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If we remove three electrons from 3 half-filled orbitals from the same p subshell, why is there a small increase in ionization energy in each of them?
Because they are at different energy levels?
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If we remove three electrons from 3 half-filled orbitals from the same p subshell, why is there a small increase in ionization energy in each of them?
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.
Like, why is there an increase in 3rd to 4th I.E. and 4th to 5th I.E?
They are degenerate orbitals ; orbitals on the same energy level.Because they are at different energy levels?
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...
He asked about the difference in I.E between the 3rd and 4 th and 4th and 5th.What?
Aren't electrons first removed from 3s in magnesium?
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...
Would you care to read it again?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?
He asked about the difference in I.E between the 3rd and 4 th and 4th and 5th.
Hydrogen bonding cannot form between H and Cl. The difference in electronegativity isn't that high to allow a very strong hydrogen bond to form. In addition to that, HCl is non-polar; it has induced dipole forces of attraction.HCL doesn't have Hydrogen bonding? Why? :/
poor HClHCL doesn't have Hydrogen bonding? Why? :/
HCL doesn't have Hydrogen bonding? Why? :/
drawing of orbitalsWhat kind of questions come in Hybridization?
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
-_- shapes of molecules!!! so with it theres always bond angle...so i wrote it -_-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.Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat? Bond angles of hybridization? What in the hell is that?
Awesome12, how much of this have we studied?
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