Search This Blog

Molecular Orbital Theory: Diatomic Molecules

Molecular Orbital Theory: Diatomic Molecules | Chemca.in

Molecular Orbital Theory (MOT)

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.

$$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \color{#d946ef}{(\pi_{2p_x} = \pi_{2p_y}) < \sigma_{2p_z}} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$$

Rule 2: Molecules with $> 14$ Electrons

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.

$$\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \color{#d946ef}{\sigma_{2p_z} < (\pi_{2p_x} = \pi_{2p_y})} < (\pi^*_{2p_x} = \pi^*_{2p_y}) < \sigma^*_{2p_z}$$
Bond Order (B.O.) = $\frac{N_b - N_a}{2}$
Where $N_b$ = Bonding Electrons, $N_a$ = Antibonding (*) Electrons
No molecules found matching your search.

Homoatomic Molecules

$$\ce{H2}$$
2 e⁻
$$\sigma_{1s}^2$$
Bond Order 1.0
Diamagnetic
$$\ce{H2+}$$
1 e⁻
$$\sigma_{1s}^1$$
Bond Order 0.5
Paramagnetic
$$\ce{H2-}$$
3 e⁻
$$\sigma_{1s}^2 \ \sigma_{1s}^{*1}$$
Bond Order 0.5
Paramagnetic
Less stable than $\ce{H2+}$ because the extra electron resides in an antibonding orbital.
$$\ce{He2}$$
4 e⁻
$$\sigma_{1s}^2 \ \sigma_{1s}^{*2}$$
Bond Order 0.0
Diamagnetic
Molecule does not exist.
$$\ce{He2+}$$
3 e⁻
$$\sigma_{1s}^2 \ \sigma_{1s}^{*1}$$
Bond Order 0.5
Paramagnetic
Unlike neutral $\ce{He2}$, this cation actually exists.
$$\ce{Li2}$$
6 e⁻
$$KK \ \sigma_{2s}^2$$
Bond Order 1.0
Diamagnetic
Exists in vapor phase. ($KK$ = closed 1s shells)
$$\ce{B2}$$
10 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \pi_{2p_x}^1 = \pi_{2p_y}^1$$
Bond Order 1.0
Paramagnetic
Paramagnetic due to two unpaired electrons in degenerate $\pi$ orbitals.
$$\ce{C2}$$
12 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \pi_{2p_x}^2 = \pi_{2p_y}^2$$
Bond Order 2.0
Diamagnetic
Highly unusual: Bond order is 2, but both are $\pi$ bonds! No $\sigma$ bond.
$$\ce{C2-}$$
13 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ \sigma_{2p_z}^1$$
Bond Order 2.5
Paramagnetic
$$\ce{N2}$$
14 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^2$$
Bond Order 3.0
Diamagnetic
Highest bond order, extremely stable. (1 $\sigma$ + 2 $\pi$)
$$\ce{N2+}$$
13 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^1$$
Bond Order 2.5
Paramagnetic
Removing an electron from a bonding orbital decreases B.O.
$$\ce{N2-}$$
15 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ \sigma_{2p_z}^2 \ \pi_{2p_x}^{*1}$$
Bond Order 2.5
Paramagnetic
Extra electron goes into antibonding $\pi^*$, decreasing B.O.
$$\ce{O2}$$
16 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*1} \!=\! \pi_{2p_y}^{*1})$$
Bond Order 2.0
Paramagnetic
MOT successfully explained the paramagnetism of liquid Oxygen, which VBT failed to do.
$$\ce{O2+}$$
15 e⁻
$$KK \ ... \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*1} \!=\! \pi_{2p_y}^{*0})$$
Bond Order 2.5
Paramagnetic
Removing an anti-bonding electron increases the Bond Order!
$$\ce{O2-} \text{ (Superoxide)}$$
17 e⁻
$$KK \ ... \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*2} \!=\! \pi_{2p_y}^{*1})$$
Bond Order 1.5
Paramagnetic
$$\ce{O2^{2-}} \text{ (Peroxide)}$$
18 e⁻
$$KK \ ... \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*2} \!=\! \pi_{2p_y}^{*2})$$
Bond Order 1.0
Diamagnetic
Isoelectronic with $F_2$.
$$\ce{F2}$$
18 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*2} \!=\! \pi_{2p_y}^{*2})$$
Bond Order 1.0
Diamagnetic
$$\ce{Ne2}$$
20 e⁻
$$... \ (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}) \ \sigma_{2p_z}^{*2}$$
Bond Order 0.0
Diamagnetic
Does not exist. $N_b = N_a$.

Heteroatomic Molecules

$$\ce{CO}$$
14 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^2$$
Bond Order 3.0
Diamagnetic
Isoelectronic with $N_2$. Extremely strong bond.
$$\ce{CN-}$$
14 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 = \pi_{2p_y}^2) \ \sigma_{2p_z}^2$$
Bond Order 3.0
Diamagnetic
Isoelectronic with $N_2$ and $CO$. Strong field ligand.
$$\ce{CN}$$
13 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ \sigma_{2p_z}^1$$
Bond Order 2.5
Paramagnetic
Cyanide radical.
$$\ce{CN+}$$
12 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2)$$
Bond Order 2.0
Diamagnetic
Isoelectronic with $C_2$.
$$\ce{NO+}$$
14 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 = \pi_{2p_y}^2)$$
Bond Order 3.0
Diamagnetic
Nitrosonium ion. Highly stable due to B.O. of 3.
$$\ce{NO}$$
15 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ \pi_{2p_x}^{*1}$$
Bond Order 2.5
Paramagnetic
Fills similarly to $O_2$ pattern. Has one unpaired electron in anti-bonding $\pi^*$.
$$\ce{NO-}$$
16 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ \sigma_{2p_z}^2 \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ (\pi_{2p_x}^{*1} \!=\! \pi_{2p_y}^{*1})$$
Bond Order 2.0
Paramagnetic
Isoelectronic with $O_2$. Possesses two unpaired electrons.
$$\ce{CO+}$$
13 e⁻
$$KK \ \sigma_{2s}^2 \ \sigma_{2s}^{*2} \ (\pi_{2p_x}^2 \!=\! \pi_{2p_y}^2) \ \sigma_{2p_z}^1$$
Bond Order 2.5 (or 3.5)
Paramagnetic
Standard MOT predicts 2.5. Actual B.O. is 3.5 (electron removed from weakly antibonding $\sigma_{2s}^*$).
Powered by

πŸ“š Also Read

Lecture Notes

Calculate Van't Hoff Factor with Degree of Dissociation | JEE & NEET - CHEMCA

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.

  1. Find the number of ions (n):
    Ionization equation: Mg3(PO4)2 → 3Mg2+ + 2PO43-
    Total ions: n = 3 + 2 = 5
  2. Apply the formula:
    Ξ± = (i - 1) / (n - 1)
    0.6 = (i - 1) / (5 - 1)
  3. 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.

© 2026 CHEMCA. All Rights Reserved. Designed for Premier Online Chemistry Education in India.

Subscribe to our YouTube channel and visit www.chemca.in for full courses.

Powered by

πŸ† Solve Another Important Question

Find Geometrical Isomers for Square Planar Complex [MABCD]

How to Calculate Ecell of a Galvanic Cell using Nernst Equation | CHEMCA

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 (cell)

Identify the Cathode (Right Side / Reduction) and the Anode (Left Side / Oxidation). Then apply:
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.
Cu2+/Cu = 0.34 V, Ag+/Ag = 0.80 V.

  1. Calculate cell:
    cell = 0.80 - 0.34 = 0.46 V
  2. 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.
  3. 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: Zn2+/Zn = -0.76 V, Cu2+/Cu = 0.34 V.

Step-by-step Solution:

  • Standard EMF: cell = 0.34 - (-0.76) = 1.10 V
  • Balanced Equation: Zn + Cu2+ → Zn2+ + Cu. Here, n = 2.
  • Nernst Equation: Ecell = 1.10 - (0.0591 / 2) × log( [Zn2+] / [Cu2+] )
  • Ecell = 1.10 - 0.0295 × log( 0.1 / 0.01 )
  • Ecell = 1.10 - 0.0295 × log(10)
  • Since log(10) = 1: Ecell = 1.10 - 0.0295 = 1.0705 V
  • Answer: 1.07 V

Question 2: Calculate the EMF of the following cell at 298K: Mg(s) | Mg2+(10-3 M) || Ag+(10-4 M) | Ag(s).
Given: cell = 3.16 V.

Step-by-step Solution:

  • Balanced Equation: Mg + 2Ag+ → Mg2+ + 2Ag. Here, n = 2.
  • Nernst Equation: Ecell = 3.16 - (0.0591 / 2) × log( [Mg2+] / [Ag+]2 )
  • Ecell = 3.16 - 0.0295 × log( 10-3 / (10-4)2 )
  • Simplify log term: log( 10-3 / 10-8 ) = log(105) = 5
  • Ecell = 3.16 - 0.0295 × 5
  • Ecell = 3.16 - 0.1475 = 3.0125 V
  • Answer: 3.01 V

© 2026 CHEMCA. All Rights Reserved. Designed for Premier Online Chemistry Education in India.

Subscribe to our YouTube channel and visit www.chemca.in for full courses.

Powered by

Find Orbital Symbol using Radial and Angular Nodes | Atomic Structure - CHEMCA

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?

  1. 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.
  2. 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
  3. 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.

© 2026 CHEMCA. All Rights Reserved. Designed for Premier Online Chemistry Education in India.

Subscribe to our YouTube channel and visit www.chemca.in for full courses.

Powered by

Featured Post

H₂O as a Ligand: Weak vs Strong Field Cases