This application question may be slightly more challenging but it has to do with chemical bonding.
If you noticed SiO2 does not exists like CO2, where we obtain an SiO2 with a pair of Si=O bond.
Interestingly, Phosphorous prefer to exists as P4, with single bonds linking the phosphorous atoms while Nitrogen exists as N2 where there is a triple bond between the two N atoms.
You may wish to read about this (an article which you will learn in JC 2). If you understand that article, you will be able to explain for the above situation.
Regards
Mr Kwok
P.S. This is an interesting question to develop thinking skills. The chemistry concept of this will be largely covered in 2010.
Suggested Answer:
In CO2, the C=O bond is made up of a sigma bond and a pi bond. If you have read the hybridisation article, you would realise that for carbon has to undergo hybridisation of the 2s and 2p orbitals to form suitable hybrid orbital for the double bond. Since CO2 contains 2 C=O, the carbon is sp hybridised. Hence, the sp hybridised orbitals overlap with the 2p orbital of oxygen to get the sigma bond, while the 2p orbital carbon overlap in a side-on manner with 2p orbital of oxygen to get the pi bond.
In SiO2, it is just sigma bonds. Hence, silicon has hybridised its 3s and 3p orbitals to give sp3 orbitals, which are used for head-on overlapping with the p orbital of oxygen.
Hence, it is quite clear that the formation of pi bond is what is different in both cases. As the formation of pi bond between C and O is favourable, hence carbon uses the sp hybridised orbitals. The reason why pi bond between C and O is favourable is because the p orbital of C and that of O are able to effectively overlap with each other.
Usually large period 3 and below elements find it harder to form pi bonding. Their p orbital usually prefer not to have side-on overlap with the 2p orbital of small atoms such as O as the overlapping may not be sufficiently effective for the particular type of hybridisation to take place.
It is this reluctance to form pi bond between atoms that results in phosphorous to exist as P4 instead of P2. The former allows each P atom to form 3 bonds (hence satisfying their octet requirements), while the latter forces each p orbtial to contribute two 3p orbitals for side-on overlapping for pi bond. This side-on overlapping is not sufficiently effective and hence the hybridisation to get sp or sp2 is not favoured. Instead a sp3 is used to obtain sigma bonds between P atoms.
Comments:
I hope that you found this entry abit interesting. It is probably a "S" paper Chemistry question, but it is really testing us whether we understand why bonding take place between two atoms and why different types of bond can occur between the two atoms.
I shall also publise all your contributions and perhaps you could take time and compare the difference in how the thought processes were formulated.
Commendable Student's answers:
Please read Johanan's and Brandon's input. Jing Yew's contribution is interesting but it did not hit what I feel is the key point. The same should be said about Lyria's answer, which was coherent but I was not convinced that she understood the key point about how easy was it for the pi bond to be formed.
23 comments:
the electronic configuration of P is 1s2 2s2 2p6 3s2 3p3. the s and p orbitals mix to form sp2 orbitals and gives it 3 bonds at 120 degrees angle, so they can only join in that shape, but not like N2's triple bond.
N is 1s2 2s2 2p3, but it doesnt undergo hybridisation so it bonds using normal p orbital bonding, and it can join in N2, but not like the P4 shape.
Observation 1: SiO2 and CO2
Covalent bonds involve the overlapping of two singly-occupied orbitals from two different atoms. Based on this definition, it does not seem possible that Carbon is capable of forming two double Covalent bonds as it only possesses two singly-occupied p orbitals, an empty p orbital and a fully occupied s orbital in its valence shall. However, in the case of CO2, one of the p orbitals mixes with the s orbital to form two hybrid orbitals, known as sp. This allows for four Covalent bonds to be formed, since the two p and two sp orbitals now contain one electron each. Due to the orientation of these orbitals with respect to each other, two sigma bonds and two pi bonds are formed, which can be observed when Carbon forms a double bond with two separate Oxygen atoms.
In the case of Si in SiO2, all three of its 3p orbitals mix with its 3s orbital, forming four hybrid orbitals, known as sp3. In order for this mixing to take place though, energy has to be used. This energy is compensated later through the formation of covalent bonds. This also holds true for the formation of sp orbitals in CO2. However, the mixing of orbitals to form sp3 orbitals needs more energy compared to the formation of sp orbitals. The reason why this extra energy can be provided is because Si can form even stronger covalent bonds than C in CO2. Therefore, the orientation of Si’s four sp3 orbitals allows it form four sigma bonds, which is observed in all four of the Si-O bonds in SiO2.
This therefore explains the difference between SiO2 and CO2.
Observation 2: N2 and P4
In Nitrogen in N2, one of its p orbitals mixes with the s orbital to form two hybrid orbitals, known as sp. This allows for threeCovalent bonds to be formed, since both the sp orbitals and one of the p orbitals now contain one electron each (the second p orbital is completely occupied). Due to the orientation of these orbitals with respect to each other, one sigma bond and two pi bonds are formed, which can be observed when Nitrogen forms a triple bond with another Nitrogen atom in N3.
In Phosphorous in P4, all three of its 3p orbitals mix with its 3s orbital, forming four hybrid orbitals, known as sp3. The formation of these sp3 orbitals requires more energy than the formation of the sp orbitals in Nitrogen. This extra energy is compensated by the stronger covalent bonds formed in Phosphorous, compared to Nitrogen. Therefore, with the four sp3 orbitals, Phosphorous can form three Covalent bonds since three of the sp3 orbitals are singly-occupied. Due to the orientation of P’s sp3 orbitals, three sigma bonds can be formed, which is observed in P4.
This therefore explains the difference between N2 and P4.
-Timothy Kwok
In order for C and Si to form double covalent bonds with each of the two O molecules, hybridization would have to occur in these atoms first between their 2s orbitals and 3 2p orbitals, forming 4 sp3 orbitals in the process. Energy, however, would be required for hybridization to occur.
As C is a smaller atom than Si, it would have more effective overlapping of its atomic orbitals with the O atoms, causing stronger covalent bonds to be formed. The formation of these stronger covalent bonds would release sufficient energy to compensate for the energy required for hybridization to occur in C. Hence, it would be possible for C to form double bonds with each of the two O atoms.
Si however, is a larger atom than C. Hence, there would be less effective overlapping of its atomic orbitals with that of the O atoms as compared with C, resulting in the formation of covalent bonds weaker than that in CO2 . Correspondingly, the energy released from the formation of these weaker covalent bonds would be lesser, and presumably, insufficient to compensate for the energy required for hybridization to occur in Si. Hence,Si does not hybridize and SiO2 does not exist as CO2 does, but rather as a giant molecular structure.
My postulate for the second part is this:
It should theoretically be easier for head on overlapping to occur between atomic orbitals as compared with side on overlapping.
Seeing that N is a much smaller atom than P, it would be much easier for both head on and especially side on overlapping between its atomic valence orbitals to occur with another N molecule, allowing sigma and pi bonds to be formed relatively easily between two N atoms. Hence, a triple bond between the two N atoms, comprising of 1 sigma bond and 2 pi bonds can be easily formed.
P, however, is a larger molecule than N. Hence, between two P atoms, it would be hard for side- on overlapping to occur between their valence p orbitals, thus making the formation of pi bonds difficult between the two atoms. This would make the formation of P2, which would require 1 sigma bond and 2 pi bonds to be formed between the two atoms, unlikely.
Hence, P would rather hybridize to form 4 sp3 orbitals using its valence s and 3 p orbitals. This would result in 3 hybrid orbitals filled with only 1 electron each and hence, capable of forming covalent bonds, with 1 hybrid orbital filled with an electron pair ( lone pair). These sp3 orbitals may only form sigma bonds.
The difficulty is then overcome. P can easily form single sigma bonds( head on overlapping of orbitals) each with 3 other P atoms using their hybrid orbitals, allowing the central P atom to fulfill its octet requirement.
Again, the formation of 3 sigma bonds with just 1 other P atom using their hybrid orbitals is highly unlikely due to the arrangement of these orbitals about the atoms and probably the large size of the atoms themselves, making it hard for 3 instances of head on overlapping of orbitals to occur between just two P atoms.
Hence, P would exist as P4.
…did the above even make any sense?
Regards,
Chia Wei Jie
Part (i):
Both silicon and carbon form four covalent bonds with oxygen atoms. In carbon, when the s and p orbitals hybridize, a certain amount of energy X is required. However, when carbon forms a sigma bond and a pi bond with an oxygen atom twice, the total energy given out by this bonding is greater than X, therefore carbon forms two double bonds with oxygen atoms. When the s and p orbitals in silicon hybridize, a certain amount of energy Y is required. When silicon bonds with oxygen, however, if silicon formed a sigma bond and a pi bond with oxygen like carbon, the energy released from this bonding would be less than Y. Since the formation of a sigma bond releases more energy than a pi bond, the silicon forms 4 sigma bonds with oxygen atoms such that an amount of energy which is greater than the energy released when silicon forms 2 sigma bonds and 2 pi bonds with oxygen atoms is released and is greater than Y, which allows for the hybridization of the s and p orbitals in silicon.
Part (ii):
Nitrogen has a greater electronegativity value than phosphorus. In addition, the 3 valence electrons in nitrogen are in the 2p shell as compared to the 3p shell in phosphorus. This means that the valence electrons in nitrogen are closer to the nucleus than in phosphorus. In phosphorus, it prefers to form 3 sigma bonds with other phosphorus atoms rather than a sigma bond and 2 pi bonds with another phosphorus atom, because the sigma bonds release more energy when formed and makes the compound more energetically stable. However, in nitrogen this does not happen because, as the valence electrons are closer to the nucleus than in phosphorus, the nitrogen nuclei are brought too closely together and repel each other. Hence, nitrogen chooses to form a sigma bond and 2 pi bonds with just one other nitrogen atom in order to minimize this repulsion. Therefore phosphorus exists as P4 while nitrogen exists as N2.
Sorry, I don’t really understand hybridization, but I’ll try to answer.
In phosphorus, there is a mixing of one s and two p obitals, thus resulting in the formation of three 3p2 obitals, leaving out one p obital that is unmixed with the rest. During bond formation, the orientation of the P atom allows for 3 sigma bonds and 1 pi-bond to be form. Thus, one P atom could be covalently bonded to 3 other P atoms to form P4.
In Nitrogen, it may because energy involved in mixing the obitals could not be compensated by the energy released from the formation of strong covalent bonds. Therefore, as the obitals are unable to be mixed, N could only form triple bond with one other N atom, unlike P4.
Both SiO2 and CO2 are oxides of group IV elements.
When carbon reacts with oxygen, hybridization occurs, moving one 2s electron to the 2p level, producing 4 unpaired electrons. By hybridizing the 2s and one 2p electron, it makes 2 sp1 orbitals of equal energy. This arrangement gives the carbon a different electron cloud which would be responsible for the bonding with oxygen with the lone pairs found in the hybrid orbitals.
Hybridization occurs in O as well, with sp2 hybrids being formed by the 2s and two of the 2p orbitals, giving 3 orbitals with same energy (degenerate). One 2p orbital is unaffected. Two of the sp2 orbitals carry the lone pair of electrons responsible for covalent bonds. When these atoms are brought together, sigma bonds form between the hybrid orbitals. As these bonds result in the atoms being close enough, sideways overlap occurs, forming pi bonds. Thus this allows for double bonds to be formed, thus making CO2 a simple discrete molecule.
For SiO2, the same hybridization applies. However, Si is a larger atom, thus Si-O bonds would be longer than C-O bonds. With these longer bonds, the p orbitals on the Si and O atoms would be too far apart for sideways overlapping to occur, thus a pi bond would not be formed. This results in Si to bond with O such that only single bonds are formed. Such single bond formation would be continuous, bring about the tetrahedral structure similar to diamond, a giant covalent structure.
For the case of bonding in Nitrogen and Phosphorus, N is a smaller atom than P found in the same group. This means more effective overlapping of orbitals between 2 N atoms as compared to 2 P atoms; through the use of triple bonds. Hence the triple bond of N would be stronger than the triple bond of P.
For this reason, when N reacts with other N atoms to complete its octet, it would share its atoms with neighbouring N atoms, resulting in a triple bond forming. However, P does not form strong triple bonds between P atoms. Thus in order to fulfil its octet, it would share a single bond with 3 other P atoms. This results in 4 sigma bonds forming, with strength greater than that of 1 sigma and 2 pi bonds from triple bonding of 2 P atoms.
Nitrogen exists as N2 where there is a triple bond (1 sigma bond, 2 pi bonds) between the 2 nitrogen atoms whereas Phosphorus exists as P4 where each P atom forms a single (sigma) bond with 3 other P atoms in a tetrahedral structure.
Nitrogen has the ability to form a stable triple bond with another N atom but Phosphorus does not. This is because nitrogen has 2 principal quantum shells only, but phosphorus has 3 principal quantum shells. Hence, the valence electrons are closer to the nucleus in nitrogen than in phosphorus. The nitrogen atom is smaller than the phosphorus atom.
The smaller atomic size of nitrogen compared to phosphorus allows the electron orbitals of nitrogen atoms to overlap more effectively with one another. Hence, the 2 pi bonds in the N2 molecule are possible because of the effective side-on overlap of the 2p orbitals not involved in the sigma bond.
However, the phosphorus atom is larger, hence the side-on overlapping of the p orbitals is less effective compared to nitrogen, thus phosphorus is less inclined to form pi bonds.
Rearragement of the electrons (Hybridisation of the 3s and 3p electron orbitals in phosphorus) occur, such that the 2 electrons in the 3s orbital are promoted to a hybrid orbital - the sp3 hybrid orbital. There are now 4 sp3 orbitals whereby 1 of the sp3 orbitals contains a lone pair of electrons and the 3 other orbitals contain 1 electron each. The 3 sp3 orbitals from each phosphorus atom overlaps head-on with the sp3 orbitals from the other three P atoms, forming strong sigma bonds, with 6 sigma bonds in total per P4 molecule.
It is energetically profitable, and therefore preferred, for phosphorus to exist as P4 because the energy released from the formation of the 6 sigma bonds in P4 compensates for the energy taken in to promote the 3s electrons to the sp3 hybrid orbital.
Carbon dioxide is a gas whereas silicon dioxide is a solid. Carbon can form simple discrete molecules with oxygen because it can form double bonds with oxygen.
When carbon forms bonds with oxygen, 4 unpaired electrons are produced in carbon by promoting one of the electrons in the 2s level into the empty 2p level. The 2s electron and one of the 2p electrons are hybridised to make two sp1 hybrid orbitals of equal energy.
Hybridisation occurs in the oxygen as well. Sp2 hybrids are formed with the s orbital and two of the p orbitals being rearranged to give 3 orbitals of equal energy - leaving a temporarily unaffected p orbital. Two of the sp2 hybrid orbitals contain lone pairs of electrons.
The hybrid orbitals of both carbon and oxygen atoms overlap to form covalent bonds (sigma bonds). The various p orbitals on the carbon and oxygen atoms then overlap sideways to form pi bonds.
The atomic radius of nitrogen is relatively small, therefore nitrogen atoms are able to come close enough together to form very strong bonds (triple bonds). On the other hand, phosphorus atoms are too big to come close together to form strong triple bonds.
N is a period 2 element with electronic configuration 1s2 2s2 2p3. As the atomic size is small, there will be effective overlapping of orbitals between its 3 valence electrons of 2 nitrogen atoms forming N2.
However, P being a period 3 element with electronic configuration 1s2 2s2 2p6 3s2 3p3 has a larger atomic size. As such, sharing of equal number of electrons does not take place. Instead, hybridisation exist which causes the phosphorous central atom to bond to 3 other phosphorous atom. 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 which results of the formation of sp3.
phosphorous's 3s and 3p electrons mix with each other to for unpaired electrons. as such. phosphorous prefers to exist with single bonds while nitrogen exists as N2 with triple bonds. the energy supplied for this mixing is due to the energy released when forming strong covalent bonds.
ps. i completely do not understand hybridisation
an addition. SiO2 does not exist like CO2 has the hybridisation occurs in oxygen and carbon in CO2 while this does not occur in SiO2
Energy must be provided before hybridisation can occur. This energy comes from the formation of strong covalent bonds. The formation of sp,sp2 and sp3 orbitals require different amounts of energy. Hence, it is the strength of the bond that determines the type of orbitals formed.
Sp3 requires most energy since more mixing occurs (3p and 1s orbital)
The Si-O bond is stronger than the C-O bond. Hence, Si-O has sufficient energy to form sp3 orbitals. However, because the C-O bond is weaker than the Si-O bond, it does not have sufficient energy to form sp3 orbitals. Hence only sp orbitals form. As different orbitals are obtained, SiO2 does not exist like CO2.
In P4, only sigma bonds are present. Hence, only sp3 orbitals are orbitals are obtained. A triple bond exists between the two Ns in N2. This means that there are 2pi and 1 sigma bond. Sp orbitals are formed. As such, Phosphorous prefers to exist as P4 while Nitrogen exists as N2.
-Abbie
SiO2 exist in the giant molecular structure form similar to the stucture of Diamond. While CO2 exist as simple discrete molecules.
The electronic configuration of Si is [Ne] 3S2 3P2.
Hence supposedly it should'e formed the same type of bond as CO2.
However hybridisation occured in SiO2, as the the S orbitals is going to mix with the P orbitals forming the 3SP3 orbitals. In which the electrons are not paired in each orbitals. Hece this allows the Si atom to create 4 covalent bonds with 4 O atoms, which leads to the formation of the tetrahedral stucture that we observe. This is possible because the energy released in the formation of the SI-O bond is large enough to actually compensae for the energy needed to mizthe S and Porbitals, allowing hybridisation to occur in this case.
While in CO2 the C=O formed does not release enough energy for hybridisation to occur hence it can only stay in the form of simple discrete molecules.
In P4 only sigma bonds are formed, while in N2 there is 1 sigma bond and also 2 phi bond being formed. Henc ethis indicates that in P4 the formation of Sp3 orbitals occured, while in N2 the formation of Sp orbitals occured leaving 2 P orbitals unmixed.
This difference in hybridisation may be caused by the fact that, in the formation of N2 bond and P4 bond, different amount of energy is released. And different levels of hybridisation also needs different level of energy to be absorbed, in order to allow the hybridisation to occur.
P4 bond formation releases more energy which is needed to form Sp3 as in i's formation more mixing occurs, hence more energy is needed. In N2, the energy released in bond formation is sufficient only to form the Sp orbitals.
Hybridisation has taken place, it is where two (at least) different types of orbitals are mixed to produce a hybrid. Hybridisation does not come free. In-order to mix the s and p orbital, there must be energy investment which results in the excitation of the electrons and orbitals. The energy used to mix the orbitals is compensated by the energy released from the formation of strong covalent bonds. Hence, multiple bonds are formed in P4 and N2 because hybridisation has taken place were there is mixing of s and p orbitals. N2 is a sp3 orbital and that is why it has triple bond within 2 N atoms and
(sorry sir i really don't understand... :()
This is my second submission.
Observation 1: SiO2 and CO2
Covalent bonds involve the overlapping of two singly-occupied orbitals from two different atoms. Based on this definition, it does not seem possible that Carbon is capable of forming two double Covalent bonds as it only possesses two singly-occupied p orbitals, an empty p orbital and a fully occupied s orbital in its valence shell. However, in the case of CO2, one of the p orbitals mixes with the s orbital to form two hybrid orbitals, known as sp. This allows for four Covalent bonds to be formed, since the two p and two sp orbitals now contain one electron each. Due to the orientation of these orbitals with respect to each other, two sigma bonds and two pi bonds are formed, which can be observed when Carbon forms a double bond with two separate Oxygen atoms.
In the case of Si in SiO2, all three of its 3p orbitals mix with its 3s orbital, forming four hybrid orbitals, known as sp3. In order for this mixing to take place though, energy has to be used. This energy is compensated later through the formation of covalent bonds. This also holds true for the formation of sp orbitals in CO2. However, the mixing of orbitals to form sp3 orbitals needs more energy compared to the formation of sp orbitals. The reason why this extra energy can be provided is because Si-O is an even stronger bond than C-O. It is the difference in strength of Si-O and C-O which results in different orbitals, sp3 and sp, being formed. The orientation of Si’s four sp3 orbitals allows it form four sigma bonds when Si bonds with other atoms, which is observed in SiO2.
Bond strength therefore explains the difference between the structures of SiO2 and CO2.
Observation 2: N2 and P4
In Nitrogen in N2, one of its 2p orbitals mixes with the 2s orbital to form two hybrid orbitals, known as sp. This allows for three Covalent bonds to be formed, since both the sp orbitals and one of the p orbitals now contain one electron each (the second p orbital is completely occupied). Due to the orientation of these orbitals with respect to each other, during bonding with another atom, one sigma bond and two pi bonds are formed. This can be observed when Nitrogen forms a triple bond with another Nitrogen atom in N3.
In Phosphorous in P4, all three of its 3p orbitals mix with its 3s orbital, forming four hybrid orbitals, known as sp3. The formation of these sp3 orbitals requires more energy than the formation of the sp orbitals in Nitrogen. This extra energy is compensated by the stronger covalent P-P bonds formed in Phosphorous, compared to N-N in Nitrogen. Therefore, with the four sp3 orbitals, Phosphorous can form three Covalent bonds since three of the sp3 orbitals are singly-occupied. Due to the orientation of P’s sp3 orbitals, three sigma bonds can be formed when P bonds with other P atoms, which is observed in P4.
The difference in the strengths of the P-P and N-N bonds therefore explains the different structures of N2 and P4.
-Timothy Kwok
(a) In CO2,the bonds present are C=O while in SiO2 the bonds present are Si-O. In CO2, hybridization takes place to allow the formation of a sigma bond between the C and O atoms which will then bring them close enough to have a side-on overlap of p orbitals to form a pi bond. The simple reason why Si=O bonds are not formed can be explained by the basics of covalent bonding. Si is bigger than C and since covalent bonding is the effective overlapping of orbitals, if Si=O were to be formed the same way as C=O, the larger size of Si would result in the Si=O bonds that do not have as effective overlap as the C=O bonds. Also, the Si and O atoms would not be able to come close enough to form a stable pi bond. Hence, Si forms only single bonds with oxygen, giving rise to its macromolecular structure.
(b) Phosphorus is a much larger atom than Nitrogen, just as Silicon is larger than Carbon. As such, while nitrogen can form strong triple bonds with another Nitrogen molecule, a phosphorus-phosphorus triple bond would be much weaker as the molecules are unable to come as close together to experience an effective overlap of orbitals. Because phosphorus does not form strong multiple bonds with itself, phosphorus consists of tetrahedral P4 molecules in which each atom forms single bonds with three neighboring atoms by hybridizing its atoms to 4 give sp3 orbitals, to allow it fill its valence shell in the most optimal manner.
In nitrogen, the element exists as diatomic gas molecules with a triple bond between every pair of N atoms. This is bacause nitrogen is a Group V element thus needing 3 more electrons to fill up the 2p orbitals to achieve its stable octet. However in the case of phosphorous, its 3p orbitals are all half-filled, thus requiring hybridisation to occur between the 3s and 3p orbitals so as to allow the 3p orbitals to be fully filled. The energy for this hybridisation to occur would come from the formation of strong covalent bonds between the phosphorous atoms.
Phosphorus has an electronic configuration of 1s2 2s2 2p6 3s2 3p3. It is observed that phosphorus has a single bond linking the phosphorus atoms. This is because each phosphorus atom uses one 3s and three 2p orbitals for form four identical sp3 hybrid orbitals. Hybridisation results in a re-ordering such that all the valence electrons are singly placed in orbitals without any pairing, thus single bonds link the phosphorus atoms.
Nitrogen has an electronic configuration of 1s2 2s2 2p3. It is observed that a triple bond exists between two N atoms because of the mixing of one s orbital and one p orbital, forming sp orbitals and leaving two p orbitals not mixed. During hybridisation of sp orbitals, multiple bonds are formed as a result if bond energy released is sufficient to compensate the energy required to mix orbitals.
CO2 if formed when C is double bonded to two O atoms covalently. that is made possible when carbon undergoes hybridisation, mixing two of its p orbitals and a s orbital for it to obtain 4 unpaired electrons to bond covalently to the two O atoms. energy required for hybridisation is accounted for by the energy that will be released with the formation of the strong ovalent bonds between C and O. however, Si does not undergo hybridisation like C because it is much larger and it is not energetically favorable for it to form dounble bonds with two O atoms when its p orbitals are far from O. the strong covalent bonds that it might form with O atoms will not release sufficient energy to compensate the energry loss for hybridisation in the first place. therefore, Si only uses 2electrons from its 3p orbitals to bond covalently and singly to two O atoms.
Sorry sir, i am not able to explain the 2nd situation because i really have not a clue on why P4 exists as so. Is it because N undergoes sp2 hybridisation while P undergoes sp3 hybridisation?
Si and P aare in period 3 while C and N are in period 2. Si and P are much larger atoms than C and N. Si and C have 4 valence electrons while P and N have 5 valence electrons.
Since the atoms are much larger, to form stable bonds between the P atoms in P4 and between Si and O in SiO2 hybridisation has to take place, where hybridized 3p and 3s orbitals will have an energy level which is between that of 3s and 3p and these will be used to form sigma bonds between the atoms. The effective overlapping of sigma bonds is more effective than that of pi bonds, so the bonding within the molecule will be more stable.
CO2 exists as simple discrete molecules, in which one carbon atom is covalently bonded with two oxygen atoms, each oxygen atom share two pairs of valence electrons with the central carbon atom. Si, being in the same group as C yet SiO2 exists as a giant molecular structure, in which there is a sharing of one pair of valence electrons between every Si and O atom. This is because in CO2, the valence electrons Carbon uses in bonding are in the second quantum shell whereas in SiO2, the valence electrons Si uses in bonding are in the third quantum shell.
One 2s and two 2p orbitals in the valence shell of C can mix to form three hybrid sp orbitals. These hybrid orbitals, together with the remaining ones, overlap with oxygen's orbitals to form double covalent bonds between C and O atoms. Si, on the other hand, has much bigger atomic radius than C, thus unable to form strong covalent bond with oxygen. The amount of energy required to mix orbitals in Si cannot be compensated by the energy released by bond formed with oxygen atoms. Hence, Si can only form single bond with oxygen.
(Sorry I know this answer of my is utterly confusing)
-----------------------------------
In order for multiple bonds (double, triple bond) to be formed, sigma bonds must be able to be formed between two atoms first. Phosphorus, having a big atomic radius, has a samll extent of overlapping of orbitals, hence the sigma bond formed between two phosphorus atoms is weaker as compared to that between two nitrogen atoms. Thus, forming more bonds between phosphorus atoms would require an amount of energy that cannot be compensate by the energy released by bond forming, hence forming multiple bonds between P atoms is not favourable. Nitrogen, being smaller, is able to perform more effective overlapping of orbitals. The energy released is enough to form two more pi bonds between N atoms (ie triple bond)
When C forms the sigma and pi bond, the energy released is sufficient to form the sp3 hybrid orbital. However, in SiO2 when Si forms a sigma and pi bond with oxygen, the energy released is inadequate to form the sp3 orbital. A sigma bond releases more energy than a pi bond when it is formed, therefore silicon will form 4 sigma bonds instead so that enough energy to form the hybrid sp3 orbital.
P will form P4 because P4 is more energetically stable because it contains only sigma bonds. While N contains both sigma and pi bonds reuslting in it being less energetically stabel and so has to form an N2 bond.
Sorry for my late reply.
My answer: Si and P are both in period 3,thus they have available empty d orbitals to form more bonds with other atoms.This is because the d orbitals are engergetically accessible so little energy is needed to utilize them.Also,with the usage of these d orbitals,more energy is needed to break the covalent bonds,thus the compounds are more stable.Si and p are also more willing to form these bonds as energy used to mix the orbitals to form bonds is compensated by the energy released from the formation of strong covalent bonds.
NOTE: Mr Kwok, my answer is based on chemical bonding alone,so I'm not sure my answer is correct.
Post a Comment