Chemistry HL/SL help

[dessskris]

Ok so there is this reaction:

KSCN + Fe(NO₃)₃ <--> K⁺ + 3NO₃^- + FeSCN²⁺

FeSCN²⁺ is a red complex ion. This reaction has reached an equilibrium. Then I added NaF, the solution turned colourless and apparently the complex ion has now totally dissociated.

Q: Fluoride ions do not react directly with the complex ions but they do react with Fe³⁺ ions. What does the effect of fluoride ions show about the reaction of Fe³⁺ ions and thiocyanate ions (SCN^-) to form the complex?

I am an SL student and I don't know much about complex ions. What does that question mean? Is there such thing like weak complex ion or something like that?

EDIT:

Does it make sense if I say Fe³⁺ ions are central ions, then S^2-, C⁴⁺ and N^3- are the ligands?

[master135]

Carbon and Nitrogen will not exist as separate ions, and are definitely not ligands separately. Cyanide (CN) or SCN are ligands on the other hand - and ligands tend to form complexes with a central metal cation (such as Fe3+).

[wombat123]

I don't understand how it is said that the only thing that can shift equilibrium is temperature. Doesn't a change in concentration also change the equilibrium??

[genepeer]

I don't understand how it is said that the only thing that can shift equilibrium is temperature. Doesn't a change in concentration also change the equilibrium??

Concentration can change the position of equilibrium by shifting it to the right or left of the equation. I believe they say that the only thing that can change the equilibrium constant is temperature.

I tried typing it here but it looks ugly.

edit: When I preview my post the image shows fine, but after I submit it I can't see anything and I get an error when I try to download it. What's wrong? Can you see the image?

Edited January 14, 2011 by Gene-Peer

[dessskris]

@Gene Peer, I can see the image, don't worry

A simple question guys, about redox. Yet I can't manage to get it.

Use the half equation method to write a balanced equation for the following reaction:

Acidified permanganate(VII) ions oxidising methanol to carbon dioxide.

The answer is:

6 MnO₄^- + 5 CH₃OH + 18 H⁺ --> 6 Mn²⁺ + 5 CO₂ + 19 H₂O

But I want the half equations. I cannot understand the above equation. Can anybody help me get the half equations? Thank you.

[Drake Glau]

MnO₄^- + CH₃OH --> Mn²⁺ + CO₂ is where you start.

Half reactions are like taking it apart, the parts that look like they go together. 2 halves make a whole

MnO₄^- --> Mn²⁺

First thing is balancing the O by adding water from the surrounding aqueous solution. (actually the FIRST thing is balancing anything but O and H by doing the normal equation balancing with coefficients, O and H are special )

MnO₄^- --> Mn²⁺ + 4H₂O

Now you need to balance the hydrogen from the water by adding H⁺

8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

Now your charges don't balance, so we need to do that. By adding some electrons. You have +7 on the left and +2 on the right, so add 5 electrons to the left.

5e^- + 8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

Ok, that one is done now.

Your other equation would be the methanol and CO₂

CH₃OH --> CO₂

Your oxygen and hydrogens are now unbalanced, and you balance oxygen with waters (they come from the aqueous solution that the reaction is happening in)

H₂O + CH₃OH --> CO₂

Now for hydrogens, 6 on the left, none on the right, so the right side gets 6 H ions.

H₂O + CH₃OH --> CO₂ + 6H⁺

Charges are now off, 0 on the left and +6 on the right so the right side gets 6 electrons

H₂O + CH₃OH --> CO₂ + 6H⁺ + 6e^-

Now we have both half reactions done, we get to combine them. When you combine the reactions the electrons MUST cancel out (this is much like how you do hess's law by the way).

H₂O + CH₃OH --> CO₂ + 6H⁺ + 6e^-

5e^- + 8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

One has 6 electrons, one has 5 electrons, least common multiple of 30 (sadly ) so your methanol reaction needs to be multiplied by 5 and your permanganate equation gets to be multiplied by 6

5H₂O + 5CH₃OH --> 5CO₂ + 30H⁺ + 30e^-

30e^- + 4818H⁺ + 6MnO₄^- --> 6Mn²⁺ + 2419H₂O

Now we start canceling stuff out just like hess's law. The electrons are 30 and 30, they're gone. Next we'll do water, 5 and 24, 24-5=19 waters on the right hand side. 48 hydrogen ions on the left, 30 on the right, so 18 are left on the left side.

So what we have left is. 6MnO₄^- + 5CH₃OH + 18H⁺ --> 5CO₂ + 6Mn²⁺ + 19H₂O

And wow the amount of bbc code in this is amazing, I'm hoping this works...

You can make steps for doing this if you have troubles on where to start, it's like a 7 step process for both reactions, just do a couple more and you'll quickly pick up on it

Edited January 20, 2011 by Walter Drake

[Meera.m]

Hey chem students,

I was wondering if any of you can tell me the concept. I'm not understanding it. I am not looking for a vivid explanation, because if I knew the concept I'll be able to figure out the rest.

Thank You

[timtamboy63]

Basically, as the two atoms come together for covalent bonding, the orbitals themselves 'merge' to form hybrid orbitals which can each only make a sigma bond. Let me know if you need a more detailed explanation.

[sid1729]

Well just as timtamboy pointed out... the orbital merge together to form hybridized orbits.. which basically are orbits of the same enegry level.

So when Carbon 1s^2 2s^2 2 p^2 ... ( which has 4 electrons in the outermost shell) reacts... it form four hybrid orbitals where , out hear it will be sp3 orbital! where all four electron have equal energy levels. If not for hybridizations the s and p orbital would have different energy level.

[Meera.m]

Basically, as the two atoms come together for covalent bonding, the orbitals themselves 'merge' to form hybrid orbitals which can each only make a sigma bond. Let me know if you need a more detailed explanation.

It is really complicated. I just have few questions to ask.

When A sigma bond is formed, how and when does the the 'pi' bond interfere?

Do the orbitals have to be identical, like for instance 1s of one atom and 1s of another?

What does Hybridisation result in?

Thanks for the offer of a more detailed explanation, I really appreciate it.

[Sandwich]

I found it easiest to think of as that the atom wants to find the most convenient position for all its electrons in the orbitals.

There are little rules you can apply -- like for instance that it's better to have an electron in each little orbital 'box' than it is for instance to have 2 (oppositely spinning, remember) electrons in 1 little orbital 'box' and 2 empty boxes (in a p orbital this is, where you have 3 boxes). They like for each box to try and contain a similar number to the other boxes on its level, and to that end they hybridise between layers

So from the example Sid gave:

Carbon 1s^2 2s^2 2 p^2

With < being an arrow indicating the presence of an electron and the direction of its spin...

That would look like

[<>] (the 1s^2 orbital)

[<>] (the 2s^2 orbital)

[<] [<] [ ] (the p^2 orbital)

The atom prefers to have an orbital full, either full of single direction spin electrons [<] or full of full of both sets [<>] but either of those beats [ ] !

So it hybridises to give an sp3 orbital:

[ < ] [ < ] [< ] [< ]

Which is basically the 2s^2 orbital merged with the p^2 orbital and the electrons redistributed between them to give a full orbital rather than a partly empty one, as you previously had with p^2 where only 2 of its boxes were full and one was blank. And of course it'd still have the full 1s^2 orbital because nothing's happened to that one.

Once you get the hang of it it's actually really straight forward!

An sp2 orbital would be that 2 of the p orbital boxes (rather than all 3) merge with an s, and sp is where just 1 of the p orbital boxes does it.

http://home.arcor.de/ctornau/hybridorbitale.png <-- that might be helpful in demonstrating it visually a little better!

[Meera.m]

I found it easiest to think of as that the atom wants to find the most convenient position for all its electrons in the orbitals.

There are little rules you can apply -- like for instance that it's better to have an electron in each little orbital 'box' than it is for instance to have 2 (oppositely spinning, remember) electrons in 1 little orbital 'box' and 2 empty boxes (in a p orbital this is, where you have 3 boxes). They like for each box to try and contain a similar number to the other boxes on its level, and to that end they hybridise between layers

So from the example Sid gave:

Carbon 1s^2 2s^2 2 p^2

With < being an arrow indicating the presence of an electron and the direction of its spin...

That would look like

[<>] (the 1s^2 orbital)

[<>] (the 2s^2 orbital)

[<] [<] [ ] (the p^2 orbital)

The atom prefers to have an orbital full, either full of single direction spin electrons [<] or full of full of both sets [<>] but either of those beats [ ] !

So it hybridises to give an sp3 orbital:

[ < ] [ < ] [< ] [< ]

Which is basically the 2s^2 orbital merged with the p^2 orbital and the electrons redistributed between them to give a full orbital rather than a partly empty one, as you previously had with p^2 where only 2 of its boxes were full and one was blank. And of course it'd still have the full 1s^2 orbital because nothing's happened to that one.

Once you get the hang of it it's actually really straight forward!

An sp2 orbital would be that 2 of the p orbital boxes (rather than all 3) merge with an s, and sp is where just 1 of the p orbital boxes does it.

http://home.arcor.de/ctornau/hybridorbitale.png <-- that might be helpful in demonstrating it visually a little better!

OMG!! You're amazing ! Thank you so much !

I understood it well.

Thank you for your time..

Thank You

[Hinuku]

Hi,

I'm having trouble with understanding hybridization - could anyone somehow explain it?

Thanks a lot,

Matt

[Guest IBkidd]

Dude I've got a test tomorrow about it

anyway, just know that it basically means mixing orbitals of an atom together.

anyway, you have to first count the bonds of a molecule and then you should also check whether it has got any lone pairs (because they count as a bond..)

(l.p.'s). umm... and= just depending on how many bonds are present in the element, there are 3 types of hybridization.

if there are 3 then its sp2 hybridized. if its 4 then its sp4 hybridized.

if there are 2 then its sp hybridized. basically, just one less.

just ignore, triple double bonds---- they all count as one bond.

i know its confusing, but im just in my first half year if ib............thats my excuse.

hope i helped

Hi,

I'm having trouble with understanding hybridization - could anyone somehow explain it?

Thanks a lot,

Matt

[Hinuku]

Thanks ^^

Still, sp4 is definetely my favorite hahah

Ive got the exact same excuse

Edited January 25, 2011 by Hinuku

[dessskris]

KIO₃ + KI + H₂SO₄ --> K₂SO₄ + H₂O + I₂

All the reactants are in aqueous form and all the products are aqueous too I guess except for water which is a liquid. However the product gave me dark brown ppt. Is that from Iodine?

[ILM]

Yes, i think it is iodine. Because iodine is solid and it do not dissolve in water. And also it has a brown color. But i want others to ensure that.

[Drake Glau]

KIO₃ + KI + H₂SO₄ --> K₂SO₄ + H₂O + I₂

All the reactants are in aqueous form and all the products are aqueous too I guess except for water which is a liquid. However the product gave me dark brown ppt. Is that from Iodine?

First off, not balanced

And it is likely the iodine, k2so4 is white when solid (according to wiki's pictures...)

[dessskris]

Oh ya forgot to balance it

KIO₃ (aq) + 5 KI (aq) + 3 H₂SO₄ (aq) --> 3 K₂SO₄ (aq) + 3 H₂O (l) + 3 I₂ (aq)

That is the equation I got through googling..but then if I₂ is aqueous, did I really see brown ppt?? As inm said, iodine is insoluble in water. If that is true, does that mean that what I saw was not precipitate? Because I saw some substance that did not dissolve in water so I thought it was ppt, but my friend just told me she didn't see any ppt :/

So if it's not ppt, what do you call it??

Also, I saw black powder after I titrated my 3 K₂SO₄ (aq) + 3 H₂O (l) + 3 I₂ (aq) with Na₂S₂O₃.5H₂O. What would that black powder be? Because at the end of the reaction the black powder was gone.

[Drake Glau]

I'm running out of ideas on this rather quickly...

The titration is a whole new reaction using the product from the original reaction and this new thing as th reactants so I don't know what the powder could have been but it could very well have nothing to do with your black ppt from before. There's always the possibility of chemical contamination. We do a lab in SL chem where we do some stuff and make a yellow ppt but half the time we end up with some orange color instead of yellow due to the chemicals being somewhat old and every year some idiot always puts stuff back in the bottles after taking the stuff out which adds stuff you don't want to your reaction =/