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Chemistry: Post your doubts here!

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Mg.gif

Like, why is there an increase in 3rd to 4th I.E. and 4th to 5th I.E?
After the removal of an electron, the already present electrons tend to get closer to each other. Thus it is becomes harder to remove another electron.
Such questions won't really come. They would rather prefer to ask simplistic statements, like why is there an abrupt increase in the ionization energy between the 2nd and 3rd I.E




Another possible explanation. In Magnesium, the p sub-shell is complete with THREE PAIRS. The first electron in the FIRST PAIR is removed at the cost of residual repulsion. This makes removing the electron easier. Then electrons get closer to each other. The other electron in the FIRST PAIR is more difficult to remove...
 
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Another possible explanation. In Magnesium, the p sub-shell is complete with THREE PAIRS. The first electron in the FIRST PAIR is removed at the cost of residual repulsion. This makes removing the electron easier. Then electrons get closer to each other. The other electron in the FIRST PAIR is more difficult to remove...

What?

Aren't electrons first removed from 3s in magnesium?
 
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After the removal of an electron, the already present electrons tend to get closer to each other. Thus it is becomes harder to remove another electron.
Such questions won't really come. They would rather prefer to ask simplistic statements, like why is there an abrupt increase in the ionization energy between the 2nd and 3rd I.E




Another possible explanation. In Magnesium, the p sub-shell is complete with THREE PAIRS. The first electron in the FIRST PAIR is removed at the cost of residual repulsion. This makes removing the electron easier. Then electrons get closer to each other. The other electron in the FIRST PAIR is more difficult to remove...

So you are saying repulsions decrease and that's the reason?
My theory is (once I thought about it) that when one electron leaves, the charge goes to +1 and thus it's the same amount of protons attracting one less electron so the bond between the nucleus and the remaining electrons gets stronger thus rising the I.E.

And Dark Destination, read the question clearly.
 
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So you are saying repulsions decrease and that's the reason?
My theory is (once I thought about it) that when one electron leaves, the charge goes to +1 and thus it's the same amount of protons attracting one less electron so the bond between the nucleus and the remaining electrons gets stronger thus rising the I.E.

And Dark Destination, read the question clearly.
Would you care to read it again?
 
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drawing of orbitals
e.c
sketchin graph for orbitals
definition simply
shapes of molecules n bond angles
attraction of electrons
polarisation of orbitals

to name a few

Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat? Bond angles of hybridization? What in the hell is that?

Awesome12, how much of this have we studied?
 
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Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat? Bond angles of hybridization? What in the hell is that?

Awesome12, how much of this have we studied?
-_- shapes of molecules!!! so with it theres always bond angle...so i wrote it -_-
chain reaction u see :p
hybridisation --> shape of molecule --> bon angle --> charge densities n all tht
 
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Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat? Bond angles of hybridization? What in the hell is that?

Awesome12, how much of this have we studied?
We have done bond angles of hybridization, but it the same as the ones in the shapes of the molecules. Like AB4 has bond angles = 109.5 degress. Thus in hybridization, it will be SP3 also having bond angles = 109.5 degrees. Nothing to worry about.
 
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