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A reddish brown compound element X on heating in air, becomes a black colored compound Y. The compound X and Y are:
A. Cu, CuO
B. Pb, PbO
C. $Fe,F{{e}_{2}}{{O}_{3}}$
D. $Cr,C{{r}_{2}}{{O}_{3}}$

Last updated date: 11th Jun 2024
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Hint: Heating of the elements in the presence of air or oxygen is called combustion. During the combustion process heat and light are going to produce. Metals generally lose their color on heating in presence of air.

Complete step by step answer:
- In the question it is given that a reddish brown color compound (X) on heating becomes black colored compound (Y).
- Coming to the given options, Option B, Pb means lead. Lead is in grey color. So, option B is wrong.
- Coming to option C, Fe means iron. Iron is in silver-grey color. So, option C is also wrong.
- Coming to option D, Cr means chromium. Chromium is steely-grey in color. So, option D is also wrong.
- Coming to option A, Cu means copper. Copper is in reddish brown in color. On heating copper forms a black color compound CuO due to oxidation.
- The chemical reaction of copper with air or oxygen present in the air is as follows.
\[\underset{X}{\mathop{2Cu}}\,+{{O}_{2}}\to \underset{Y}{\mathop{2CuO}}\,\]
- In the above reaction two moles of copper reacts with one mole of oxygen present in the air and forms 2 moles of copper oxide as the product.
- Copper oxide is black in color.
- Therefore, the X is copper and Y is CuO.

- So, the correct option is A.

Note: The formed black color compound (CuO) reacts with hydrochloric acid and converts into blue color due to the formation of calcium chloride. The chemical reaction of copper oxide with hydrochloric acid is as follows.
\[CuO+2HCl\to CuC{{l}_{2}}+{{H}_{2}}O\]