Molecular Orbital Theory

MOLECULAR ORBITAL THEORY

In Molecular Orbital Theory (MOT) the atoms in a molecule are supposed to loose their individual control over the electrons. The nuclei of the bonded atoms are considered to be present at equilibrium inter-nuclear positions. The orbitals where the probability of finding the electrons is maximum are multicentred orbitals called molecular orbitals extending over two or more nuclei.

In MOT the atomic orbitals loose their identity and the total number of electrons present are placed in Mo's according to increasing energy sequence (Auf Bau Principle) with due reference to Pauli's Exclusion Principle and Hund's Rule of Maximum Multiplicity.

When a pair of atomic orbitals combine they give rise to a pair of molecular orbitals, the bonding and the anti-bonding. The number of molecular orbitals produced must always be equal to the number of atomic orbitals involved. Electron density is increased for the bonding MO's in the inter-nuclear region but decreased for the anti-bonding MO's, Shielding of the nuclei by increased electron density in bonding MO's reduces inter nuclei repulsion and thus stabilizes the molecule whereas lower electron density even as compared to the individual atom in anti-bonding MO's increases the repulsion and destabilizes the system.

In denotation of MO's, $\sigma$ indicates head on overlap and $\pi$ represents side ways overlap of orbitals. In simple homonuclear diatomic molecules the order of MO's based on increasing energy is

This order is true except B₂, C₂ & N₂. If the molecule contains unpaired electrons in MO's it will be paramagnetic but if all the electrons are paired up then the molecule will be diamagnetic.

Bond order

= $\frac{no.of{e}^{-}soccupyingbondingM{O}^{\prime }s-no.of{e}^{-}soccupyingantibondingM{O}^{\prime }s}{2}$

Application of MOT to homonuclear diatomic molecules.

H₂ molecule : Total no. of electrons = 2

Arrangement : ${\sigma }_{1s}^2$

Bond order : ½ (2 – 0) = 1

$H_2^{+}$ molecule : Total no. of electrons = 1

Arrangement : ${\sigma }_{1s}^1$

Bond order : ½ (1 – 0) = 1/2

He₂ molecule : Total no. of electrons = 4

Arrangement : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$Bond order:½ (2 – 2) = 0\He₂ molecule does not exist.$H{e}_2^{+}$molecule:Total no. of electrons = 3Arrangement:${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 1}$

Bond order : ½ (2 – 1) = 1/2

So $H{e}_2^{+}$ exists and has been detected in discharge tubes.

Li₂ molecule : Total no. of electrons = 6

Arrangement : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$

Bond order : ½ (4 – 2) = 1

No unpaired e's so diamagnetic

Be₂ molecule : Total no. of electrons = 8

Arrangement : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$Bond order:½ (4 – 4) = 0No unpaired e^–s so diamagneticB₂molecule:Total no. of electrons = 10Arrangement:${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${{\sigma }^2}_{2p_x}$Bond order:½ (6 – 4) = 1 \diamagneticBut observed Boron is paramagneticC₂molecule:Total no. of electrons = 12Arrangement:${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${\sigma }_{2p_x}^2$$\left\{\begin{array}{c}{\pi }_{2p_y}^1\\ {\pi }_{2p_z}^1\end{array}\right\}$

Bond order : ½ (4 – 0) = 2

It is paramagnetic

But observed C₂ is diamagnetic

N₂ molecule : Total no. of electrons = 14

Arrangement : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^2$${\sigma }_{2px}^2$$\left\{\begin{array}{c}{\pi }_{2p_y}^2\\ {\pi }_{2p_z}^2\end{array}\right\}$Bond order:½ (6 – 0) = 3$\therefore$It is diamagnetic O₂molecule:Total no. of electrons = 16Arrangement:${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${\sigma }_{2p_x}^2$$\left\{\begin{array}{c}{\pi }_{2p_y}^2\\ {\pi }_{2p_z}^2\end{array}\right\}$$\left\{\begin{array}{c}{\pi }_{2p_y}^{\ast 1}\\ {\pi }_{2p_z}^{\ast 1}\end{array}\right\}$

Bond order : ½ (6 – 2) = 2

It is paramagnetic

F₂ molecule : Total no. of electrons = 18

Arrangement : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${\sigma }_{2p_x}^2$$\left\{\begin{array}{c}{\pi }_{2p_y}^2\\ {\pi }_{2p_z}^2\end{array}\right\}$$\left\{\begin{array}{c}{\pi }_{2py}^{\ast 2}\\ {\pi }_{2pz}^{\ast 2}\end{array}\right\}$${\sigma }_{2px}^{\ast }$Bond order:½ (6 – 4) = 1It has been seen that in case of B₂, C₂ & N₂ the order of filling the e’s is different from the normal sequence.B₂:${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^{\ast 2}$$\left\{\begin{array}{c}{\pi }_{2p_y}^1\\ {\pi }_{2p_z}^1\end{array}\right\}$

It is paramagnetic

C₂ : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$$\left\{\begin{array}{c}{\pi }_{2p_y}^2\\ {\pi }_{2p_z}^2\end{array}\right\}$

It is diamagnetic

N₂ : ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${\sigma }_{2s}^{\ast 2}$.\left\{ {\begin{array}{*{20}{c}} {\pi _{2py}^2} \\ {\pi _{2pz}^2}\end{array}} \right\}\sigma _{2px}^2

It is diamagnetic

Example 1. Compare the bond energies of O₂, $O_2^{+}$&$O_2^{-}$

Solution: Higher the bond order greater will be the bond energy.

Now configuration of O₂ = ${\sigma }_{1s}^2$${\sigma }_{1s}^{\ast 2}$${\sigma }_{2s}^2$${\sigma }_{2s}^2$${\sigma }_{2s}^{\ast 2}$${\sigma }_{2p_x}^2$$\left\{\begin{array}{c}{\pi }_{2p_y}^2\\ {\pi }_{2p_z}^2\end{array}\right\}$$\left\{\begin{array}{c}{\pi }_{2p_y}^{\ast 1}\\ {\pi }_{2p_z}^{\ast 1}\end{array}\right\}$

Now formation of $O_2^{+}$means to remove an electron from anti-bonding one, which means increase in B.O.

B.O. of $O_2^{+}$ = ½ (6-1) = 2.5

$O_2^{-}$means introduction of an e^– in the anti-bonding thereby reducing the bond order.

Bond order of $O_2^{-}$ = ½ (6 – 3) = 1.5

So bond energy of $O_2^{+}$> O₂ > $O_2^{-}$

M.O. of Some Diatomic Heteronuclei Molecules

The molecular orbitals of heteronuclei diatomic molecules should differ from those of homonuclei species because of unequal contribution from the participating atomic orbitals. Let's take the example of CO.

The M.O. energy level diagram for CO should be similar to that of the isoelectronic molecule N₂. But C & O differ much in electronegativity and so will their corresponding atomic orbitals. But the actual MO for this species is very much complicated since it involves a hybridisation approach between the orbital of oxygen and carbon.

HCl Molecule: Combination between the hydrogen 1s A.O's. and the chlorine 1s, 2s, 2p & 3s orbitals can be ruled out because their energies are too low. The combination of H 1s¹ and $3p_x^1$ gives both bonding and anti-bonding orbitals, and the 2 electrons occupy the bonding M.O. leaving the anti-bonding MO empty.

NO Molecule: The M.O. of NO is also quite complicated due to energy difference of the atomic orbitals of N and O.

As the M.O.'s of the heteronuclei species are quite complicated, so we should concentrate in knowing the bond order and the magnetic behaviour.

INERT PAIR EFFECT

Heavier p-block and d-block elements show two oxidation states. One is equal to group number and second is group number minus two. For example Pb(5s25p2) shows two OS, +II and +IV. Here +II is more stable than +IV which arises after loss of all four valence electrons. Reason given for more stability of +II O.S. that 5s2 electrons are reluctant to participate in chemical bonding because bond energy released after the bond formation is less than that required to unpair these electrons (lead forms a weak covalent bond because of greater bond length).