MCQ
The electron affinity of $Be$ is similar to :-
- ✓$He$
- B$B$
- C$Li$
- D$Na$
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The product formed is
The compound $A$ is a above reaction is
$Fe|F{e^{ + 2}}\left( {0.2\,M} \right)||A{u^{ + 3}}\left( {0.02\,M} \right)|Au$, If $E_{F{e^{ + 2}}/Fe}^o = - 0.44\,V$ and $E_{Au/A{u^{ + 3}}}^o = - 1.50\,V$
