Question

A simplified application of MO theory to the hypothetical ‘molecule' OF would give its bond order as:
A.2
B.1.5
C.1.0
D.0.5

Answer 1

Hint:Bond order: Bond order is defined as a molecule or ion, the number of bonds present between the two atoms is called bond order. For example, the bond order of few molecules like H-H is one, O=O is two and N≡N is three. The isoelectronic species has the same bond order like $F{}_{2}$ and $O{}_{2}^{2\_}$ both have 18 electrons and the bond order is one.

Complete step by step answer:
First we will discuss about the MO theory which is molecular orbital theory,
-Molecular orbital theory is the method for describing the electronic structure of molecules using quantum mechanics. It was proposed early in the 20th century by Hund and Mullikan in 1932.
-According to this theory, when two atomic orbitals combine or overlap, they lose their identity and form new orbitals and the new orbital is known as molecular orbital.
-Now according to question, we will calculate the bond order of hypothetical molecule ‘OF',
Each oxygen atom has 2+6 = 8 electron
Each Fluorine atom has 2+7 = 9 electron
-To calculate the Bond order,
Bond order = number of bonding molecules – number of antibonding molecules / 2
= $\dfrac{10-7}{2}$ = $\dfrac{3}{2}$ =1.5
This OF has 1.5 bond order.
Hence, the correct option is (B).

Note:
According to the Lewis Langmuir concept of covalent bond, greater the bond order greater is the stability of the bond that is greater is the bond enthalpy and also shorter is the bond length. The odd electron molecule, three electron bond is considered as equivalent to half covalent bond.