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$(a)$ $Be _{2} C$
$(b)$ $CaC _{2}$
$(c)$ $Mg _{2} C _{3}$
$(d)$ $Al _{4} C _{3}$
$\mathop {{H_3}C\mathop C\limits^ \ominus {H_2}}\limits_{(i)} \,,$$\mathop {{H_2}C = \mathop C\limits^ \ominus H}\limits_{(ii)} $ and $\mathop {H - C \equiv \mathop C\limits^ \ominus }\limits_{(iii)} $
is in the order of
$F{e^{ + 3}}\left( {aq} \right) + {e^ - } \to F{e^{2 + }}\left( {aq} \right);{E^o} = + 0.77\,V$
$A{l^{ + 3}}\left( {aq} \right) + 3{e^ - } \to Al\left( s \right);{E^o} = - 1.66\,V$
$B{r_2}\left( {aq} \right) + 2{e^ - } \to 2B{r^ - };{E^o} = + 1.09\,V$
Considering the electrode potentials, which of the following represents the correct order of reducing power?