Facebook Twitter Instagram
    ScienceMotive
    • Class 9
      • Matter in Our Surroundings
      • Is Matter Around Us Pure
      • Atoms and Molecules
      • Structure of the Atom
      • The Fundamental Unit of Life
    • Class 10
      • Chemistry
    • Class 11
      • Chemisrty
        • Chapter – 1 Some Basic Concepts of Chemistry
        • Chapter – 2 Structure Of Atom
        • Chapter – 3 Classification of Elements and Periodicity in Properties
        • Chapter – 4 Chemical Bonding and Molecular Structure
        • Chapter – 5 States of Matter
        • Chapter – 6 Thermodynamics
        • Chapter – 7 Equilibrium
        • Chapter – 8 Redox Reaction
        • Chapter – 10 s-Block Elements
        • Chapter – 13 Hydrocarbons
    • Class 12
      • Chemistry
        • The Solid State
        • Solutions
        • Electrochemistry
        • Chemical Kinetics
        • Surface Chemistry
        • p – Block Elements
        • d & f Block Elements
        • Coordination Compounds
        • Haloalkanes and Haloarenes
        • Alcohols, Phenols and Ethers
        • Aldehydes, Ketones and Carboxylic Acids
        • Amines
        • Biomolecules
        • Polymers
        • Chemistry in Everyday Life
    • Practice Questions
      • +1
      • +2
    • Test Series
      • Class 9 Test Series
      • Class 10 Test Series
      • Class 11 Test Series
      • Class 12 Test Series
    • World
      • Current Affairs
      • General Knowledge
    ScienceMotive
    Home » Bond order

    Bond order

    Dr. Vikas JasrotiaBy Dr. Vikas JasrotiaAugust 7, 2021No Comments
    Share
    Facebook WhatsApp Telegram Twitter Email

    Bond Order

    It was introduced by Linus Pauling. Bond order is defined as half the difference between the number of bonding electrons and antibonding electrons.
    In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. The bond number gives an indication of the stability of a bond. Isoelectronic species have the same bond number.

    Bond Order Formula:

    The Bond Order Formula can be defined as half of the difference between the number of electrons in bonding orbitals and antibonding orbitals.
    B.O = ½ [Nb – Na]
    Where,
    Nb is the number of bonding electrons
    Na is the number of antibonding electrons
    If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have resonance, the B.O. does not need to be an integer.

    Bond Order in Molecular Orbital Theory:
    Let us take few examples to find bond order using the Molecular orbital energy level diagram and configuration.
    1. Dihydrogen (H2): One hydrogen atom contains 1 electron, as we have two hydrogen atoms in dihydrogen so we have to write a configuration for two electrons.
    Configuration: (σ1s)2
    Molecular Orbital Energy Level Diagram:

    B.O. = ½ [Bonding Electrons – Antibonding electrons]
    As we have 2 electrons in bonding only
    Therefore B.O. = ½ [2 – 0] = 1

    2. Helium (He2): One He atom contains 2 electrons, as we have two He atoms in He2 so we have to write configuration for 4 electrons (i.e 2 + 2).
    Configuration of He2 = (σ1s)2  (σ*1s)2
    Molecular Orbital Energy Level Diagram:

    B.O. = ½ [Bonding Electrons – Antibonding electrons]
    As we have 2 electrons in bonding and two electrons in antibonding
    Therefore B.O. = ½ [2 – 2] = 0
    As B.O is zero the He2 molecule does not exist

    3. Carbon (C2): One Carbon atom contains 6 electrons, as we have two carbon atoms in C2 so we have to write a configuration for 12 electrons (i.e 6 + 6).
    Configuration of C2 = (σ1s)2  (σ*1s)2  (σ2s)2  (σ*2s)2   [л(2px)2 = л(2py)2]
    Molecular Orbital Energy Level Diagram:

    B.O. = ½ [Bonding Electrons – Antibonding electrons]
    As we have 8 electrons in bonding and 4 electrons in antibonding
    Therefore B.O. = ½ [8 – 4] = 2
     

    4. Oxygen (O2): One Oxygen atom contains 8 electrons, as we have two Oxygen atoms in O2 so we have to write a configuration for 16 electrons (i.e 8 + 8).
    Configuration of O2 = (σ1s)2  (σ*1s)2  (σ2s)2  (σ*2s)2  (σ2s)2  [л(2px)2 = л(2py)2]
    [л*(2px)1 = л *(2py)1]
    Molecular Orbital Energy Level Diagram:


    B.O.= ½ [Bonding Electrons – Antibonding electrons]
    As we have 10 electrons in bonding and 6 electrons in antibonding
    Therefore B.O. = ½ [10 – 6] = 2
     
     
     

     

    Advertisement
    Bond Order
    Share. Facebook Twitter Pinterest LinkedIn Tumblr Email
    Dr. Vikas Jasrotia
    • Website

    Related Posts

    How to Find the Neutrons

    February 2, 2023

    Value-Based Questions Class 11 Chemistry Chapter 7

    February 1, 2023

    Sample Paper Class 11 Chemistry

    January 31, 2023

    Leave A Reply Cancel Reply

    READ ALSO

    Periodic Classification of Elements Questions PDF (Answers)

    March 1, 2023

    Periodic Classification of Elements Questions

    February 28, 2023

    Carbon and Its Compounds Class 10 Solutions of Practice Questions

    February 22, 2023

    Carbon and Its Compounds Class 10 Practice Questions

    February 21, 2023
    Class 10 Test Series

    Periodic Classification of Elements Questions PDF (Answers)

    By Dr. Vikas JasrotiaMarch 1, 2023

    Periodic Classification of Elements Questions PDF (Answers) Periodic Classification of Elements Questions PDF Ans 1.…

    Class 10 Test Series

    Periodic Classification of Elements Questions

    By Dr. Vikas JasrotiaFebruary 28, 2023

    Periodic Classification of Elements Questions Periodic Classification of Elements Questions Que 1. What is the…

    Class 10 Test Series

    Carbon and Its Compounds Class 10 Solutions of Practice Questions

    By Dr. Vikas JasrotiaFebruary 22, 2023

    Carbon and Its Compounds Class 10 Solutions of Practice Questions Carbon and Its Compounds Class…

    Advertisement
    Advertisement
    Facebook Twitter Instagram Pinterest YouTube
    • Disclaimer
    • Contact Us
    • Privacy Policy 
    • Terms and Conditions
    © 2023 All Rights Reserved ScienceMotive.

    Type above and press Enter to search. Press Esc to cancel.