sblock Elements - Result Question 72
####25. Arrange the following in increasing order of basic strength :
$MgO, SrO, K _2 O, NiO, Cs _2 O$
(1991, 1M)
Show Answer
Solution:
- PLAN This problem can be solved by using concept of oxidant and reductant.
Oxidant Oxidant increases the oxidation number of the species with which it is reacted.
Reductant Reductant decreases the oxidation number of the species with which it is reacted.
$H _2 O _2$ reacts with $KIO _4$ in the following manner:
On reaction of $KIO _4$ with $H _2 O _2$, oxidation state of $I$ varies from +7 to +5 , i.e. decreases. Thus, $KIO _4$ gets reduced hence, $H _2 O _2$ is a reducing agent here.
With $NH _2 OH$, it given following reaction:
$$ \stackrel{-1}{N} H _2 OH+H _2 O _2 \longrightarrow \stackrel{+3}{N _2} O _3+H _2 O $$
In the above reaction, oxidation state of $N$ varies from -1 to +3 . Here, oxidation number increases, hence $H _2 O _2$ is acting as an oxidising agent here.
Hence, (a) is the correct choice.
