Resonance and Resonance Hybrid

In this post we will try to find out the correct structure and right way to represent the fractional bond of (CO₃­)^-2ion. When we draw Kekulé structure of (CO₃­)^-2 ion, we show double bond between C and O atom and single bonds in between C and O^-ion. But in MOT we discovered that the bonds between C and O are neither double nor single; instead, their bond order is 1(1/3).

In MOT we have seen that the pi bond is delocalised over all atoms. Here all O atoms demonstrate their comradeship and one by one share the negative charge and pi bond as well. Let’s see how.

These structures don’t exist separately but they contribute to form the actual structure. The real structure is a combination of all these structures. These structures all called resonating structures and they combine to form a resonance hybrid which represents the actual structure. This phenomenon of delocalisation of charge and pi bond is known as resonance.

Resonance has a great significance in stability of a number of molecules like (NO₃)^- ion which has bond order 1(1/3) and (NO₂)^- ion which has bond order 1.5.

Resonance plays an important role in determining the stability of Benzene (C₆H₆). In the whole aromatic chemistry we study Benzene and its derivatives. Stability of Benzene was a big question because it is a cyclic molecule and has 3 double bonds. It is expected to be 1(1/2) very unstable because of the above mentioned qualities. MOT determines its bond order to be , that means double bonds are delocalised over all the 6 C atoms.

The higher number of resonating structures gives extra stability to Benzene molecule. In resonating hybrid you can see 3 delocalised pi bonds which are represented by rings above and below the plane hexagon.

You have learned about ionic and covalent bonds. But if you try to guess the bonding in O₃ molecule you need a third type of bond covalent co-ordinate bond. In the next post we will discuss it in detail.     

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Delocalisation of pi π bond

MOT revealed one of the most admiring qualities of bonded atoms. In a molecule atoms are ready to share the burden of other atom. You will understand it better by an example of (CO_3­)^-2 ion. This ion has got extra stability by team work of its atoms. Let’s see how these atoms work together to gain extra stability.

When you work out the Lewis dot structure for (CO_3­)^-2ion you will find that two of the three Oxygen atoms are negatively charged (O^-1). Write the configuration of C and O atoms.

Electronic configuration of ⁶C: 1s², 2s², 2p²

Electronic configuration of ⁸O: 1s², 2s², 2p⁴

When O atom gains one electron it will form O^-1 ion so the

Electronic configuration of O^-1: 1s², 2s², 2p⁵.

Thus 6+8+9+9=32 electrons are involved in the formation of (CO_3­)^-2 ion. With the help of VSEPR we can predict the shape of the molecule.

Excited state configuration of central atom C is: 1s², 2s¹, 2p³

It opts sp² hybridization and uses three sp² hybridised orbitals for making three sigma (σ) bonds and unhybridised pz orbital for making one pi (π) bond. That’s why its shape is triangular planer. You can see here that one of the three O atoms has double bond and two O atoms have to bear negative charge. But MOT finds out the real picture of (CO_3­)^-2ion.

We have counted that (CO_3­)^-2 ion has 32 electrons. We are not able to distribute them in MOs because the energy order of MOs for (CO_3­)^-2ion isn’t known exactly. That’s why MOT suggests a bypass to calculate the bond order of such multi cantered (atomic) molecules. This procedure involves the following steps:

1. Find out the shape by using VSEPR: triangular planer
2. Add up the total number of valence electrons: C has 4 valance electrons, O has 6 and O-1 has 7, thus, the total 4+6+7+7=24
3. Determine the number of electrons that participate in the formation of (π) bonds: = Total number of valence electrons – (total number of electrons involved in (σ) bonds)+(lone pair of electrons):  (CO_3­)^-2has 3 (σ) bonds which involved 6 electrons and each O has two sets of lone pairs thus 6 sets that means 12 electrons, now we have to subtract 6+12=18 electrons from total number of valence electrons (24), so we get 24-18= 6 electrons which will participate in π bonding.
4. Count the number of atomic orbitals which can take part in π bonding: C has one 2pz orbital and each of three O has one 2pz orbital thus 4 AO are available for π bonding.
5. Combine these AO and find out the number of MO which will be formed:  4 AOs combined to form 4 four centered MOs as (CO_3­)^-2ion is a four centered molecule. That means MOs are spread over all the four atoms.  
6. Decide which MO are bonding, antibonding and non bonding: lowest energy MO will be bonding and highest energy MO will be antibonding and remaining two will be non bonding and also of same energy. Non bonding MO is intermediate state of bonding and antibonding.
7. Find the number of π bonds: on distributing 6 electrons in MO you will find that 2e are filled in bonding MO and 4e are filled in nonbonding MO. On applying the formula used to calculate bond order we will get (2 bonding e) - (0 antibonding e) /2  = 1 π bond.

By VSEPR we know that (CO_3­)^-2ion has 3 (σ) bonds and MOT gives the number of π bond formed. But the important discovery from MOT is about the distribution of π bond. MOT says that this π bond is equally distributed in between all atoms. That means each of the three C-O bonds has got 1/3 share of π bond. Thus each C-O bond will possess the (1+1/3) bond order.

That means the pair of electrons of π bond is not localised, it is delocalised over all bonds. So, all the bonds share the burden and give extra stability to the molecule. But how can we draw the bond of fractional bond order? In the next post we will try to solve this problem.

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

MOT for Heteronuclear Molecules

We have learnt in MOT that atomic orbitals of similar energy and similar symmetry get overlapped to form molecular orbitals. This concept is well accepted when we work on molecules of same element but there is some controversy when we talk about molecules of different elements. Since atomic orbitals belonging to different elements have different energy levels, so in that case we cannot get molecular orbitals by linear combination of atomic orbitals.

Although LCAO can be applied for heteroatomic molecules formed by elements which are closely placed in periodic table. Let’s take an example of NO molecule.

Because the energy levels of MOs are not known exactly, we have to try which order will be well fitted. Let’s take the energy order we have used for heavier atoms like O₂.

NO molecule has got 15 electrons, 7 electrons from N and 8 electrons from O atom. Thus we get electronic configuration:

σ1s², σ*1s², σ2s², σ*2s², σ2px², {π2py², π2pz²}, {π*2py¹, π*2pz⁰}

The NO molecule is paramagnetic since it contains an unpaired electron and the bond order will be (10-5)/2=2.5.  Here you might get surprised by fractional bond order. It means the bond will be somewhat in between double and triple bond.

The fractional bond order has great significance in the world of molecules. This is the second most important discovery of MOT. This incomplete bond gives extra stability to the multi atomic molecules. In the next post we will explore it in detail.

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Why is O2 Paramagnetic while N2 Diamagnetic?

I hope you have seen the video of the link I had given you in previous post. Did you notice that Oxygen somehow dances between the poles while Nitrogen escapes? This strange behaviour can be explained by MOT.

You can see Oxygen gets attracted toward the magnetic field while Nitrogen repels it. In both cases N₂ and O₂ behave like a magnet. You may get surprised here and ask, how can I relate a molecule to a magnet? You must have learnt in physics that magnetic field is associated with moving charged particle. Similarly in molecules moving negatively charged electrons generate a magnetic field. Thus, the magnetic behaviour of an atom or a molecule is related to the orbital and spin motion of its electrons. Quantum number m_l and m_s represent the magnetic factor of an electron.

That means the reason behind their strange behaviour lies in their electronic arrangement. So, have a look on electronic configuration of N₂ and O₂again.

Electronic configuration of N₂: σ1s², σ*1s², σ2s², σ*2s², {π2py², π2pz²}, σ2px²

Electronic configuration of O₂: σ1s², σ*1s², σ2s², σ*2s², σ2px², {π2py², π2pz²}, {π*2py¹, π*2pz¹}

You can see that in N₂ all electrons are paired while O₂ has 2 unpaired electrons. These unpaired electrons are responsible for the magnetic nature of O₂.

These unpaired electrons of O₂ move around in their orbits. Their orbital motion generate loop of current which produces magnetic field. You may think that both unpaired electrons spin in clockwise direction so their magnetic field will add to give a strong resultant magnetic field which makes O₂ a powerful magnet. But it doesn't happen because these electrons are randomly arranged in a molecule so they cancel each-other's magnetism and very little magnetism is left within a molecule.

Paramagnetism

As we apply external magnetic field these tiny magnets get aligned in the same direction as the external magnetic field thus they produce induced magnetism in the direction of applied field which is proportional to the applied field. That's why O₂ get attracted towards external magnetic field. This type of magnetism is called paramagnetism.

In case of N₂molecule, all electrons are paired. That means half of the electrons spin clockwise and half of the electrons spin anticlockwise. Because of their opposite spins they produce magnetic field in opposite direction thus the resultant magnetism becomes zero. When such molecules are placed in an external magnetic field they produce induced magnetic field in opposite direction and that's why they repel the magnetic field. This type of magnetism is called diamagnetism.

Diamagnetism

Actually all atoms/molecules are diamagnetic inherently but the presence of unpaired electrons produces some magnetism in the atoms/molecules and make them paramagnetic. Paramagnetic molecules get attracted towards external magnetic field and diamagnetic repel the external magnetic field.

Diamagnetic and Paramagnetic

You can easily predict the magnetic nature of any molecule/atom by its electronic configuration. If it has any unpaired electrons it will be paramagnetic and otherwise it will be diamagnetic. In the next post we will see how MOT deals with the molecules formed by two different elements.

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Bond Order Calculation by MOT

In the previous post you have seen how MOT helps us to guess the existence of a molecule. In this post we will explore some more applications of Molecular orbital theory. Let’s take O₂  molecule as an example and see what information can be drawn by MOT.

Oxygen ⁸O has 8 electrons. Its electronic configuration is 1s², 2s², 2p⁴. When two O atoms come closer to form O₂ molecule their atomic orbitals get overlapped and form molecular orbitals. Each O atom has 5 atomic orbitals thus they combine to form 10 MO. The energy order of these MO is as follows:

σ1s, σ*1s, σ2s, σ*2s, σ2p_x, {π2p_y, π2p_z}, {π*2p_y, π*2p_z}, σ*2p_x

This kind of energy order is usually applicable for Oxygen and other heavier elements.

O₂ molecule has got 16 electrons. When you fill these electrons in respective MOs you will get the following configuration:

σ1s², σ*1s², σ2s², σ*2s², σ2p_x², {π2p_y², π2p_z²}, {π*2p_y¹, π*2p_z¹}

By MOT we can calculate the bond order of the molecule. Bond order defines the number of bonds formed in between the bonded atoms in that molecule.

Bond Order = (Number of e^-occupying bonding MO) – (Number of e^- occupying antibonding MO)

         2

Let’s calculate it for O₂  molecule. It has 10 electrons in bonding MO and 6 electrons in antibonding MO. Thus its bond order will be (10-6)/2= 2, which corresponds to the double bond.

Now take another example of N₂ molecule. Nitrogen 7N has 7 electrons. Its electronic configuration is 1s², 2s², 2p³. When two N atoms come closer to form N₂ molecule their atomic orbitals get overlapped and form molecular orbitals. N₂ molecule has got 14 electrons, 7 electrons from each N atom. Each N atom has 5 atomic orbitals thus they combine to form 10 MO. The energy order of these MO is slightly different from the energy order given for O₂  molecule. This kind of energy order is applicable for the lighter elements like Nitrogen, Carbon and Boron:

σ1s, σ*1s, σ2s, σ*2s, {π2p_y, π2p_z}, σ2p_x, σ*2p_x, {π*2p_y, π*2p_z}

When we fill the electrons in MOs, we will get the following configuration:

σ1s², σ*1s², σ2s², σ*2s², {π2p_y², π2p_z²}, σ2p_x²

Let’s calculate its bond order; it has 10 electrons in bonding MO and 4 electrons in antibonding MO, so the bond order will be (10-4)/2 = 3, which corresponds to the triple bond.

Nitrogen and Oxygen both are gases but they behave differently when magnetic field is applied to them. I recommend you to see this video to understand what I am going to say. As you can see in the video that liquid Nitrogen is unaffected by external magnetic field and falls right through it while liquid Oxygen is affected by magnetic field and sort of dances between the poles and stays there for a while. This amazing nature of Oxygen can be explained by MOT, in the next post we will see how MOT reveals the secret of Oxygen. 

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Applications of MOT

MOT helps us to solve a number of mysteries about molecules. We will explore its applications one by one. Now you must be able to interpret the nature of element and predict the type of bond which it would prefer to form. You are familiar with Hydrogen and Oxygen, we have studied a number of molecules which contains H and O. Both of them exist in diatomic state which means Hydrogen exists as H₂ molecule and Oxygen exists as O₂ molecule. Do you know the first element of 18^thgroup is Helium? It is a noble gas. Does Helium exist as He₂ molecule like H₂ and O₂ or not?

MOT can prove that Hydrogen exists as H₂ molecule and Oxygen exists as O₂ molecule and also can solve the question about He₂ molecule. Let’s see how MOT solves these mysteries.

Let’s take an example of H₂ molecule. Its electronic configuration is 1s¹. How do two atoms of H combine to form H₂molecule? 1s orbitals of both H atoms get overlapped and form one bonding MO (σ1s) and one antibonding MO (σ^*1s). Now we have to fill electrons in these MOs. Filling of electron in MO is just similar to filling electrons in atomic orbitals. H₂ molecule gets one electron from each H atom, thus it has 2 electrons. Bonding MO (σ1s) has lower energy so it has to be filled first.

In the last post we have seen that bonding MO is more stable because it has lesser energy than parent AO. In H₂ molecule both electrons are filled in the bonding MO. If stabilization energy for one electron is ΔE, then it will be 2ΔE for two electrons. That means H₂ molecule is stabilised by 2ΔE as compared to the H atom.

Do you know the Darwin’s theory of evolution? “Survival of the fittest”; similar concept is applicable in the world of chemistry. Only those molecules exist which have lowest energy. That’s why H prefers to form H₂ molecule for stable existence.

Let’s examine Helium in the light of MOT. Its electronic configuration is 1s². In order to form He₂ molecule, 1s atomic orbitals of both Helium atoms will have to overlap and form one bonding MO (σ1s) and one antibonding MO (σ^*1s). If this happens, the resulting He₂ molecule will get two electrons from each Helium atom, thus it will have 4 electrons. Bonding MO (σ1s) has lower energy so it has to be filled first. 2 of the 4 electrons will be filled in σ1s and remaining 2 will be filled in σ^*1s. 

As a result, the stabilization energy gained by 2 electrons will be cancelled by destabilization of 2 electrons of σ^*1s. As we can see that Helium atom is not benefited by formation of He₂ molecule and prefers to stay alone as Helium atom. That’s why He₂ molecule doesn’t exist.

In the next post we will discuss the formation of O₂ molecule and see how MOT helps us to find the number of bonds formed by O atoms in O₂ molecule.​

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Linear Combination of Atomic Orbitals px, py and pz

In the previous post we have seen how two s atomic orbitals get overlapped and form one bonding and one antibonding molecular orbitals. In this post we will see how other atomic orbitals take part in the formation of molecular orbitals.

If atom A and atom B have three p orbitals p_x, p_y and p_z each, how can we decide which two orbitals will get overlapped? In LCAO, it is a rule that two atomic orbitals must have similar energies. That means p_x will overlap with p_x, similarly p_y with p_yand p_z with p_z.

Let’s see how p_x orbital gets overlapped with other p_x orbital. They can only meet head-on to overlap as they are parallel to the inter-nuclear axis. The bond formed by their head-on overlapping is called sigma bond. But there are two ways this can be done, either both overlapping lobes have similar sign or have opposite sings.

When both p_x orbitals have same signs they form sigma (σ) bonding molecular orbital. In this situation electron density is increased between the two nuclei. And if both p_x orbitals have opposite signs they form sigma star (σ^*) antibonding molecular orbital. Here electron density is decreased between the nuclei and node is formed. Node is a place where probability of finding electron is zero.

In the last post we have discussed about the symmetry of MO, either they are gerade or ungerade. Let’s examine σ and σ^*and try to find out which one is gerade or ungerade.

The signs of the lobes of σ MO remain the same on rotating it along the line joining the two nuclei and then about a line perpendicular to this line, which means it is gerade. But when we repeat similar procedure with σ* MO, the signs of the lobes change, which means it is ungerade.

Lets discuss the overlapping of p_y orbitals. They are perpendicular to the nuclear axis, that’s why they can overlap only in sidewise manner and the bond is called pi (π) bond. Here also two possibilities arise, either both lobes of overlapping orbitals have similar signs or one orbital get inverted at the time of overlapping resulting in oppositely signed lobes.

When both p_y  orbitals have similar signs they form π bonding MO and in the other case they form π^* antibonding MO. You can clearly see that in π MO electron density is increased above and below the inter-nuclear axis and node is formed at the inter-nuclear axis. While in π^* MO electron density is decreased everywhere and node is formed.

Let’s examine their symmetry. When we rotate π MO along the line joining the two nuclei and then about a line perpendicular to this line, the signs of the lobes change, which means it is ungerade. And when we repeat similar procedure with π ^* MO, the signs of both lobes changes in an identical manner that’s why it is gerade.

Next is p_z orbital. Two p_z orbital share the same axis so they can overlap sidewise and form pi (π) bond. They also form πbonding MO and in other case they form π^* antibonding MO.

What happens when two d orbitals get overlapped? They also make sidewise overlapping but the difference is that the bond formed is called the δ bond. When both orbitals have similar signs they form δ bonding MO and in other case they form δ ^*antibonding MO. If you try to find out their symmetry it becomes difficult because the sign on lobes changes four times on rotating them about the inter-nuclear axis.

I hope you have learned how two atomic orbitals combine to form molecular orbitals. But what is its significance? What information can we draw by this Molecular Orbital Theory? In the next post we will explore the applications of MOT.

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

Linear Combination of Atomic Orbitals

Now we know that electrons exhibit dual nature. When we consider an electron as a particle, we are able to give its address in an atom by a set of quantum numbers n, l, m_l and m_s. When we consider an electron as a wave, we describe it by wave function. A wave function also includes these quantum numbers except m_s, because as a wave we can only define orbital (a space where is the maximum probability of finding an electron).

Wave function has two parts one is Radial function which depends on quantum numbers n and l and the other is Angular wave function which depends on quantum numbers land m_l.

Radial function gives the number of degenerate orbitals for corresponding l value. Otherwise it has no physical meaning. Angular wave function is of much importance because it is used to draw polar diagram of orbital. In LCAO this polar diagrams are used to define overlapping of bonding orbitals.

Angular wave function depends on Polar coordinates r, θ and Φ. Let’s see how these polar coordinates are related to the Cartesian coordinates.

When you try to calculate angular wave function for orbital p_z you will find that the value of cosθ is positive for upper lobe and negative for lower lobe, that’s why in polar diagram upper lobe of p_zorbital has positive sign and lower lobe has negative sign. Similarly, polar diagrams for each type of orbitals have been developed by Schrödinger wave equation.

The signs of polar diagram play an important role in bonding. I’ll give you an example to explain it, suppose two persons extend their right hands forward to shake hands, they can easily and effectively do so. But if one person extends his left hand and the other forwards his right hand, shaking hands will not be efficient. 

Similarly in LCAO when two orbitals get overlapped they have two possibilities, either they are facing similar signs or they are facing opposite signs. When they are facing similar signs, their wave functions add to get more enhanced wave function. It results in increased electron density in between the nuclei which in turn leads to strong bonding of orbitals and formation of bonding molecular orbital.

When they are facing opposite signs, their wave functions get cancelled by each other and results in zero electron density between the nuclei which leads to the formation of antibonding molecular orbital.

Thus each time when two atomic orbitals get combined, they produce one bonding molecular orbital ψ_(g) and one antibonding molecular orbital ψ_(u).

ψ_(g)  = N { ψ_(A)  + ψ_(B) }

ψ_(u)  = N { ψ_(A)  + (- ψ_(B)) } ⋍ N { ψ_(A)  - ψ_(B) }

Bonding molecular orbital's wave function is denoted by ψ_(g); g stands for gerade that means when you rotate the MO about the line joining the two nuclei and then about a line perpendicular to this line, the signs of lobes remain the same. Antibonding molecular orbital's wave function is denoted by ψ_(u); u stands for ungerade that means odd. When you rotate the MO about the line joining the two nuclei, the sign of lobes changes. In short g and u tells us the symmetry of orbitals about its centre.

​

How will you define an established person? Someone who has fewer worries, less anxiety and more security in his life is to be considered as an estabilised person. Similarly in the world of atoms stability means less stress and a cool calm life. If a molecular orbital is formed by the overlapping of similar lobes it forms a strong bond by collaboration of atomic orbitals and gets stabilised. Two atomic orbitals which have opposite sign are not compatible with each other and when they get combined they form an antibonding MO. Because of less compatibily of AOs, antibonding MO has much stress and it becomes destabilised.

This stability of MO’s can be measured by amount of their energy. Minimum energy is a sign of stabilization. Stable MO has lesser energy than their parent AOs and unstable MO has more energy. 

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.

MOT: Linear Combination of Atomic Orbitals (LCAO)

Molecular orbital theory is quite successful in solving the mystery of molecules. In VBT we have learnt that only valence electrons and their corresponding orbitals of different atoms take part in bonding and form a molecule. The MOT suggests that all atomic orbitals of different atoms come together and produce molecular orbitals, thus whole atoms participate in bonding to form a molecule. And these molecular orbitals are associated with all the nuclei in the molecule (nucleus of all bonded atoms).

Do you ever imagine how do any two atoms combine to form a molecule? Two different atoms come closer to each other, but their nuclei repel each other and their electrons are attracted by other's nuclei. Both attractive force and repulsive force work between them, but at certain distance these forces balance each other. This optimum distance, where both forces are in equilibrium with each other, is called the bond length. It means that at this distance these atoms can make a stable bond together.

As I have told you that MOT is based on the dual nature of electron. So here we will describe an electron occupying an atomic orbital by its wave function ψ and we will see how MOT finds the wave function of a molecular orbital.

But we will try ourselves first. If we have 2 atoms A and B, each has 2 orbitals and all of them will combine to produce molecular orbitals. So we have to produce 2+2=4 molecular orbitals. Let’s suppose these orbitals are made up of clay. a₁, a₂ orbitals and atom B has b₁, b₂ orbitals. We can take half of the clay from each orbital from each atom and mix them to form one molecular orbital like:

½(a₁) + ½(b₁) = MO₁,

 ½(a₁) + ½(b₁) = MO₂,similarly we can get other 2 MOs.

Or we can proceed through another way. We can mix the clay from all the atomic orbitals and then make new MOs. Like:

 a₁ + a₂ + b₁ + b₂  = MO₁ + MO₂ + MO₃+ MO₄

MOT also has two procedures to get wave function of molecular orbital. One way is similar to our first procedure, in which it suggests that atoms come to the equilibrium distance and mix their atomic orbitals to produce molecular orbitals. This approach is called the ‘Linear Combination of Atomic Orbitals’ (LCAO).

Other way is similar to our second procedure, in which MOT suggests that atoms superimpose each other and produce new molecular orbitals and then move apart to their equilibrium distance. This approach is called the ‘United Atom Method’.

LCAO is the most accepted approach of MOT. Let’s see what is LCAO. Consider two atoms A and B which have atomic orbitals described by wave function ψ_(A) and ψ_(B). If these two atoms come to the equilibrium distance, their electron clouds overlap with each other and the wave function of molecular orbital can be obtained by linear combination of atomic orbitals ψ_(A) and ψ_(B).

ψ_(AB)  = N { c₁ψ_(A)  + c₂ ψ_(B) }  

Where ψ_(AB) is the wave function of molecular orbitals of the molecule AB. N, the normalization constant chosen to get the probability of finding an electron in a space is unity. And c₁ and c₂are constants for energy.

In the next post we will learn how do real orbitals (s, p, d) combine to form molecular orbitals.​

This work is licensed under the Creative Commons Attribution- NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.