DonExodus

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Catalysts lower activation energy of the reactants, they do not increase the free energy, correct?

Also, I have a question about atomic size:
"Rank the following atoms and ions in order of increasing atomic size?
S^2- , Cl- , Ar , K+ ?

I thought it should go Ar < Cl- < S2- < K+ ,
but the book had it as S2- < Cl- < Ar < K+

Which is correct? If I was incorrect, why? More e-, less protons = larger radius....

Also, were there any Electrochemical calculations on the test? i.e. E= +2.41V, etc?

In an atom with the e- configuration of 1s2-2s2-2p6-3s2-3p2, how many unpaired e-'s are there? I chose 4 (silicon), but the key said 2, why?

Finally, are heats of fusion and vaporiazation, the specific heat of water, etc included, or do we have to memorize those as well?

Thanks!
-Don
 

dat_student

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DonExodus said:
Catalysts lower activation energy of the reactants, they do not increase the free energy, correct?
correct

DonExodus said:
Also, I have a question about atomic size:
"Rank the following atoms and ions in order of increasing atomic size?
S^2- , Cl- , Ar , K+ ?

I thought it should go Ar < Cl- < S2- < K+ ,
but the book had it as S2- < Cl- < Ar < K+

Which is correct? If I was incorrect, why? More e-, less protons = larger radius....
compare (positive charge)/(negative charge) ratios & number of shells.

DonExodus said:
Also, were there any Electrochemical calculations on the test? i.e. E= +2.41V, etc?

Finally, are heats of fusion and vaporiazation, the specific heat of water, etc included, or do we have to memorize those as well?

Thanks!
-Don
ADA doesn't say whether you need to remember those numbers. I didn't memorize them.
 
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DonExodus

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compare (positive charge)/(negative charge) ratios & number of shells.
The shell number is the same, sans K. What about the ratios? Something negative will have a larger radius, correct?
 
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dental#1

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The ionic radius is estimated from the distance between cations and anions that are adjacent in ion crystals. It is assumed that the internuclear distance between two ions in the ionic structure is equal to the sum of the radius of those ions, though the ionic radius is not actually constant because it depends on the ionic valency, the coordination number, and the state of spin of the ion. Generally, the ionic radius of the negative ion becomes large and that of the positive ion becomes small compared with the former neutral atom. Ionic radius also refers to the size of an atom after losing all valence electrons (the outer electron shell).

0.37 Å 37 pm Sulfur S 16
1.81 Å 181 pm Chlorine Cl 17
1.38 Å 138 pm Potassium K 19

In the periodic table, atomic radii increase down a group as new electron shells are added, and decrease left-to-right as the nuclear charge (or number of protons) is increased - an important exception are the noble gases. They do not form bonds.

Catalyst have no effect on gibbs free energy.
 

dat_student

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DonExodus said:
The shell number is the same, sans K. What about the ratios? Something negative will have a larger radius, correct?
when shell number is the same >> higher +/- (i.e. proton/electron) ratio = smaller
 

dental#1

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Thats right the larger the negative charge on the ion the larger the radius the more positive the ion the smaller the radius compared with the former neutral atom . As far as the unfilled shells the answer is two. You have two electrons in two different orbitals with no buddy to hang out with.
 
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DonExodus

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So then I was right in my thinking? S2- IS larger than Cl- ?

Also,
Also, were there any Electrochemical calculations on the test? i.e. E= +2.41V, etc?

In an atom with the e- configuration of 1s2-2s2-2p6-3s2-3p2, how many unpaired e-'s are there? I chose 4 (silicon), but the key said 2, why?
Thoughts?

Thanks for all your help.
 

dental#1

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Yes S^2-< Cl-< Ar< K+........ Largest,2nd largest, 2nd smallest and smallest. They are in the same period and both have negative charges so the larger negative charge prevails and the K+ ends up in the same period.
 

dat_student

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dental#1 said:
Yes S^2-< Cl-< Ar< K+........ Largest,2nd largest, 2nd smallest and smallest. They are in the same period and both have negative charges so the larger negative charge prevails and the K+ ends up in the same period.
DonExodus said:
...
Also, I have a question about atomic size:
"Rank the following atoms and ions in order of increasing atomic size?
S^2- , Cl- , Ar , K+ ?

I thought it should go Ar < Cl- < S2- < K+ ,
but the book had it as S2- < Cl- < Ar < K+
...
arrows are pointing in the wrong direction ;)

S^2- (largest) > Cl-> Ar > K+ (smallest)
 

slayerdeus

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I'm confused. Can someone explain ionic radius in relation to the question that was originally asked, ie giving an example? I don't know what you guys are talking about when you say ratio to positive and negative. Thanks.
 

dental#1

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dat_student said:
arrows are pointing in the wrong direction ;)

S^2- (largest) > Cl-> Ar > K+ (smallest)

Oops!!!!!!!!!!!!!!!! I am working QR at the same time :scared:
 
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DonExodus

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I compensated lol, it said rank them in increasing size, but had the choices with the > the wrong way... i.e.
small > medium > large lol
 

slayerdeus

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Okay for anyone that was confused like I was. The answer shoudl have been, when ordering the ions in increasing ionic radius:

(smallest) K+ < AR < Cl- < S2- (largest)
 

jaybo

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In answer to your electron question -

Electrons will always find their own orbital inside of the shell before they start pairing up. In your case, all shells are filled except for the very last p shell. A p shell can hold 6 electrons, and therefore must have only 3 orbitals in the shell. Since electrons will decide to get their own orbital if there is one vacant, the first 3 p electrons will always be unpaired (because there are 3 orbitals possible). Also, notice that only 3 p electrons can possibly be unpaired. Since you have 2 p electrons, they both will get their own orbital and remain unpaired. Because all other shells are filled, those are the only electrons in question, making 2 the answer.

Hope this helps! Let me know if you need further clarification!
 
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DonExodus

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jaybo said:
In answer to your electron question -

Electrons will always find their own orbital inside of the shell before they start pairing up. In your case, all shells are filled except for the very last p shell. A p shell can hold 6 electrons, and therefore must have only 3 orbitals in the shell. Since electrons will decide to get their own orbital if there is one vacant, the first 3 p electrons will always be unpaired (because there are 3 orbitals possible). Also, notice that only 3 p electrons can possibly be unpaired. Since you have 2 p electrons, they both will get their own orbital and remain unpaired. Because all other shells are filled, those are the only electrons in question, making 2 the answer.

Hope this helps! Let me know if you need further clarification!
Thanks a lot, that does help- so if it was 3p5, then there would be 1 unpaired electron, correct?

Were there any Electrochemical calculations on the test? i.e. E= +2.41V?
 

gwest

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DonExodus said:
Thanks a lot, that does help- so if it was 3p5, then there would be 1 unpaired electron, correct?

Were there any Electrochemical calculations on the test? i.e. E= +2.41V?
Right. Since there r 3 orbitals in the p subshell, 2 of the orbitals would b paired while the other orbital would b unpaired.

Probably just b familiar with the formula deltaG= -nEF; n being the number of moles of e-'s transferred & F being Faraday's constant (96500 C/mole e-)
and that the E is always positive in a Galvanic cell, while being negative in an electrolytic cell.
 
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