Why H$_2$Te is a good reducing agent than H$_2$O .
I think it should be opposite as in water , oxygen has only tendency to release electrons .
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ Are you sure about the statement about oxygen's affinity for electrons? $\endgroup$– airhuffCommented Jan 26, 2017 at 8:44
-
$\begingroup$ @airhuff yes it is given in my book $\endgroup$– user123733Commented Jan 26, 2017 at 8:45
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
According to this Wikipedia entry:
"[electronegativity] is a chemical property that describes the tendency of an atom to attract electrons (or electron density) towards itself.
then according to an electronegativity scale called the Allen electronegativity:
"...neon has the highest electronegativity of all elements, followed by fluorine, helium, and oxygen."
By any scale, oxygen is among the most electronegative elements. More to the point, electronegativity decreases as you go down a period, so $\ce{Te}$ is much less electronegative, much more able to give up its electrons, than oxygen. Thus, $\ce{H2Te}$ makes a better reducing agent than $\ce{H2O}$.
