Question

An ore found in nature contains two elements A and B from the $4^{th}$ period of the periodic table. Both elements belong to the same group in the list of basic radical analysis. An orange compound of element A, on reacting with $KCl$ and concentrated $H_{2}S{O}_4$, produces deep red vapours. These deep red vapours on passing through aqueous $H_{2}S{O}_4$ solution acquire an orange colour. If barium chloride is added to this solution along with barium hydroxide, a yellow ppt is obtained.

On the basis of this information, the element B can be ________ .

• A. Ca
• B. Fe
• C. Hg
• D. Al

Answer 1

Answer: (B)

Fe

The element A is Cr. The ore is chromite ore $FeC{r}_2{O}_4$.

The orange compound is $K_{2}C{r}_2{O}_7$, which reacts with $KCl$ and conc. $H_{2}S{O}_4$ to produce deep red vapours of chromyl chloride $\left(Cr{O}_{2}C{l}_2\right)$. The reaction is as follows :

$K_{2}C{r}_2{O}_7+4KCl+3H_{2}S{O}_4\to 2Cr{O}_{2}C{l}_2+3K_{2}S{O}_4+3H_{2}O$

$\left(Cr{O}_{2}C{l}_2\right)$ is passed through aqueous $H_{2}S{O}_4$ to form a orange colored compound $H_{2}C{r}_2{O}_7$.

The reaction is as follows :

$Cr{O}_{2}C{l}_2+H_{2}S{O}_4\to H_{2}C{r}_2{O}_7$

This orange coloured compound reacts with barium hydroxide and then with barium chloride to form a yellow ppt of $BaCr{O}_4$.

The reactions are as follows :

$H_{2}C{r}_2{O}_7+4NaOH\to 2N{a}_{2}Cr{O}_4+3H_{2}O$

$N{a}_{2}Cr{O}_4+BaC{l}_2\to BaCr{O}_4\downarrow \left(yellow\right)+2NaCl$

The other element belongs to the same group as Cr, i.e., either Fe or Al. Also, it belongs to the $4^{th}$ period. Hence, the element B is Fe.

Therefore, B is correct option.