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Schrodinger Equation & Quantum Numbers – JEE Guide

Schrodinger Equation & Quantum Numbers – JEE Guide

Schrodinger Equation – Importance & How Quantum Numbers Arise (JEE & NEET Complete Guide)

1. Introduction – Why Schrodinger Equation is the Heart of Atomic Structure

The Schrodinger equation is the foundation of modern quantum mechanics. Every concept you study in atomic structure — orbitals, quantum numbers, electronic configuration, periodic properties — ultimately comes from this equation.

In JEE Main, JEE Advanced, and NEET, you are not required to derive it mathematically, but you must understand:

  • What the equation represents
  • What the wave function (ψ) means
  • How quantum numbers originate
  • Why orbitals have specific shapes
Chemca Tip: If you understand Schrodinger equation conceptually, atomic structure becomes logical — not memory-based.

2. Limitations of Bohr’s Model – Why Schrodinger Was Needed

Bohr model explained hydrogen spectrum but failed for:

  • Multi-electron atoms
  • Fine spectral lines
  • Wave nature of electrons
  • Heisenberg uncertainty principle compatibility

This led to development of wave mechanics by Erwin Schrodinger.

3. The Schrodinger Wave Equation

3.1 Time-Independent Schrodinger Equation

For hydrogen atom:

Ĥψ = Eψ

Where:

  • Ĥ = Hamiltonian operator (total energy operator)
  • ψ = Wave function
  • E = Energy of system

Expanded form (hydrogen atom):

− (h² / 8π²m) ∇²ψ − (e² / 4πε₀r) ψ = Eψ

You are NOT expected to memorize this equation in JEE. You must understand its meaning.

4. Physical Meaning of Wave Function (ψ)

The wave function ψ itself has no direct physical meaning.

But ψ² gives probability density.

Probability of finding electron in a small region of space.

This replaces Bohr’s fixed orbit concept with orbital concept.

Chemca Concept: Orbit = fixed circular path (Bohr).
Orbital = region of high probability (Quantum Mechanics).

5. How Quantum Numbers Arise from Schrodinger Equation

When we solve Schrodinger equation for hydrogen atom in spherical coordinates, the solution separates into three parts:

  • Radial part
  • Angular part (θ)
  • Azimuthal part (φ)

Mathematically solving it gives three quantum numbers naturally.

5.1 Principal Quantum Number (n)

Arises from boundary conditions applied to radial part.

  • n = 1, 2, 3, 4...
  • Determines energy and size of orbital

Energy of hydrogen atom:

Eₙ = −13.6 / n² eV

5.2 Azimuthal Quantum Number (l)

Arises from angular solution.

  • l = 0 to (n−1)
  • Determines shape of orbital
l Value Subshell Shape
0 s Spherical
1 p Dumbbell
2 d Cloverleaf
3 f Complex

5.3 Magnetic Quantum Number (ml)

Arises due to spatial orientation solutions.

  • ml = −l to +l
  • Total orbitals = 2l + 1
Example:
  • For p (l=1): ml = −1, 0, +1 → 3 orbitals

5.4 Spin Quantum Number (ms)

Not directly from Schrodinger equation but introduced later to explain spectral splitting.

  • ms = +½ or −½

6. Total Number of Orbitals from Quantum Mechanics

  • Total orbitals in nth shell = n²
  • Total electrons = 2n²

This is directly derived from allowed quantum numbers.

7. Importance of Schrodinger Equation in JEE & NEET

  • Explains atomic orbitals
  • Explains quantum numbers logically
  • Justifies electronic configuration
  • Explains shapes of orbitals
  • Basis of periodic trends

Internal Linking Suggestion:

  • Link to “Electronic Configuration of Elements”
  • Link to “Periodic Properties of Elements”

8. Graphical Interpretation – What Diagrams Should Show

Create the following diagrams:

  • 1s orbital probability density (spherical cloud)
  • 2p orbital dumbbell shape
  • Radial distribution curve for 1s and 2s
  • Nodes in 2s orbital
JEE Advanced Alert: Questions on radial nodes and angular nodes are common.

9. Nodes – Direct Outcome of Wave Equation

  • Total nodes = n − 1
  • Angular nodes = l
  • Radial nodes = n − l − 1
Example:

For 3p:

  • n = 3
  • l = 1
  • Total nodes = 2
  • Angular nodes = 1
  • Radial nodes = 1

10. Common Mistakes Students Make

  • Thinking orbitals are circular paths
  • Confusing orbit and orbital
  • Not understanding origin of quantum numbers
  • Memorizing without conceptual clarity

11. Quick 10-Point Revision

  1. Schrodinger equation describes wave nature of electron.
  2. Ĥψ = Eψ is core form.
  3. ψ² gives probability density.
  4. Orbitals are probability regions.
  5. n determines size & energy.
  6. l determines shape.
  7. ml determines orientation.
  8. ms determines spin.
  9. Total orbitals in shell = n².
  10. Total nodes = n − 1.

12. Practice Questions (JEE/NEET Level)

MCQ 1

Maximum number of orbitals in n=4 shell?

Answer: 16

MCQ 2

Angular nodes in 3d orbital?

Answer: l = 2 → 2 angular nodes

Integer Type

Total radial nodes in 4f?

n=4, l=3 → 4−3−1 = 0

13. FAQs (Schema Ready)

FAQ 1: What is Schrodinger equation in simple terms?

It is a mathematical equation that describes wave behavior of electrons in atoms.

FAQ 2: Does JEE require derivation?

No, only conceptual understanding is required.

FAQ 3: How do quantum numbers arise?

They arise naturally when solving Schrodinger equation with boundary conditions.

FAQ 4: What does ψ² represent?

Probability density of finding electron.

FAQ 5: Why is Schrodinger equation important?

It explains orbitals, quantum numbers and electronic structure of atoms.

Conclusion

The Schrodinger equation is the mathematical backbone of atomic structure. It gives birth to quantum numbers, orbitals, and ultimately explains the structure of the periodic table.

Do not fear the equation — understand its meaning.

Explore more free chemistry resources at www.chemca.in

Bookmark this page and revise before every mock test.

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Electronic Configuration of Elements – JEE & NEET Guide

Electronic Configuration of Elements – JEE & NEET Guide

Electronic Configuration of Elements – Complete Guide for JEE & NEET

Primary Keyword: Electronic configuration of elements

1. Introduction – Why Electronic Configuration is Crucial for JEE & NEET

The electronic configuration of elements is the foundation of modern chemistry. If you understand this topic deeply, you automatically gain clarity in:

  • Periodic properties
  • Chemical bonding
  • Magnetism
  • Oxidation states
  • Color of transition metals
  • Coordination chemistry

In JEE Main, JEE Advanced, and NEET, questions are rarely direct. Instead, they are concept-based — especially from d-block configuration, electronic configuration exceptions, and electronic configuration of ions.

Chemca Tip: If you master this chapter conceptually, you can easily score 2–3 guaranteed questions in JEE and 1–2 in NEET.

2. Basics Required Before Learning Electronic Configuration

2.1 Atomic Structure Recap

Atoms consist of:

  • Protons
  • Neutrons
  • Electrons

Electrons revolve around the nucleus in specific energy levels called shells and subshells.

Subshells:

  • s (2 electrons)
  • p (6 electrons)
  • d (10 electrons)
  • f (14 electrons)

2.2 Quick Revision of Quantum Numbers

Electronic configuration depends on quantum numbers:

Quantum Number Symbol What It Represents
Principal n Shell number
Azimuthal l Subshell (s,p,d,f)
Magnetic ml Orientation
Spin ms Spin of electron

Internal Link Suggestion: Link this section to your detailed “Quantum Numbers” article on Chemca.

3. Rules for Writing Electronic Configuration

3.1 Aufbau Principle

The Aufbau principle states that electrons fill orbitals in increasing order of energy.

Energy increases according to the n + l rule.

Diagram Instruction: Create an energy level diagram showing 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f → 5d → 6p → 7s → 5f → 6d → 7p.

Chemca Tip: Lower (n+l) value fills first. If equal, lower n fills first.

3.2 Pauli Exclusion Principle

The Pauli exclusion principle states:

  • No two electrons in an atom can have the same four quantum numbers.
  • Each orbital can hold maximum 2 electrons with opposite spins.

3.3 Hund’s Rule

Hund’s rule states that electrons fill degenerate orbitals singly first before pairing.

Example: Nitrogen (Z = 7)

1s² 2s² 2p³

In 2p:

↑ ↑ ↑ (not ↑↓ ↑ _)

This explains paramagnetism.

4. Order of Filling Orbitals (n + l Rule Table)

Orbital n l n+l Order
1s1011
2s2022
2p2133
3s3034
3p3145
4s4046
3d3257
4p4158
5s5059
4d42610
5p51611
6s60612
4f43713
5d52714
6p61715
7s70716
5f53817
6d62818
7p71819

Memory Trick: Use diagonal arrow rule diagram.

5. Writing Electronic Configuration Step-by-Step

5.1 Simple Elements (Z < 20)

Example: Calcium (Z = 20)

1s² 2s² 2p⁶ 3s² 3p⁶ 4s²

5.2 Transition Elements

Example: Iron (Z = 26)

[Ar] 4s² 3d⁶

5.3 Noble Gas Configuration (Electronic Configuration Shortcut)

The Noble gas configuration simplifies writing long configurations.

Example:

Fe = [Ar] 3d⁶ 4s²

Chemca Tip: Always use noble gas shorthand in JEE to save time.

6. Exceptions in Electronic Configuration

Chromium (Z = 24)

Expected: [Ar] 3d⁴ 4s²

Actual: [Ar] 3d⁵ 4s¹

Copper (Z = 29)

Expected: [Ar] 3d⁹ 4s²

Actual: [Ar] 3d¹⁰ 4s¹

Reason: Extra stability of half-filled and fully filled subshells.

Conceptual Explanation:

  • Exchange energy
  • Symmetry
  • Reduced electron repulsion

7. Electronic Configuration of Ions

Important Rule: Electrons are removed from highest n value first.

Fe = [Ar] 4s² 3d⁶

Fe²⁺ = [Ar] 3d⁶

Fe³⁺ = [Ar] 3d⁵

Even though 4s fills before 3d, it is removed first.

Common JEE Trap: Students remove from 3d first. That is wrong.

8. d-block and f-block Configurations

d-block Configuration

General configuration:

(n-1)d¹–¹⁰ ns¹–²

f-block Configuration

(n-2)f¹–¹⁴ (n-1)d⁰–¹ ns²

Brief Note: Lanthanide contraction affects atomic size.

9. Common Mistakes Students Make

  • Ignoring electronic configuration exceptions
  • Removing wrong electrons in ions
  • Confusing 4s and 3d order
  • Not applying Hund’s rule correctly
  • Forgetting spin while calculating unpaired electrons

10. Quick Revision – 10 Point Summary

  1. Electrons fill lowest energy orbital first.
  2. Follow n + l rule.
  3. If same n+l, lower n fills first.
  4. Maximum 2 electrons per orbital.
  5. Opposite spin required.
  6. Degenerate orbitals fill singly first.
  7. Half-filled and fully-filled subshells are stable.
  8. 4s fills before 3d.
  9. 4s empties before 3d.
  10. Use noble gas shorthand in exams.

11. Practice Questions (JEE/NEET Level)

MCQ 1

Which element shows anomalous electronic configuration?

a) V b) Cr c) Mn d) Ti

Answer: b) Cr

MCQ 2

Number of unpaired electrons in Fe³⁺?

Solution: 3d⁵ → 5 unpaired electrons

Integer Type

Find total electrons in 4th shell of Ca.

Solution: 4s² → 2

12. FAQs (Schema Ready)

FAQ 1: What is electronic configuration of elements?

It is the distribution of electrons in various orbitals of an atom.

FAQ 2: Why does 4s fill before 3d?

Because 4s has lower (n+l) value than 3d.

FAQ 3: Why are Cr and Cu exceptions?

Due to extra stability of half-filled and fully filled subshells.

FAQ 4: How to remove electrons in ions?

Remove from highest n shell first.

FAQ 5: Is electronic configuration important for NEET?

Yes, especially for periodic trends and magnetism questions.

Conclusion

The electronic configuration of elements is not a memory-based chapter. It is a logic-based system governed by quantum mechanics principles like the Aufbau principle, Hund’s rule, and Pauli exclusion principle.

If you master this topic conceptually, you unlock half of inorganic chemistry.

Explore more free chemistry resources at www.chemca.in

Bookmark this page and revise before every mock test.

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Simultaneous Solubility: Complete Guide & Solved Problems

Simultaneous Solubility: Complete Guide & Solved Problems | ChemCa
ChemCa

Simultaneous Solubility

A comprehensive guide to solving equilibrium problems involving multiple sparingly soluble salts. Master the Common Ion Effect and complex stoichiometry.

Ionic Equilibrium Advanced Chemistry

What is Simultaneous Solubility?

Simultaneous solubility refers to the condition where two or more electrolytes (usually sparingly soluble salts) are present in the same solution and dissolve until equilibrium is established.

The defining characteristic of these problems is usually the presence of a Common Ion. According to Le Chatelier's Principle, the presence of this common ion shifts the equilibrium backward, decreasing the solubility of both salts compared to their solubility in pure water.

"The solubility of a salt is suppressed in the presence of another salt containing a common ion."

Governing Principles

  • 1 Constraint: All Solubility Product ($K_{sp}$) expressions must be satisfied simultaneously.
  • 2 Summation: The concentration of the common ion is the sum of contributions from all sources.
  • 3 Electroneutrality: Total positive charge must equal total negative charge in the solution.

Case Study 1: Simple Stoichiometry (1:1)

Problem Statement

Find the simultaneous solubility of Silver Chloride ($AgCl$) and Silver Bromide ($AgBr$) in water.
Given: $K_{sp}(AgCl) = 10^{-10}$, $K_{sp}(AgBr) = 5 \times 10^{-13}$.

Step 1: Set up variables

Let solubility of $AgCl$ be $x$ and $AgBr$ be $y$. The common ion is $Ag^+$.

$$ \begin{align} [Ag^+]_{total} &= [Ag^+]_{from AgCl} + [Ag^+]_{from AgBr} = x + y \\ [Cl^-] &= x \\ [Br^-] &= y \end{align} $$

Step 2: Formulate Equations

Apply the $K_{sp}$ equation for both salts using the total silver concentration.

$$ \begin{align} (1) \quad K_{sp}(AgCl) &= [Ag^+_{total}][Cl^-] = (x+y)(x) = 10^{-10} \\ (2) \quad K_{sp}(AgBr) &= [Ag^+_{total}][Br^-] = (x+y)(y) = 5 \times 10^{-13} \end{align} $$

Step 3: Solve

Divide equation (1) by equation (2) to find the relationship between $x$ and $y$. This eliminates the common term $(x+y)$.

$$ \frac{(x+y)x}{(x+y)y} = \frac{10^{-10}}{5 \times 10^{-13}} \implies \frac{x}{y} = \frac{1000}{5} = 200 \implies x = 200y $$

Substitute $x = 200y$ back into Equation (1):

$$ \begin{align} (200y + y)(200y) &= 10^{-10} \\ (201y)(200y) &= 10^{-10} \\ 40200y^2 &= 10^{-10} \\ y^2 &\approx 2.48 \times 10^{-15} \\ y &= \sqrt{24.8 \times 10^{-16}} \approx 5 \times 10^{-8} \text{ M} \end{align} $$
Final Answer
  • Solubility of AgBr (y) = 5 × 10⁻⁸ M
  • Solubility of AgCl (x) = 1 × 10⁻⁵ M

Case Study 2: Complex Stoichiometry

Problem Statement

Calculate simultaneous solubility of Silver Chromate ($Ag_2CrO_4$) and Silver Chloride ($AgCl$).
Given: $K_{sp}(Ag_2CrO_4) = 1.1 \times 10^{-12}$, $K_{sp}(AgCl) = 1.8 \times 10^{-10}$.

Analysis

Salt 1: Ag₂CrO₄ (Solubility x)
$$ [Ag^+] = 2x, \quad [CrO_4^{2-}] = x $$
Salt 2: AgCl (Solubility y)
$$ [Ag^+] = y, \quad [Cl^-] = y $$
Warning: The total silver concentration is the sum: $[Ag^+]_{total} = 2x + y$

Equations

$$ \begin{align} (1) \quad [Ag^+]^2[CrO_4^{2-}] &= (2x+y)^2(x) = 1.1 \times 10^{-12} \\ (2) \quad [Ag^+][Cl^-] &= (2x+y)(y) = 1.8 \times 10^{-10} \end{align} $$

Strategic Solution

Direct division is difficult here because of the square term. A better strategy is to check if one concentration is negligible. Let's calculate individual solubilities ($S$) in pure water first.

$$ S_{AgCl} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} $$ $$ S_{Ag_2CrO_4} = \sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \approx 6.5 \times 10^{-5} $$

Since $6.5 \times 10^{-5} > 1.34 \times 10^{-5}$, the chromate contributes significantly more silver ions. However, they are close enough that we should try to solve exactly or use the division method carefully.

The Division Trick: Divide Eq(1) by the Square of Eq(2).

$$ \frac{K_{sp1}}{(K_{sp2})^2} = \frac{(2x+y)^2(x)}{[(2x+y)(y)]^2} = \frac{(2x+y)^2 x}{(2x+y)^2 y^2} = \frac{x}{y^2} $$ $$ \frac{x}{y^2} = \frac{1.1 \times 10^{-12}}{(1.8 \times 10^{-10})^2} = \frac{1.1 \times 10^{-12}}{3.24 \times 10^{-20}} \approx 3.39 \times 10^7 $$ $$ x = 3.39 \times 10^7 y^2 $$

Now substitute $x$ back into the simpler equation (Eq 2) or solve iteratively.

Summary of Formulas

Case Type Salt Types Key Relation (Approx)
Type AB + Type AC AgCl + AgBr x/y = Ksp1 / Ksp2
Type A₂B + Type AC Ag₂CrO₄ + AgCl x/y² = Ksp1 / (Ksp2)²
Strong Electrolyte AgCl + AgNO₃ (C) s = Ksp / C
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Some Basic Concepts of Chemistry: Guide for Class 11, JEE & NEET | Chemca

Some Basic Concepts of Chemistry: Guide for Class 11, JEE & NEET | Chemca

Some Basic Concepts of Chemistry: The Ultimate Guide for Class 11, JEE, and NEET

Mastering chemistry starts with a rock-solid foundation. Whether you are a Class 11 student just beginning your journey or a JEE/NEET aspirant looking to polish your fundamentals, the chapter "Some Basic Concepts of Chemistry" is the gateway to the entire subject.

In this guide, we break down the essential topics, from the laws of chemical combination to the ever-important Mole Concept, and provide you with exclusive resources to ace your exams.

Why is this Chapter Important?

This chapter isn't just a requirement for your syllabus; it’s the language of chemistry. It introduces the quantitative side of science. Without understanding molarity, stoichiometry, or atomic masses, Physical Chemistry becomes an uphill battle.

If you are looking for a deep dive into the syllabus-specific notes, check out our Detailed Guide on Basic Concepts of Chemistry for JEE & NEET.

Core Topics You Must Master

1. Matter and Its Classification

Matter is anything that has mass and occupies space. Understanding the difference between mixtures (homogeneous and heterogeneous) and pure substances (elements and compounds) is fundamental to identifying how chemicals react.

2. The Laws of Chemical Combination

Chemistry is governed by rules. These laws explain how elements combine to form compounds:

  • Law of Conservation of Mass: Mass is neither created nor destroyed.
  • Law of Definite Proportions: A compound always contains the same elements in the same proportion by mass.
  • Law of Multiple Proportions: Formulated by Dalton, this explains how elements like Carbon and Oxygen can form both \(CO\) and \(CO_2\).

3. Dalton’s Atomic Theory

Dalton proposed that matter consists of indivisible atoms. While we now know about subatomic particles, his theory laid the groundwork for the concept of atomic mass and chemical reactions.

4. Atomic and Molecular Masses

  • Atomic Mass: Expressed in \(u\) (unified mass), it is relative to \(1/12^{th}\) of the mass of a Carbon-12 atom.
  • Molecular Mass: The sum of atomic masses of the elements in a molecule. For example, for \(H_2O\): \((2 \times 1.008) + 16.00 = 18.016 \, u\).

The Heart of Chemistry: The Mole Concept

The Mole is the bridge between the microscopic world of atoms and the macroscopic world of the laboratory. One mole of any substance contains exactly \(6.022 \times 10^{23}\) particles. This number is known as Avogadro’s Number (\(N_A\)).

\[\text{Number of Moles (n)} = \frac{\text{Given Mass (m)}}{\text{Molar Mass (M)}}\]

Struggling with the math? Don't worry! We have prepared a dedicated resource for you. Download our Chemca Formula Sheet: Mole Concept & Stoichiometry to keep all the vital equations at your fingertips.

Empirical and Molecular Formulas

  • Empirical Formula: Represents the simplest whole-number ratio of various atoms present in a compound.
  • Molecular Formula: Shows the exact number of atoms of different elements present in a molecule.

Relationship: \(\text{Molecular Formula} = n \times (\text{Empirical Formula})\), where \(n\) is a simple whole number.

Stoichiometry and Limiting Reagents

Stoichiometry deals with the calculation of masses and volumes of reactants and products involved in a chemical reaction.

Pro Tip: Always identify the Limiting Reagent—the reactant that is completely consumed in a reaction and limits the amount of product formed.

Boost Your Preparation with Chemca Resources

Final Thoughts

"Some Basic Concepts of Chemistry" is more than just Chapter 1; it is the foundation of your career in science. Focus on units, practice numericals, and ensure your basics are clear.

Happy Learning with Chemca!

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Ionic Equilibrium - JEE Advanced Numerical Challenge | Chemca.in

Ionic Equilibrium - JEE Advanced Numerical Challenge | Chemca.in

Ionic Equilibrium Challenge

30 Advanced Numerical Problems for JEE | www.chemca.in

Course Progress 0/30 Solved

Targeting JEE Advanced?

Analyze every solution carefully to understand common-ion effects, salt hydrolysis, and complexation concepts.

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