In the reactions $I$ and $II$,the covalencies of $Be$ and $Al$ in $X$ and $Y$ are respectively:
$I$ $Be(OH)_2 + NaOH \text{ (excess)} \rightarrow X$
$II$ $Al(OH)_3 + NaOH \text{ (excess)} \rightarrow Y$
$I$ $Be(OH)_2 + NaOH \text{ (excess)} \rightarrow X$
$II$ $Al(OH)_3 + NaOH \text{ (excess)} \rightarrow Y$
- A$4, 6$
- B$4, 4$
- C$6, 4$
- D$3, 6$
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