Destroyer- CHEM #107

[keliao]

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more CHEM q. 107
which specie is paramagnetic?
a. Ti +4 has[Ar] configuration
b. Al 3+ has [Ne] config
c. Ne is noble gas
d. F- has [Ne] config
e. Fe +2 3d6 (d subshell holds 10 e-, and the electrons are unpaired therefore is paramagnetic) ANS.

i dont k now why Fe2+ doesnt have Cr configuration instead???on the solution they said Fe2+ has Fe2+

Cr will still be unpaired.

 

[keliao]

some1??????????? please.....

 

[polarmolar]

Yes some1 please explain that and what makes something paramagnetic, thanks!!

 

[Albuterol]

paramagnetic is when not all electrons are paired.
diamagnetic all electrons are paired.

honestly when i did this question i just did POE and now im confused too. my reasoning would be that it would lose e- from the dshell and s-shell to get it 4s1 3d5 bc thats a really stable config and paramagnetic. but im not sure 100%

 

[JBarr29]

ok so this is a tough problem and really tests your knowledge of writing out configurations. First step is to draw the ground state of Fe, clearly shown in the book. Now it has a 2+ configuration right? So when you remove electrons, you remove them from the valence shell. which would be furthest away from the center. ground state has blah blah 4s2 3d6. removing 2 electrons from the valence shell will knock out the 4s2, and just leave you with 3d6. remember the highest n value (4 in this case) will be your valence shell. this is paramagnetic since there are 4 unpaired electrons in the D-shell. hope it helps. post back if you still have questions.

 

[Albuterol]

nice !

 

[keliao]

there's Paramag and Diamagnetic
Paramagnetic means unpaired electrons on the subshells..so lets say
Ni = [Ar] 4s2 3d8
n=3
ml= 5 at d-orbital. because at ml( s=1)(p=3)(d=5) (f=7)
so you draw 5 lines, each line fits 2 electrons. e e e e e after filling 5 of them in order , got 3 more to go. so ee ee ee e e here you see two unpaired therefore is paramagnetic. [i used e instead of up/down arrow cause' i dont have the arrows on my keyboard]

d-orbital holds total of 10 electrons

 

[keliao]

ok so this is a tough problem and really tests your knowledge of writing out configurations. First step is to draw the ground state of Fe, clearly shown in the book. Now it has a 2+ configuration right? So when you remove electrons, you remove them from the valence shell. which would be furthest away from the center. ground state has blah blah 4s2 3d6. removing 2 electrons from the valence shell will knock out the 4s2, and just leave you with 3d6. remember the highest n value (4 in this case) will be your valence shell. this is paramagnetic since there are 4 unpaired electrons in the D-shell. hope it helps. post back if you still have questions.

what i am confused is....on the solution they said Fe2+ is at 3d6...
why not Chromium which is 4s1 3d5. you see the 2+ on Fe?? arent you supposed to subtract two elements back to be on Cr???

 

[JBarr29]

no ml= +/- 2 not 5.
read up on some of the stuff. There are five orbitals available to hold a total of 10 e-s, but it it does not have a mL number of 5.

 

[JBarr29]

i see where you are confused bud. let me try and explain it to thebest of my ability. i could show ya in two seconds if you were here haha. ok so yes i agree with you, if you have Fe+2, you will have the same number of electrons as Cr. But you MUST draw the ground state first. if an element loses electrons, it will lose them from its valence shell, which is the 4s2, which is why the answer is what it is. Cr and Cu are two exceptions with the electron config. if you first draw the ground state for Cr, it is 4s2 3d4. but it is energetically favorable for an electron to fill the last open d shell, so one electron jumps from the 4s2 into the 3d4, resulting in 4s1 3d5. Always draw the ground state first, meaning if no electrons were to be gained or lost, then take into account how many you lose or gain. Fe2+ will lose 2 electrons from its valence shell which is key to understanding this problem. did that help? haha

 

[keliao]

let say if you have Zn 3+ will the config be Cobalt?? should be..rite...

thank you so much.

 

[Albuterol]

Since Zn+3 is impossible lol but Zn+2 should be 3d10 and Zn+ should be 4s1 3d10 ..right?

 

[keliao]

haha...true.......oops..u know what i was trying to imply.
in this case.. Zn2+ would be 3d8 and Zn+ would be the same 3d10

 

[JBarr29]

yeah there ya go. hope i helped a lil. i know it was confusing its just real hard to explain in text vs visual ya know?

 

[Albuterol]

haha...true.......oops..u know what i was trying to imply.
in this case.. Zn2+ would be 3d8 and Zn+ would be the same 3d10

yo man i dont think that is right...

Zn 4s2 3d10
+2 would be 3d10
+1 would be 4s1 3d10

 

[keliao]

yo man i dont think that is right...

Zn 4s2 3d10
+2 would be 3d10
+1 would be 4s1 3d10

think again!. on +2 should be config of Nickel 4s2 3d8
+1 is rght.

 

[Albuterol]

yea but on +2 u are removing from the outer shell which is 4s..if u remove from the 3d shell ur making it unstable again.

http://wiki.answers.com/Q/Electron_configuration_of_Zn_plus_2_ion
i know not the most credible source lol

 

[keliao]

http://wiki.answers.com/Q/Electron_c..._Zn_plus_2_ion whattt!!! is this right?


some1 correct us please.!!

 

[Albuterol]

i see where you are confused bud. let me try and explain it to thebest of my ability. i could show ya in two seconds if you were here haha. ok so yes i agree with you, if you have Fe+2, you will have the same number of electrons as Cr. But you MUST draw the ground state first. if an element loses electrons, it will lose them from its valence shell, which is the 4s2, which is why the answer is what it is. Cr and Cu are two exceptions with the electron config. if you first draw the ground state for Cr, it is 4s2 3d4. but it is energetically favorable for an electron to fill the last open d shell, so one electron jumps from the 4s2 into the 3d4, resulting in 4s1 3d5. Always draw the ground state first, meaning if no electrons were to be gained or lost, then take into account how many you lose or gain. Fe2+ will lose 2 electrons from its valence shell which is key to understanding this problem. did that help? haha

4s2 loses e- first.

 

[pbure9]

yo man i dont think that is right...

Zn 4s2 3d10
+2 would be 3d10
+1 would be 4s1 3d10

that looks right to me... the 4s loses electrons first in transition metals