This is the second question.
Suggest an explanation to why XeF2 exists but NeF2 does not. [Hint: You are required to use the concept of atomic structure and chemical bonding in your answer]
Answer to Question 2
Xe is a Period 5 element and although it has 8 valence electrons, it has available empty d-orbital which can be used to expand the octet for additional bonding. While Ne is a Period 2 element and does not have available d-orbtial which can be used to expand the octet.
In additional, to expand the octet in Xe, it requires the unpairing of the valence electrons and exciting them to the d orbital, since in a same shell the d orbital has a higher energy than the s orbital and the p orbital. Thus, this rearrangement of electronic configuration requires energy. Noticeably, this energy required is compensated by the Xe-F bond that is formed (bond formation releases energy). Therefore, XeF2 exist.
Even if Ne can expand its octet, Ne's valence electrons are so close to the nucleus and thus experience strong electrostatic attraction. Thus to unpair them and move it to a higher energy orbital, you will need so much energy that even the formation of the Ne-F bond is unable to compensate the energy required use to rearrange the electrons configuration.
Comments for Question 2
1. If you have submitted the work, a mark is awarded to you. If you have mentioned expansion of octet due to available d orbitals or available orbitals a second mark is given to you. However, if you mentioned 3d orbital, you will not get the mark.
2. All of you had problems in getting the 3rd and 4th mark. I guess one has to realise that in order to expand the octet for bonding, there is a need to rearrange the electronic configuration of Xe, to make the electrons unpaired. This gives the 3rd mark. One of the unpaired electrons will occupy a higher energy orbital such as the d orbital and the unpair electron awaits to be bonded with F. (Note that formation of Xe-F bond makes use of equal sharing of electrons, where Xe and F each contribute one electron. Hence, you will need Xe to unpair its electrons pair.)
3. Finally, because the Xe-F bond is formed (bond formation is exothermic or energy releasing - this was mentioned in an earlier blog post), the energy release can compensate the energy required to rearrange the electronic configuration. This is the 4th mark.
4. Hence, to expand the octet these are the available reasons. Hence, sometimes when we want to consider about expansion of octet, the type of bond formed will determine if the expansion is favourable. Hence, if a strong covalent bond can be formed, we will get expansion of the octet. For this reason, XeBr2 does not exist as the Xe-Br bond is not sufficiently strong.
Both xenon and neon are in group 0 of the periodic table. This means that they are non-metals. As such, they are more likely to form covalent bonds with other non-metals and ionic bonds with metals. This is because they would rather accept more electrons to fill available subshells rather than lose an entire valence shell of electrons to form cations. Hence they would form anions when bonding with metals and would share electrons if possible when bonding with non-metals.
ReplyDeleteXenon is in period 5 of the periodic table. It has 54 electrons with the electronic configuration 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2, 5p6. Hence, it has empty 4f and 5d orbitals. It can accept more electrons to fill these orbitals. Therefore it can form covalent bonds with fluorine and accept two more electrons into these empty orbitals to form XeF2. In fact, xenon can even react with more fluorine to form XeF4.
However, neon is in period 2 of the periodic table. It has 10 electrons with an electronic configuration of 1s2, 2s2, 2p6. It does not have an empty orbital to accept more electrons, as the 2d orbital does not exist. It can accept electrons into a 3s subshell but this is very unlikely. If fluorine were to react with neon, neon would have 2 more electrons as a result of the two covalent bonds formed with fluorine. Hence, as neon cannot accept any more electrons to form bonds with fluorine, NeF2 does not exist.
Answer to question 2:
ReplyDeleteBoth Xenon and neon are noble gases, meaning that their atoms have already had the octet structure. However, xenon is in period 5, it has empty electron orbitals, allowing expansion of its octet structure (having more than 8 valence electrons). This allow xenon to form covalent bond with fluorine to form XeF2, in which there are five valence electron pairs surrounding the xenon atom.
Neon, on the other hand, is in period 2. It does not have empty valence electron orbitals to accept more electron pairs, hence cannot form covalent bond with fluorine.
Both Ne and Xe are noble gases found in period 2 and period 5 respectively. As Ne is found in period 2, we can deduce it has an electronic configuration of 1s2 2s2 2p6 . Xe would thus have an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6. As Ne does not have an empty d orbital, it would not be able to accept electrons any further, thus being unable to form a covalent bond with fluorine. On the other hand, Xe has an empty d orbital, thus it would be able to accept electrons and able to expand its octet. Thus in suitable conditions, Xe would be formed as Xe would donate an electron each to each of the fluorine. Moreover, this occurs because Fluorine is a highly electronegative element, thus having a great affinity for electrons (Such reactions would not occur with atoms of low electronegativity.)It would then draw the valence electrons of Xe towards it, thus forming the covalent bonds and the hypervalent molecule of XeF2
ReplyDelete2. Firstly, Xe and Ne are noble gases in Group 0 of the periodic table, meaning that they have stable electronic configuration and exist as monoatomic gases. However, Ne is from period 2 while Xe is from period 5. This means that Xe is able to expand its octet due to empty d orbitals to recieve additional electrons, thus allowing it to have 10 electrons in its valence shell when covalently bonded to 2 flourine atoms. Ne would thus be unable to form NeF2 as it lacks empty d orbitals, preventing it from expanding its octet.
ReplyDeleteFirstly, let us establish that the intramolecular bonds present in XeF2 and NeF2 are covalent bonds. Covalent bonds are formed by the overlapping of atomic orbitals of atoms, causing a sharing of electrons to take place between the atoms.
ReplyDeleteF is a very,very electronegative atom. This allows F to attract electrons to it easily.
Xe and Ne both possess an octet structure. F requires 1 more electron to fulfil its octet structure.
Xenon is a large atom. Hence, it has a large electron cloud, causing the electrons in its valence shell to be held by relatively weak electrostatic forces of attraction to the Xe nucleus.
When Xe is brought close to F under certain conditions, allowing the atomic orbitals of Xe and F to come close to each other, F’s electronegativity would allow it to strip an electron away from Xe, forcing a sharing of electrons to take place and forming a covalent bond, fulfilling F’s octet requirement in the process. This is possible due to the aforementioned weakness of the attraction between Xe valence electrons to its nucleus.
As Xe possesses an energetically available d orbital, it has the capability to expand its octet structure. Hence, it has the ability to accommodate the extra electrons after forming covalent bonds with 2F, allowing XeF2 to exist.
NeF2 cannot exist for two reasons. Firstly, Ne is a small atom. Hence, it would be nigh impossible for even F to strip an electron away from Ne as the Ne valence electrons would experience strong electrostatic forces of attraction to the Ne nucleus.
Secondly, even if it was possible for F to attract an electron away from Ne, Ne does not possess an energetically accessible d orbital and hence, cannot expand its octet structure. This would prevent it from being able to accommodate any extra electrons obtained from the sharing of electrons with F, hence preventing NeF2 from existing.
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Homework...overhwhelming...
Regards,
Chia Wei Jie
-under chemical bonding...)
ReplyDeleteXeF2 and NeF2 do not exist as an ion because both Xe and Ne are noble gases and have a full outer shell already. there is no need to change the number of electrons to get a noble gas structure because they are after all noble gases with the noble gas structure. therefore with no ions formed, no ionic bonding can take place.
metallic bonding is out because it only occurs in metals. none of the elements mentioned are metals.
-under atomic structure...)
XeF2 therefore can only exists as a covalently bonded molecule. there is empty d orbital in Xe which are available to accept more electrons and expand octet, therefore allowing it to be covalently bonded, and hence XeF2 exists.
but NeF2 is not exist even as a covalent molecule because it is in period 2 and has no empty d orbital to accept more electrons and is at its maximun number of electrons already. this prevents any bonding from taking place, so Ne can only exist by itself, not bonded to anything, hence NeF2 does not exist.
Ne is a period 2 element. It therefore has only 2 quantum shells that are completely filled up with electrons. As such, it cannot form covalent bonds with F as the Ne does not have extra orbitals to hold the 2 extra electrons it would gain from forming covalent bonds with the two F atoms.
ReplyDeleteHowever, Xe is a Period 5 element. It therefore has five quantum shells, meaning that it has extra orbitals. As Xe has extra obitals, it can therefore expend its octet structure to include extra electrons, allowing it to form covalent bonds with other elements. Thus, it is able to form covalent bonds with 2 F atoms, forming XeF2.
2) Xe and Ne each possess 8 valence electrons. Therefore, in order for them to bond covalently with F2 to form XeF2 or NeF2, they would have to expand their octet. This is possible for Xe because it is a Period 5 element, which has empty d orbitals to accept electrons from F. However, Ne is a Period 2 element, which does not have any energetically vacant d orbitals. It is therefore unable to expand it octet, and NeF2 does not exist as a result.
ReplyDelete- Timothy Kwok
Neon(1s2 2s2 2p6) is an element in period 2. Xenon(1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6) is an element in period 3.
ReplyDeleteTherefore, Xenon can expand its octet since it has vacant 3d orbitals which can accommodate the extra electrons. As such, Xe despite having an octet valence shell is able to share 1 electron with 2 fluorine atoms, having 10 electrons in its valence shell.
On the other hand, Ne is not able to accommodate more than 8 electrons in its valence shell since it does not have the vacant 3d orbitals and hence NeF2 does not exist.
XeF2 thus exists as a simple discrete molecule as the central atom Xenon is covalently bonded to 2 Fluorine atoms.
Ne's has 10 electrons. Its electronic configuration is 1s2, 2s2, 2p6. Xe has 54 electrons, and its electronic configuration is 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2, 4p6, 4d10, 5s2, 5p6.
ReplyDeleteXeF2 exists because Xe is able to expand its octet due to the fact that it has vacant d orbitals to accommodate the extra electrons. Hence, 2 electrons of Xe form 2 single covalent bonds with 2 F to form XeF2. This result in Xe having 10 valence electrons.
NeF2 does not exist because Ne cannot form covalent bonds with F. Ne is in Period 2, it does not have vacant d orbitals to accommodate extra electrons, hence it cannot expand its octet.
I think the answer is because F is a very electronegative atom which means that it has a very high polarising power. Hence even Xe which has already attained the noble gas structure due to it's large size of electron cloud, since it is located on period 5. Hence it's electron cloud is able to be polarized by F who is v electronegative forming XeF2. However this can't happen with Ne as Ne is located at period 2. Thus it has a small electron cloud size which located near the nucleus hence experiencing a high attractive forces from the nucleus. This fact combined with the fact that it has already attain a noble gas configuration caused it's electron not to be polarized even by the most electronegative atom which is F.
ReplyDeleteHence XeF2 can exist while NeF2 can't exist
Xe, an element that is below period 3, is able to expand its octet because it has empty d-orbitals which are available to accept more electrons. On the other hand, F is willing to donate its electrons to Xe to form dative covalent bonds.
ReplyDeleteNe is found on period 2, therefore it is unable to expand its octet to accept more electrons.
Answer:
ReplyDeleteIn atomic structure, as the number of protons increases, the elements have more electron shells.This is because elements are of neutral charges.Thus,we can infer that down the noble gas group, the noble gases will get heavier,resulting in more electron shells.Also, Xe is on period 5,unlike Ne which is in period 2.Thus, XeF2 is possible as elements from period 3 onwards can utilize their available 3d orbitals, thus expanding their octet.
From chemical bonding, fluorine is the most electronegative element,thus it will easily attract electrons to itself. On the other hand,there are many electron shells in the Xe atom because of its large proton number.Electrons, being of similar negative charges, will repel each other.Thus,the outermost electrons will experience repulsion from the inner electrons.As a result, the outermost electrons are less attracted to the positively-charged nucleus.Thus, they are able to form compounds with the most electronegative elements, for example, fluorine.
XeF2 and NeF2 are covalent compounds that are exceptions to the octet rule as there are more than 8 valence electrons around the central atom. However, while XeF2 exists, NeF2 does not. This is because Xenon which is in period 5 contains d and f orbitals which it can make use of to expand its octet, accepting more electrons to form a covalent bond with Fluorine. On the other hand Neon, in period 2 does not have any empty d-orbitals and thus, cannot accept additional electrons to form covalent bonds with Fluorine. Hence, XeF2 exists while NeF2 does not.
ReplyDeleteXe is much larger than Ne, so its valence electrons are more easily polarized by a highly electronegative atom, e.g F, such that 2 F atoms can form single covalent bonds with Xe's d-orbital electrons.
ReplyDeleteXeF2 exist because Xe is a Period 5 element and hence can expand its octet due to its empty 5d orbital. Thus, although xenon already has 8 valence electrons, it can expand its octet to accomodate the 2 extra electrons shared by the 2 fluorine atoms. The xenon is now covalently bonded to 2 fluorine atoms, forming XeF2.
ReplyDeleteNeF2 does not exist because it already has a completely filled valence electron shell with 8 electrons; but unlike xenon, it is from Period 2 and hence does not possess an empty 3d orbital to accomodate any more electrons. Thus, it is unable to expand its octet, that is, to accomodate the extra electrons shared by the fluorine atoms during covalent bonding.
Covalent bonds result from the sharing of a pair of electrons between two atoms. The orbitals overlap.
ReplyDeleteIn the case of XeF2, as Xe is in the 5th period, it can expand its octet and have more than 8 electrons in the valence shell as it has vacant 4d orbitals which can accommodate extra electrons. Together with the fact that F is willing to donate its electrons, the orbitals between Xe and F would overlap. This results in covalent bonds being formed. Hence, XeF2 exists.
In the case of NeF2, Ne is in period 2. Therefore, it cannot expand its octet as it does not have empty orbitals. Therefore, the maximum number of valence electrons it can have is 8. Ne is a noble gas as it has already achieved stable octet structure. It does not have empty orbitals to accept any more electrons. Hence, the orbitals of Ne and F would not overlap. Covalent bonds will not be formed. Thus NeF2 does not exist.
-Abbie
Xe and Ne are both Group 0 elements, each with 8 valence electrons, and are very stable noble gases. Ne lies in Period 2 while Xe lies in Period 5. Even though both alreay fulfil the octet rule with 8 electons on their outermost shell, Xe is able to expand its octet and accept more electrons due to the presence of its electronically accessible d orbitals. Xe can thus react with 2 F atoms to form XeF2, with its outermost shell now containing 10 electrons. However, Ne, having no electronically accessible d orbitals, cannot break the octet rule of 8 electons on its outermost shell, and thus cannot react with 2 F atoms to form NeF2.
ReplyDeleteFirstly, it is because Xe has a highly energetic and accessible 3d orbital that will allow it to expand its octet. Hence, XeF2 can be formed and not NeF2 due to its absence of 3d orbital in the Ne atom.
ReplyDeleteSecondly, there is a bigger difference in electronegativity between Xe and F and hence they can form a compound and NeF2 cannot.
Non-metals from period 3 onwards are able to utilize their valence d-orbitals to form bonds. Xenon is in period 5 and thus is able to under go hybridization, involving its d-orbitals in bonding with Florine. However, Neon is in period 2 and does not have any d-orbitals in its atomic structure to enable it to bond with Florine.
ReplyDeleteXeF2 would have a total of 10 valence electrons around the Xe atom which is more than the required octet, it is the expanded octet model which requires accommodation by the d-orbitals. However, this bonding that occurs between Florine and Xenon do not violate the octet rule as the normal covalent bonds between them do not exceed its original number of valence electrons.
Therefore, because Ne is in period 2, it is unable to form bonds with F to form the compound NeF2.