Question

Which of the following has minimum bond length?
A.${{O}_{2}}^{+}$
B.${{O}_{2}}^{-}$
C.${{O}_{2}}^{2-}$
D.${{O}_{2}}$

Answer 1

Hint: Bond length can be defined as the experimentally determined average distance between two bonded atoms.

Complete answer:
VSEPR theory is defined as the electron pairs surrounding the atom present in the centre that must be arranged in space in such a distance so as to minimize the electrostatic repulsion experienced between them.
The most important rule of the VSEPR theory states that the bond angles about a central atom are those that minimize the overall repulsion which is experienced between the Electron pairs in the valence shell of the atom.
As the interior angle increases, the repulsion forces decrease sharply. As the value of electronegativity of an atom forming a molecule gets increased, the influence of a bonding electron pair decreases.
Multiple bonds behave similar to a single electron pair for the sake of VSEPR bond theory.
Hybridization of an atom can be found out by calculating the summation of lone pairs and sigma bonds.
If a sum of sigma bonds and lone pairs is 2, we have sp hybridization
If a sum of sigma bonds and lone pairs is 3 , we have $s{{p}^{2}}$ hybridization
If a sum of sigma bonds and lone pairs is 4 , we have $s{{p}^{3}}$ hybridization
If a sum of sigma bonds and lone pairs is 5, we have $s{{p}^{3}}d$ hybridization
If a sum of sigma bonds and lone pairs is 6, we have $s{{p}^{3}}{{d}^{2}}$hybridization.
Now bond length is inversely proportional to bond order. Now ${{O}_{2}}^{+}$ has a bond order of 2.5 which has a maximum bond order . So it has a minimum bond length.

So the correct option is A.

Note:
Bond order is defined as the (total no of electrons present in bonding orbital - total no of electrons present in antibonding electrons ) divided by 2.