Match the following lists:List-IList-II(A) Be | vedclass

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(A) The correct matching is:$(A)$ Benzene matches with $4$. $(4n + 2) \pi$ electrons: Benzene has $6 \pi$ electrons,which follows the Huckel rule $(4n

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$A-4, B-3, C-2, D-1$

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(A) The correct matching is:$(A)$ Benzene matches with $4$. $(4n + 2) \pi$ electrons: Benzene has $6 \pi$ electrons,which follows the Huckel rule $(4n + 2) \pi$ electrons.$(B)$ Ethylene matches with $3$. Mustard gas: Ethylene reacts with $S_2Cl_2$ to form mustard gas.$(C)$ Acetaldehyde matches with $2$. Silver mirror: Acetaldehyde gives a positive silver mirror test with Tollen's reagent.$(D)$ Chloroform matches with $1$. Phosgene: Chloroform on oxidation gives phosgene,which is a poisonous gas.Therefore,the correct sequence is $A-4, B-3, C-2, D-1$.

Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

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List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

Match the following lists:List-$I$List-$II$$(A)$ Benzene$1$. Phosgene$(B)$ Ethylene$2$. Silver mirror$(C)$ Acetaldehyde$3$. Mustard gas$(D)$ Chloroform$4$. $(4n + 2) \pi$ electrons$5$. Carbylamine