Question

Explain the diamagnetic behaviour of $F_{2}$ molecular on the basis of molecular orbital theory.

Answer 1

Important features of “Molecular orbital” theory:

1) The molecular orbital obtained by the “additive interaction” of “atomic orbitals” is called “bonding molecular” orbital.

2) The “molecular orbital” obtained by the “subtractive interaction” of “atomic orbitals” is called “bonding molecular” orbital.

3) In “boding molecular” orbital, the “electron density” is “concentrated” in-between the “nuclei”, hence, greater attraction and lower energy.

4) Nucleus, hence, less attraction and high energy.

5) “Bond order” is defined as the half of the distance between the “number of electrons” in “bonding molecular” orbital and anti-bonding molecular orbital.

6) $Bond\quad order$ $=\quad \frac { 1 }{ 2 } \left( Number\quad of\quad bonding\quad orbitals-Number\quad of\quad anti-bonding\quad orbitals \right)$

7) “Bond order” is “directly proportional” to the “bond dissociation” energy and inversely proportional to the “bond length”.

8) Bond order is equal to the number of “covalent bonds” in molecule or ion.

9) If there are unpaired electrons in a molecule or ion it will be paramagnetic.

The Molecular orbital diagram of ${ F }_{ 2 }$ is shown below.

$Electronic\quad configuration\quad ({ F }_{ 2 })\quad =\quad \sigma { 1s }^{ 2 }\quad { \sigma }^{ * }{ 1s }^{ 2 }\quad \sigma 2{ p }_{ z }^{ 2 }\quad \pi 2{ p }_{ x }^{ 2 }\quad \pi 2{ p }_{ y }^{ 2 }\quad { \pi }^{ * }2{ p }_{ x }^{ 2 }\quad { \pi }^{ * }2{ p }_{ y }^{ 2 }$

$Bond\quad order$ $=\quad \frac { 1 }{ 2 } \left( Number\quad of\quad bonding\quad orbitals-Number\quad of\quad anti\quad bonding\quad orbitals \right)$

$=\quad \frac { 1 }{ 2 } \quad \left( 8\quad -\quad 6 \right)$

Bond order = 1

Bond order of ${ F }_{ 2 }$ is one therefore, two fluorine atoms are bonded by single bond and it does not have unpaired electrons hence, ${ F }_{ 2 }$ is diamagnetic.