Master the electronic configurations of diatomic molecules. Learn how s-p mixing changes the energy level diagram and predict bond order and magnetic behavior instantly.
Rule 1: Molecules with $\le 14$ Electrons
For molecules like $B_2$, $C_2$, and $N_2$, s-p mixing occurs. This pushes the $\sigma_{2p_z}$ orbital higher in energy, placing it above the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals.
For heavier molecules like $O_2$ and $F_2$, the energy gap between 2s and 2p is large. No s-p mixing occurs, meaning $\sigma_{2p_z}$ drops back down to its normal, lowest-energy position.
Calculate Van't Hoff Factor with Degree of Dissociation | JEE & NEET - CHEMCA
How to Calculate Van't Hoff Factor from Degree of Dissociation (Ξ±)
Published by Abhishek Sengar | CHEMCA India
In the Physical Chemistry chapter of Solutions and Colligative Properties, dealing with electrolytes means you have to account for the Van't Hoff factor (i). While strong electrolytes fully dissociate, weak electrolytes only partially break down. This partial breakdown is measured by the Degree of Dissociation (Ξ±).
If you are aiming for JEE Main, JEE Advanced, or NEET, knowing the exact formula connecting i and Ξ± is non-negotiable. Let's break down a numerical problem to understand its application.
Video Tutorial: Step-by-Step Calculation
Watch Abhishek Sengar sir from CHEMCA solve a classic question: "What will be the Van't Hoff factor of Mg3(PO4)2 if Ξ± is 0.6?"
The Master Formula
To find the Van't Hoff factor for a solute undergoing dissociation, we use the following standard relation:
Ξ± = (i - 1) / (n - 1)
Ξ± (Alpha): Degree of dissociation (given as 0.6 in our example).
i: Van't Hoff factor (what we need to find).
n: Total number of ions produced by the dissociation of one molecule of the electrolyte.
Pro-Tip from Sir: If the problem states the solute is an electrolyte but doesn't mention the value of Ξ±, you must assume it is a strong electrolyte with 100% dissociation. In that case, Ξ± = 1, which simplifies the formula directly to i = n.
Solved Example: Magnesium Phosphate
Let's solve the exact question from the video: Find i for Magnesium Phosphate Mg3(PO4)2 given Ξ± = 0.6.
Find the number of ions (n):
Ionization equation: Mg3(PO4)2 → 3Mg2+ + 2PO43-
Total ions: n = 3 + 2 = 5
Apply the formula: Ξ± = (i - 1) / (n - 1) 0.6 = (i - 1) / (5 - 1)
Solve for i: 0.6 = (i - 1) / 4 0.6 × 4 = i - 1 2.4 = i - 1 i = 3.4
Notice that because dissociation is partial (60%), the Van't Hoff factor (3.4) is greater than 1, but less than the theoretical maximum of 5.
Practice Questions for JEE & NEET
Test your understanding with these exam-level questions. Click the button to verify your calculations!
Question 1: Calculate the Van't Hoff factor for an aqueous solution of Aluminium Sulphate Al2(SO4)3 if it is 80% dissociated.
Step-by-step Solution:
Given: Ξ± = 80% = 0.8
Ionization: Al2(SO4)3 → 2Al3+ + 3SO42-
Number of ions (n) = 2 + 3 = 5
Formula: Ξ± = (i - 1) / (n - 1)
0.8 = (i - 1) / (5 - 1)
0.8 = (i - 1) / 4
3.2 = i - 1
Answer: i = 4.2
Question 2: An electrolyte A2B3 is dissolved in water. If no information about its degree of dissociation is given, what will be the value of its Van't Hoff factor (i) used in colligative property formulas?
Answer: 5
Reasoning: As Abhishek Sir pointed out, if Ξ± is not mentioned, we assume the compound is a strong electrolyte undergoing 100% dissociation. Therefore, Ξ± = 1.
The compound A2B3 breaks into 2 A ions and 3 B ions. Thus, n = 2 + 3 = 5.
When Ξ± = 1, the formula Ξ± = (i - 1) / (n - 1) simplifies to i = n. Therefore, i = 5.
How to Calculate Ecell of a Galvanic Cell using Nernst Equation | CHEMCA
How to Calculate the EMF (Ecell) of a Galvanic Cell
Published by Abhishek Sengar | CHEMCA India
The Nernst Equation is one of the most critical formulas in the Electrochemistry chapter. It allows us to calculate the cell potential (Ecell) under non-standard conditions—meaning when the concentration of the electrolyte solutions is not exactly 1 Molar or the temperature is not 298K.
While the formula itself is straightforward, students frequently make a critical error when formulating the reaction quotient. Let's learn the foolproof way to solve these numericals for JEE Main, JEE Advanced, and NEET.
Video Tutorial: Solving a Copper-Silver Cell
Watch Abhishek Sengar sir from CHEMCA break down a standard Galvanic cell problem involving Copper and Silver electrodes. Pay close attention to the second step!
The 3-Step Strategy
Step 1: Find Standard Cell Potential (E°cell)
Identify the Cathode (Right Side / Reduction) and the Anode (Left Side / Oxidation). Then apply: E°cell = E°cathode - E°anode
Crucial Step 2: Write the Balanced Chemical Equation!
Do NOT jump straight to the Nernst Equation. You must write the balanced cell reaction to find the correct number of transferred electrons (n) and the correct stoichiometric coefficients to square/cube your concentration terms. If you skip this, your answer will likely be wrong.
Step 3: Apply the Nernst Equation (at 298K)
Ecell = E°cell - (0.0591 / n) × log10(Q)
Where Q is the Reaction Quotient: [Products] / [Reactants] raised to their stoichiometric powers.
Solved Example from the Video
Given: A cell with Cu | Cu2+(0.01M) || Ag+(0.1M) | Ag. E°Cu2+/Cu = 0.34 V, E°Ag+/Ag = 0.80 V.
Calculate E°cell: E°cell = 0.80 - 0.34 = 0.46 V
Balanced Equation:
Cu gives 2 electrons, but Ag accepts 1. We must multiply Ag by 2. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Electrons transferred (n) = 2.
Nernst Equation: Ecell = 0.46 - (0.0591 / 2) × log( [Cu2+] / [Ag+]2 ) Ecell = 0.46 - (0.0591 / 2) × log( 0.01 / (0.1)2 )
Since (0.1)2 = 0.01, the log term becomes log(0.01 / 0.01) = log(1) = 0. Final Answer: Ecell = 0.46 V
Practice Questions for JEE & NEET
Try these hand-picked numericals. Remember to balance the equation first!
Question 1: Calculate the EMF of the Daniel Cell at 298K: Zn(s) | Zn2+(0.1M) || Cu2+(0.01M) | Cu(s).
Given: E°Zn2+/Zn = -0.76 V, E°Cu2+/Cu = 0.34 V.
Find Orbital Symbol using Radial and Angular Nodes | Atomic Structure - CHEMCA
How to Find the Orbital Symbol using Radial & Angular Nodes
Published by Abhishek Sengar | CHEMCA India
In the Atomic Structure chapter, understanding quantum numbers and nodes is essential for both JEE and NEET exams. A "node" is a region in space around the nucleus where the probability of finding an electron is exactly zero.
Examiners love to test this concept in two ways: either they give you the orbital symbol (like 3p or 4d) and ask for the number of nodes, OR they give you the number of nodes and ask you to reverse-engineer the exact orbital symbol. Let's learn how to do both!
Video Tutorial: Reverse-Engineering the Orbital
Watch Abhishek Sengar sir from CHEMCA instantly solve a question asking for the orbital symbol given 3 radial nodes and 2 angular nodes.
The Core Formulas
To solve any question regarding nodes, you only need to memorize two simple formulas relying on the Principal Quantum Number (n) and the Azimuthal Quantum Number (l):
Angular Nodes= l
Radial (Spherical) Nodes= n - l - 1
Quick Tip: Remember your azimuthal quantum numbers (l):
s-orbital: l = 0 | p-orbital: l = 1 | d-orbital: l = 2 | f-orbital: l = 3
Solved Example from the Video
Question: What will be the symbol of an orbital that has 3 radial nodes and 2 angular nodes?
Find the Azimuthal Quantum Number (l):
We know: Angular Nodes = l
Given Angular Nodes = 2. Therefore, l = 2.
An l value of 2 corresponds to a d-orbital.
Find the Principal Quantum Number (n):
Formula: Radial Nodes = n - l - 1
We know Radial Nodes = 3, and we just found l = 2.
Substitute the values: 3 = n - 2 - 1 3 = n - 3 n = 3 + 3 = 6
Combine for Final Symbol:
With n = 6 and orbital type d, the exact symbol is 6d.
Practice Questions for JEE & NEET
As Abhishek Sir mentioned, exam questions can be asked from both directions. Try these two variations!
Question 1 (Forward Type): Calculate the number of radial nodes, angular nodes, and total nodes for a 5p orbital.
Step-by-step Solution:
For a 5p orbital, the quantum numbers are: n = 5 and l = 1 (since it's a p-orbital).
Angular Nodes = l = 1
Radial Nodes = n - l - 1 = 5 - 1 - 1 = 3
Total Nodes = Angular + Radial = 1 + 3 = 4(Alternatively, Total Nodes = n - 1 = 5 - 1 = 4)
Question 2 (Reverse Type): Find the symbol of the orbital which has exactly 2 radial nodes and 1 angular node.
Step-by-step Solution:
Find l: Angular node = l = 1. This means it is a p-orbital.
Find n: Radial nodes = n - l - 1 = 2.
Substitute l = 1 into the equation: n - 1 - 1 = 2.
n - 2 = 2 ⟶ n = 4.
Answer: Since n = 4 and l = 1, the orbital symbol is 4p.