Inter-molecular Hydrogen Bonding

In this post we will study the molecules which possess Hydrogen bonds. First we will list out the specifications needed for making Hydrogen bond. It is formed between partially negative atom and partially positive Hydrogen. That means molecule must have:

• Hydrogen as one of the bonded atoms.
• Considerable electronegativity difference between Hydrogen and the atom directly bonded to it. That means covalent bond between Hydrogen and other atom must have polar character.

Molecules in which Hydrogen is directly bonded to Nitrogen, Oxygen or Fluorine are likely to form Hydrogen bond. Let’s take examples of such molecules.

In Hydrogen fluoride HF, Hydrogen is directly bonded to Fluorine. There is considerable electronegativity difference between H and F which makes their bond polar and develops partially negative charge on F and partially positive charge on H. Fluorine has 3 sets of lone pairs and so it shares its one set of lone pairs with electron deficient Hydrogen and makes a Hydrogen bond.

In ammonia, NH₃, more electronegative N pulls bonding pairs from all three bonded Hydrogen atoms and makes them electron deficient which develops partially positive charge on H atoms and partially negative charge on N atom. You may think that in NH₃ molecule N of one molecule can form Hydrogen bond with H of three different molecules. In NH₃ molecule three electron deficient Hydrogen atoms are available for bonding but N is running short of lone pairs of electrons because it has only one set of lone pair so it can bond with only one H.

Now you can understand why hydrogen bond gives a number of unique qualities to water. Because in water molecule Oxygen has two sets of lone pair and also have two electron deficient Hydrogen atoms available for bonding. It is the perfect arrangement for hydrogen bonding.

In the above examples hydrogen bond is formed between two or more molecules, that’s why it is called Intermolecular hydrogen bond. Is it possible for a molecule to form Hydrogen bond even if it doesn’t have H directly bonded to electronegative element? In the next post we will try to solve this mystery.

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Hydrogen Bond

Hydrogen bond is not a real bond, actually it is a type of electrostatic attraction. It plays very important role in the case of water. So let’s learn more about it with the example of water molecule. You have studied bonding and hybridization of H₂O molecule.

H₂O is a bent shaped molecule. There is a considerable electronegativity difference between H and O atoms which makes the H-O bond polar.  More electronegative O pulls bonding pair of electrons and acquires a partial negative charge while Hydrogen develops a partial positive charge.

When two molecules of water come closer, the electrostatic force comes in action. Partially negative charged Oxygen of one molecule attracts partially positive charged Hydrogen of another molecule by electrostatic attraction. Electron rich Oxygen shares its lone pair of electron with electron deficient Hydrogen atom and forms an invisible bond of attraction which is called Hydrogen bond. This electrostatic attraction isn’t sufficiently strong to form an ionic bond and the electrons are not shared enough to make it a coordinate covalent bond, but this attraction is somehow capable of keeping the molecules together. Hydrogen bond is represented by a dotted line.

It is weaker than ionic and covalent bond. But it is solely responsible for the amazing nature of water. Let’s see how it makes water so wonderful.

Hydrogen bonds make a network of water molecules which is responsible for the liquid state of water. When we try to evaporate water into vapours, we need to break a large number of hydrogen bonds to let water molecule free from the network. And it requires a large amount of heat to break multiple Hydrogen bonds that’s why water remains liquid in large range of temperature and boils at higher temperature (100°C).

Do you know why sweating lowers your body temperature? Sweat is also formed by water and for evaporation it uses your body heat to break Hydrogen bonds. Now you can guess why temperature is lowered near the water bodies. Definitely because water molecules utilise heat present in the atmosphere which results in lowering of temperature.

Why ice floats in water? Have you ever thought about it? You will be surprised if I say Hydrogen bond is the culprit again. Water molecules are bent shaped and when hydrogen bond is formed between these molecules it forms a network. When we freeze water into ice, this network hinders the close packing of molecules and air gets trapped in the vacant space which lowers the density of ice. That’s why ice floats in water.

Hydrogen bonding is not exclusive only for water molecules but it isn’t present in every other molecule. In the next post we will learn about the molecules which possess Hydrogen bonds and try to find why they are chosen for hydrogen bonding.

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Co-Ordinate Covalent Bond

You have learnt two types of bonds ionic bond and covalent band. But when we try to draw the structure of O₃molecule we find these two bonds are not capable to define the bonding involved in O₃ molecule.

The coordinate covalent bond is like covalent bond but the only difference is in the way of sharing of electrons. You have seen that in covalent bond both bonded atoms share single electron, this shared pair of electrons makes a bond and both atoms get the equal share of it.

The co-ordinate covalent bond is a kind of partnership where the whole investment is done by one atom while other partner invests nothing but gets the equal share of the company.

In the previous post we have studied NH3and BF₃ molecules. In NH₃ molecule N has a lone pair of electron and in BF₃ molecule B has an incomplete octet. So they choose such type of joint-venture where NH₃ invests its lone pair of electrons with BF₃. In this way NH₃ helps BF₃to achieve octet. Now the bonded pair of electron is equally shared between N and B.

Once the co-ordinate covalent bond is formed it is identical to the normal covalent bond. Let’s try to solve the mystery of O₃ molecule. We have studied O₂ molecule, it has a double bond and two lone pair of electrons. It gives one of the lone pair to the third O atom. Thus, O₃ is formed by a double bond a co-ordinate covalent bond.

If you apply LCAO treatment for O₃molecule you will find its bond order is 1.5 which shows that each O-O bond is formed by 1 sigma and half pi bond. The pi bond is delocalised over all three O atoms. So the structure of O₃ is best explained by resonating hybrid.

This type of co-ordinate covalent bond is mostly formed by d block elements. We will study about such molecules, when we study “Co-ordination chemistry”.

There is one more type of bond which is called Hydrogen Bond. We will learn about this in our next post.

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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.     

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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.

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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.

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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.

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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. 

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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.​

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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.

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