Oxidation States

[King E]

How is the oxidation state in covalent compounds calculated ?  For example, in  H₂S, why does hydrogen have an oxidation state of +1 and sulphur has oxidation state of -2 as the electrons are being shared ? Can hydrogen have oxidation state of -1 and sulphur +2 (because the electrons are shared) ?

Edited October 6, 2020 by King E

[OldChemE]

Oxidation states relate to the electron structure of the given atom, generally taking into account the octet rule.  The Octet rule basically says that atoms "prefer" to have 8 electrons in their outer shells (there are exceptions and it is not very rigorous-- but it works).    The outer shells of Sulfur are 3s2 3p4, for a total of 6 electrons.  By 'sharing' 2 more electrons (from the hydrogen atoms) the atom fulfills the octet rule, but that gives it an electric charge of -2 (2 extra electrons).  As these two electrons are from the two Hydrogen atoms, they are now are each +1.  The extent to which the electrons are only shared, as opposed to one atom taking and varies.  There is a lot more technical detail to that.  You should note that the arrangement of the Periodic Table is ties into this, as elements that usually have the same oxidation states are all in the same column.

[Asheekay]

Oxidation states have a lot to do with the electronegativity value of an element, regardless of whether the compound is covalent or ionic.

For example, in your case of $H_2S$, hydrogen has an electronegativity of 2.20 while sulphur has a value of 2.58, being slightly more electronegative than hydrogen.

Hence in hydrogen sulphide the shared electron pair spends slightly more time with sulphur than they do with hydrogen. That is why hydrogen gets a partial negative charge while sulphur gets a partial positive charge.

Edited October 7, 2020 by Asheekay

[King E]

7 hours ago, Asheekay said:

Oxidation states have a lot to do with the electronegativity value of an element, regardless of whether the compound is covalent or ionic.

For example, in your case of $H_2S$, hydrogen has an electronegativity of 2.20 while sulphur has a value of 2.58, being slightly more electronegative than hydrogen.

Hence in hydrogen sulphide the shared electron pair spends slightly more time with sulphur than they do with hydrogen. That is why hydrogen gets a partial negative charge while sulphur gets a partial positive charge.

What about in case of O₂ ? Or H₂ ? The electronegativity is same. Which atom will get partial negative charge and which will get partial positive charge in the above mentioned cases?

15 hours ago, OldChemE said:

Oxidation states relate to the electron structure of the given atom, generally taking into account the octet rule.  The Octet rule basically says that atoms "prefer" to have 8 electrons in their outer shells (there are exceptions and it is not very rigorous-- but it works).    The outer shells of Sulfur are 3s2 3p4, for a total of 6 electrons.  By 'sharing' 2 more electrons (from the hydrogen atoms) the atom fulfills the octet rule, but that gives it an electric charge of -2 (2 extra electrons).  As these two electrons are from the two Hydrogen atoms, they are now are each +1.  The extent to which the electrons are only shared, as opposed to one atom taking and varies.  There is a lot more technical detail to that.  You should note that the arrangement of the Periodic Table is ties into this, as elements that usually have the same oxidation states are all in the same column.

Each hydrogen atom in H2S , no doubt gives 1 electron to sulphur but also takes 1 electron from sulphur. Similarly the sulphur atom takes 1 electron from each hydrogen atom but also gives 1 electron to each hydrogen atom. So why isn't the oxidation state of each hydrogen atom -1 and that of sulphur +2 in H2S ?

[studiot]

On 10/6/2020 at 2:31 PM, King E said:

How is the oxidation state in covalent compounds calculated ?  For example, in  H₂S, why does hydrogen have an oxidation state of +1 and sulphur has oxidation state of -2 as the electrons are being shared ? Can hydrogen have oxidation state of -1 and sulphur +2 (because the electrons are shared) ?

The following summary of rules and examples may help you here.

Hydrogen is a good atom to follow because it has only one electron.

Thus it can only ever loose 1 and reach the +1 oxidation number.

Note that the text shares the electrons equally in the case of two bonded atoms being identical.

Thus two hydrogen atoms in a hydrogen molecule have one each of the bonding pair.
Therefore neither atom has lost or gained an electron so they both start of with an oxidation number of zero for an unbonded atom and continue at zero when bonded into a hydrogen molecule.

Note this example was not in the text, but the rules I have used are.

 

[Asheekay]

In the case of homogeneous biatomic molecules, H2, O2 etc, none of the atoms gets any oxidation state. Oxidation states are reached when there is at least some polarity in the compound. Perfectly nonpolar compounds will not have any oxidized atoms at all.

[studiot]

1 hour ago, Asheekay said:

In the case of homogeneous biatomic molecules, H2, O2 etc, none of the atoms gets any oxidation state. Oxidation states are reached when there is at least some polarity in the compound. Perfectly nonpolar compounds will not have any oxidized atoms at all.

What do you understand an oxidation state of zero to mean ?

 

[Asheekay]

14 hours ago, studiot said:

What do you understand an oxidation state of zero to mean ?

 

That there is no partiality at all in the sharing of electrons. In a perfectly nonpolar covalent compound, the oxidation state of all the atoms should be zero.

Atoms or groups of atoms get oxidized or reduced when there is at least some disparity in the destination of electrons which are exchanged/shared during the chemical reaction.

For example in the case of NaCl, the oxidation states are highly conspicuous, due to the nature of the compound (ionic). Na gets a +1 while Cl gets a -1. In the case of covalent compounds, the oxidation states become relative (as compared to absolute, for ionic compounds). Take the case of water, which is polar covalent. Each H gets a +1 (less electronegative) and the O gets a -2 (more electronegative).

However, this simple logic becomes a nightmare with organic compounds. Take methane $$CH_4$$ for example. While carbon is more electronegative than hydrogen (2.55 vs 2.20), it is written before hydrogen. So theoretically carbon should have oxidation state of -4 while the hydrogens have +1. But for reasons unknown and probably lame, carbon is written at the cation side (positive oxidation state) while hydrogen is written at the anion side (negative oxidation state).

[studiot]

44 minutes ago, Asheekay said:

However, this simple logic becomes a nightmare with organic compounds. Take methane

CH4

for example. While carbon is more electronegative than hydrogen (2.55 vs 2.20), it is written before hydrogen. So theoretically carbon should have oxidation state of -4 while the hydrogens have +1. But for reasons unknown and probably lame, carbon is written at the cation side (positive oxidation state) while hydrogen is written at the anion side (negative oxidation state).

 

 

Seems to me that an oxidation number of -4 is allocated to the carbon and +1 to each hydrogen halfway down page 422 of my attachment.

 

I agree that some oddities occur, for instance when they lead to fractional oxidation numbers.
For this reason IUPAC has withdrawn their use.

However they are still used to bamboozle schoolboys.

Oxidation numbers are a model of what is happening and like all models are limited in situations where they match what theya re modelling.

[John Cuthber]

16 hours ago, studiot said:

What do you understand an oxidation state of zero to mean ?

 

It's an interesting question;
https://en.wikipedia.org/wiki/Nickel_tetracarbonyl

has Ni in the zero oxidation state.

And in some cases the oxidation state is pretty much a matter of opinion.