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  4. The higher oxide of an element ( E ) has the formula E O 3. Its hydride contains 2.47 % hydrogen, the element is

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Q. The higher oxide of an element ( $E$ ) has the formula $E O _{3}$. Its hydride contains $2.47 \%$ hydrogen, the element is

Classification of Elements and Periodicity in Properties

Solution:

Since the element $(E)$ forms a higher oxide with the

formula, $E O _{3}$, therefore, it must belong to group $16$ .

The hydrides of 16 group are $H _{2} S , H _{2} Te , H _{2} Se$.

$\%$ of hydrogen in $H _{2} S =\frac{2}{34} \times 100=5.88$

$\%$ of hydrogen in $H _{2} Se =\frac{2}{81} \times 100=2.47$

$\%$ of hydrogen in $H _{2} Te =\frac{2}{129.6} \times 100=1.54$

As such Si which lies in group $14$ and would form $SiO _{2}$ as the higher oxide stands rejected.