Dear CH304,
This week's question is as following observations:
(i) Explain why the bond energy of O-F is smaller than the bond energy of O-Br. (The former is 212 kJmol-1 and the latter is 217 kJmol-1.)
(ii)As working, draw the Lewis diagram of N2 and the CN- ion. You do not need to submit your Lewis diagram. (Clue: Draw HCN first)
Explain why the triple bond in N2 is stronger than the triple bond in CN-. (The former's bond energy is 944 kJmol-1 and the latter is 890 kJmol-1
This week's question is as following observations:
(i) Explain why the bond energy of O-F is smaller than the bond energy of O-Br. (The former is 212 kJmol-1 and the latter is 217 kJmol-1.)
(ii)As working, draw the Lewis diagram of N2 and the CN- ion. You do not need to submit your Lewis diagram. (Clue: Draw HCN first)
Explain why the triple bond in N2 is stronger than the triple bond in CN-. (The former's bond energy is 944 kJmol-1 and the latter is 890 kJmol-1
Answers
The following pictures illustrate and explain the questions.
Although, O-Br bond also have lone pair on both O and Br, the repulsion is much less because of Br's large size.In addition, the size of the atom of the element decreases as we go along the period. While the size of the atom of the element increases when we go down the group. Hence, using this concept, we are able to conclude that the C triple bond N is weaker than the N triple bond N.
CommentsFor (i)
Generally (i) has better answers. But, many missed out the key phrase - lone pair on O and the lone pair repulsion on F repel each other because they are so close together (due to O and F being small sized), hence weakening the O-F bond. - I think Samuel's answer for (i) is nice. Succinct, not trying to explain too much and doesn't complicate the situation further.
In addition, using electronegativity to explain (i) is NOT correct.
Interestingly, only Nicholas mentioned that overlapping between small atoms is weak because orbtial overlap is not effective. This is highly incorrect.
For (ii)
I think the biggest problem is that many of you tried to explain that lone pair repulsion, hence making the bond weaker. That is not true as illustrated by the picture above, where both CN- and N2 have lone pairs of electrons.
Hence, in short. The general strategy to explain strength of covalent bond is as follow.
(A) Direct application of the definition of covalent bonds.
If the two examples you are given cannot be concluded using (A), then either one of the two exceptions will be employed.
(B1) A polar covalent bond. A partial positive and partial negative charge, this is an added electrostatic attraction that strengthens the bond between the two atoms.
(B2) Lone pair repulsion. This causes the bond to be weaker.
Generally (i) has better answers. But, many missed out the key phrase - lone pair on O and the lone pair repulsion on F repel each other because they are so close together (due to O and F being small sized), hence weakening the O-F bond. - I think Samuel's answer for (i) is nice. Succinct, not trying to explain too much and doesn't complicate the situation further.
In addition, using electronegativity to explain (i) is NOT correct.
Interestingly, only Nicholas mentioned that overlapping between small atoms is weak because orbtial overlap is not effective. This is highly incorrect.
For (ii)
I think the biggest problem is that many of you tried to explain that lone pair repulsion, hence making the bond weaker. That is not true as illustrated by the picture above, where both CN- and N2 have lone pairs of electrons.
Hence, in short. The general strategy to explain strength of covalent bond is as follow.
(A) Direct application of the definition of covalent bonds.
If the two examples you are given cannot be concluded using (A), then either one of the two exceptions will be employed.
(B1) A polar covalent bond. A partial positive and partial negative charge, this is an added electrostatic attraction that strengthens the bond between the two atoms.
(B2) Lone pair repulsion. This causes the bond to be weaker.
(i) Both O-F and O-Br are bonded covalently. The atomic radii of F is smaller than the atomic radii of Br, so by right, there is a larger extent of overlapping between atomic orbitals between O and F and therefore a stronger covalent bond. However, F being small and containing lone pair of electrons, these lone pair of electrons are nearer to the nucleus of the O atom than those of Br and therefore causes repulsion to occur between the the lone pair of electrons on the F and those on O, making it unfavourable for O and F to come close together, therefore the covalent bond is weaken.
ReplyDelete(ii) Both triple bond in CN- and N2 involves the same amount of electrons in bonding. Hence we cannot evaluate the strength of bonds by comapring the number of bonding electrons. The CN- has a polar covalent bond as the N atom is more electronegative therefore having a slightly more negative charge.The C atom is slightly positive and therefore results in a polar covalent bond. Since CN- has a dipole moment and N2 does not, therefore N2 should be a stronger bond with zero dipole moment.
i) Due to the F and O atoms’ small size, they repel each other, weakening the O-F bond. Therefore, the bond energy of O-F is smaller than that of O-Br.
ReplyDeleteii) C in CN- has accepted an extra electron, causing it to repel N. This weakens the C-N triple bond, therefore, the N-N triple bond in N2 is stronger.
-Timothy Kwok
i) Assuming that there wasn’t an error with the question…
ReplyDeleteTheoretically, the bonds between O-F and O-Br are covalent bonds, which are strong electrostatic forces of attraction formed between the positive nuclei of the two atoms involved and the bonding electrons shared between them. A covalent bond can only be formed when there is an overlapping of atomic orbitals.
O-F is a non-polar compound while O-Br is a polar compound. The electronegativity difference between O-F is insufficient for F, the more electronegative element, to polarise the bonding electrons towards itself.
The difference in electronegativity between O and Br, however, allows O, the more electronegative element, to polarise the bonding electron pair towards itself. This gives the bond between O and Br some ionic character, which the bond between O and F lacks. This results in additional electrostatic forces of attraction present between O and Br, resulting in more energy being required to break the bond between O and Br as compared with that between O and F, hence the higher bond energy of the former.
ii)Now for my incredibly insane/stupid explanation as to why the triple bond in N2 is stronger than the triple bond in CN- .
H has a high charge density, due to its small size and relatively large charge.
H, bonded to C in CN- would be able to polarise the electron cloud (ie. the triple bond ) towards H.
This results in the triple bond being polarised more towards between H and C as compared with between C and N. Due to this, the bond between C and N would become weaker.
Hence, the triple bond in CN- would be weaker than the triple bond in N2 , despite C having approximately the same atomic size as N and both molecules having the same number of bonding electrons. This results in less energy being required to break the triple bond between CN- as compared with the triple bond in N2, hence the lower bond energy of the former.
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Regards,
Chia Wei Jie
Disclaimer: The above postulates were made under the influence of sleep-inducing flu medicine. No offense to the sensibilities of chemists was intended.
(i) Both O-F and O-Br are covalent bonds. In O-F bond, both atoms contain valence shell lone electron pairs which repel each other, hence weaken the covalent bond formed. In O-Br bond, although there are also lone pairs of electron in both atoms' valence shell, the oxygen atom is more electronegative than bromine atom, hence it will attract more towards itself, overcoming the repulsion of electron lone pairs. Hence, O-F bond has smaller energy than O-Br bond.
ReplyDelete(ii)
In N2, the two nitrogen atoms have one valence shell lone pair of electron each, thus experiencing repulsion between N atoms.
In CN-, the carbon and nitrogen also have one valence shell lone pair of electron each, hence also experience the reulsion as in N2. However, CN- ions are formed when HCN dissociates to form H+ ions in water. When hydrogen leaves the molecule, it leaves its electron for the carbon atom, making the carbon atom negatively charged. Since in CN- the nitrogen atom has a valence shell lone pair of electron, repulsion occurs between the negative carbon and the lone pair, thus weakening the covalent bond between C and N (in addition to the repulsion between the lone pairs). As a result, the triple bond in N2 is stronger than the triple bond in CN-.
[i]
ReplyDeleteboth O-F and O-Br are covalently bonded. however F is a smaller atom, thus there is little overlapping of orbitals with O. Br on the other hand is a large atom, with a lot of overlapping with O. since O-Br overlapping is more effective, more energy is needed to break the bond, thus the bond energy is higher than in O-F.
[ii]
in N2, the bonds are all normal covalent bonds.
in CN-, the bonds are two normal covalent bonds and one dative bond (donator is N).
Dative bonds are weaker than normal covalent bonds. so, N2 need more energy to break its bonds, and thus N2 is stronger than CN-.
(i) By definition, since the size of O and F atoms are smaller than the size of the Br atom, the overlapping of the atomic orbital of O and F should be more effective than O and Br, thus forming a stronger covalent bond than O-Br. However, despite that, the bond energy of O-F is still lower than that of O-Br. This is because the O and F atoms each contain lone pairs of electrons, which repel each other. This repulsion cases it to be unfavorable for the O and F to come close together, thus weakening the bond between O and F. Thus, O-Br has larger bond energy than O-F.
ReplyDelete(ii) In the CN- ion, the extra electron that came from the H is found in the C atom. Therefore, the C atom and the N atom both contains lone pairs of electrons. As a result, the lone pair of electrons repel each other, weakening the bonds between the C and the N. As the C atom is smaller than the N atom, there is a smaller distance between the C-N triple bond as compared to the N-N triple bond. As a result, the repulsion between the lone pair of electrons between C and N is greater than the repulsion of lone pair of electrons between N-N. Thus, this causes the bond between the CN- ion to be weaker than that of N-N triple bond. Therefore, N-N triple bond have a larger bond energy than the CN- triple bond.
(i) Firstly, both of these bonds are strong as O is a very electronegative element.However,there are other factors that affect the strength of the bond.In O-F bond, the elements are of similar electronegativity,thus there is no charge separation in these molecules, hence no dipole moment.On the other hand, in O-Br, there is a electronegativity difference between the elements,thus causing the O atom to acquire a partial negative charge,and the Br atom acquires a partial positive charge.This polarization of covalent bond gives additional strength to the covalent bond.Thus, more energy is required to break the bond between O and Br,resulting in a higher bond energy.
ReplyDelete(ii) In nitrogen. The nitrogen atoms will share 3 electrons with the other atom to achieve the stable octet structure.This is a equal sharing of electrons.In Cyanide(CN- ), N, being more electronegative, will accept the valence electron. Then, nitrogen atom will then form 2 normal covalent bonds and a dative covalent bond with the carbon atom.Because of the dative covalent bond,there is an unequal sharing of electrons,with nirigen contributing more electrons.Added onto the fact is that nitrogen atom is more electronegative than carbon,thus it is more unwilling to give/share its electrons.Thus, not much energy is required for the CN- bond to break, as nitrogen is willing to retain its electrons.
(i) The stronger the covalent bond, the more energy required to break the covalent bond and hence the higher the bond energy. However, in this case, because the F atom is very small and has lone pairs of electrons, when it comes together with the O atom there will be electronic repulsion due to the lone pairs, hence weakening the O-F bond. Therefore, the bond energy of O-F is smaller than the bond energy of O-Br.
ReplyDelete(ii) C atom has a lone pair of electrons in CN-, when it comes together with the N atom there will be electronic repulsion due to the lone pairs, hence weakening the triple bond in CN-. However, in N2, there is minimal electronic repulsion and therefore, the triple bond in N2 is stronger than the triple bond in CN-.
(i) Bond in OF has less energy than bond of OBr, because both O and F are very electronegative atoms, hence their difference in electronegativity is not that high, while OBr has higher difference in electronegativity, hence bond in OBr has higher energy compared to bond in OF. As with higher electronegativity difference, this will cause the bond to become more polar in nature as the more electronegative atom would pull the electron cloud towards it,hence the bond will be harder to break/ requires more energy to break.
ReplyDelete(ii)I am not really sure how to answer this question, and rather than writing rubbish i prefer to just admit that i have no idea >.<
i) In O-F, both the atoms are of similar size; both being relatively small unlike in O-Br where the Br atom is bigger compared to an F atom. However, although the atoms in the O-F bond experience a more effective overlapping of atomic orbitals compared to that in O-Br, both the O atom and the F atom are too close to each other in the O-F bond, resulting in their lone pairs repelling each other, weakening the bond, thus resulting in a lower bond energy compared to O-Br.(This also explains for the slight difference in bond enrgy between the two bonds.)
ReplyDeleteii) Firstly, the CN- ion is the cyanide ion which is a nucleophile, meaning that it contains an active lone pair of electrons. Also, when bonded to a H atom, the compound exhibits acidic properties albeit being weak. Thus, the CN- ion does not lose its H atom easily, and when it does, the H atom leaves behind one electron to form the active lone pair on the carbon atom, resulting in a negative ion. As such, with the active lone pair on the carbon atom, the triple bond in the CN- ion is weaker compared to that in nitrogen as the bond is 'stretched' because of the tendency for the carbon atom to be attracted to positive species in the solution.
Fluorine is found in period 2 of group VII while Bromine is in period 4 of the same group. Thus both of them have 7 valence electrons and form a single bond (sigma bond) with oxygen. Both types of bonds also possess a 3 lone pairs of electrons.
ReplyDeleteHowever, as bromine has a larger atomic radius due to having more quantum shells, the lone pairs of electrons are found further away from each other. This results in weaker forces of repulsion between these sets of lone pair electrons.
On the other hand, fluorine has a smaller atomic radius, with lesser quantum shells. This results in the lone pairs being found nearer each other. Thus the forces of repulsion between the lone pairs of O and lone pairs of F are greater than the forces of repulsion in the first case.
This difference in strength of repulsion results in different bond energy as the greater the repulsion, the lesser attractive forces between the 2 atoms, making it easier to break the bond.
In CN- the carbon atom is negatively charged while N is of neutral charge. Despite the fact that both atoms have a lone pair of electrons, the negative charge of the carbon atom + the negative charge of the electrons on C would repel the lone pair of electrons in N.
In N-N, both N atoms have a lone pair of electrons. This lone pair thus possesses a negative charge resulting in repulsive forces.
As the repulsion in CN- is greater than that in N2, the bond energy of CN- is lesser than that of N2 as this repulsion would reduce the strength of the triple bond.
(ii) I want to give it at try for the 2nd question. Is it because of the fact that CN- has 1 less lone pair compared to the N2,hence there will be less repulsion between the bonding pair and the lone pairs, causing the bonding pairs to be very close with each other just like what happen in N2.
ReplyDeleteBecause in N2, since there are 2 lone pairs, it repels the bonding electrons that they were located closer the each other. hence the repulsion between the 3 bonding electrons weakened the bond.
Whilefor CN- since it only has 1 bond pair and the other one is later used as the electrons to bond with H, (in HCN) hence, there will be less repulsion exerted by the lone pair to the bonding electrons. This causes the bonding electrons not to be to close with each other, hence less repulsion happened between the bonding electrons, and therefore the bond at CN- is stronger compared to the bond at N2.
{i} Difference in electro-negativity between O&Br is greater than O&F. Thus, O-Br forms a greater polar bond, which is why bond energy of O-Br is greater than O-F
ReplyDelete(ii) C takes up the extra electron to form a negative CN ion? Then, because N is the more electronegative atom, it forms the partial negative charge, whereas C forms the partial positive charge. yet because of the extra electron its partial positive charge is less strong and hence the attraction between both atoms isnt as strong as N2
(i) The bond length of O-F is smaller than that of O-Br. As a result, the electrons would repel each other. As such, the electrons would be found further away from the nucleus. Hence its bonds would be easier to break than those in O-Br. Hence the bond energy in O-F is smaller than that of O-Br.
ReplyDelete(ii) In CN-, since C has a negative charge, it would repel from the lone pair of electrons from N. As a result, there is a stronger repulsion in CN-. Hence, C and N would be less attracted to each other compared to N2. Hence the the triple bond in N2 is stronger than the triple bond in CN-.
-Abbie
(i)There is a larger electronegativity between O and Br than O and F. Hence, there is a greater displacement of the electron cloud, leading to a more polar molecule and stronger attraction between the dipoles, making the O-Br bond harder to break.
ReplyDeleteAlternatively, i'm not sure if this is true but perhaps: Since the fluorine atom is smaller than the bromine atom, it should be able to achieve a greater extent of overlap with the orbitals of the oxygen atom. While both fluorine and bromine have a lone pair of electrons that will repel the lone pair on oxygen, due to fluorine's smaller size, it would be able to come closer to the oxygen atom, leading to a greater degree of repulsion between the two atoms, making the O-F bond weaker than the O-Br bond as the Br atom would experience less repulsion compared to the F atom.
(ii) N2 is a non polar molecule while CN- is polar. Hence, in CN- the electrons would be pulled towards the N atom, giving it some ionic character. Because of this unequal electron distribution, it would be easier to break the triple bond in CN- compared to N2 as it would be similar to cleaving an ionic bond.
O and F contain lone pair of electrons. These lone pair of electrons on O and F will repel each other, making it unfavourable for O and F to come close together to form a covalent bond. The covalent bond between O and F is weaken, resulting in a smaller bond energy.
ReplyDeleteIn CN- ion, C is negatively charged when it accepts an electron from H. Hence, the electron repulsion between the more negatively charged C and the lone pairs of electrons of the N will be more than the electron repulsion between the lone pairs of electrons from 2 N atoms in N2 molecule. The covalent bonding in CN- ion is weakened more than the covalent bonding in N2.
(i) Even though F is smaller and has a larger electronegativity difference compared with O than Br, which would normally lead to a stronger bond formed with O, O-F bond is weaker than O-Br bond because F being so small that when it comes together with the nucleus of the other atom (O) the lone pairs on F repel the lone pairs on the O, therefore weakening the bond between them which leads to O-F having a smaller bond energy.
ReplyDelete(ii) I honestly do not know the answer. Sorry :/
(i): F is smaller than Br. Hence the lone pairs of electrons in the valence shell of F are closer together than the lone pairs of electrons in the valence shell of Br. Hence those electrons in the valence shell of F exert a repulsive force on the other electron pairs, including the bonding pair of electrons. This force is greater than that exerted by the electrons in the valence shell of Br. Therefore the bond pair of electrons will be 'pushed' away from the nucleus. This repulsive force is greater in F than in Br. As a result the bond energy of O-F is smaller than the bond energy of O-Br.
ReplyDelete(ii): In N2 there are two nitrogen atoms which share three electrons each and have another lone pair of electrons. In CN-, the carbon atom forms three bonds with nitrogen, of which one is a dative covalent bond. Both carbon and nitrogen have three bonding pairs of electrons and one lone pair of electrons. However, in carbon the positive nuclear charge is 6 while in nitrogen the positive nuclear charge is 7. This means that the attractive force between carbon and electrons is weaker than nitrogen since carbon has a lower charge. This also means that the forces of attraction between the carbon nucleus and the bonding pair of electrons is weaker than that between the nitrogen nucleus and the bonding pair of electrons. Carbon and nitrogen also have the same number of bonding and lone pairs of electrons. Hence for nitrogen in N2 and CN- the attraction between the nucleus and electrons are stronger than in carbon in CN-. As a result the triple bond in N2 is stronger than the triple bond in CN-.
i) By right, the smaller the atomic size, the larger the extent of overlapping between atomic orbitals, the stronger the covalent bond and so theoretically, the bond energy of O-F should be bigger than the bond energy of O-Br because F is a smaller ion than Br. However, Flourine ion is an exception as it is very small, and there are many lone pairs of electrons present. Hence, the bond energy of O-F is smaller than the bond energy of O-Br.
ReplyDeleteii) In CN-, CN is negatively charged, and it will repel the lone pair in N, hence, the bond energy in CN- is smaller than the N2 as there is also no negatively charged particle in N2.
(i) the bond pairs for the covalent bond formed between O and F is further from the nuclei as F is a larger atom compared to Br. As such, the attraction between the nuclei would be weaker.
ReplyDelete(ii)the C atom in CN- is negatively charged. Therefore, it would repel the lone pairs in the N atom. As such, the bond energy of CN- is weaker than N2.
i)
ReplyDeleteBoth F and Br are in GroupVII and are covalently bonded to oxygen. The bond energy of O-F is smaller than the bond energy of O-Br because of the larger difference in electronegativity between O-Br as compared to O-F, even though F is smaller than Br and would allow for larger overlap of atomic orbitals and a stronger bond. This is due to the fact that the higher the electronegativity value between two atoms covalently bonded together, the stronger the respective δ+ charge(more electronegative atom) and δ- charge(less electronegative atom), the more polar the bond, the greater the ionic character which gives additional attractions to strengthen the bond, and therefore the O-Br bond will be stronger and have a higher bond energy than O-F.
ii)
The triple bond in N2 is stronger than the triple bond in CH- due to N2 having a lone pair in each of its 2 atoms. This is compared to the CH- bond where there is only one lone pair on N which is not even the central atom(C) and so does not distort the shape of the central molecular ion. The lone pair repulsion in N2 cause the bond angle between bond-pair bond-pair of electrons to be smaller due to its proximity to the nucleus as it is only attracted by one nucleus whereas the bond pairs are attracted by 2. As such, the bond angle between the bond pairs in N2 will be smaller than the bond angles in CN-, and so the N2 bond will be stronger.
i) The bond energy of O-F is smaller than that of O-Br because the difference in electronegativity of O and F (0.5) is smaller than the difference in electronegativity of O and Br (0.7). As such, there are more polarisation between the covalent bond of O and Br, resulting in a stronger bond than O-F.
ReplyDeleteii) N, being more electronegative than C, uses only 2 of its valence electrons, while C uses all of its valence electrons, to form CN-. On the other hand, in N2, both nitrogen atoms use 3 of its valence electrons to form a covalent triple bond. Thus, the attraction between C and N in CN- is not as strong as that in N2.
(i) Both O-F and O-Br form polar covalent bonds due to the small difference in electronegativity between the two atoms in each of the compounds. F is more electronegative than Br, attracting the bonding electrons more strongly than in Br; thus the polar covalent bond in O-F is stronger than in O-Br, and this weakens the covalent bond. Thus the O-F bond energy is smaller than the O-Br bond energy.
ReplyDelete(ii) I'm not very sure about this answer, but I'm guessing it has something to do with the lone pairs. One of the electrons from CN- is gained from hydrogen in order to achieve a stable noble gas configuration, whereas the nitrogen atoms in N2 do not gain but share electrons. The gained electron probably weakens the electrostatic forces of attraction between the shared electrons and nuclei in CN-, thus making its triple bond weaker than that of N2's.
(i)Due to the F atom being smaller in size than the Br atom, the F atom is nearer to the O atom that the BR atom is to its own O atom. This causes O-F to repel from each other more than O-Br, resulting in a larger magnitude of weakening of the O-F bond than the O-Br bond. As such, Bond energy of O-F is smaller than that of O-Br.
ReplyDelete(ii) The C in CN- has accepted an extra electron, which causes a larger magnitude of repulsion of the N, weakening the CN- triple bond more than the N-N triple bond, which experiences a lower magnitude of repulsion between its bond due to not having an extra electron.