From the above illustration it is shown that element phosphorous exists as P4 and element nitrogen exists as N2.In addition, the following are relevant bond energy data:
- Single bond of nitrogen is 160 kJ mol-1,
- triple bond of nitrogen is 994 kJ mol-1,
- single bond of phosphorous is 201 kJ mol-1,
- triple bond of phosphorous is 488 kJ mol-1.
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UPDATE(26th Jan 2009):
Comments to answers
The answers to this question is posted as a comment by me. Please check your answers with the comments and do clarify with me if need to. This question requires you to make use of hybridisation theory and chemical bonding to explain for the difference in observation.
I must add that many of you have good arguments, incooperating energetics (by calculating the enthalpy changes of formation), fewer actually relate it to the hybrid orbitals. Unfortunately, there were not any which actually mentions about bond energy helping the choice of hybridisation.
Some glaring errors include:
(1) counting the number of single bonds available wrongly. There are 6 covalent bonds in a P4 or N4 structure. My sense is that the error can also be attribute to you not using the equation 4N (g) -> N4 (g).
(2) Electronic configuration is quite unnecessary in this question.
(3) Not stating the correct hybrid orbital used. E.g sp3 for a tetra-atomic molecule and sp for a bi-atomic molecule.
(4) Sweeping comments on whether the bond can be formed. This question requires you to show actual calculation with the assumption that these bonds actually will be formed and then you argue for the corresponding situation which is more favourable.
The 2s and p orbitals(core orbitals) of Nitrogen is nearer to its nucleus compared to the 3s and p orbitals of Phosphorous. Hence, the excitation of Phosphorous orbitals would have a higher energy level and it would be easier to form hybrids. Moreover, being a larger atom, a phosphorous atom would have a smaller percentage area of overlapping when it is bonded to another single atom in comparison to a nitrogen atom bonded to another single nitrogen atom; this causes the P2 bond to be weaker than N2 bond. Therefore, a phosphorous atom would prefer to increase its overlapping bonded area by forming three other single bonds. Moreover, the lower the potential energy of an orbital, the more stable the bond. Since the formation of a nitrogen triple bond gives -994kj/mol of energy, it will be more stable than the formation of N4 which has -640kj/mol. Similarly, P4 molecule would be more stable with -804kj/mol than P2 of -488kj/mol.
ReplyDeleteFor N2:
ReplyDeleteOne of the 2s electron of Nitrogen(N) will undergo excitation and this electron will combine with 1 2p orbital to form sp hybridisation. One of the sp hybrid orbitals overlaps head-on with that of the adjacent N atom to give one N-N sigma bond. The two other unhybridised 2p orbitals will overlap side-on to from 2 pi bonds which lie at right angles to each other. This will form the triple bonds of N, which will give out 994 kJ mol-1 of energy.
For N4:
The Nitrogen(N) atom has 3 electrons in each of its 2p orbitals and each of the 2p orbital will overlap head-on with the 2p orbitals of the other 3 nitrogen atoms to form 3 N-N sigma bonds. The formation will give out 160x4=640 kJ mol-1 of energy, which is lesser than the 994 kJ mol-1 of energy that N2 gives out. As N2 gave out more energy, it has a stronger bond and is more stable than N4. Therefore, N will combine to form N2 instead of N4 as N2 is more electrostatically stable.
For P2:
One of the 3s electron of Phosphorus(P) will undergo excitation and this electron will combine with 1 3p orbital to form sp hybridisation. One of the sp hybrid orbitals overlaps head-on with that of the adjacent P atom to give one P-P sigma bond. The two other unhybridised 3p orbitals will overlap side-on to from 2 pi bonds which lie at right angles to each other. This will form the triple bonds of P, which will give out 488 kJ mol-1 of energy.
For P4:
The Phosphorus atom has 3 electrons in each of its 3p orbitals and each of the 3p orbital will overlap head-on with the 3p orbitals of the other 3 phosphorus atoms to form 3 P-P sigma bonds. This formation of compound will give out 804 kJ mol-1 of energy, which is more than the 488 kJ mol-1 of energy that P2 gives out. As P2 gave out less energy, it has a weaker bond and is less stable than P4. Therefore, P will combine to form P4 instead of P2 as P4 is more stable.
From the data, we can see that the bond energy of a single P-P bond is 201 kJ mol-1 Thus the bond energy of P4 is 3 x 201 = 603 kJ mol-1 and the bond energy of P2 is 488 kJ mol-1 .
ReplyDeleteP2 molecule formation is not feasible due to its last shell of valence electrons. This shell of electrons is relatively far away from the nucleus of the Phosphorus atom. Additionally, the electrons in the 3rd quantum shell of P are sp3 hybridised, with one orbital containing a pair of electrons while the other 3 are unpaired. Hence the electrons have a higher energy level and are even further away from the nucleus; hence a triple bond between 2 P atoms is difficult to form due to the weak electrostatic forces of attraction between the nucleus of an atom of P and the bonding electrons in the other atom of P. From this, there is little area of effective overlap between the electron clouds of the 2 P atoms, making the P-P triple bond weak (488 kJ mol-1 ) and unstable. At room temperature, there is enough energy to break the P-P triple bond if it is formed. Therefore, formation of P2 is not feasible.
However, the formation of P4 is possible as single bonds formed between the P atoms are strong and stable, as seen from the bond energy value of 603 kJ mol-1 , which is 115kJ mol-1 stronger than that of a P-P triple bond. At room temperature, there is insufficient energy to break the P4 bonds.
From the data, we can calculate that if nitrogen atoms were to form N4 , the bond energy for N4 will be 3 x 160 = 480 kJ mol-1 , while the bond energy of an N2 molecule is 994 kJ mol-1 .
The N atoms are similar to Phosphorus such that the valence electrons its 2nd quantum shell are sp3 hybridised as well, with one orbital of paired and 3 orbitals of unpaired electrons. However unlike P, the N atom is relatively smaller. Hence, when an N-N triple bond is formed, the area of effective overlap of the electron clouds is large relative to its small size. The nuclei of both atoms are brought closer to the valence electrons due to the head-on and side-on sigma and pi bonds respectively. Hence, there is strong electrostatic attraction between the nucleus of an N atom and the bonding electrons of the other N atom, making the N-N triple bond strong and the N2 molecule stable. At room temperature, there is not enough energy to break the bonds in N2 , hence it is feasible.
For the N4 bond, however, it is not feasible. To form N4 would require 4 N atoms to come together to form 3 single N-N bonds each. As the N atoms are small, when these 3 bonds are formed, the nuclei of each atom would be in close proximity to each other. Hence in the case of N4 , the nucleus of an N atom will be repelled by the nuclei of the other 3 N atoms, hence making it difficult to form 3 N-N single bonds with 3 N atoms. Hence, even if N4 were formed, its bonds would be weak (480 kJ mol-1 vs. 994 kJ mol-1 in N2 ) and the molecule will be unstable such that at room temperature, there will be enough energy present to break the bonds in N4 , hence formation of N4 is not feasible.
Based on the electronic configuration of phosphorus, 1s2 2s2 2p6 3s2 3p3, and nitrogen, 1s2 2s2 2p3, the phosphorus atom is bigger than the nitrogen atom. The atomic radius of phosphorus atom is larger due the increase in the screening effect that outweighs the increase in the nuclear charge. Due to the large phosphorus orbital, it is unable to form Pi bonds after the Sigma bond is formed. Therefore there is a combination of the s and different number of p orbitals which yield degenerate hybrid orbitals. The hybrid orbitals in the phosphorus are used for bonding through head-on overlapping with three other atoms of phosphorus and unable to form side-on overlapping of p orbitals. However nitrogen is a smaller atom as compared to phosphorus. Therefore after the head-on overlapping of orbitals between 2s orbital of the nitrogen, it is able to form Pi bonds between the 2p orbitals.
ReplyDeleteThe bond energy to form P4 is 804 kJ mol-1 and P2 is 488 kJ mol-1. As the 3 single bonds are more exothermic than the triple bond, P4 is more stable than P2 in standard conditions. The bond energy to form N2 is 994 kJ mol-1 and N4 is 480 kJ mol-1, thus the triple bond in N2 is more exothermic than N4 which makes N2 more stable than N4 in standard conditions.
Therefore it is not feasible for P4 to exist as P2 and N2 to exist as N4 in standard conditions.
The more exothermic a reaction is, the more stable is the product of that reaction.
ReplyDeleteFormation of 6 single bond for P4 releases 6*201=1206KJ of energy while formation of 1 triple bond for P2 releases only 488KJ of energy. Since the formation of P4 releases more energy than the formation of P2, P4is more stable. Thus, phosphorous exists as P4
Formation of N2 releases 994KJ of energy while formation of N4 releases only 6*160=960KJ of energy. Thus, N@ is more stable and nitrogen exists as N2.
Both nitrogen and phosphorous form sp3 hybrid obitals with only 3 half-filled orbitals which form covalent bonds and a lone pair of electrons. So, they can both form N2, P2, N4 and P4. however, formation of N4 and P2 are not energically favoured. hence, they exist as P4 and N2.
Total energy of bonds of N2 molecule=994kJmol^-1
ReplyDeleteTotal energy of bonds of theoretical N4 molecule=4(160)=640kJmol^-1
Total energy of bonds of theoretical P2 molecule=488kJmol^-1
Total energy of bonds of P4 molecule=4(201)=804kJmol^-1
Reason why N forms diatomic molecules while P forms tetra atomic molecules:
N atoms are more likely to be found in the N2 diatomic state because N2 has a larger total bond energy and more energy is required to break it up after formation as compared to N4.
P atoms are more likely to be found in the P4 state because P4 has a larger total bond energy and more energy is required to break it up after formation as compared to P2.
Reason why P forms weaker triple bonds:
N's triple bond(994kJmol^-1) is proportionately stronger than that of P(488kJmol^-1) when compared to their respective single bond energies. This trend of their triple bond energies can be attributed to the fact that P is a larger atom than N. When 2 P atoms form pi bonds, the side-on overlap of its pi orbitals is less extensive(because of P's larger atomic radius, causing the bond length to be larger) and is therefore weaker. As P2 has a lower total bond energy than N2, P2requires less energy to break up than N2.
The formation of sigma and pi bonds will create stronger bonds between atoms and thus N2 is formed instead of N4. As for phosphorus, its atom is too large for the efficient overlapping of p orbitals, thus P4 is formed instead of P2.
ReplyDeleteThe higher the bond energies, the more kinetically stable the compound is. The energy released by the formation of N2 is 994 KJ/mol while that of N4 is 640 KJ/mol. Hence, a N2 molecule is more kinetically stable and will be formed instead of N4. The energy released by the formation of P4 is 804 KJ/mol while that of P2 is 488 KJ/mol. Hence, a P4 molecule is more kinetically stable and will be formed instead of P2.
Phosphorus (P)
ReplyDeleteP has 15 electrons and therefore a electronic configuration of 1s2 2s2 2p6 3s2 3p3. With 3 valence electrons, it is feasible for P to bond with 3 other P atoms, each via a single bond, to form P4.
There is no need for any form of hybridization as there is already a maximum separation of the valence electrons in the 3p3 orbital.
At the same time, the bond strength of 4 single bonds of P (201x4 = 804kj/mol) to form a P4 molecule, is larger than that of a single triple P bond (488kj/mol) to form a P2 molecule.
Hence, the formation of P4 is more feasible than that of P2.
The case of Nitrogen (N) is similar to that of P.
N has 7 electrons, and therefore an electronic configuration of 1s2 2s2 2p3. Likewise to P, N also has 3 valence electrons in the 2p3 orbital, which are already as far apart from each other (maximum separation), rendering the possibility of any form of hybridization unnecessary.
As such, N should form an N2 molecule via a single triple bond.
In addition, the bond strength of a single triple bond of N (994kj/mol) forming a N2 molecule, is significantly larger than that of 4 single bonds of N (160x4 = 640kj/mol) forming a N4 molecule.
Hence, the formation of N2 is more feasible than that of N4.
In order to form N4, the valence shell of each nitrogen atom would have to occupy more than 8 valence electrons. Nitrogen being in period 2 would require so much energy to expand its octet or to excite its 2p3 electrons that it would not be feasible to form N4.Moreover, a single triple nitrogen bond forming N2(994kj/mol) is stronger than that of 4 single nitrogen bonds forming N4(640kJ/mol) hence N2 is more stable and feasible because more energy is needed to overcome the strong covalent bonds between nitrogen atoms.
ReplyDeleteAs phosphorous has 15 electrons,its last electronic configuration is 3p3 and therefore has 3 valence electrons.Hence there is no need for hybridization and therefore not feasible to form P2.Moreover, a single triple phosphorous bond forming P2 (448kJ/mol) is weaker than that of 4 single phosphorous bonds forming P4 (804kJ/mol) hence forming P4 is more feasible because more energy is needed to overcome the strong covalent bonds between Phosphorous atoms and therefore harder to break down as it forms a more stable structure.
Hi Mr Kwok,
ReplyDeleteNitrogen is a smaller atom than Phosphorous, thus when it forms a triple bond with other nitrogen atoms, there would be a greater extent of overlapping of orbitals as compared to nitrogen forming 3 single bonds with another Nitrogen atom. Thus, Nitrogen would prefer to form N2 instead of N4 due to N2 being more stable than N4 as seen from the energy needed to break a triple bond (994 kJ mol-1 ) compared to 3 single bonds ( 480 kJ mol-1).For Phosphorous, the degree of overlapping of orbitals is greater when it forms 3 single bonds as compared to a triple bond due to the size of the atom. Thus, Phosphorous would tend to form P4 than P2 as P4 is more stable. This can be seen from the energy needed to break the triple bond of P2 ( 488 kJ mol-1 ) as compared to the energy needed to break the 3 single bonds of P4 ( 603 kJ mol-1).As a result, P4 would be formed instead.
The N atoms are small enough to engage in pπ-pπ overlap, allowing for the formation of triple bond.
ReplyDeleteThe P atom, on the other hand, which is much larger, experiences a much higher amount of internuclear repulsion than N does, so it cannot form multiple bonds.
The bond energy of N2 (994 kJ mol-1) is larger than that of N4 (480kJ mol-1) hence it would be less feasible for nitrogen atoms to form N4 since the bond between the N atoms in N2 is stronger.
However, in P4 (603 kJ mol-1), the bond energy is stronger than that of P2 (488 kJ mol-1). Therefore, it is less feasible for P to exist as P2.
*cheryl*
The formation of pi bonds in conjunctions with sigma bonds would form stable and strong multiple bonds, stronger than the sigma bond by itself. However, in P, the P atom is too large for 2 P atoms to come close enough to allow p-orbitals to overlap efficiently to form pi bonds as the 3p orbitals are much more extended in space. Hence, P can't form triple bonds, thus, unable to exist as P2.
ReplyDeleteKinetically, we can calculate the amount of energy given off during the formation of a N2 molecule as compared to a hypothetical N4 molecule. During the formation of a mole of N2 molecules, the energy given off is 994 kJ. While that from a mole of N4 molecules is only 640 kJ. Clearly the formation of a N2 molecule is more kinetically stable as the formation of N2 is more exothermic.
As for a P4 vs. P2 molecule, the energies given off are, 804 kJ/mol, and 488 kJ/mol. The P4 molecule is more kinetically stable and it would require more energy to break the bonds of P4 than P2. Hence, P exists as P4 rather than P2.
The energy of a sigma bond in phosphorus is 201 kJ/mol. Since a triple bond comprises of a sigma bond and 2 pi bonds, each pi bond is 143.5 kJ/mol. Hence, it is not difficult for phosphorus to form P4 as the energy level of the 3s and 3p are very close so it is easy for sp3 hybridisation of the orbitals to occur. This is why P2 do not form since there is another alternative.
ReplyDeleteHowever, for nitrogen, each sigma bond has an energy of 160 kJ/mol and each pi bond, 417 kJ/mol. The pi bonds have much higher energy than the sigma bonds. Hence, it is difficult for the hybridisation of all orbitals to take place. It is easier for some pi bonds to remain, resulting in an sp hybridisation of nitrogen. Hence, it is not feasible for N4 to form as it is difficult to yield degenerate hybrid orbitals.
It is not very likely for phosphorus to form pi-bonds as its p-orbitals are far to large to overlap side on, instead it will form sigma bonds, each P atom will form 3 single bonds^* hence forming P4. In a P4 atom 6 single bonds are present hence amounted to a total bond energy is 1206KJ mol^-1 while a triple bond in P2 will have a bond energy of 488KJ mol^-1, which is relatively less stable. Hence it would be more likely for a phosphorus molecule is exist as P4 rather then P2.
ReplyDelete* the three unpaired electrons in the three 3p orbitals combine with the two electrons in the 3s orbital to form three electron pairs of opposite spin, available for the formation of three bonds. The remaining hybrid orbital contains two paired non-bonding electrons, which show as a lone pair in the Lewis structure. Hence forming sp3 hybiridization.
As the p-orbitals of N are small enough, they are able to overlap side-on and form pi-bonds. Hence it would be possible to form a triple bond with another N atom to form N2. Although it would be possible to from 3 single bonds to form an N2 it would be less feasible as the N2 structure would be more stable then the N4. A single N-N triple bond would have a bond energy of 994 KV mol^-1 whereas the 6 single bonds will have a bond energy of 960 KV mol^-1. Hence it would be more likely for the nitrogen molecule to exist as N2 rather then N4
The triple bond in molecular nitrogen (N2) is very strong with its stable octet structure.
ReplyDeleteN2 has a triple bond = 994 kJ mol-1
If it was to form N4, then it has 6 single bonds in total = 160 x 6 = 960 kJmol-1 which is of lower energy level and it is less stable than N2. Hence, nitrogen prefers to form N2 than N4 due to its difficulty of being converted into other compounds and its ease of converting nitrogen compounds into elemental N2 with an associated high energy release.
To form N2, it involves 1 sigma and 2 pi bonds wherby pi bonds are high in energy level. Hence, the high bond energy release by the formation of strong covalent bonds in N2 is able to compensate the energy used to mix the orbitals in hybridisation. However, to form N4, nitrogen undergoes sp2 hybridisation such that it will have 3 sp2 hyrbird orbitals and a lone pair of electrons in the p orbitals.
It will have 3 sigma bonds with other nitrogen atoms. However, this molecule is unstable and hence easily being converted.
The diphosphorous allotrope, P2, is stable only at high temperatures.
Phosphorous prefers a tetrahedral form P4 because P-P pi-bonds are high in energy.
P2 = 488 kJmol-1
P4 = 201 x 6 = 1206 kJmol-1
Hence, phosphorus prefers to form P4 than P2 since P4 is more stable and dissociates more energy.
Phosphorous undergoes sp2 hybridisation such that it will be able to have 3 sp2 hybrid orbitals and a lone pair of electrons in the p orbitals. As a result, it will be able to form 3 sigma bonds with the other phosphorous atoms. Hence, it will form P4. It is unable to form P2 because it will involve 1 sigma and 2 pi bonds whereby pi bonds are high in energy level and the low bond energy release by the formation of strong covalent bonds in P2 is unable to compensate the energy used to mix the orbitals in hybridisation.
By: Li Ling
(1) Using the hybridisation theory, when P4 undergoes sp2hybridisation, it will form 3 equivalent sp2 hybrid orbitals with 3 unpaired electrons inside(electrons ready for bonding) and 1 3p orbital with a pair of electrons inside(2 unpaired valence electrons). Hence, as seen from the diagram, each P atom will be linked to 3 other P atoms via a single bond each. Moreover, the sp2 orbital of phosphorus is too large to overlap side on to form pi bond which will give P2. Hence, it can only overlap head on to form sigma bond which will give P4.
ReplyDeleteWhen nitrogen undergoes sp hybridisation, it will form 2 equivalent sp hybrid orbitals with 2 unpaired electrons and 2 2p orbitals with 3 electrons. Hence, as can be seen from the diagram, the 3 electrons which makes up 1 sigma and 2 pi bonds will form the triple bond with another nitrogen atom to form N2. Nitrogen will exist as N2 instead of N4 because the sp orbital is small enough to overlap side on to form the pi bond.
(2) Using the idea of bond energy, phosphorus will prefer to exist as P4 instead of P2 because total bond energy required to form 6 single bonds for P4 is (+201 X 6 = +1206KJ) while total bond energy required to form a triple bond in P2 is +488 KJ. Since energy required to form P4 is more than that to form P2, P4 will be a more stable compound because more energy will be required to break the compound in the reverse process. Similarly, total bond energy required to form a triple bond in N2 is +994KJ which is more than that of forming 6 single bonds in N4 which requires +160 X 6 = +960KJ of energy.Hence, Nitrogen will exist as N2 instead of N4.
It is not feasible for the P atoms to form triple bonds and thus existing as P2 as the P atoms have large electron cloud size. Hence, the valence electrons from the P atoms can only overlap each other head-on, forming sigma bonds and is not possible for the P atoms to come into close proximity for them to overlap side on and forming pi-bonds. Also, it is more stable for P atoms to exits as P4 as the total bond energy is 201X6 = 1206kj/mol which is higher than that of an assuming existable P2 molecule in which its bond energy is only 408kj/mol.
ReplyDeleteIt is not feasible for N atoms to exist as N4 molecules as N atoms are able to form triple bonds with each other due to its small electron cloud size which are far stronger than six sigma bonds joining four N atoms together to form N4. This is also proven as the bond energy of a triple bond between two N atoms is 994kj/mol which is higher than that of six sigma bonds, 160 X 6 = 960kj/mol.
Phosphorus exists as a sp3 atom. While nitrogen exists as a sp atom.
ReplyDeleteFor it to exist as p2, phosphorus needs to form 2 sigma bonds and 1 pi bond with another phosphorus atom.
However, since pi bonds can only be formed when the atoms are close to each other, phosphorus is not able to form pi bonds since its 3p orbital is too big. Therefore, P2, which requires pi bonds, is not possible in nature.
For nitrogen, the 2p orbital is smaller in comparison and thus it is able to form pi bonds and thus form N2 instead of N4.
It forms N2 instead of N4 because N2 is more stable then N4. The bond energy for N2 is 994 kj per mole as compared to N4 which is 6(160)=960 kj per mole.
One reason why phosphorous cannot exist as P2 and nitrogen as N4 is due to the diffrent bond energies they have. A single bond of nitrogen is weaker than a single bond of phosphorus while its triple bond is stronger than phosphorus. This causes the nitrogen atoms the form triple bonds with each other so that they could become more stable than when they have 3 single bonds. Phosphorus atoms on the other hand forms 3 single bonds with each other for it to become more stable compared to having just a triple bond. Also, it is also due to the differences in character between the valence 2p and valence 3p orbitals of nitrogen and phosphorus, respectively. The 2s and 2p orbitals of first row atoms are localized in roughly the same region of space, while the 3p orbitals of phosphorus are much more extended in space, allowing the phosphorus atom to bond with more of its own atoms compared to a nitrogen atom.
ReplyDeletePhosphorous
ReplyDeleteFor phosphorous to be P2, it needs to form a triple bond which is 1 sigma bond and 2 pi bonds. However, the 3p orbitals of phosphorous are too large to overlap sideways with one another. Thus, phosphorous is highly unlikely to form P2.
Also, the bond energy of P2 is 488kJmol-1 while the bond energy of P4 is 6x201=1206kJmol-1. Hence P4 is more stable as compared to P2 as more energy is required to break the bonds.
Therefore, it is more likely for phosphorous to form P4 than P2.
Nitrogen
As for nitrogen, the 2p orbitals are able to overlap sideways with one another as the 2p orbitals are smaller as compared to 3p orbitals. Thus, it is able to form pi bonds and forms a triple bond with another nitrogen atom. Hence, N2 is formed.
Although it is also feasible for N4 to be formed as 4 single bonds are also possible, but N2 is more stable than N4.
This is because the bond energy of N2 is 994kJmol-1 while N4 is 6x160=960kJmol-1. Thus, N2 is more stable as compared to N4 as more energy is required to break the bonds too.
Therefore, it is more likely for nitrogen to form N2 than N4.
Phosphorus cannot exist as P2, as its bond energy of 488 kJmol-1 is very much lower than that of its actual state P4 which is 1206 kJmol-1. This is makes P4 more stable than P2, therefore phosphorous exists as P4.
ReplyDeleteAlso, the 2s and 2p orbitals are nearer together as compared to the 3s and 3p orbitals, therefore it makes hybridisation for the 2s and 2p orbitals easier compared to the 3s and 3p orbitals. Being further away from the 3s orbital, the 3s orbital is harder to excite to the 3sp state. Therefore, phosphorous cannot exist as P2, as hybridisation is not very viable.
Nitrogen cannot exist as N4, as it is not chemically as stable as N2. N4 has a bond energy of 960 kJmol-1, which is lower than that of N2, which possesses a bond energy of 994 kJmol-1, making it more stable than N4. Therefore, nitrogen exists as N2, and not N4.
Nitrogen is able to hybridise as it hybridises at the n=2 shell. However, in the case of N2, it does need to hybridise in order to triple bond.
Phosphorous: 1s2 2s2 2p6 3s2 3p3
ReplyDeleteNitrogen: 1s2 2s2 2p3
Both Phosphorous and Nitrogen have not obtained octect configuration. To do so, they would undergo covalent bonds to obtain octect configuration. In forming these covalent bonds, the overlapping of orbitals are involved.
For Phosphorous, it forms sigma bonds in its existing state, P4. However, if Phosphorous were to exist as a P2 molecule, it would form a triple bond like in N2, which forms one sigma bond, and a pie bond. The sigma bond will first undergo head-on overlapping. The pie bond then undergoes side-on overlapping. However, because the 3p-orbital is used for side-on overlapping for the P2 molecule, it would not be feasible as the 3p-orbital would be too large. Because pi-bonds produce electron density above and below the line joining the two neuclei, a large atomic radius will widen the distance between both nuclei, thus it would be difficult for the two Phosphorous atoms to carry out side-on overlapping.
Energy required to break P4 bonds = 6(201) = 1206kJ mol-1
Energy required to break P2 bonds = 1(488) = 488 kJ mol-1
As the energy released when P4 bonds are broken is larger than the amount that the P2 bonds release, the P4 molecule will have a smaller energy after breaking of bonds. With a lower energy level, the P4 molecule is more stable than the P2 molecule. Therefore the P2 molecule is less feasible than the P4 molecule.
For Nitrogen, it forms both sigma and pi bonds in its existing state, N2.
Energy required to break N2 bonds = 1(994) = 994 kJ mol-1
Energy required to break N4 bonds = 6(160) = 960 kJ mol-1
As the energy released when N2 bonds are broken is larger than the amount that the P2 bonds release, the N2 molecule will have a smaller energy after the breaking of bonds. With a lower energy level, the N2 molecule will be more stable than the N4 molecule. Therefore the N4 molecule is more feasible than the N2 bond.
Both nitrogen and phosphorus exist as simple discrete covalent molecules.
ReplyDeleteNitrogen is group V period 2 while phosphorus is group V period 3.
The electronic configurations of nitrogen and phosphorus atoms are 1s2 2s2 2p3 and 1s2 2s2 2p6 3s2 3p3 respectively. Although they are from the same group and have many similar properties, nitrogen exists as N2 while phosphorus exists as P4 naturally at standard conditions. This is due to the different bonds energies of the nitrogen-nitrogen bonds and phosphorus-phosphorus single and triple bonds.
Bond Energy/ kJ mol-1
N–N 160
N≡N 994
P–P 201
P≡P 488
2N(g) → N2 (g)
∆Hr = -∆H (N≡N)
= -994 kJ mol-1
4N(g) → N4 (g)
∆Hr = -6∆H (N–N)
= -6(160) kJ mol-1
= -960 kJ mol-1
2P(g) → P2 (g)
∆Hr = -∆H (P≡P)
= -488 kJ mol-1
4P(g) → P4 (g)
∆Hr = -6∆H (P–P)
= -6(201) kJ mol-1
= -1206 kJ mol-1
Energy is released when bonds are formed. The more negative the ∆Hr, the more stable the product formed as compared to the reactants.
As can be seen from above, energy released while forming N2 is greater (more negative) than that while forming N4 (-994 kJ mol-1 < -960 kJ mol-1) Therefore, it is energetically more preferable for nitrogen to exist as N2 as it is more stable than N¬4. Moreover, if N4 were to be formed, more energy has to be lost in order to compensate for the extra energy needed to hybridise the 3s and 3p orbitals, so there is no incentive for forming N4.
As for phosphorus, it is evident that P4 is more stable than P2 because the energy released when forming P4 is greater(a lot more negative) than the energy released when forming P2 (-1206 kJ mol-1 < -488 kJ mol-1). Although energy will be needed to excite the electrons to form hybrid orbitals, it can be compensated with the very much larger amount of energy released upon forming P4.
From this data, we can conclude that nitrogen exists as N2 and phosphorus as P4, and not N4 and P2.
In the N2 molecule, there will be 1 σ-bond and 2 π-bonds between the nitrogen atoms (N≡N). There is no hybridisation of 2s and 2p orbitals.
In the P4 molecule, phosphorus atoms undergoes sp3 hybridisation in the 3rd quantum shell, forming four sp3 orbitals, with one of them already filled with two electrons. The remaining three sp3 orbitals form a σ-bond with one other phosphorus atom each in the P4 molecule, resulting in a tetrahedral structure.
ANSWER TO HOMEWORK
ReplyDelete(1) 2N (g) -> N2(g)
enthalpy change = -994
4N (g) -> N4(g)
enthalpy change = 6 x (-160) = -960
Therefore, based on enthalpy changes. It is more favourable to form N2.
(2) 2P (g) -> P2(g)
enthalpy change = -488
4P (g) -> P4(g)
enthalpy change = 6 x (-201) = -1206.
Therefore, it is more favourable to form P4.
(3) In the formation of a triple bond, energy is require to excite the orbital to hybridise to form sp orbital, with two p orbital available.
In the case of N2, the formation of the triple bond is favourable for the hybridisation to take place - giving the sp hybrid.
While for P2, the large 3p orbital results in poor side-on overlapping between the orbital, hence the pi bonds formed are not as strong.
Therefore, hybridisation to give sp orbital is not so favourable.- since energy release from formation of the triple bond does not compensate energy required to hybridise the orbital.
(4) Hence, P4 is favourable for phosphorous as the energy for formation of 6 P-P single bonds compensate for the energy required formation of sp3 hybridised orbital.
While, N4 is unfavourable for Nitrogen as the energy for the formation of 6 N-N single bonds probably does not compensate the energy require to form the sp3 hybrid orbitals.