Question

The energy-level diagram shown here can be continued to higher energies. The next few orbitals, in order of increasing energy, are sigma15, pattern is repeated. Determine the bond order o f a ?g2 molecule. Predict the stability of a ?g2 molecule

Answer 1

Concepts and reason

The bonding and anti – bonding interaction of the molecules can be explained with the help of molecular orbital diagram. It also gives detailed description of bonding in molecules.

Bond order represents the number of bonds present between two bonded atoms. It also explains the stability of bond.

Fundamentals

Molecular orbital diagram:

First, the electrons fills the bonding molecular orbital and then anti – bonding molecular orbital. If the electrons in molecular orbitals are paired, then it is said to be diamagnetic. If the electrons in molecular orbitals are unpaired, then it is called paramagnetic.

Bond order can be calculated by:

${\rm{Bond}}\,\,{\rm{order}}\,\,{\rm{ = }}\,\,\frac{{\left( \begin{array}{l}\\{\rm{Number}}\,\,{\rm{of}}\,\,{\rm{electrons}}\,\,{\rm{in}}\,\\\\{\rm{bonding}}\,\\\\{\rm{molecular}}\,\,{\rm{orbital}}\\\end{array} \right)\, - \,\,\left( \begin{array}{l}\\{\rm{Number}}\,\,{\rm{of}}\,\,{\rm{electrons}}\,\\\\\,{\rm{in}}\,\,{\rm{antibonding}}\\\\{\rm{molecular}}\,\,{\rm{orbital}}\\\end{array} \right)}}{{\rm{2}}}$

The stable and unstable compounds can be determined with the help of bond order. If the bond order is 0, then the molecule becomes very unstable and it disappears.

${\rm{Electronic configuration of Mg:}}\,\,{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{6}}}{\rm{3}}{{\rm{s}}^{\rm{2}}}$ :

$\begin{array}{l}\\{\rm{Bond}}\,\,{\rm{order}}\,\,{\rm{ = }}\,\,\frac{{\left( \begin{array}{l}\\{\rm{Number}}\,\,{\rm{of}}\,\,{\rm{electrons}}\,\,\\\\{\rm{in}}\,{\rm{bonding}}\,\\\\{\rm{molecular}}\,\,{\rm{orbital}}\\\end{array} \right)\, - \,\,\left( \begin{array}{l}\\{\rm{Number}}\,\,{\rm{of}}\,\,{\rm{electrons}}\\\\\,\,{\rm{in}}\,\,{\rm{antibonding}}\\\\{\rm{molecular}}\,\,{\rm{orbital}}\\\end{array} \right)}}{{\rm{2}}}\\\\\,{\rm{Bond}}\,{\rm{order}}\,\,{\rm{ = }}\,\,\frac{{{\rm{12}}\, - \,{\rm{12}}}}{{\rm{2}}}\\\\\,{\rm{Bond}}\,{\rm{order}}\,\,{\rm{ = }}\,\,{\rm{0}}\\\end{array}$

For the stable molecule, the molecule should contain any value for the bond order. For the given molecule, the bond order is 0. Therefore, the molecule is not stable.

For the given molecule, the bond order is 0 which implies that the molecule is unstable.

Ans:

The bond order of ${\rm{M}}{{\rm{g}}_{\rm{2}}}$ molecule is 0.

Stability of ${\rm{M}}{{\rm{g}}_{\rm{2}}}$ molecule:

O	Unstable
O	Stable