# Redox reaction

1. Nov 3, 2004

### dibilo

redox reaction << need help badly

hi all. i need help on a few problems regarding redox.

1) what is the oxidation state of Cr in Cr(CO)6
in this case i only know that the oxidation state of oxygen is 2-. i need to know carbon's oxidation state to solve..

2) oxidation state of Ag in AgNO3.. same problem as above

3)K2C2O4+KMnO4+H2SO4 >>>> K2SO4+MnSO4+H2O+CO2
from here, i know that carbon is oxidised
C3+ >>> C4+
manganese is reduced
Mn(VII)7+ >>> Mn(II)2+

then where do i go from here.. pls help thx

Last edited: Nov 3, 2004
2. Nov 3, 2004

### chem_tr

1. Carbonyl in CO is neutral, so if the complex is also neutral, you can conclude something.
2. Nitrogen in NO3- is 5+, as oxygen is almost always 2- (except the compound $OF_2$ in which oxygen is forced to take 2+ oxidation state.
3. In oxalate titration with permanganate, you can (and must) calculate the oxidation states, and you've done this correctly. Then use half redox reactions to find how many molecules must undergo redox (i.e., one electron is given away from carbon, but five electrons must go into manganese, so if an equilibrium with no electrons free outside is set, you must balance electron counts.

3. Nov 3, 2004

### dibilo

"Carbonyl in CO is neutral, so if the complex is also neutral, you can conclude something."

ermm don't understand...what does it mean by carbonyl is neutral? my guess is C is contributing 1- in this case but i don't know why

4. Nov 3, 2004

### chem_tr

Yes, carbonyl is neutral; even if it seems confusing, there is a negative sign on carbon while a positive sign on oxygen; this makes the molecule neutral. The carbon has three bonds with oxygen, and one non bonding electron pair.

Carbonyl complexes are extremely stable and have extraordinary valences. There are even 1- charged complexes. We explain this phenomena with "back-donation"; and its extreme result is that a double-bond-like bonding is formed between carbon and metal. I don't want to go into more detail; give some feedback and we'll discuss it.

5. Nov 4, 2004

### garytse86

yeah i am interested actually - would u want to explain that? Thanks

6. Nov 4, 2004

### chem_tr

If you are not at least undergraduate students, you may encounter some difficulty in understanding what I'll write here. The reason why carbonyl is so reactive but neutral can be explained with molecular orbital theory. In this theory, atomic orbitals are thought to change upon building the molecular orbital. CO molecule is composed of two atomic orbitals, one of them is C, and the other O. When you write the electronic configurations of these two atoms, the 1s orbitals may be omitted since their contribution to overall molecular orbital bonding is very limited. We can consider the contributions of 2s and 2p orbitals for CO, and this is enough. As oxygen is more electronegative than carbon, oxygen has a lower energy and its 2s-2p energy gap is higher than that of carbon.

There is hybridizations of 2p and 2s orbitals, and two hybrid orbitals with different s (and p) percentages; h1 and h2 hybrid orbitals denote approximately 70% and 30% of s orbital density, respectively (note that the percentages are only for understanding purposes). As p orbitals are triplicate, two of them will remain intact; these will be used for $\pi$ bonding.

The unsymmetrical nature of these two atomic orbitals makes the h1 hybrid orbital of oxygen remains unbonded; this is the non-bonding orbital of oxygen. h1 of carbon and h2 of oxygen are of similar energy, so they combine to form a sigma bond. h2 of carbon cannot find a suitable atomic orbital to bond; so this is the second non-bonding orbital.

Note that bonding of orbitals is a energy-giving process, so the sigma bond has the lowest energy among all orbitals. Then comes n orbitals of oxygen.

$\pi$ bonds are formed with involvements of two p orbitals of each atom; here, as a new molecular bond is formed, its energy must be lower than the two p orbitals. The others form an anti-bonding counterpart (as in the first example with sigma bond; two s orbitals give $\sigma$, while two p orbitals give $\pi$ molecular bonds with their anti bonding orbitals, $\sigma^*$ and $\pi^*$, respectively).

As the sigma bond has the lowest energy potential, its counterpart, sigma star orbital has to be the highest (there are some exceptions, however, pi star orbital is also sometimes the highest one). When we begin to put the electrons on hand from bottom to top, we see that the highest occupied molecular orbital (HOMO; the most available molecular orbital for reactions) is the non-bonding orbital of carbon; that is why the molecule prefers to bind from carbon; and there is a three-bond-profile in CO molecule.

I will try to find the molecular orbital scheme for CO and include this thread.

Please look at this link for an asymmetric MO diagram based on those information I gave above.

You may want to view the following web pages:

Last edited by a moderator: Apr 21, 2017 at 9:57 AM