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Topic: Oxidation states  (Read 7113 times)

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Offline Schrödinger

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Oxidation states
« on: September 25, 2009, 12:52:33 AM »
I have a basic doubt regarding oxidation states.

As far as i know, oxidation states are only for polar covalent bonds(even if weakly polar). And i also know how to calculate them.(i think )

Consider the MnO4- ion. When we say that the oxidation state of Mn = 7, what does it actually mean?

Like... does it mean that Mn is in its +7 ionized state? i.e, have 7 electrons been removed due to the 4 oxygen atoms?
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Offline AWK

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Re: Oxidation states
« Reply #1 on: September 25, 2009, 02:17:10 AM »
Oxidation state of Mn in manganate(VII) [ MnO4- ] is +7
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Offline Borek

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Re: Oxidation states
« Reply #2 on: September 25, 2009, 02:49:23 AM »
When we say that the oxidation state of Mn = 7, what does it actually mean?

Nothing.

There is no measurable physical property that can be attributed to oxidation state (or oxidation number) of an atom (in other words: oxidation state can be not determined experimentally). Oxidation state is just a way of counting electrons.
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Offline Schrödinger

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Re: Oxidation states
« Reply #3 on: September 26, 2009, 05:09:23 AM »
Thanks Borek.

Now...let me use this thread do get some of my other doubts clarified as well...

How do you compare the relative strengths of ionic, covalent and polar covalent bonds?
Which one is usually stronger ? i.e, which one's breaking usually involves a greater energy?
 
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Offline Borek

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Re: Oxidation states
« Reply #4 on: September 26, 2009, 05:12:11 AM »
That should be explained in every textbook.
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Offline Schrödinger

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Re: Oxidation states
« Reply #5 on: September 26, 2009, 05:22:01 AM »
every textbook does explain it. But i came across certain questions that did not agree with the theory.

So, i thought i'd ask you guys on the forum. So, the exceptions can be overlooked right? That is, i don't have to bother about them?
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Offline Borek

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Re: Oxidation states
« Reply #6 on: September 26, 2009, 05:59:24 AM »
Give examples of what you mean.
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Offline renge ishyo

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Re: Oxidation states
« Reply #7 on: September 26, 2009, 11:49:07 AM »
The bond strength outside of water is:

ionic > polar covalent > covalent

The reason for this is that the charge separation adds an additional electrostatic attraction between the two atoms that helps hold them together. This confuses a lot of students because most of the chemistry you study takes place in water where the apparent (i.e. not correct) strength of bonds is:

covalent > polar covalent > ionic

Actually, the bond strength in an of itself is the same as it was outside of water. What is different now is that water is able to solvate ionic compounds easier than covalent ones. By surrounding the ions, the water pulls the ions apart a greater distance...weakening both the original bond and the electrostatic attraction and making it easy to break ionic bonds in water. In contrast non-polar covalent compounds are not solvated in water efficiently so those bonds remain just as hard to break in water as they are outside of it.

I have only seen one textbook adequately explain this, and that would be Linus Pauling's "The Nature of the Chemical Bond."

Offline Schrödinger

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Re: Oxidation states
« Reply #8 on: September 29, 2009, 11:49:32 AM »
@ Borek : Let us say we have two molecules (crystals) with us. Say C1 and C2. Suppose from Fajan's rules, I find that the covalent character of C2>C1, can I jump to the conclusion that bond strength in C1>C2, as ionic bonds are usually stronger ??
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Offline MrTeo

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Re: Oxidation states
« Reply #9 on: September 30, 2009, 02:23:16 PM »
I'll try to answer you ;)
As renge ishyo correctly pointed out there's no real strength scale for the chemical bonds: it depends on the medium you're working with (e.g. water or unpolar substances) and also (obviously) from the bonds formed between the atoms.
There is no real division between ionic and covalent bonds: only homopolar covalent bonds (between two atoms of the same element, such as H2 or Cl2) have an ionic percentage of 0% but all the other bonds (e.g. C-H ones, even if in this case the ionic effect is really tiny) have got both a covalent and an ionic part of the bond working together to make the link stronger. In your case the ionic bond will be (maybe, as we don't know the real molecule so all we can say are hypotheses) stronger in air but probably weaker in water because of the polar effect of the solvent (think at the dissolution of salts).

Hope it's all clear now  ;D
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