MCQ
End product $(C)$ in above, reaction is


- ✓

- B

- C

- D








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| Period number | Group number | |
| $Q$ | $2$ | $15$ |
| $P$ | $3$ | $2$ |
Then formula of the compound formed by $P$ and $Q$ element is
$F{e^{3 + }}(aq) + {e^ - } \to F{e^{ - 1}}(aq);\,{E^o} = + 0.77\,V$
$A{l^{3 + }}(aq) + 3{e^ - } \to Al(s);\,{E^o} = - 1.66\,V$
$B{r_2}(aq) + 2{e^ - } \to 2B{r^ - }(aq);\,{E^o} = + 1.08\,V$
Based on the data given above, reducing power of $F{e^{2 + }},\,Al$ and $B{r^ - }$ will increase in the order