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Bond Order

  1. Dec 8, 2003 #1
    Does anyone know what border is.
  2. jcsd
  3. Dec 8, 2003 #2
    I have a structure of KNO3, and a question I have asks for the NO bond number. The answer, aparently, is 4/3. The equation I am trying to use Bond order = 0.5(Bonde electrons - antibonding electrons. I simply can't get 4/3 though.
  4. Dec 9, 2003 #3


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    The existence of electrons in a antibonding orbital tends to repel two atoms apart while electrons in a molecular bonding orbital tends to stablize the two atoms and thus forming a bond. In order for any bond to exist the number of electrons in the bonding orbitals must be greater than the electrons in the antibonding orbitals. Noble gases rarely exist as diatomic molecules (through intramolecular bonding) simply because equal numbers of electrons would have to occupy both antibonding and bonding orbitals.

    The nitrate ion has delocalization of electrons through a Pz orbital and will exist as a hybrid of 4 resonance structures. 3 of the 4 structures (structures with double bonding) predominate or are more stable than the 4th structure. Since a bond order of 2 would indicate a double bond we can understand why each bond has a order of 4/3. Simply because pi bond is "distributed" through each of the three bonds (1 bond divided by three). In order to see which molecular orbitals are involved in bonding, simply draw the full diagram and find which specific orbital(s) has a unequal number of antibonding and bonding orbital.

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