Electronegativity of $\ce{H}$ is greater than $\ce{P}$, but according to my book in $\ce{PH3}$, $\ce{P}$ has oxidation number of -3. How is it possible? Is any of the data wrong?


Here is a way to get the answer. If you place a reagent with oxidation number of H = -1 (hydride) it would produce dihydrogen gas. For example: $\ce{KH + H2O-> KOH + H2}$.

If you place a molecule with oxidation number of $\ce{H} = +1$ dihydrogen gas is NOT formed. $\ce{NH3 + H2O -> NH4+ + OH-}$. Phosphine doesn't form dihydrogen gas on contact with water so $\ce{H}$ is in oxidation state +1 and $\ce{P}$ is in oxidation state -3. Reaction with strong acids leads to $\ce{PH4+}$ which has similar structure to the ammonium ion. For these reasons even for $\ce{As}$ in $\ce{AsH3}$ the oxidation state of $\ce{As = -3}$.

The problem with electronegativity is that we are trying to assign a single number, while we actually need a function. We take average of electronegativity of $\ce{H}$ in $\ce{HF, NaH, H2O, CH4, NH4+}$ and come up with a single number. This works well if you need to guesstimate the behavior, but it is not accurate. It doesn't even account for softness/hardness of a partner.

Analysis "will $\ce{PH3 + H2O}$ form $\ce{H2}$?" (no) vs "will $\ce{PH3 + P2O3}$ form $\ce{P + H2O}$?" (yes) is more accurate way to assign formal charges.

  • $\begingroup$ But H is more electronegative than P...how do you explain that? $\endgroup$ – user14857 Jul 23 '16 at 16:03
  • $\begingroup$ @ZOZ please see updated version. $\endgroup$ – sixtytrees Jul 23 '16 at 16:14
  • 2
    $\begingroup$ @ZOZ As a rule of thumb, H will pretty much always be in oxidation state +1. You can work your way through many redox reactions starting with this assumption! Only in some rarer cases like metal hydrides (the KH example above) will H be have the oxidation number -1. $\endgroup$ – tipavi Jul 23 '16 at 17:29

Electronegativity is a mere indication of what is happening and you will find many different values for the electronegativity of H and P (and all the other elements).

If you have a look at the C-H bond, electronegativities are 2.55 for carbon and 2.2 for hydrogen, i.e. the difference is much larger than for phosphorus (2.19) and hydrogen (2.2). However, C-H bonds are generally regarded as non-polar.

What really makes phosphorus different in its reactivity is that it is extremely oxophilic, i.e. it will try to form P-O or P=O bonds on every occasion. This is why PH3 is pyrophoric, for example.


Oxidation numbers are generally used to determine if a molecule acts as a reducer or oxidiser in a given reaction. The same molecule can be a reducer in one reaction and oxidiser in another if it is an intermediate oxidation state. Knowing the electronegativity (however defined) does not help in this instance.

If the oxidation number of a particular atom is not known it may be deduced from the known oxidation numbers of other atoms associated with it in a molecule. For example, the oxidation state of Pu in PuO$_2^{2+}$ is not known. It is calculated as

charge on Pu in PuO$_2^{2+}$ + charges on two O's (or 4-) = charge on PuO$_2^{2+}$ or 2+

thus Pu has an oxidation number of 6+.

In some cases (S$_4$O$_6^{2-}$, Fe$_3$O$_4$) the oxidation number can be fractional. Thus oxidation numbers can be negative, positive, zero or fractional. Of course the same atom can have different oxidation numbers in different molecules. An example is shown below for nitrogen

-3 in NH$_3$, 0 in N$_2$ , +1 in N$_2$O, +2 in NO, +3 in HNO$_2$, +4 in NO$_2$, +5 in HNO$_3$

where NH$_3$ is only a reducer and HNO$_3$ only an oxidiser, but the other compounds can be reducers or oxidisers. Note that it is zero in N$_2$.

A molecule is an oxidiser of its oxidation state decreases, for example iodine oxidises the strong reducer H$_2$SO$_3$ e.g.

I$_2$ + H$_2$SO$_3$ +H$_2$O $\rightarrow$ 2HI+H$_2$SO$_4$,

the iodine oxidation state changes from 0 to -1. In a reduction the oxidation state increases.

You can see that in your compound the phosphorous' oxidation state is -3.


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