Question

The higher oxide of an element ( $E$ ) has the formula $E{O}_3$. Its hydride contains $2.47\%$ hydrogen, the element is

• A. $Te$
• B. $Se$
• C. $S$
• D. $Si$

Answer 1

Answer: (B)

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 $Si{O}_2$ as the higher oxide stands rejected.