question (1 Viewer)

[toknblackguy]

why are transition metals in high oxidation states strong oxidants?

 

[mercury]

high oxidation state means a high postive charge like Mn7+
eg. 8H+ + MnO4- - +5e- > Mn2+ + 4H2O
they have great attraction for electrons, which means, they can scab electrons very easily as an oxidant.
remember oxidant gets reduced - ie. grabbing electrons.

so for transition metals in high oxidation states, like Cr(VI), Mn(VII), they are bullies in the competition for electrons.

 

[toknblackguy]

i think that's wrong
high oxidation state doesn't mean high charge
the MnO4- has a charge of only -1, but the element Mn has an oxidation state of 7+ not a charge of 7

 

[mercury]

soz, me referring not to overall charge of molecule

as the oxidation state of a metal in a compound increases, that metal's gonna become more positively charged, so its electronegativity increases as its attraction for electrons increases.

Just think of it this way, transition metals with high oxidation states like +7 can't get oxidised any further, the tendency for them is to get reduced as far as possible.

and also notice that as u go from left to very right of transition metal series, high ox states are not observed.

 

[Gruma]

dis wat i got on it...


the max oxidation state equals the total number of 4s and 3d electrons in the atom. The higer oxidation states tend to be when a transition metal is bonderd to a more electronegative atom such as oxygen or with the additional loss of 3d electrons forming simple charged ions. Transition metal ions that have in which the metal has a high oxidation number tend to be strong oxidising agents

half eqns (there pretty messy)

Cr2O7 2- + 14H + + 6e- (orange) = 2Cr 2+ + 7water (green)

MnO4 - + 8H + + 5e - (purple) = Mn 2+ + 4water (colourless)


Finally, the strength of the oxidizing agent depends on the ease of which it will accept electrons and hence be reduced.

 

[toknblackguy]

i thought Cr2+ was blue, and Cr3+ was green...

 

[Gruma]

sorry that 2+ is a typo it is suposed to be 3+

 

[toknblackguy]

cool
and do ppl know why Mn2+ is colourless/
it's got electrons in it's d orbitals, so why no colour?

 

[toknblackguy]

perhaps the energies between it's d orbitals are too small or too large resulting in photons outside the visible spectrum

 

[phenol]

not totally right

Mn 2+ has a stable electronic configuration

the explanation goes into crystal field theory

it has 5 d electrons, i.e. 2 in its e[g] set and 3 in its t[2g] set.
As both sets are half filled and hence very stable (there are quantum mechanical explanations for this).

a t[2g] to e[g] transition in the d orbital will require a photon of very large frequency - most likely outside visible (and infact it is)

 

[toknblackguy]

ok phenol...but why are high oxidation state things powerful oxidants?

 

[phenol]

well you can think of it this way

do you think gold is stable? yes, it is very inert

what if we take 30 electrons off it? (an exaggerated example)

it'll be VERY unstable, it'll want those electrons back

similar for high OS transition metals. the most stable form usually 2+ or 3+ for first row transition metals and if it is in a very high OS then it'll have tendency to get those electrons back

 

[toknblackguy]

ok then
that seems to be the consensus
there's no real techie explanaiton
oh well
hopefully it won't even come up in the exam
thanks

 

[phenol]

you're the first person i know on these boards that actually enjoy techie explanations

that's very nice to know!

hope you are doing science in uni!

 

[sneaky pete]

Originally posted by toknblackguy
ok then
that seems to be the consensus
there's no real techie explanaiton
oh well
hopefully it won't even come up in the exam
thanks

The "instability" of a high oxidation state can be associated with the enormous energy that would be required to form the highly charged cation with the charge defined by the oxidation state. The total energy that would be required to form the true Fe6+ cation in the gas phase, for example, would be 27,407 kJ mol-1 ! To stabilise such a species, "charge return" in some form is necessary and one way to do this is to form a "coordinate" bond, since the donation of a lone pair by the ligand amounts to donating one electron to the Lewis acid (assuming equal "sharing" in the bond, which may, of course be far from the truth). The Lewis base must be a rather good donor, since the charge demand is so high, but it must also be rather resistant to oxidation in order to prevent the metal simply "stripping away" some electrons.

Tech enough for you? lol..
http://www.chem.uwa.edu.au/enrolled_students/Chem100/Solutions/soln6.html
Prob. Set 6, Q1b

 

[phenol]

that still doesnt really explain WHY (though it seems common sense enough)

and the ligand example is somewhat dodgy - what are you trying to show with it?

 

[toknblackguy]

hahahaha
don't worry bout it m8
it's a small dot point which i am welland truly overcomplicating!!
i believe it's a mere "account for.."

don't worry
the explanation before will easily suffice
thanks

 

[sneaky pete]

lol, I don't know
I thought only the first sentence answered his question, but chose to keep the remaining of the (irrelevant) information with it as to increase the techness of the response.