Question
Question: A certain compound ( \(X\) ) shows the following reaction: (i) When \(KI\) is added to an aqueous ...
A certain compound ( X ) shows the following reaction:
(i) When KI is added to an aqueous solution of X containing acetic acid, Iodine is liberated.
(ii) When CO2 is passed through an aqueous suspension of X , the turbidity transforms into a precipitate .
(iii) When a paste of X in water is heated with ethyl alcohol, a product of anaesthetic is obtained.
Identify X
A.CaOCl2
B.PbCl2
C.CaCl2
D.None of these
Solution
The reaction between KI and X which liberates Iodine is an oxidation reaction. Hence X must have oxygen as an atom in the molecular structure.
-Formation of turbidity with CO2 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 KI on treatment with X liberates Iodine gas.
Iodine is in a −1 oxidation state in KI and the oxidation state changes to 0in 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 should be formed along with chlorine gas to be liberated.
CaOCl2+2CH3COOH(CH3COO)2Ca+Cl2+H2O
This Chlorine Gas liberates, displaces iodine from potassium iodine to liberate iodine gas.
Cl2+2KI2KCl+I2
Turbidity formation by adding CO2 is due to the formation of metal carbonate salt. In water,
X gets converted into its hydroxide form and reacts with carbon dioxide to form metal carbonate.
CaOCl2+H2OCa(OH)2+Cl2
Then this hydroxide reacts with the Carbon Dioxide to warm metal carbonate salt.
Ca(OH)2+CO2CaCO3+Cl2
The white precipitate is of Calcium Carbonate.
In the third reaction , X react with water to form metal hydroxide
CaOCl2+H2OCa(OH)2+Cl2
Then the product formed reacts with ethyl alcohol to form:
CH3CH2OH+Cl2+Ca(OH)2CHCl3+(HCOO)2Ca+CaCl2+H2O
Here, CHCl3 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 CaOCl2 can form chloroform with ethyl alcohol as it’s a mixture of lime CaCO3 and calcium chloride CaCl2.