What are the oxidation numbers in [Re2Cl9]2- and how can it exist?

  • Thread starter kankerfist
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    Oxidation
In summary: For example, in the molecule H2S, the atoms have a number of protons in addition to the number of electrons. But even though it has a number of protons, it isn't a gas because the number of electrons (in the form of a negative electron) is too high.
  • #1
kankerfist
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Sorry about no latex, when I try using latex my formulas just wind up as a link
that says "LaTeX graphic is being generated. Reload this page in a moment."


Anyway, I was trying to balance a redox equation and I am completely stumped
as to how:

[Re2Cl9]^2-

can be possible. I'm aware that some arrangements can cause the same atoms
in an ion to have different oxidation numbers, but I can't seem to figure out how
that molecule can exist under any arrangement. Any hints would be appreciated
 
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  • #2
kankerfist said:
Sorry about no latex, when I try using latex my formulas just wind up as a link
that says "LaTeX graphic is being generated. Reload this page in a moment."


Anyway, I was trying to balance a redox equation and I am completely stumped
as to how:

[Re2Cl9]^2-

can be possible. I'm aware that some arrangements can cause the same atoms
in an ion to have different oxidation numbers, but I can't seem to figure out how
that molecule can exist under any arrangement. Any hints would be appreciated

To help with latex, I've seen that message when i try to preview a post but if i actually make a post then it will show up. Sometimes I make an error in my latex and i can just edit it real fast. Just try to post some latex and not preview see if that works.
 
  • #3
unless you get this error then you really did something wrong
[tex] \LATEX [/tex]
 
  • #4
Ok, I'll test it by quoting someone that used it sucessfully.

[tex]^{33}_{17}Cl^{16}\xrightarrow{n,n}~^{31}_{15}P^{16 }+~^4_2He^2[/tex]
 
  • #5
There You Go. Now see if you can edit your first post with some good code!
Im looking forward to your work!

Edit: You the man
 
Last edited:
  • #6
[tex]{ReCl}_5} + {{H_2}O}[/tex]
-->
[tex]{Re_2}{Cl_9}^{2-} + {ReO_4^-} + {Cl^-} + {H^+}[/tex]
 
Last edited:
  • #7
kankerfist said:
Anyway, I was trying to balance a redox equation and I am completely stumped
as to how:

[Re2Cl9]^2-

can be possible. I'm aware that some arrangements can cause the same atoms
in an ion to have different oxidation numbers, but I can't seem to figure out how
that molecule can exist under any arrangement. Any hints would be appreciated

It's possible for atoms to have a charge that isn't a whole number, or the number can be very different from what is expected. For example, what is carbon monoxide, CO? Does the oxygen have 4 bonds? Does the carbon have 2 bonds? Do they compromise to make 3 bonds each? (edit: I looked it up and yes they make 3 bonds each).
The number of bonds between atoms or groups of atoms is called the Bond Order.
From that link "A good example of this is bonds between carbon in the molecule benzene where the delocalized molecular orbitals contain 6 pi electrons over six carbons essentially yielding half a pi bond. Together with the sigma bond the bond order is 1.5"

Is it possible that Re has a bond order of 3.5? Sure, why not? :smile:
 
Last edited:
  • #8
kankerfist said:
[tex]{ReCl}_5} + {{H_2}O}[/tex]
-->
[tex]{Re_2}{Cl_9}^{2-} + {ReO_4^-} + {Cl^-} + {H^+}[/tex]

Ok, finally figured it out above. I am trying to balance this equation, and in order to break it into half reactions, I need to know the ox #'s of each atom to see which is the reduction half and which is the ox half. If the Re is 3.5 like you say, then am I on the right track that the oxidation half reaction is:
2 ReCl5 + 3e- --> [tex]{Re_2}{Cl_9}^{2-}[/tex]

edit: sorry, I mean reduction half
 
  • #9
The best approach is to assume that oxidation numbers are just an accounting device and they don't refer to any real chemical property.
 

Related to What are the oxidation numbers in [Re2Cl9]2- and how can it exist?

1. What is the oxidation number of the metal in [Re2Cl9]2-?

The oxidation number of the metal in [Re2Cl9]2- is +4.

2. How do you determine the oxidation number in [Re2Cl9]2-?

To determine the oxidation number in [Re2Cl9]2-, you must first assign a variable to the unknown oxidation number (x). Then, use the formula: (oxidation number of metal) x 2 + (oxidation number of chlorine) x 9 = charge of the ion (-2). This will give you the equation: 2x + 9(-1) = -2. Solving for x, you will get an oxidation number of +4 for the metal.

3. Why is the oxidation number of the metal in [Re2Cl9]2- +4?

The oxidation number of the metal in [Re2Cl9]2- is +4 because the overall charge of the ion is -2, and there are two metal atoms present. Therefore, each metal atom must have an oxidation number of +2, and since there are two of them, the total oxidation number for the metal is +4.

4. Is the oxidation number of chlorine in [Re2Cl9]2- the same as in HCl?

No, the oxidation number of chlorine in [Re2Cl9]2- is different from that in HCl. In [Re2Cl9]2-, each chlorine atom has an oxidation number of -1, while in HCl, the oxidation number of chlorine is -1 for one chlorine atom and 0 for the other.

5. Can the oxidation number of the metal in [Re2Cl9]2- vary?

No, the oxidation number of the metal in [Re2Cl9]2- cannot vary. It is always +4 because the compound has a fixed chemical formula and the oxidation numbers of each element must balance out to give a total charge of -2 for the ion.

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