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Mole Concept

An ore found ...

Question

An ore found in nature contains two elements A and B from the $4^{t h}$ 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 $K C l$ 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 water, a yellow ppt is obtained.
The element B in its stable and highest oxidation state gives blue colour with:

N H_{3}

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K_{3} [F e {(C N)}_{6}]

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K_{4} [F e {(C N)}_{6}]

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N a S C N

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Solution

The correct option is D $K_{4} [F e {(C N)}_{6}]$
The element A is Cr. The ore is chromite ore $F e C r_{2} O_{4}$ .

The orange compound is $K_{2} C r_{2} O_{7}$ , which reacts with $K C l$ and conc. $H_{2} S O_{4}$ to produce deep red vapours of chromyl chloride $(C r O_{2} C l_{2})$ . The reaction is as follows:

$K_{2} C r_{2} O_{7} + 4 K C l + 3 H_{2} S O_{4} \to 2 C r O_{2} C l_{2} + 3 K_{2} S O_{4} + 3 H_{2} O$

$(C r O_{2} C l_{2})$ 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:

$C r O_{2} C l_{2} + H_{2} S O_{4} \to H_{2} C r_{2} O_{7}$

This orange coloured compound reacts with baryta water $(N a O H)$ and then with barium chloride to form a yellow ppt of $B a C r O_{4}$ .
The reactions are as follows:

$H_{2} C r_{2} O_{7} + 4 N a O H \to 2 N a_{2} C r O_{4} + 3 H_{2} O$
$N a_{2} C r O_{4} + B a C l_{2} \to B a C r O_{4} ↓ (y e l l o w) + 2 N a C l$

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

Its stable and highest oxidation state is +3.

In this state, it gives blue colour with excess of $K_{4} [F e {(C N)}_{6}]$ .

The reaction is as follows:

$4 F e^{3 +} (a q) + 3 K_{4} [F e (C N)_{6}] (a q) \Leftrightarrow F e_{4} [F e (C N)_{6}]_{3} (b l u e) + 12 K^{+}$

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Similar questions

4^{t h}

K C l

H_{2} S O_{4}

H_{2} S O_{4}

S_{1}

K_{2} C r_{2} C_{7}

H_{2} S O_{4}

S_{2}

S_{3}

N a N O_{2}

K_{4} [{F e (C N)}_{6}]

{H N O}_{3}

K_{3} [{F e (C N)}_{6}]

A g {N O}_{3}

N a O H

K_{2} H g l_{4}

B a C l_{2}

H C l

K_{2} C r_{2} O_{7}

H_{2} S O_{4}

(A)

(B)

(F e C r_{2} O_{4})

N a_{2} C O_{3}

H_{2} S O_{4}

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