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Q. Given the following reactions involving $A , B , C$ and $D$
(i) $C+B^{+} \rightarrow C^{+}+B$
(ii) $A^{+}+D \rightarrow$ No reaction
(iii) $C^{+}+A \rightarrow$ No reaction
(iv) $D+B^{+} \rightarrow D^{+}+B$
The correct arrangement of $A , B , C , D$ in order of their decreasing ability as reducing agent

Delhi UMET/DPMTDelhi UMET/DPMT 2011

Solution:

$C+B^{+} \rightarrow C^{+}+B ;$
$ \therefore C > B$ (reducing power)
$A^{+}+D \rightarrow$ No reaction;
$\therefore A>D$ (reducing power)
$C^{+}+A \rightarrow$ No reaction;
$\therefore C>A$ (reducing power)
$D+B^{+} \rightarrow D^{+}+B ; $
$\therefore D>B$ (reducing power)
Hence, order of reducing power
$=C>A>D>B$