mass and it depends on nature of substance
which vary from one condition to another condition.
n-factor Calculation
When there is No
reaction given
For Redox For Non-Redox
Reaction Reactions
n- factor
ACID Number of H+
BASE Number of OH–
ION Magnitude of Charge on Ion
z SALT Total Positive/Negative Charge
furnished by 1 molecule of Salt
Number of electrons lost or gained
OA/RA
by 1 molecule of OA or RA
n- factor of acid
n-factor of acid = basicity of the acid
Basicity : Number of replaceable H+ ion.
Ex. n factor of HCl = 1
n factor of CH3COOH = 1
n factor of H2SO4 = 2
n- factor of base
n factor of base = acidity of the base
Acidity: Number of replaceable OH- ion.
Ex. n factor of NaOH = 1
n factor of Ca(OH)2 = 1
n factor of Al(OH)3 = 3
N factor of B(OH)3 = 1 (because it is a monobasic acid)
n- factor of salt
n factor Salt = Total number of positive or negative charge
Ex. n factor of NaOH = 1
n factor of Na2SO4 = 2
n factor of K2SO4.Al(SO4)3. 24H2O = 8
In case of redox reaction
From oxidation number:
n-factor of oxidising or reducing agent = change in oxidation number per molecule.
n- factor
n- factor
n- factor
n- factor
n- factor
n- factor
Vedantu courses and vedantu Programs
❏ Long term courses
❏ Curriculum Structure
❏ Microcourses
Ex. Find the n factor of KMnO4 is different medium.
Ex. Find the n factor of KMnO4 is different medium.
Sol. (i)in acidic medium
KMnO4 ⟶Mn+2
Change in oxidation number of Mn = +7-2 = 5
∴ the n factor of KMnO4 = 5 Ans.
Ex. Find the n factor of KMnO4 is different medium.
(ii) basic medium
KMnO4 ⟶K2MnO4
n factor of KMnO4 = +7-6 = 1 Ans.
Ex. Find the n factor of KMnO4 is different medium.
(iii) neutral medium
KMnO4 ⟶ MnO2
n factor of KMnO4 = +7-4 = 3 Ans.
Calculate n-factor
CaOCl2 + H2O + 2KI → Ca(OH)2 +I2 + 2KCl
Calculate n-factor
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Calculate n-factor
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
The equivalent weight of HCl in the given reaction is :
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 3Cl2 + H2O
A 16.25
B 36.5
C 73
D 85.1
The equivalent weight of HCl in the given reaction is :
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 3Cl2 + H2O
A 16.25
B 36.5
C 73
D 85.1
Solution:
14 mole HCl- loses 6 mole e-;
6
∴ 1 mole of HCl loses mole e-
14
∴ eq. Wt. of HCl = M
6
14
36.5 × 14
⇒ = 85.1
6
When BrO3- ion reacts with Br- in acid medium, Br2 is liberated. The equivalent
weight of Br2 in this reaction is:
5M 5M
A B
8 3
3M 4M
C D
5 6
When BrO3- ion reacts with Br- in acid medium, Br2 is liberated. The
equivalent weight of Br2 in this reaction is:
5M 5M
A B
8 3
3M 4M
C D
5 6
Solution:
2BrO3- + 12H+ + 10Br- → 6Br2 + 6H2O
10 mole required for formation of 6 moles of Br2
10 5
∴ n-factor of Br2 = =
6 3
Mol. wt. m 3M
Eq. wt. = = =
n 5/3 5
MASTER CLASS
TEACHER TEACHER
Enroll into Multiple course
JEE 2023
Benefits of Pro Pro Lite Pro Classic Pro Plus
Subscription ₹2,700/- ₹3,600/- ₹5,400/-
LIVE Interactive Online Classes
Test series and analysis
Assignments and notes
Doubt solving during class
Doubt Solving on mobile app
Personal mentor
COUPON CODE:
MRPRO ₹53,999/- ₹67,499/- ₹1,07,999/-
How to Avail Discount ?
Special Discount for this class
Apply Coupon Code:
MRPRO
Visit: https://vdnt.in/YTPRO
Link in Description
Join Vedantu JEE
Telegram channel
NOW!
Assignments
Notes
Daily Update
https://vdnt.in/JEEVedantu
