Question

The calculated bond order in $H_ {2} ^ {-}$ ion is:
(A) $12$
(B) $\dfrac {1}{2} $
(C) $\dfrac {-1}{2} $
(D) $1$

Answer 1

Hint: Bond order, as introduced by Linus Pauling, is defined as the difference between the number of bonds and antibonds. The bond number itself is the number of electron pairs between a pair of atoms.

Complete step by step solution:
In molecular orbital theory, electrons in a molecule are not assigned to individual chemical bonds between atoms but are treated as moving under the influence of the atomic nuclei in the whole molecule. Quantum mechanics describes the spatial and energetic properties of electrons as molecular orbitals that surround two or more atoms in a molecule and contain valence electrons between atoms.
Now, using molecular orbital theory:
The valence electrons in $H_ {2} ^ {-} $ion is: $2+1=3$,
So, the configuration would be: $1{{s}^{2}}2{{s}^ {1}} $,
Or we can write as: ${{\left (\sigma 1s \right)} ^ {2}} {{\left (\sigma *1s \right)} ^ {1}} $,
Now we will be calculating the bond order:
Bond order is defined as half the difference between the number of electrons in bonding molecular orbital (Nb​) and the number of electrons in the antibonding molecular orbitals (Na​).
So, the formula of bond order is: $\dfrac {1}{2} \left({{N}_{b}}-{{N}_{a}} \right) $,
So, keeping the values in the above formula of bond order: $\dfrac {1}{2} \left (2-1 \right) $,
Solving it further we would get: bond order: \[\dfrac {1}{2} =0.5\],

Therefore, we can say that option (B) is correct.

Note: Molecular orbital theory is more powerful than valence-bond theory because the orbitals reflect the geometry of the molecule to which they are applied. But this power carries a significant cost in terms of the ease with which the model can be visualized.