How do you find the n-factor in an intramolecular redox reaction?

The n-factor for a compound undergoing intramolecular redox is equal to either the total number of moles of electrons lost by the oxidized element OR the total number of moles of electrons gained by the reduced element. Do not add them together.

What is the n-factor of KClO3 in its thermal decomposition?

The n-factor of KClO3 is 6. The chlorine atom gains 6 electrons (reducing from +5 to -1), and the three oxygen atoms collectively lose 6 electrons (oxidizing from -2 to 0).

What is the n-factor of ammonium dichromate during decomposition?

The n-factor of (NH4)2Cr2O7 is 6. The two nitrogen atoms lose a total of 6 electrons (oxidizing from -3 to 0), while the two chromium atoms gain a total of 6 electrons (reducing from +6 to +3).

What is the n-factor of ferrous oxalate (FeC2O4)?

The n-factor of ferrous oxalate is 3. The iron (Fe) oxidizes from +2 to +3 (losing 1 electron), and the two carbon atoms in oxalate oxidize from +3 to +4 (losing 2 electrons total). The total electrons lost per molecule is 1 + 2 = 3.

How do you calculate the n-factor when multiple elements oxidize?

When multiple elements in the same compound are oxidized, you calculate the total number of electrons lost by each element (multiplying by their stoichiometric subscript) and add them together to get the total n-factor of the compound.

What is the n-factor of Cuprous Sulfide (Cu2S) when reacted with KMnO4?

The n-factor of Cu2S is 8. The two copper atoms oxidize from +1 to +2 (losing 2 electrons total), and the single sulfur atom oxidizes from -2 to +4 in SO2 (losing 6 electrons). The total is 2 + 6 = 8.

n-Factor in Intramolecular Redox Reactions | Chemca

Advanced Redox Concepts

n-Factor in Intramolecular Redox Reactions

What do you do when one element in a compound gets oxidized and another element in the same compound gets reduced? Master this advanced equivalent weight concept for JEE and NEET.

By Abhishek Sengar 10 Min Read

Standard redox reactions usually involve two distinct reactants: an oxidizing agent and a reducing agent. But sometimes, a single chemical compound acts as both. When different elements within the same molecule undergo simultaneous oxidation and reduction, the process is called an Intramolecular Redox Reaction.

Calculating the equivalent weight of such compounds in JEE Main, JEE Advanced, and NEET frequently causes panic because students don't know which element to use for the n-factor calculation.

The Golden Rule for n-Factor

In any balanced redox reaction, the total number of electrons lost must equal the total number of electrons gained. Therefore, for an intramolecular redox compound, the n-factor is:

n-factor = | Total e^- lost by oxidized element |
OR
n-factor = | Total e^- gained by reduced element |

CRUCIAL: You calculate one OR the other. You NEVER add them together!

Case Study 1: Decomposition of Potassium Chlorate (KClO₃)

This is a classic NTA favorite. When KClO₃ is heated, it decomposes to form potassium chloride and oxygen gas.

Step-by-Step Calculation

KClO₃ → KCl + 3/2 O₂

The Reduction Part

Element: Chlorine (Cl)
Initial State: +5 (in KClO₃)
Final State: -1 (in KCl)
Change per atom: Gains 6 e^-
Atoms per molecule: 1
Total e^- gained = 6

The Oxidation Part

Element: Oxygen (O)
Initial State: -2 (in KClO₃)
Final State: 0 (in O₂)
Change per atom: Loses 2 e^-
Atoms per molecule: 3
Total e^- lost = 3 × 2 = 6

n-factor of KClO₃ = 6

Case Study 2: Decomposition of Ammonium Dichromate ((NH₄)₂Cr₂O₇)

This reaction (the "chemical volcano") is the ultimate test of your understanding. Here, Nitrogen is oxidized, and Chromium is reduced.

Step-by-Step Calculation

(NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4 H₂O

The Oxidation Part

Element: Nitrogen (N)
Initial State: -3 (in NH₄⁺)
Final State: 0 (in N₂)
Change per atom: Loses 3 e^-
Atoms per molecule: 2
Total e^- lost = 2 × 3 = 6

The Reduction Part

Element: Chromium (Cr)
Initial State: +6 (in Cr₂O₇^2-)
Final State: +3 (in Cr₂O₃)
Change per atom: Gains 3 e^-
Atoms per molecule: 2
Total e^- gained = 2 × 3 = 6

n-factor of (NH₄)₂Cr₂O₇ = 6

The Deadly Addition Trap

In both examples above, the electrons lost (6) perfectly match the electrons gained (6), confirming the reaction is balanced internally.

Equivalent Weight Calculation

Equivalent Weight = Molecular Weight (M)

n-factor

WRONG Answer

Eq. Wt = M / 12

(Student added 6 lost + 6 gained)

CORRECT Answer

Eq. Wt = M / 6

(Student took the magnitude of transfer)

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Lecture Notes

n-Factor When Multiple Elements are Oxidized: Ferrous Oxalate & Cu2S | Chemca

Advanced Equivalent Concept

n-Factor: When Multiple Elements are Oxidized in the Same Compound

What happens when an oxidizing agent attacks two different elements in the same molecule? Master the "Addition Rule" to conquer the hardest titration questions in JEE and NEET.

By Abhishek Sengar 10 Min Read

In standard redox reactions, usually, only one element in a compound undergoes a change in oxidation state. However, in advanced titration problems, you will encounter compounds where a strong oxidizing agent (like KMnO₄ or K₂Cr₂O₇) simultaneously oxidizes two or more elements within the same reactant molecule.

If you don't calculate the total electron transfer correctly, your equivalent weight calculation will be completely wrong.

The Addition Rule for n-Factor

Unlike intramolecular redox (where one element oxidizes and another reduces, and we DO NOT add them), here both elements are undergoing the same process (oxidation). Therefore, the molecule as a whole is acting as a massive electron donor.

Total n-factor = Σ (Moles of e^- lost by Element 1) + (Moles of e^- lost by Element 2) ...

Rule of Thumb: When elements suffer the same fate (both oxidized or both reduced), ADD their individual n-factors.

Case Study 1: Ferrous Oxalate (FeC₂O₄)

This is arguably the most frequently asked question in JEE physical chemistry. When Ferrous Oxalate reacts with an acidic solution of KMnO₄, the permanganate oxidizes both the Ferrous ion and the Oxalate ion.

Step-by-Step Breakdown

FeC₂O₄ → Fe³⁺ + 2 CO₂

Element 1: Iron (Fe)

Process: Oxidation
Initial State: +2 (Ferrous)
Final State: +3 (Ferric)
Change: Loses 1 e^- per atom
Atoms per molecule: 1
e^- lost by Fe = 1 × 1 = 1

Element 2: Carbon (C)

Process: Oxidation
Initial State: +3 (in Oxalate)
Final State: +4 (in Carbon Dioxide)
Change: Loses 1 e^- per atom
Atoms per molecule: 2
e^- lost by C = 2 × 1 = 2

Total n-factor = 1 (from Fe) + 2 (from C) = 3

Therefore, Equivalent Weight of FeC₂O₄ = M / 3

Case Study 2: Cuprous Sulfide (Cu₂S)

This is a slightly more advanced example often seen in JEE Advanced. When Cuprous Sulfide is treated with a strong oxidizing agent, both Copper and Sulfur are oxidized to their higher stable states.

Step-by-Step Breakdown

Cu₂S → 2 Cu²⁺ + SO₂

Element 1: Copper (Cu)

Initial State: +1 (Cuprous)
Final State: +2 (Cupric)
Change: Loses 1 e^- per atom
Atoms per molecule: 2
e^- lost by Cu = 2 × 1 = 2

Element 2: Sulfur (S)

Initial State: -2 (Sulfide)
Final State: +4 (in Sulfur Dioxide)
Change: Loses 6 e^- per atom
Atoms per molecule: 1
e^- lost by S = 1 × 6 = 6

Total n-factor = 2 (from Cu) + 6 (from S) = 8

Therefore, Equivalent Weight of Cu₂S = M / 8

The Subscript Trap

The number one mistake students make is calculating the change in oxidation state correctly but forgetting to multiply by the subscript of the element in the parent molecule.

Take Ferrous Oxalate (FeC₂O₄) as an example:

The Mistake

Fe loses 1.

Carbon loses 1.

n-factor = 1 + 1 = 2

(Forgot there are TWO carbons!)

The Correct Way

Fe loses 1.

Carbon loses 1 × 2 = 2.

n-factor = 1 + 2 = 3

(Multiplied by the subscript C₂)

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Lecture Notes

The energy required to break one mole of Cl-Cl bonds in Cl2 is 242 kJ mol-1 | Chemca Solution

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Question

The energy required to break one mole of Cl-Cl bonds in Cl₂ is 242 kJ mol^-1. The longest wavelength of light capable of breaking a single Cl-Cl bond is:

Given: \( c = 3 \times 10^8 \text{ m s}^{-1} \) and \( N = 6.02 \times 10^{23} \text{ mol}^{-1} \)

(a) 494 nm

(b) 594 nm

(d) 700 nm

Detailed Step-by-Step Solution

To find the longest wavelength of light capable of breaking a single Cl-Cl bond, we need to determine the energy required to break just one bond, and then use the Planck-Einstein relation to find the corresponding wavelength.

Step 1: Calculate energy required for one single bond

We are given the energy for one mole of bonds:

E_{\text{mole}} = 242 \text{ kJ mol}^{-1} = 242 \times 10^3 \text{ J mol}^{-1}

To find the energy for a single bond (\(E_{\text{bond}}\)), we divide by Avogadro's number (\(N_A\)):

E_{\text{bond}} = \frac{E_{\text{mole}}}{N_A} \) \( E_{\text{bond}} = \frac{242 \times 10^3 \text{ J}}{6.02 \times 10^{23}} \) \( E_{\text{bond}} \approx 4.0199 \times 10^{-19} \text{ J}

Step 2: Relate Energy to Wavelength

According to the Planck-Einstein relation, the energy of a photon is related to its wavelength (\(\lambda\)) by the formula:

E = \frac{hc}{\lambda}

Rearranging this formula to solve for wavelength (\(\lambda\)):

\lambda = \frac{hc}{E}

Where Planck's constant \( h = 6.626 \times 10^{-34} \text{ J s} \) and the speed of light \( c = 3 \times 10^8 \text{ m s}^{-1} \).

Step 3: Calculate the Wavelength

Substitute the values into our rearranged equation:

\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m s}^{-1})}{4.0199 \times 10^{-19} \text{ J}} \) \( \lambda = \frac{19.878 \times 10^{-26}}{4.0199 \times 10^{-19}} \text{ m} \) \( \lambda \approx 4.944 \times 10^{-7} \text{ m}

To convert meters to nanometers (nm), we multiply by \(10^9\):

\lambda = 4.944 \times 10^{-7} \times 10^9 \text{ nm} = 494.4 \text{ nm}

Conclusion: The calculated wavelength is approximately 494 nm. Therefore, the correct option is (a) 494 nm.

Enhance Your Chemistry Preparation

The problem discussed above is a classic application of the photoelectric and quantum mechanical principles. It is crucial for competitive exams like JEE and NEET. To deeply understand the relationship between photons, energy, and electromagnetic radiation, we highly recommend reviewing our detailed notes on the Structure of Atom Class 11 Chemistry.

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Lecture Notes

A gas absorbs a photon of 355 nm and emits at two wavelengths | Chemca Solution

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Question 43

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at:

(a) 1035 nm

(b) 325 nm

(d) 518 nm

Detailed Step-by-Step Solution

This problem is based on the Law of Conservation of Energy. According to this principle, the total energy of the absorbed photon must be equal to the sum of the energies of the emitted photons.

Step 1: Set up the energy equation

Let \( E_{\text{abs}} \) be the energy absorbed, and \( E_{1} \) and \( E_{2} \) be the energies of the two emitted photons. We can write:

E_{\text{abs}} = E_{1} + E_{2}

We know from Planck's quantum theory that energy \( E = \frac{hc}{\lambda} \). Substituting this into our equation gives:

\frac{hc}{\lambda_{\text{abs}}} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}

Step 2: Simplify and plug in the values

We can cancel out Planck's constant (\(h\)) and the speed of light (\(c\)) from both sides of the equation. This simplifies to a relationship between the wavelengths:

\frac{1}{\lambda_{\text{abs}}} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}

Given the values:

Absorbed wavelength (\( \lambda_{\text{abs}} \)) = \( 355 \text{ nm} \)
First emitted wavelength (\( \lambda_{1} \)) = \( 680 \text{ nm} \)
Second emitted wavelength (\( \lambda_{2} \)) = ?

\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_{2}}

Step 3: Solve for the unknown wavelength (\(\lambda_2\))

Rearrange the equation to isolate \( \frac{1}{\lambda_2} \):

\frac{1}{\lambda_{2}} = \frac{1}{355} - \frac{1}{680}

Find a common denominator or use cross-multiplication:

\frac{1}{\lambda_{2}} = \frac{680 - 355}{355 \times 680} \) \( \frac{1}{\lambda_{2}} = \frac{325}{241400}

Now, invert the fraction to solve for \( \lambda_2 \):

\lambda_{2} = \frac{241400}{325} \) \( \lambda_{2} \approx 742.76 \text{ nm}

Rounding off to the nearest whole number gives us \( 743 \text{ nm} \).

Conclusion: The wavelength of the other emitted photon is approximately 743 nm. Therefore, the correct option is (c) 743 nm.

Mastering Atomic Structure Concepts

Questions involving photon absorption and multi-step emission rely heavily on Planck's Quantum Theory and the principles of energy conservation. If you find numericals like these tricky, we recommend brushing up on your foundational concepts. You can read our comprehensive, easy-to-understand notes on the Structure of Atom Class 11 Chemistry.

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Lecture Notes

The ionization enthalpy of hydrogen atom is 1.312 x 10^6 J mol-1 | Chemca Solution

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Question

The ionization enthalpy of hydrogen atom is \( 1.312 \times 10^6 \text{ J mol}^{-1} \). The energy required to excite the electron in the atom from \( n = 1 \) to \( n = 2 \) is:

(a) \( 9.84 \times 10^5 \text{ J mol}^{-1} \)

(b) \( 8.51 \times 10^5 \text{ J mol}^{-1} \)

(d) \( 7.56 \times 10^5 \text{ J mol}^{-1} \)

Detailed Step-by-Step Solution

To solve this problem, we will use Bohr's model of the hydrogen atom. The ionization enthalpy is the energy required to remove an electron completely from the ground state (\(n=1\)) to infinity (\(n=\infty\)).

Step 1: Understand the Energy Formula

The energy of an electron in the \(n^{\text{th}}\) orbit of a hydrogen atom is given by:

E_n = -\frac{\text{Ionization Energy}}{n^2}

We are given the Ionization Energy (I.E.) = \( 1.312 \times 10^6 \text{ J mol}^{-1} \).

Step 2: Calculate Energy of the Initial and Final States

For the ground state (\(n=1\)):

E_1 = -\frac{1.312 \times 10^6}{1^2} = -1.312 \times 10^6 \text{ J mol}^{-1}

For the first excited state (\(n=2\)):

E_2 = -\frac{1.312 \times 10^6}{2^2} = -\frac{1.312 \times 10^6}{4} \) \( E_2 = -0.328 \times 10^6 \text{ J mol}^{-1}

Step 3: Calculate the Excitation Energy (\(\Delta E\))

The energy required to excite the electron from \(n=1\) to \(n=2\) is the difference between their energies:

\Delta E = E_2 - E_1 \) \( \Delta E = (-0.328 \times 10^6) - (-1.312 \times 10^6) \) \( \Delta E = 1.312 \times 10^6 - 0.328 \times 10^6 \) \( \Delta E = 0.984 \times 10^6 \text{ J mol}^{-1}

To match the options, we can rewrite this in standard scientific notation:

\Delta E = 9.84 \times 10^5 \text{ J mol}^{-1}

Conclusion: The energy required to excite the electron is \( 9.84 \times 10^5 \text{ J mol}^{-1} \). Therefore, the correct option is (a).

Deepen Your Knowledge of Atomic Structure

Understanding the energy levels of hydrogen-like species is a fundamental part of the Bohr Model. This concept frequently appears in both board exams and medical/engineering entrance tests. For a deep dive into atomic models, quantum numbers, and electron excitation, study our comprehensive guide on the Structure of Atom Class 11 Chemistry.

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n-Factor in Intramolecular Redox Reactions | Chemca

The Golden Rule for n-Factor

Case Study 1: Decomposition of Potassium Chlorate (KClO3)

Step-by-Step Calculation

The Reduction Part

The Oxidation Part

Case Study 2: Decomposition of Ammonium Dichromate ((NH4)2Cr2O7)

Step-by-Step Calculation

The Oxidation Part

The Reduction Part

The Deadly Addition Trap

Frequently Asked Questions

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n-Factor When Multiple Elements are Oxidized: Ferrous Oxalate & Cu2S | Chemca

The Addition Rule for n-Factor

Case Study 1: Ferrous Oxalate (FeC2O4)

Step-by-Step Breakdown

Element 1: Iron (Fe)

Element 2: Carbon (C)

Case Study 2: Cuprous Sulfide (Cu2S)

Step-by-Step Breakdown

Element 1: Copper (Cu)

Element 2: Sulfur (S)

The Subscript Trap

Frequently Asked Questions

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The energy required to break one mole of Cl-Cl bonds in Cl2 is 242 kJ mol-1 | Chemca Solution

The energy required to break one mole of Cl-Cl bonds in Cl2 is 242 kJ mol-1. The longest wavelength of light capable of breaking a single Cl-Cl bond is:

Detailed Step-by-Step Solution

Step 1: Calculate energy required for one single bond

Step 2: Relate Energy to Wavelength

Step 3: Calculate the Wavelength

Enhance Your Chemistry Preparation

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A gas absorbs a photon of 355 nm and emits at two wavelengths | Chemca Solution

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at:

Detailed Step-by-Step Solution

Step 1: Set up the energy equation

Step 2: Simplify and plug in the values

Step 3: Solve for the unknown wavelength (\(\lambda_2\))

Mastering Atomic Structure Concepts

Get new posts by email

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The ionization enthalpy of hydrogen atom is 1.312 x 10^6 J mol-1 | Chemca Solution

The ionization enthalpy of hydrogen atom is \( 1.312 \times 10^6 \text{ J mol}^{-1} \). The energy required to excite the electron in the atom from \( n = 1 \) to \( n = 2 \) is:

Detailed Step-by-Step Solution

Step 1: Understand the Energy Formula

Step 2: Calculate Energy of the Initial and Final States

Step 3: Calculate the Excitation Energy (\(\Delta E\))

Deepen Your Knowledge of Atomic Structure

Get new posts by email

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Featured Post

H₂O as a Ligand: Weak vs Strong Field Cases

Case Study 1: Decomposition of Potassium Chlorate (KClO₃)

Case Study 2: Decomposition of Ammonium Dichromate ((NH₄)₂Cr₂O₇)

Case Study 1: Ferrous Oxalate (FeC₂O₄)

Case Study 2: Cuprous Sulfide (Cu₂S)

The energy required to break one mole of Cl-Cl bonds in Cl₂ is 242 kJ mol^-1. The longest wavelength of light capable of breaking a single Cl-Cl bond is: