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Question: A certain compound ( \(X\) ) shows the following reaction: (i) When \(KI\) is added to an aqueous ...

A certain compound ( XX ) shows the following reaction:
(i) When KIKI is added to an aqueous solution of XX containing acetic acid, Iodine is liberated.
(ii) When CO2C{O_2} is passed through an aqueous suspension of XX , the turbidity transforms into a precipitate .
(iii) When a paste of XX in water is heated with ethyl alcohol, a product of anaesthetic is obtained.
Identify XX

A.CaOCl2CaOC{l_2}
B.PbCl2PbC{l_2}
C.CaCl2CaC{l_2}
D.None of these

Explanation

Solution

The reaction between KIKI and XX which liberates Iodine is an oxidation reaction. Hence XX must have oxygen as an atom in the molecular structure.
-Formation of turbidity with CO2C{O_2} in solvent is a characteristic test used to determine the presence of Calcium in the solvent.

Complete step by step solution:
Let us go over the reactions one by one and eliminate the options simultaneously by the data we collect.
In the first reaction we have been given information, that KIKI on treatment with XX liberates Iodine gas.
Iodine is in a 1 - 1 oxidation state in KIKI and the oxidation state changes to 00in Iodine Gas.
This means there is an oxidation reaction taking place, or in other words, there is an oxidising agent in the reaction. On reaction with acetic acid, Metal Anhydride (CH3COO)2M{(C{H_3}COO)_2}M should be formed along with chlorine gas to be liberated.
CaOCl2+2CH3COOH(CH3COO)2Ca+Cl2+H2OCaOC{l_2} + 2C{H_3}COOH\xrightarrow{{}}{(C{H_3}COO)_2}Ca + C{l_2} + {H_2}O
This Chlorine Gas liberates, displaces iodine from potassium iodine to liberate iodine gas.
Cl2+2KI2KCl+I2C{l_2} + 2KI\xrightarrow{{}}2KCl + {I_2}
Turbidity formation by adding CO2C{O_2} is due to the formation of metal carbonate salt. In water,
XX gets converted into its hydroxide form and reacts with carbon dioxide to form metal carbonate.
CaOCl2+H2OCa(OH)2+Cl2CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{(OH)_2} + C{l_2}
Then this hydroxide reacts with the Carbon Dioxide to warm metal carbonate salt.
Ca(OH)2+CO2CaCO3+Cl2Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + C{l_2}
The white precipitate is of Calcium Carbonate.
In the third reaction , XX react with water to form metal hydroxide
CaOCl2+H2OCa(OH)2+Cl2CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{(OH)_2} + C{l_2}
Then the product formed reacts with ethyl alcohol to form:
CH3CH2OH+Cl2+Ca(OH)2CHCl3+(HCOO)2Ca+CaCl2+H2OC{H_3}C{H_2}OH + C{l_2} + Ca{(OH)_2}\xrightarrow{{}}CHC{l_3} + {(HCOO)_2}Ca + CaC{l_2} + {H_2}O
Here, CHCl3CHC{l_3} is the anaesthetic, it is called chloroform and was used in surgeries before modern anaesthetics were developed.
Hence Option 1 is correct.

Note:
-Lead anhydride does not form at normal temperature conditions, since lead is a weak metal, it needs an external energy to help form hydroxide, like temperature and catalyst.
-Only Calcium Hypochlorite CaOCl2CaOC{l_2} can form chloroform with ethyl alcohol as it’s a mixture of lime CaCO3CaC{O_3} and calcium chloride CaCl2CaC{l_2}.