Missradioactive
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why is the answer to this B ?

Pigster
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antibody123
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11578 - 2475 = biggest difference

so if you leap from from no1 to no2 to no3 to 11578 it's the 3rd jump so it's group 3...
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Missradioactive
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(Original post by antibody123)
11578 - 2475 = biggest difference

so if you leap from from no1 to no2 to no3 to 11578 it's the 3rd jump so it's group 3...
Why 3rd jump = group 3, exactly ? I'm so confused !
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antibody123
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(Original post by Missradioactive)
Why 3rd jump = group 3, exactly ? I'm so confused !
the biggest difference between the two numbers tells you the group number
the biggest difference is between no3 and no4 it is the 3rd group

no1 to no2 = 1st 'leapfrog'
no2 to no3 = second 'leapfrog'
no3 to no4 being the biggest difference in ionisation energies and it being the 3rd 'leapfrog' gives you group 3

im not the best at explaining stuff but thats how i do it sorry
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Missradioactive
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(Original post by Missradioactive)
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why is the answer to this B ?

Pigster
IE1 = 578, i.e. is doesn't take much energy to remove, i.e. the 1st e- isn't too hard to remove.
IE2 = 1817, i.e. about 3x more energy (but you'd expect that as the p+:e- ratio has gone up), i.e. the 2nd e- isn't too hard to remove.
IE3 = 2745, again not too big a jump, the 3rd e- is also not too hard to remove.
IE4 = 11578, OH MY LORDY, that is a bit increase!, the 4th e- is MUUUUUCH harder to remove. What could have caused that? Niels Bohr suggested that a massive jump like this would be caused by the existence of "shells". The 4th e- is found in a shell that is closer to the nucleus and hence requires far more energy to remove than one in a higher shell.

In conclusion, the 1st 3 e-, being much easier to remove than the 4th are found in one shell. The 4th is found in a shell closer to the nucleus.

Ergo, there are 3 e- in the outer shell. Hence valence and therefore group = 3.
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Missradioactive
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(Original post by Pigster)
IE1 = 578, i.e. is doesn't take much energy to remove, i.e. the 1st e- isn't too hard to remove.
IE2 = 1817, i.e. about 3x more energy (but you'd expect that as the p+:e- ratio has gone up), i.e. the 2nd e- isn't too hard to remove.
IE3 = 2745, again not too big a jump, the 3rd e- is also not too hard to remove.
IE4 = 11578, OH MY LORDY, that is a bit increase!, the 4th e- is MUUUUUCH harder to remove. What could have caused that? Niels Bohr suggested that a massive jump like this would be caused by the existence of "shells". The 4th e- is found in a shell that is closer to the nucleus and hence requires far more energy to remove than one in a higher shell.

In conclusion, the 1st 3 e-, being much easier to remove than the 4th are found in one shell. The 4th is found in a shell closer to the nucleus.

Ergo, there are 3 e- in the outer shell. Hence valence and therefore group = 3.
Wow !

Thank you so much, you're a lifesaver
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Missradioactive
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(Original post by Pigster)
IE1 = 578, i.e. is doesn't take much energy to remove, i.e. the 1st e- isn't too hard to remove.
IE2 = 1817, i.e. about 3x more energy (but you'd expect that as the p+:e- ratio has gone up), i.e. the 2nd e- isn't too hard to remove.
IE3 = 2745, again not too big a jump, the 3rd e- is also not too hard to remove.
IE4 = 11578, OH MY LORDY, that is a bit increase!, the 4th e- is MUUUUUCH harder to remove. What could have caused that? Niels Bohr suggested that a massive jump like this would be caused by the existence of "shells". The 4th e- is found in a shell that is closer to the nucleus and hence requires far more energy to remove than one in a higher shell.

In conclusion, the 1st 3 e-, being much easier to remove than the 4th are found in one shell. The 4th is found in a shell closer to the nucleus.

Ergo, there are 3 e- in the outer shell. Hence valence and therefore group = 3.
One more question:

We know that the hydroxide ion contains 10 electrons.

Then why are there only 8 electrons shown in its dot and cross diagram?
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Ailurophile03
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(Original post by Missradioactive)
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why is the answer to this B ?

Pigster
Which year is this
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Missradioactive
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(Original post by Ailurophile03)
Which year is this
January 2019, unit-1
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Ailurophile03
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(Original post by Missradioactive)
January 2019, unit-1
Thank you
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Pigster
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(Original post by Missradioactive)
One more question:

We know that the hydroxide ion contains 10 electrons.

Then why are there only 8 electrons shown in its dot and cross diagram?
The dot and cross diagram only shows the outer shell electrons, i.e. it missed off the 2x 1st shell e- on the O.
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Deborah247
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(Original post by Missradioactive)
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why is the answer to this B ?

Pigster
What exam board is this?
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Ailurophile03
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(Original post by Deborah247)
What exam board is this?
Edexcel
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Missradioactive
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(Original post by Pigster)
The dot and cross diagram only shows the outer shell electrons, i.e. it missed off the 2x 1st shell e- on the O.
Hmm.

Also, there's this sentence in my book: "Carbocations are more stable when there are more electron-releasing alkyl groups attached to the carbon with the positive charge."

Why are alkyls referred to as electron-releasing here ?
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Pigster
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(Original post by Missradioactive)
Hmm.

Also, there's this sentence in my book: "Carbocations are more stable when there are more electron-releasing alkyl groups attached to the carbon with the positive charge."

Why are alkyls referred to as electron-releasing here ?
Consider the simplest alkyl group: CH3.

C is more electronegative than H.

C will be delta -ve and H will be delta +ve.

If a delta negative C (in CH3) is next to a +ve C atom (your carbocation), some of the extra electron density on the delta +ve C will be pushed onto the +ve C atom, reducing the size of the +ve charge, hence increasing the stability of the ion.
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Missradioactive
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(Original post by Pigster)
Consider the simplest alkyl group: CH3.

C is more electronegative than H.

C will be delta -ve and H will be delta +ve.

If a delta negative C (in CH3) is next to a +ve C atom (your carbocation), some of the extra electron density on the delta +ve C will be pushed onto the +ve C atom, reducing the size of the +ve charge, hence increasing the stability of the ion.
Thank you so much, that was really helpful
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Missradioactive
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How to do this? I don't get it :bawling: Pigster


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Pigster
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(Original post by Missradioactive)
How to do this? I don't get it :bawling: Pigster
You have identified two of the values from the table and assigned them to two of the arrows on the cycle.
There are two other values given to you and two arrows they can be assigned to.
Either you 50:50 it, or you work out which arrow deserves which value.
Then you Hess the hell out of it.
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Missradioactive
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(Original post by Pigster)
You have identified two of the values from the table and assigned them to two of the arrows on the cycle.
There are two other values given to you and two arrows they can be assigned to.
Either you 50:50 it, or you work out which arrow deserves which value.
Then you Hess the hell out of it.
I still don't get it :bawling:


Also
is THIS one correct ?

In THIS one, they said to think about the gaseous carbon so is the Hess law the way to do so ?
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