Question
You have learnt about the oxidising agent KMnO4. This reagent is made up of potassium and permanganate ions. In an ion of MnO4-, there is covalent bond between Mn and O. By using your knowledge from these posts, (i) suggest plausible reason(s) to why covalent bond exists between Mn and O and (ii) to why a dimer or polyer is not formed (which is what happen in AlCl3 and BeCl2 respectively.).
Suggested Answers
(i) If ionic bonding occurs in MnO4-, the Mn must exist as a Mn7+. This cation will be very small and will have a very large change, hence highly polarising. O2- is generally quite small, but when placed next to a Mn7+, it will polarise the electron cloud of O and hence resulting in an overlapping of electron cloud between Mn and O. Therefore, covalent bond exists between Mn and O instead.
(ii) Usually such structures, we will expect to see that the metal atom will contain empty orbitals while the non-metal atom will contain available lone pair of electrons of donation. This results in a formation of dimer (Al2Cl6) or a polymer (e.g. BeCl2). In the situation of MnO4-, it is possible to suggest that O being a highly electronegative element is reluctant to donate its lone pair of electrons to form dative bond. Hence, there is no dimer formed nor polymer.
Common errors and comments
There some nice answers given by your classmates. I reckon that Jing Yew provides the best answer. Joey gave me a very detailed analysis to why the bonding between Mn and O is covalent for MnO4-. Wei Jie also provided me with a good answer to why the bond between Mn and O is covalent.
Basically, I required you to make use of the ideas to suggest why a metal and non-metal produces a covalent bond between them. In addition, to account for formation of dative bond. If you notice, generally covalent compounds are metal and non-metal combination, it is capable of dative bond. - This is not the case for MnO4-
1. Suggesting that a Mn2+ will exist in MnO4-. This is not true.
2. Mn has no empty orbital - Actually I do not think I can conclude that. I think the d orbital is available. And since Mn has the electronic configuration of 1s22s22p63s23p63d54s2, it should still have empty orbital available for lone pairs of electrons donated to it.
3. Some fail to conceptualise why AlCl3 exists as a dimer. That idea is discussed in tutorial and explained over here! There is a weakness in conceptualising the phenomenon of AlCl3.
4. I don't understand what has octet got to do with the type of bond formed between Mn and O.
5. The only two ways to account for why MnO4- has covalent bonds between Mn and O: (i) The above answer. (ii) Using IE. It takes up too much energy to form Mn7+, hence the ion prefers to have covalent bond instead.
6. One of you mentioned about orbital overlap, since Mn 3d and 4s are relatively close in energy. I would think that by saying these orbitals overlap presupposes that covalent bond is formed. I am actually curious then to why this doesn't occur to the other metals.
7. Some mentioned that O lacks lone pair. That is not true.
8. Mass of the atoms is irrelevant to the charge density.
i)
ReplyDeleteThe Mn7+ ion has a charge of +7. this is a very large charge. therefore even at a relative atomic mass of 59.9, the ion has a high charge density. this gives it a lot of polarising power.
On the other hand, the O2- ion has a charge of -2. it also has a small electron cloud. but the Mn ion really has too much polarising power, so it distorts the electron cloud of O ion anyway.
This leads to overlapping of valence orbitals, which is defined as a covalent bond. no ionic bond is formed at any point.
ii)
There are no free orbitals in any of the atoms in the MnO4- ion. and the only atoms with lone pairs are the O atoms, which tend to not donate their lone pairs. thus dative bonding cannot take place, meaning that dimers and polymers cannot be formed.
(i)A covalent bond exists between Mn and O because they would rather form a covalent bond compared to an ionic one due to the impossibility of Mn giving 7 electrons to 4 O atoms to allow the O atoms to achieve octet configuration. it would be much more feasible for a double covalent bond to form between Mn and O. Also as O is much more electronegative compared to Mn, a dipole would be induced with the polarisation of the covalent bond due to the O atom attracting more elctrons to itself.
ReplyDelete(ii) A dimer is not formed because Mn lacks an empty orbital and O lacks an extra lone pair (6 valence electrons) thus a dative covalent bond cannot be formed.
Kevin
(i) The electronic configuration of Manganese is 1s2 2s2 2p6 3s2 3p6 3d5 4s2. We know that the energy levels of the 3d and the 4s subshells are very similar. Hence, bonding in manganese can use its 4s orbitals or both 4s and 3d subshells. Thus, this is why Manganese can form compounds with oxidation states from +2 to +7. Assuming that the manganese loses 7 electrons to give an ion of Mn7+, that would definitely have an extremely high polarising power and will polarise the electron clouds of the oxygen ions to form a covalent bond. However, it seems more likely that at low oxidation numbers, manganese exists as an ion (e.g. Mn2+ ) This is because like most ions, the first few ionisation energy are low. However, at higher oxidation states (e.g. +7) losing 7 electrons would take a very large amount of energy to remove the electrons. Thus, it forms covalent bonds by sharing electrons instead as a simpler method of bonding.
ReplyDelete(ii) In order for a dative covalent bond to be formed, there must be an atom with a lone pair as well as another with an empty orbital. Aluminum and Beryllium both have empty orbitals that can accept the lone pair. However, in the case of Manganese, all of its d-orbitals are half filled already and hence it cannot form dative covalent bonds. Thus, there is no dimmer or polyer formed.
Okay, this assignment totally drived me crazy. Lots of research done, but to no avail.Nonetheless, let me try to answer the question with my very best.
ReplyDeleteFirst, I would like to address an issue on your post that I looked up on the net. "...oxidising agent KMnO4. This reagent is made up of potassium and manganate ions." I spent alot of time checking up on it. If I'm not mistaken, it should have been permanganate ions instead of manganate ions. According to Wiki(if its right), permanganate ions refer to MnO4- ions with Mn having an oxidation state of +7. On the other hand, manganate ions refer to MnO4 2-, with Mn having an oxidation state of +6.
That aside, I'm having trouble drawing the diagram for MnO4-, which is driving me nuts. But I'll answer the question by theory, nonetheless.
Oxygen is the second most electronegative element. Which means that it has a greater tendency to attract bonding electrons to itself in a chemical bond. As such, 4 O2 will attract the Manganese atom to the centre and share 2 electrons per oxygen atom. Since Manganese lies in period 4, it is able to have more than 8 valence electrons die to empty 3d orbitals that can accommodate the extra electrons and hence expand the octet.
Perhaps it is also exactly like AlCl3 in the sense of overlapping electron clouds. Mn is probably of high charge density and this being highly polarising. And oxygen is an anion that is relatively large and therefore being polarisable. As such, Mn polarises the O electron clouds and the elctron clouds of the particles overlap with each other. Therefore sharing of electrons occur and a covalent bond is formed.
For the second part, let us take note that AlCl3 and BeCl2 is of neutral charge. As such, it is right for them to be a dimer and polymer respectively. However, the permanganate ion (MnO4-) has a negative charge. How can a dimer or polymer be formed when the molecule already gained an electron to form an anion?
Okay, this assignment is tough. But yea, done =)
Part (i) answer:
ReplyDeleteManganese in MnO4 has a very large charge of 7+, hence if MnO4 was an ionic compound, the Mn7+ cation would be willing to polarise the electron cloud of the anion.
O2- is a small anion. However, it has a charge of 2- making it easier to polarise than an anion of similar size with a charge of 1-. Hence, when O2- and Mn7+ come into close proximity with each other, the electron cloud of O2- would be attracted to Mn7+, causing a distortion in the electron cloud of O2-. This distortion results in an overlapping of the electrons cloud of both O2- and Mn7+. When that happens it mimics the overlapping of orbitals that characterize covalent bonds, hence a covalent bond is formed.
Part(ii) answer:
For a dimer or polymer to be formed, the atoms/ions of the compounds in question must contain a free orbital to accomodate more electron pairs. There must also be another atom/ion in another compound which has free electron pairs and is willing to share these electrons.
In MnO4-, Mn can accept more electron pairs as it can hold more than 8 electrons. Oxygen in MnO4- also has lone pairs of electrons. However, oxygen is very electronegative and as such is unwilling to share these electrons with another Mn in another MnO4- ion. Therefore a dative covalent bond cannot be formed, and hence MnO4- does not form dimers or polymers.
i) Even though Mn is a metal and O is a non-metal, Mn and O do not form an ionic compound. Mn forms a small cation of high charge (+7), while O is a large anion, whose electron cloud is easily polarised.
ReplyDeleteHence the cation (Mn7+) is willing to polarise the electron cloud of the anion (O2-).
When O2- is placed next to Mn7+, the electron cloud of O2- is distorted by Mn7+. The two electron clouds of both ions overlap, which is similiar to the overlapping of orbitals. As such, a covalent bond is formed between Mn and O.
ii) Mn and O do not form a dimer or a polymer as there is no formation of dative covalent bond. Dative covalent bond is formed when either Mn or O have a pair of non-bonding electrons for donation and the other atom has an empty orbital to accept the electron pair. However, this is not the case. As such, dative covalent bonds cannot be formed between two or more molecular ions of MnO4-.
Mn is a transition metal with an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d5 4s2
ReplyDeleteWhen Mn reacts with O, it does not just use its 2 valence electrons for covalent bonding. Instead, it would use the electrons in the 3d and 4s orbitals to form the covalent bonds. This would be possible due to the fact that the 3d and 4s orbitals are close and have almost similar energy levels. Thus when the 4s orbital of Mn and 2p orbital of O overlap, the 3d orbital would also be able to overlap with the 2p orbital of O. (this would be due to the fact that the nucleus would not have sufficient forces of attraction to keep these electrons from being shared.) This would therefore allow the O atoms to gain a stable structure, and the one with one less electron to cause the ion to be negatively charge. Moreover, Mn is in period 3, thus allowing for it to make use of its energetically accessible d orbitals in bonding. Hence it would accommodate the extra electrons contributed by O and expand its octet.
It does not exist as a dimer or polyer due to the fact that the central atom (Mn) has already expanded its octet to accommodate the extra electrons contributed by O. Thus it would not be electron deficient and not need to accept electrons. Therefore it would not form dimers or polyers.
(i) In MnO4-, oxidation number of Mn is +7 and of O is -2. If ionic bonds are to be formed between Mn and O, the Mn atom would have to lose 7 electrons which requires a lot of energy and cannot be replaced enough by the bond energy. Hence it's unlikely that Mn would transfer 7 electrons to O atoms to form ionic bonds. Moreover, suppose ionic bonds are formed between Mn7+ ion and O2- ions, the positive charged of managanese would be so great that it attracts the electron clouds of the oxygen ions. The distortion of electron clouds is so severe that it results in overlapping of orbitals of manganese and oxygen ions. Consequently, covalent bonds are formed.
ReplyDelete(ii) In MnO4-, each Mn atom forms double covalent bonds with 3 oxygen atoms and one single covalent bond with another oxygen atom. So each Mn atom has 7 bonding electron pairs around it, in addition to 8 non-bonding pairs in its third quantum shell. There is no empty 3d orbitals to accept electron pairs for dative covalent bond. Hence, dimer or polyer is not formed.
i) The Mn2+ ion has a considerably large charge of +2 and is small, and has a high polarizing power. The O2- ion possesses a considerably large charge of -2 and is polarisable.
ReplyDeleteWhen both ions are placed together, Mn2+ distorts the electron cloud of O2- , causing the electron clouds of both ions to overlap. This mimics the overlapping of electron orbitals, and a covalent bond between O2- and Mn2+ is formed.
ii) Unlike Al3+ in AlCl3, Mn2+ does not have any empty electron orbitals to accept electrons from O2- . Therefore, a dative covalent bond cannot be formed between the two ions and the dimerization of MnO4- does not occur.
-Timothy Kwok
Mn is a large cation that is moderately charged. Hence, it is highly polarizable. O is a small anion that is highly charged—it has high electro negativity. Hence, it is highly polarizing. When the two are mixed, the O anion polarizes the electron cloud of the Mn cation. Thus, the electron cloud of both particles overlaps with each other resulting in both particles sharing electrons. Thus, a covalent bond is formed.
ReplyDeleteThe shape of the molecule plays a role in determining the intermolecular forces that exist between the molecules. MnO4- exists in a tetrahedral formation. In the Mn-O bonds, the O particle has higher electro negativity than the Mn particle. Thus, the electrons are found closer to the O particles, causing the o to carry a slight Negative charged and the MN to carry a slight positive charged. Since the O carries a slight negative charged, it is unable to form dimmers with other O particles of other MnO4 particles, neither can it form polymers with the Mn particle as the Mn has a full octet structure.
(i) Suggest plausible reason(s) to why covalent bond exists between Mn and O.
ReplyDeleteManganese has 7 valence electrons (electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d5 4s2) and oxygen has 6 valence electrons (electronic configuration (1s2 2s2 2p4).
Each of the 4 oxygen atoms in MnO4- has an empty p-orbital which can accept 2 electrons to achieve a stable octet electronic configuration. Mn on the other hand has 3 lone pairs of electrons + 1 electron.
The positive nuclei of the oxygen atoms attract the negatively charged lone electron pairs from Mn. However, the Mn does not readily give up its electrons completely to the O atoms but is willing to share its 3 lone pairs of electrons to fill the empty orbitals in 3 of the oxygen atoms. Thus 3 dative covalent bonds are formed between Mn and 3 oxygen atoms when Mn shares 3 pairs of its electrons with the 3 O atoms.
The fourth oxygen atom accepts an electron from another source and its nucleus also attracts the single unbonded electron from Mn while the nucleus of Mn also attracts an electron from the O atom. However, neither atom is willing to give up its electron completely. Thus, Mn and O share a pair of electrons to achieve a stable octet electronic configuration and the 4th covalent bond is formed.
(ii) Why a dimer or polyer is not formed (which is what happen in AlCl3 and BeCl2 respectively.).
A dimer or polyer is not formed between MnO4- firstly because MnO4- is negatively charged and hence repulsion occurs when MnO4- anions are brought together, thus they will not come close together and no bond can be formed between any two of the MnO4- anions as a result.
Secondly, there are no empty orbitals in the Mn or O atoms in MnO4- to accept the lone pair of electrons donated by the oxygen atoms, so no dative covalent bond can be formed and hence dimers/polyers of MnO4- do not exist.
(i) In my opinion.the reason why a covalent bond exists between Mn & O is similar to AlCl3. Splitting up the Mn & O as ions, the Mn7+ cation is small, yet it has a large charge of +7. Hence, Mn7+ is able to easily polarise the electron cloud of O2- .On the other hand,O2- is similar to Cl-.It is a large anion, thus it can be easily polarized..When it is together with Mn7+ the electron cloud of oxygen is distorted by Mn7+.This causes the overlapping of both electron clouds,which leads to the overlapping of orbitals. Together with the similar electronegativity of both elements,thus resulting in the formation of a covalent bond, instead of an ionic bond.
ReplyDelete(ii)A dimer nor polymer does not form after bonding, as Mn has more than sufficient electrons(10 electrons). Moreover, Mn does not have a pair of non-bonding electrons for donation and the oxygen particles do not contain empty orbitals to accept electron pairs.
My guesses would be…
ReplyDeletei) To attain a stable octet structure, O would need to gain 2 electrons. The alternative of losing 6 electrons would be unfeasible due to the large amount of energy required.
Mn, however, is in a quandary. Losing 7 electrons to attain its octet structure would be unfeasible due to the phenomenal amount of energy required. Neither can it gain electrons from the reaction as O would be most unwilling to give away its electrons.
Hence, in a bid to fulfil their octet requirements, Mn would form covalent bonds (ie. share electrons) with O. By forming double bonds with 3 O atoms, the octet structure of 3 O atoms would be fulfilled. Mn would then form a single bond with the final O atom as it would only have 1 lone pair of electrons left. The more electronegative O atom would pull the bonding pair towards itself, hence the “-“ in MnO4-. An ionic bond would then be formed between O and K to fulfil both their octet requirements.
In conclusion, a covalent bond would be formed between Mn and O as it is the energetically most feasible method for them to attempt to gain their octet structure. While Mn would still be unable to gain an octet structure, it is nevertheless willing to do so because it would be closer to attaining it should it react than if it didn’t. That’s my postulate anyway….
---------
ii)
As to why dimers and polymers are not formed…
This is because the only lone pairs of electrons available in an MnO4-
ion are those in the O atoms. O, however, being much more electronegative than Mn, would be most unwilling to form a dative covalent bond with Mn, even though the latter possesses the free orbitals required for it to receive electrons.
Hence, as no dative covalent bond can be formed, dimers or polymers cannot be formed.
-----
Regards,
Chia Wei Jie
Each O atom has 6 valence electrons while Mn atom has 7 electrons.
ReplyDeleteFirstly, Mn forms a double covalent bond with 2 of the O atoms so that the 2 O atoms obtain the stable octet electronic configuration. By doing so, Mn will have 11 valence electrons. Since Mn atom's electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s2, besides the 4 electrons shared with 2 O atoms, it can still gain 1 more electron from the K cation in order to fill up its 3d orbital completely. By doing so, Mn will have 2 lone pairs of electrons.
In order for the remaining 2 O atoms to achieve the stable octet electronic configuration, Mn forms dative covalent bonds with the remaining 2 O atoms by contributing each lone pair of electrons to each O atom.
No dimers can be formed as all the orbitals are completely filled with electrons and there are no empty orbitals left.
Mn has 7 valence electrons. It forms a single covalent bond with O-, and has 3 lone pairs. These 3 lone pairs form dative covalent bonds with 3 other O atoms (Mn donate to O), and thus form MnO4-
ReplyDeleteAlso, because all its lone pairs are donated to form dative covalent bonds with the 3 O atoms, it has none to form dimers like AlCl3.
My answer :
ReplyDelete1. Mn's 4s and 3d orbitals are very close in energy level, hence, the valence electron in Mn, is not only from the outer 4s shell which is 2 electrons, but also from the 3d shell which consist of 5 electrons.
Hence it has 7 valence electrons.
2. Then Mn is located on period 3 and elements from period 3 and below have the ability to expand it's octet.
3. Mn formed double bonds with 3 oxygen molecules and 1 single bond with 1 dative covalent bond with the last oxygen molecule. which is why we have the -1 ion of MnO4. This is possible because of the fact that Mn may be able to expand it's octet due to the fact that it has energetically accessible d orbitals, where the electrons can go.
4. Finally it does not form a dimer, as it's shells are already full as it is because of the fact that it had already expanded it's octet to form the 3 double bonds, one single bond and one dative bond, with the 4 Oxygen atoms.
(sry, if there are any mistakes >.<)
Ren_aldy
(i) Ionic bonds are formed when there is a transfer of electrons from a cation to an anion. Mn has an oxidation number of +7. If it were to form an ionic bond with Oxygen, it would have to loose 7 electrons. However, this would be unlikely to happen as loosing 7 electrons require a large amount of energy. Hence, covalent bonds are formed instead. Electrons are shared between O and Mn.
ReplyDelete(ii) In the case of AlCl3 , the compound usually exists as dimer as Cl has a lone pair which it is willing to donate, while Al has an empty orbital willing to accept the electrons. The Cl donates its lone pair of electrons to an Al of another molecule. However, in the case of Mn, the 3d subshell is half filled. There are no empty orbital’s to receive electrons. Hence a dimmer is not formed.
-Abbie
(i) Elements from period 3 onwards are able to expand its octet as they can utilize their energetically accessible d orbitals in bonding. Since Mn is from period 4, it is able to receive 3 pairs of electrons from O and expand its octet, hence forming three dative covalent bonds between O and Mn instead of ionic bonds.
ReplyDelete(ii)MnO4 is result of one covalent bond between oxygen and Mn and donation of three pairs of electrons by oxygen to Mn to form three dative covalent bonds. Hence, Mn04 does not form a dimer because O although there are 3 dative covalent bonds, Mn of another molecule does not have an empty orbital to receive electrons, hence unable to form a dimer.
Mn has an oxidation state of +7 in KMnO-. Logically, it woud take a large amount of ionisation enerhy to remove 7 electrons. As such, Mn would rather share it electrons with O to form an oxidation state of +7.
ReplyDeleteall the Mn d-orbitals are more than half-filled. simarily, oxygen does not have any lone pairs or empty orbitals (1s2 2s2 2p4). As such, there are no lone pairs or an empty orbital for dative bonds to form. thus it does not form dimers or polymers.
i) If MnO4 was an ionic compound, the Mn7+ cation is big and has a large charge, hence this cation is willing to polarise the electron cloud of the anion.
ReplyDeleteO2- is a small anion whose electron clouds are easily polarised. Polarisablity refers to the ease of distorting the electron cloud. When placed next to the Mn7+, the electron cloud of O2- is distorted by O2-. This severe distortion allows for both electron clouds to overlap. when that happens it mimics the overlapping of orbital hence a covalent bond is formed.
ii) Manganese is an element in period 3, which means it can utilize their energetically accessible d orbitals in bonding. The vacant 3d orbitals can be used to accomodate the extra electrons and hence expand its octet. Thus, dimers and polymers are not formed .
(i) O has the configuration of 1s2 2s2 2p4. In order to be a stable structure, it needs 2 electrons to fill the p-subshells. Mn on the other hand, has the configuration of 1s2 2s2 2p6 3s2 3p6 3d5 4s2, it is highly unlikely and improbable for it to lose 7electrons from its 4s and 3d orbitals to form a stable octet structure. Therefore, by combining with the 4Os, it will gain electrons to fill its 3d-subshells and become more stable. Since both elements require more electrons, and neither one is willing to give it up, they have to both share their electrons. Also, Mn and O have both relatively large electron clouds that can form decent overlapping to give rise to covalent bonds too. The fact that Mn is in period 4 allows it to have more than 8electrons in its valence shell due to the energetically accessible d-orbitals.
ReplyDelete(ii) In order for a dimer to be formed, one of the atoms must possess at least a pair of lone electrons and the other atom must have at least one empty orbital in its outer shell to accept the electron pair. There are lone pairs of electrons in the donor atom but there the acceptor atom, in this case Mn is not deficient in electrons. Hence, a dimer nor polymer is formed.
Covalent bonds exist between Mn and O in MnO4- due to the electrostatic forces of attraction between shared electrons and nuclei of the atoms of Mn and O. The shared electrons are brought about when the atomic orbitals of Mn and O overlap. Each atom in sharing the electrons achieves the nearest noble gas electronic configuration and gains stability.
ReplyDeleteWhen a covalent bond is formed, the substance can either exist as a simple discrete molecule or a macromolecule. In AlCl3 and BeCl2, a dimer and polymer are formed respectively. This causes them to have infinite dot and cross diagrams. MnO4- exists as a simple discrete molecule, and thus its dot and cross diagram is finite; a dimer or polymer is not formed.
(i) Under normal circumstances a metal such as Mn would form an ionic bond with a non-metal like O. However, Mn2+ has a high polarizing power, due its small size and relatively large charge of +2. In addition, O2- is very polarisable, have a large negative charge of 2-. When these 2 ions are placed together, Mn2+ distorts the electron cloud of O2-. This results in both electron clouds overlapping each other, identical to the overlapping of orbitals, similar to a covalent bond. Therefore, a covalent bond is formed between Mn2+ and O2-.
ReplyDelete(ii) Unlike Al3+ in AlCl3, the Mn2+ ion has no empty electron orbitals to accept a pair of electrons from O2-. A dative covalent bond cannot be formed and thus the dimerisation/polymerisation of
-
MnO is not possible
4