sblock Elements - Result Question 72
25. Arrange the following in increasing order of basic strength :
$MgO, SrO, K_2O, NiO, Cs_2O$
(1991, 1M)
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Solution:
- PLAN This problem can be solved by using the concept of oxidizing agent and reducing agent.
Oxidant increases the oxidation number of the species with which it is reacted.
Reductant decreases the oxidation number of the species with which it is reacted.
$H_2O_2$ reacts with $KIO_4$ in the following manner:
On reaction of $KIO_4$ with $H_2O_2$, oxidation state of $I$ varies from +7 to +5, i.e. decreases. Thus, $KIO_4$ gets reduced hence, $H_2O_2$ is a reducing agent here.
With $NH_2OH$, it gives the following reaction:
$$ \stackrel{-1}{N} H _2 OH + H _2 O _2 \longrightarrow \stackrel{+3}{N} O _3 + H _2 O $$
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 a reducing agent here.
Hence, (a) is the correct choice.
BETA
