
As compared to pure atomic orbitals, hybrid orbitals have
A) Low energy
B) Same energy
C) High energy
D) None of these
Answer
232.8k+ views
Hint: We know that, in the process of hybridisation, atomic orbitals composed of similar or close to similar energy undergo the intermixing of orbitals to produce a group of orbitals. Here, we have to determine the energy of the hybrid orbitals.
Formula used: \[H = \dfrac{{V + X + A - C}}{2}\]; Here, H means hybridisation, X is the count of monovalent atoms, A stand for anionic charge, and C stands for cationic charge
Complete Step by Step Answer:
Let's understand some characteristics of hybridization.
1) The orbitals belonging to the same atom possessing similar energies undergo hybridization.
2) The count of hybrid orbitals produced equates with the count of orbitals undergo hybridization.
3) There is a comparable shape and energy of the hybrid orbitals.
4) In hybrid orbitals, more stable bonds are generated than in the pure atomic orbitals.
5) The aim of hybridization is to form equivalent symmetrical orbitals.
Let's understand the energy of the formed hybrid orbitals. The energy of the hybrid orbitals is less as compared to the pure atomic orbitals. The reason is because of the change in the energy level of the orbitals because of hybridization. The energy possessed by 2s orbital of a Carbon atom is lower than the 2p orbitals as it is of more penetrating nature. After the process of hybridization, the energy content of all the hybrid orbitals has the same energy, lower than orbitals of p but higher than s- orbital.
Hence, option A is right.
Note: It is to be noted that, there are many types of hybridization namely, sp, \[s{p^2}\] , \[s{p^3}\] , \[s{p^3}{d^2}\] etc. Hybridisation determines the molecular shape of a molecule. If no lone pair is present, the \[s{p^3}\]hybridization indicates a tetrahedral-shaped molecule.
Formula used: \[H = \dfrac{{V + X + A - C}}{2}\]; Here, H means hybridisation, X is the count of monovalent atoms, A stand for anionic charge, and C stands for cationic charge
Complete Step by Step Answer:
Let's understand some characteristics of hybridization.
1) The orbitals belonging to the same atom possessing similar energies undergo hybridization.
2) The count of hybrid orbitals produced equates with the count of orbitals undergo hybridization.
3) There is a comparable shape and energy of the hybrid orbitals.
4) In hybrid orbitals, more stable bonds are generated than in the pure atomic orbitals.
5) The aim of hybridization is to form equivalent symmetrical orbitals.
Let's understand the energy of the formed hybrid orbitals. The energy of the hybrid orbitals is less as compared to the pure atomic orbitals. The reason is because of the change in the energy level of the orbitals because of hybridization. The energy possessed by 2s orbital of a Carbon atom is lower than the 2p orbitals as it is of more penetrating nature. After the process of hybridization, the energy content of all the hybrid orbitals has the same energy, lower than orbitals of p but higher than s- orbital.
Hence, option A is right.
Note: It is to be noted that, there are many types of hybridization namely, sp, \[s{p^2}\] , \[s{p^3}\] , \[s{p^3}{d^2}\] etc. Hybridisation determines the molecular shape of a molecule. If no lone pair is present, the \[s{p^3}\]hybridization indicates a tetrahedral-shaped molecule.
Recently Updated Pages
JEE Main Course 2026 - Important Updates and Details

JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

Chemistry Question Pattern for JEE Main & Board Exams

Chemistry Question Paper PDF Download (2025, 2024) with Solutions

JEE Main Books 2026: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions (2025-26)

Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 4 The d and f Block Elements (2025-26)

Biomolecules Class 12 Chemistry Chapter 10 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules (2025-26)

