Pb + text{dil. } HNO_3 xrightarrow{text{Warm} | vedclass

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(C) The reaction is: $3Pb + 8HNO_3 	o 3Pb(NO_3)_2 + 2NO \uparrow + 4H_2O$.Here,$Q$ is $NO$ (Nitric oxide).$NO$ has $15$ electrons,making it a paramag

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It combines with $Fe_2(SO_4)_3$

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(C) The reaction is: $3Pb + 8HNO_3 	o 3Pb(NO_3)_2 + 2NO \uparrow + 4H_2O$.Here,$Q$ is $NO$ (Nitric oxide).$NO$ has $15$ electrons,making it a paramagnetic,colourless gas.$NO$ reacts with $O_2$ in air to form $NO_2$ $(2NO + O_2 	o 2NO_2)$,which is a paramagnetic,brown-coloured gas.$HNO_2$ undergoes disproportionation: $3HNO_2 	o HNO_3 + 2NO + H_2O$,producing $NO$.$NO$ combines with $FeSO_4$ to form a brown ring complex $[Fe(H_2O)_5(NO)]SO_4$,but it does not combine with $Fe_2(SO_4)_3$ (where $Fe$ is in $+3$ oxidation state).Therefore,the statement in option $C$ is incorrect.

List-$I$ (Family)	List-$II$ (Symbol of Element)
$A$. Pnictogen (group $15$)	$I$. $Ts$
$B$. Chalcogen	$II$. $Og$
$C$. Halogen	$III$. $Lv$
$D$. Noble gas	$IV$. $Mc$

$Pb + \text{dil. } HNO_3 \xrightarrow{\text{Warm}} P + Q \uparrow + H_2O$
The incorrect statement for $Q$ is:

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Match the List-$I$ with List-$II$.
List-$I$ (Family) List-$II$ (Symbol of Element)
$A$. Pnictogen (group $15$) $I$. $Ts$
$B$. Chalcogen $II$. $Og$
$C$. Halogen $III$. $Lv$
$D$. Noble gas $IV$. $Mc$

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$Pb + \text{dil. } HNO_3 \xrightarrow{\text{Warm}} P + Q \uparrow + H_2O$The incorrect statement for $Q$ is: