| (i) \[C{{l}^{-}}\]can give up an electron more easily than\[{{F}^{-}}.\] |
| (ii) \[C{{l}^{-}}\]is a better reducing agent than \[{{F}^{-}}.\] |
| (iii) \[C{{l}^{-}}\]is smaller in size than \[{{F}^{-}}.\] |
| (iv) \[{{F}^{-}}\]can be oxidized more readily than \[C{{l}^{-}}.\] |
A) (i) and (ii)
B) (i), (ii) and (iv)
C) (iii) and (iv)
D) only (i)
Correct Answer: D
Solution :
[d] The halide ions act as reducing agents. F ion does not show any reducing nature but \[C{{l}^{-}},\] \[B{{r}^{-}}\]& \[{{I}^{-}}\] ion act as reducing agents and their reducing nature is in increasing order \[\xrightarrow[\text{Reducing}\,\,\text{nature}\,\,\text{increases}]{C{{1}^{-}}B{{r}^{-}}{{I}^{-}}}\]
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