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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 $C{O_2}$ 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 $X$

D.None of these

Last updated date: 20th Jun 2024
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Hint: 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 $C{O_2}$ 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 $KI$ on treatment with $X$ liberates Iodine gas.
Iodine is in a $ - 1$ oxidation state in $KI$ and the oxidation state changes to $0$in 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 ${(C{H_3}COO)_2}M$ should be formed along with chlorine gas to be liberated.
$CaOC{l_2} + 2C{H_3}COOH\xrightarrow{{}}{(C{H_3}COO)_2}Ca + C{l_2} + {H_2}O$
This Chlorine Gas liberates, displaces iodine from potassium iodine to liberate iodine gas.
$C{l_2} + 2KI\xrightarrow{{}}2KCl + {I_2}$
Turbidity formation by adding $C{O_2}$ 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.
$CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{(OH)_2} + C{l_2}$
Then this hydroxide reacts with the Carbon Dioxide to warm metal carbonate salt.
$Ca{(OH)_2} + C{O_2}\xrightarrow{{}}CaC{O_3} + C{l_2}$
The white precipitate is of Calcium Carbonate.
In the third reaction , $X$ react with water to form metal hydroxide
$CaOC{l_2} + {H_2}O\xrightarrow{{}}Ca{(OH)_2} + C{l_2}$
Then the product formed reacts with ethyl alcohol to form:
$C{H_3}C{H_2}OH + C{l_2} + Ca{(OH)_2}\xrightarrow{{}}CHC{l_3} + {(HCOO)_2}Ca + CaC{l_2} + {H_2}O$
Here, $CHC{l_3}$ is the anaesthetic, it is called chloroform and was used in surgeries before modern anaesthetics were developed.
Hence Option 1 is correct.

-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 $CaOC{l_2}$ can form chloroform with ethyl alcohol as it’s a mixture of lime $CaC{O_3}$ and calcium chloride $CaC{l_2}$.