past paper confusion

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 !

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.

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.

why is the answer to this B ?

@Pigster

Which year is this

January 2019, unit-1

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?

why is the answer to this B ?

@Pigster

What exam board is this?

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 ?

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.

How to do this? I don't get it @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.