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In the reactions $\mathrm{I}$ and $\mathrm{II}$ the covalencies of $\mathrm{Be}$ and $\mathrm{Al}$ in $\mathrm{X}$ and $\mathrm{Y}$ are respectively
I $\mathrm{Be}(\mathrm{OH})_2+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{X}$ (excess)
II $\mathrm{Al}(\mathrm{OH})_3+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{Y}$
Options:
I $\mathrm{Be}(\mathrm{OH})_2+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{X}$ (excess)
II $\mathrm{Al}(\mathrm{OH})_3+\underset{\text { (excess) }}{\mathrm{NaOH}} \rightarrow \mathrm{Y}$
Solution:
1726 Upvotes
Verified Answer
The correct answer is:
4,4
The covalencies of $\mathrm{Be}$ and $\mathrm{Al}$ in $\mathrm{X}$ and $\mathrm{Y}$ respectively are 4,4 .
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