First slide

Hydrogen Peroxide

Question

The substance which can not be used for preparing H₂O₂ solution is

Easy

A

Na₂O₂
B

BaO₂
C

PbO₂
D

2–ethyl anthraquinone

Solution

⇒
$\large N{a_2}{O_2}\; + \;\mathop {{H_2}S{O_4}\;}\limits_{(dil)} \to \;N{a_2}S{O_4}\; + \;{H_2}{O_2}$

⇒
$\large Ba{O_2}\; + \;\mathop {{H_2}S{O_4}\;}\limits_{(dil)} \to \;BaSO_4^{ - 2}\; + \;{H_2}O_2^{ - 1}$
⇒
⇒ compound which contain a peroxy bond will have the calculated oxidation number {assuming Ox.st of all "o" is (-2)] of central atom will be grater than group no, which is impossible such compound will have peroxy bond.
⇒ Calculated OX.no of Na in Na₂O₂ +2 group no . of Na →1
So Na₂O₂ has a peroxy bond.
Both the oxygen atom are peroxy oxygens with (-1) state.
Thus actual Oxidation state of Na is (+1)
⇒ In case of PBO₂ the calculated oxidation state of "Pb" is "+4" it is equal to its group no. Therefore it can't contain peroxy linkage and hence on reaction with dilute acids it can't give Hydrogen peroxide.
⇒ BaO₂ also possess peroxy bond and hence it also give H₂O₂ on treatment with dilute acids.
⇒ Where as PbO₂ does not possess peroxy bond and hence it does not give H₂O₂ with dilute acids.

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