It was given in my text book that reducing character of group 15 hydrides increases down the group. I am a bit confused about this. For $\ce{NH3}$ this is fine as I assume that:

$$\ce{NH3 -> N + H^+ + 3e^-}$$

My doubt arises in the higher elements of the group. In $\ce{SbH3}$, $\ce{H}$ is present as $\ce{H^{-}}$ and $\ce{Sb}$ as $\ce{Sb^3+}$. So would $\ce{Sb^3+}$ be oxidised to $\ce{Sb^5+}$ or would $\ce{H^{-}}$ be oxidised to $\ce{H2}$? I know that for $\ce{Sb}$, $\ce{Sb^3+}$ is more stable than $\ce{Sb^5+}$ . So would $\ce{H^{-}}$ be oxidised to $\ce{H2}$ ?

Also electronegativity of $\ce{As}$ and $\ce{H}$ is given tobe equal in my textbook. So what would be oxidation product of $\ce{AsH3}$ ?

For $\ce{BiH3}$ it seems like:

$$\ce{Bi^3+ -> Bi^5+ + 2e^-}$$

is more feasible as $\ce{Bi^5+}$ is more stable. Can someone please tell me the oxidation products of group 15 hydrides? Also can someone tell $\ce{E°_(H^{-}/H2)}$ value as I couldn't find it on net?


1 Answer 1


When your book says that reducing character increases down the group, it PRIMARILY meant in gaseous state. What you have described above are all ionic reactions, i.e., in some solution.

NOW in gaseous state, as we go down the group, the bond energy of hydrides reduces (due to increasing atomic size as we go down the group). So they can be more easily broken to release nascent hydrogen (see my note at the bottom), which is the cause of reducing nature. So reducing ability increases down the group in gaseous state.

NOTE : Nascent hydrogen is referred to the hydrogen which has just evolved post some reaction, and has not yet formed its molecule. This is not the same as hydrogen atom so keep that in mind.

Reducing property in a solution

This is the catch. In aqueous solution, reduction takes place due to the group 15 ion, and not hydrogen ion. So it depends on how easily the group 15 ion can lose an electron (reduction is gain of electrons) Down the group, electronegativity decreases and hence it becomes easier to remove electrons for reduction. So reduction ability decreases down the group in solutions too.

Now to answer your question, Sb(3+) will be oxidised to Sb(5+) as it is what is causing the reduction. You should be able to answer your following question now.

Btw, the E0 value for (H-/H2)is 0.00. Note that all oxidation reduction reactions of hydrogen gas have this value as 0.00

  • 1
    $\begingroup$ $E^\circ$ for $\ce{H-}/\ce{H2}$ is most certainly not 0.00. $\endgroup$ Feb 2, 2017 at 15:57
  • $\begingroup$ Forgive me if my electrochemistry is weak. Should you come to know about the value for the reaction, please do let me know. Thanks $\endgroup$
    – CupC_56
    Feb 2, 2017 at 15:58
  • $\begingroup$ But what about the $H^-$ produced in aqueous solution? Would it combine with $H^+$ and form $H_2$ ? $\endgroup$ Feb 2, 2017 at 17:47
  • $\begingroup$ I do not know why your book says that Hydrogen anion is produced in the case of SbH3. I am pretty sure that H+ is produced even in this case. $\endgroup$
    – CupC_56
    Feb 4, 2017 at 6:55
  • $\begingroup$ Check this link for yourself - books.google.co.in/… $\endgroup$
    – CupC_56
    Feb 4, 2017 at 6:56

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