chemistry multiple choice question
Hi, please could I have help on these two questions?
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
Original post by anonymous56754
Hi, please could I have help on these two questions?
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
im really sorry i tried to make sense of it, I did this diagram with the double bond and monomers labelled
The I tried to draw the graphical formulas from the short formulas and tried to work out which fitted, I thought C seemed more likely cos I thought there was only 1 double bond per monomer?? However if the correct answer is D I apologise for this as its probs very unhelpful? Anyway sorry..
Ill have a look at the second one now..
Okay so with the second one I got a little confused cos usually I would calculate the amount of electrons each one had and then work out the size of the radius from that..but obviously they all had the same amount of electrons so I used this diagram
https://images.app.goo.gl/ToAnRzF4F8zAr3tAA
From this..you would be right in saying its B i believe!
(edited 1 month ago)
Original post by anonymous56754
Hi, please could I have help on these two questions?
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
1. I don't understand how to work out the monomer. I worked out it was D but I'm confused by the placement of the double bonds, there is a c=c and c=ch2 and c=ch2?
This is the question: https://ibb.co/GTZKwbf
2. For this question, I thought the answer would be B? Neon would have the smallest atomic radius because atomic radius decreases across the period. There is an increase in atomic radius when you move to the next group, but magnesium will have a smaller atomic radius than sodium because again, atomic radius decreases across the period?
This is the question: https://ibb.co/TM3mHhp
Thanks!!
For Q1, I agree that it’s D. There must be two C=C bonds, since the new bonds formed upon “polymerisation” haven’t formed at the same part of the molecule each time and there must be a C(CH3) and a CH somewhere within the structure. This precludes all other options, since A and C only have one double bond and whilst B has CH and C(CH3)’s forming part of it’s structure alongside the two required C=C bonds, the C(CH3) forms the end of the molecule, so the methyl group cannot be correctly placed.
For Q2, I disagree entirely. The trends you have mentioned only apply for when you have a series of neutral atoms. It says the three species are isoelectronic, so they have the same electronic structures, but crucially, the charges differ and this means how tightly bound the electrons are is different. Mg^2+ is the smallest as it has the most positive charge and so has the tightest binding of the electrons and Ne is the largest.
Original post by TypicalNerd
For Q1, I agree that it’s D. There must be two C=C bonds, since the new bonds formed upon “polymerisation” haven’t formed at the same part of the molecule each time and there must be a C(CH3) and a CH somewhere within the structure. This precludes all other options, since A and C only have one double bond and whilst B has CH and C(CH3)’s forming part of it’s structure alongside the two required C=C bonds, the C(CH3) forms the end of the molecule, so the methyl group cannot be correctly placed.
For Q2, I disagree entirely. The trends you have mentioned only apply for when you have a series of neutral atoms. It says the three species are isoelectronic, so they have the same electronic structures, but crucially, the charges differ and this means how tightly bound the electrons are is different. Mg^2+ is the smallest as it has the most positive charge and so has the tightest binding of the electrons and Ne is the largest.
For Q2, I disagree entirely. The trends you have mentioned only apply for when you have a series of neutral atoms. It says the three species are isoelectronic, so they have the same electronic structures, but crucially, the charges differ and this means how tightly bound the electrons are is different. Mg^2+ is the smallest as it has the most positive charge and so has the tightest binding of the electrons and Ne is the largest.
Thank you!
1.
I see what you mean, is the reason there are 2 C=C bonds because of the different bonds forming as I mentioned earlier such as c=c and c=ch2 so there can be a break in either one of the C=C double bonds?
2.
I understand this, but just to clarify, is the greater positive charge indicating that there are less electrons than before and the same number of protons so there is a stronger force of attraction?
Original post by StudentAC
im really sorry i tried to make sense of it, I did this diagram with the double bond and monomers labelled
The I tried to draw the graphical formulas from the short formulas and tried to work out which fitted, I thought C seemed more likely cos I thought there was only 1 double bond per monomer?? However if the correct answer is D I apologise for this as its probs very unhelpful? Anyway sorry..
Ill have a look at the second one now..
Okay so with the second one I got a little confused cos usually I would calculate the amount of electrons each one had and then work out the size of the radius from that..but obviously they all had the same amount of electrons so I used this diagram
https://images.app.goo.gl/ToAnRzF4F8zAr3tAA
From this..you would be right in saying its B i believe!
The I tried to draw the graphical formulas from the short formulas and tried to work out which fitted, I thought C seemed more likely cos I thought there was only 1 double bond per monomer?? However if the correct answer is D I apologise for this as its probs very unhelpful? Anyway sorry..
Ill have a look at the second one now..
Okay so with the second one I got a little confused cos usually I would calculate the amount of electrons each one had and then work out the size of the radius from that..but obviously they all had the same amount of electrons so I used this diagram
https://images.app.goo.gl/ToAnRzF4F8zAr3tAA
From this..you would be right in saying its B i believe!
Tysm, for the first question I thought the same as the image u drew out but I think typical nerd explained it below 🙂
Original post by anonymous56754
Thank you!
1.
I see what you mean, is the reason there are 2 C=C bonds because of the different bonds forming as I mentioned earlier such as c=c and c=ch2 so there can be a break in either one of the C=C double bonds?
2.
I understand this, but just to clarify, is the greater positive charge indicating that there are less electrons than before and the same number of protons so there is a stronger force of attraction?
1.
Indeed, in theory either carbon-carbon π-bond can break to form new carbon-carbon σ-bonds, joining the monomers.
2.
Essentially yes. More protons per electron means more things in the nucleus pulling the electrons in. Charges just are a very quick way to rationalise this.
Original post by TypicalNerd
1.
Indeed, in theory either carbon-carbon π-bond can break to form new carbon-carbon σ-bonds, joining the monomers.
2.
Essentially yes. More protons per electron means more things in the nucleus pulling the electrons in. Charges just are a very quick way to rationalise this.
Great, thank you 🙂
Original post by StudentAC
im really sorry i tried to make sense of it, I did this diagram with the double bond and monomers labelledThe I tried to draw the graphical formulas from the short formulas and tried to work out which fitted, I thought C seemed more likely cos I thought there was only 1 double bond per monomer?? However if the correct answer is D I apologise for this as its probs very unhelpful? Anyway sorry..
Ill have a look at the second one now..
Okay so with the second one I got a little confused cos usually I would calculate the amount of electrons each one had and then work out the size of the radius from that..but obviously they all had the same amount of electrons so I used this diagram
https://images.app.goo.gl/ToAnRzF4F8zAr3tAA
From this..you would be right in saying its B i believe!
Ill have a look at the second one now..
Okay so with the second one I got a little confused cos usually I would calculate the amount of electrons each one had and then work out the size of the radius from that..but obviously they all had the same amount of electrons so I used this diagram
https://images.app.goo.gl/ToAnRzF4F8zAr3tAA
From this..you would be right in saying its B i believe!
Just for the sake of your learning, I'll make the following points:
-If there was only one C=C bond per monomer, it would have been used up entirely in polymerisation - you wouldn't see any C=C bonds in the product. Because the product contains C=C bonds, there must be more than one per monomer.
-You have the right idea with the monomers question in that splitting the molecule into equal sections can make it easier to work out the structure of the monomer. I think you just missed that where the new C-C bonds have formed, a C=C bond must have existed previously.
-Bear in mind that the size of a monatomic particle is affected by both its electronic structure and the size of the nucleus. If an ion is positive, it has more protons per electron and so the nuclear pull is stronger, which draws the electrons in and so decreases the radius. The trend you have referred to only works when you are comparing a series of monatomic particles of zero (or equal) charge.
(edited 1 month ago)
Original post by TypicalNerd
Just for the sake of your learning, I'll make the following points:
-If there was only one C=C bond per monomer, it would have been used up entirely in polymerisation - you wouldn't see any C=C bonds in the product. Because the product contains C=C bonds, there must be more than one per monomer.
-You have the right idea with the monomers question in that splitting the molecule into equal sections can make it easier to work out the structure of the monomer. I think you just missed that where the new C-C bonds have formed, a C=C bond must have existed previously.
-Bear in mind that the size of a monatomic particle is affected by both its electronic structure and the size of the nucleus. If an ion is positive, it has more protons per electron and so the nuclear pull is stronger, which draws the electrons in and so decreases the radius. The trend you have referred to only works when you are comparing a series of monatomic particles of zero (or equal) charge.
-If there was only one C=C bond per monomer, it would have been used up entirely in polymerisation - you wouldn't see any C=C bonds in the product. Because the product contains C=C bonds, there must be more than one per monomer.
-You have the right idea with the monomers question in that splitting the molecule into equal sections can make it easier to work out the structure of the monomer. I think you just missed that where the new C-C bonds have formed, a C=C bond must have existed previously.
-Bear in mind that the size of a monatomic particle is affected by both its electronic structure and the size of the nucleus. If an ion is positive, it has more protons per electron and so the nuclear pull is stronger, which draws the electrons in and so decreases the radius. The trend you have referred to only works when you are comparing a series of monatomic particles of zero (or equal) charge.
ooooohhh thank you that makes much more sense!
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