Ionization Energy

[The Phlebotomist]

Correct explanations for the increases and decreases in ionization energies between elements between atomic numbers 2 and 10 (and 11 and 18) include:

I. There is repulsion of paired electrons in the P^4 configuration.

II. The electrons in a filled S orbital are more effective at shielding the electrons in the p orbitals of the same n than each other.

III. Filled orbitals and subshells are more stable than unfilled orbitals and subshells.

A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III

To me this question was awfully stated. Let me take you through my thinking process.

I) P^4 configuration looks like: AB A A (A for up arrow B for down arrow) P has 3 orbitals.

Since B is an opposite spin its unstable and takes more energy to be in such orientation. Thus there is repulsion of paired electrons: This explains why group 6 (O,S,As) have less ionization energy than group 5. This repulsion force requires extra energy for the atom to spend to be in such orientation it would rather lose the "B arrow" and become stable A A A thus requiring less energy.

II. Please some one explain this to me, I have not learned this.

III. Looks Kinda true.

I would also like to mention:

a Greater effective nuclear charge explains why the radius shrinks in a period. The protons are able to hold on those electrons "tighter" or stronger. Because as the number of protons increase in a period no new shells are formed. This explains why the radius decreases from left to right. Also it shows that it requires more ionization energy to remove an electron from the atom.

My answer Choice was E:

But the answer is D.

Can you please help me?

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[ayocaptain628]

so ok
I. is defo right b/c as you said p^4 means that one p orbital will have two e- repulsing each other.

II. is right
it literally makes no sense for s electrons to "shield" each other from the nuclear change b/c they're in the same shell.
shielding should be associated with an inner shell that has already been filled

the s shell is filled first, and b/c of that, an inner shell has been filled which makes the pull of the nuclear change less on the e- in outer shells thus generating zeff, greater effective nuclear charge.

III. i think this answer is weird b/c it seems to state the answer as a definitive fact.
unfilled orbitals can actually be more stable such as in the case when u have one e- in each orbital where they are all half filled.

thats why we have weird e- configurations for like chromium b/c it likes 4s1^ 3d^5 as opposed to 4s^2 3d^4.

but now that i think of it, unfilled could also mean literally have no e- in it.
either way, filled orbitals would have repulsion of e- whereas unfilled or half filled do not have that repulsion .:. more stable

 

[The Phlebotomist]

so ok
I. is defo right b/c as you said p^4 means that one p orbital will have two e- repulsing each other.

II. is right
it literally makes no sense for s electrons to "shield" each other from the nuclear change b/c they're in the same shell.
shielding should be associated with an inner shell that has already been filled

the s shell is filled first, and b/c of that, an inner shell has been filled which makes the pull of the nuclear change less on the e- in outer shells thus generating zeff, greater effective nuclear charge.

III. i think this answer is weird b/c it seems to state the answer as a definitive fact.
unfilled orbitals can actually be more stable such as in the case when u have one e- in each orbital where they are all half filled.

thats why we have weird e- configurations for like chromium b/c it likes 4s1^ 3d^5 as opposed to 4s^2 3d^4.

but now that i think of it, unfilled could also mean literally have no e- in it.
either way, filled orbitals would have repulsion of e- whereas unfilled or half filled do not have that repulsion .:. more stable

Thanks