Question

The calculated bond order in ${{H}_{2}}^{-}$ ion is:
   (A) 0
   (B) $\dfrac{1}{2}$
   (C) $-\dfrac{1}{2}$
    (D) 1 $\dfrac{1}{2}$

Answer 1

Hint: Bond order is said to be the difference between number of electrons in bonding and antibonding molecular orbitals. In the given ion, the total number of electrons is three and thus by finding the arrangement of these electrons in the orbital we could find the bond order.


Complete step by step solution:
  - Let's begin with the concept of bond order. It refers to the kind of bond between two atoms in a molecule. In other words, bond order can be explained as the number of bonding pairs of electrons between two atoms.
  - If we are considering a covalent bond between two atoms, a single bond is said to have a bond order of one, a double bond will have a bond order of two, and a triple bond will have a bond order of three, and it goes on like this.
  - More attraction between electrons leads to high bond order. It also means that higher bonds bond the atoms together more tightly. Similarly, if the bond order is low, then there would be less attraction between electrons and thus the atoms would be held loosely.
  - Bond order also gives us an idea about the stability of the bond. As we mentioned, higher the bond order, higher would be the holding of atoms between each other and thus it leads to higher stability of bond.
 - The bond order can be found by using molecular orbital theory. By MO theory, bond order can be defined as the difference between the number of bonding electrons and number of antibonding electrons. Thus, it can be written as follows
Bond order = $\dfrac{1}{2}\left[ {{N}_{b}}-{{N}_{a}} \right]$
 Where ${{N}_{b}}$ is the number of bonding electrons in bonding molecular orbital
   ${{N}_{a}}$is the number of the antibonding electrons.
 - In ${{H}_{2}}^{-}$ ion the total number of electrons is three. Hence its electronic configuration can be written as $\sigma 1{{s}^{2}}{{\sigma }^{*}}1{{s}^{1}}$.Thus it contains two electrons in the bonding molecular orbital and one electron in the antibonding molecular orbital. Thus, the equation for bond order can be written as
  Bond Order = $\dfrac{1}{2}\left[ 2-1 \right]$
      = $\dfrac{1}{2}$

Therefore, the answer is option (B). That is the bond order for ${{H}_{2}}^{-}$ is $\dfrac{1}{2}$.



Note: Do not confuse bond order with bond energy and bond length. As we mentioned, bond order is the difference between the number of bonds and antibonds. Whereas bond energy can be defined as the energy required to break all the covalent bonds of a specific type in one mole of a chemical substance and bond length can be described as a measure of the distance between the nuclei of two chemically bonded atoms in a molecule.