The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule is _________.
Correct Answer: 8
Solution and Explanation
The molecular orbitals formed from 2s and 2p atomic orbitals are as follows:
\(• Two\ molecular\ orbitals\ of\ 2s\ and\ σ ∗2s\)
\(• Six\ molecular\ orbitals\ of\ 2p: σ2pz, σ ∗2pz, π2px, π2py, π ∗2px, π ∗2py.\)
Thus, the total number of molecular orbitals formed is 8.
Top Questions on Molecular Orbital Theory
Which of the following statement is true with respect to H\(_2\)O, NH\(_3\) and CH\(_4\)?
(A) The central atoms of all the molecules are sp\(^3\) hybridized.
(B) The H–O–H, H–N–H and H–C–H angles in the above molecules are 104.5°, 107.5° and 109.5° respectively.
(C) The increasing order of dipole moment is CH\(_4\)<NH\(_3\)<H\(_2\)O.
(D) Both H\(_2\)O and NH\(_3\) are Lewis acids and CH\(_4\) is a Lewis base.
(E) A solution of NH\(_3\) in H\(_2\)O is basic. In this solution NH\(_3\) and H\(_2\)O act as Lowry-Bronsted acid and base respectively.
- JEE Main - 2025
- Chemistry
- Molecular Orbital Theory
Which of the following linear combinations of atomic orbitals will lead to the formation of molecular orbitals in homonuclear diatomic molecules (internuclear axis in z-direction)?
(1) \( 2p_z \) and \( 2p_x \)
(2) \( 2s \) and \( 2p_x \)
(3) \( 3d_{xy} \) and \( 3d_{x^2-y^2} \)
(4) \( 2s \) and \( 2p_z \)
(5) \( 2p_z \) and \( 3d_{x^2-y^2} \)
- JEE Main - 2025
- Chemistry
- Molecular Orbital Theory
- In \( \text{O}_2^- \), \( \text{O}_2 \), and \( \text{O}_2^{2-} \) molecular species, the total number of antibonding electrons respectively are:
- MHT CET - 2024
- Chemistry
- Molecular Orbital Theory
- In which of the following sets, the sum of bond orders of the three species is maximum?
- AP EAMCET - 2024
- Chemistry
- Molecular Orbital Theory
- The total number of anti bonding molecular orbitals, formed from 2s and 2p atomic orbitals in a diatomic molecule is _____________.
- JEE Main - 2024
- Chemistry
- Molecular Orbital Theory
Questions Asked in JEE Main exam
- In YDSE, light of intensity \( 4I \) and \( 9I \) passes through two slits respectively. Difference of maximum and minimum intensity of interference pattern is:
- JEE Main - 2025
- Wave optics
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
- JEE Main - 2025
- Differential Equations
Find the IUPAC name of the compound.
- JEE Main - 2025
- Aromaticity & chemistry of aromatic compounds
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32
- JEE Main - 2025
- Limits and Exponential Functions
- The integral \[ \int_0^\pi \frac{8x}{4\cos^2 x + \sin^2 x} \, dx \text{ is equal to:} \]
- JEE Main - 2025
- Trigonometric Functions