Assertion (A) :- Fe^{+3} is not used in the b | vedclass

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(D) The brown ring test is used to detect the presence of nitrate ions $(NO_3^{-})$.In this test,freshly prepared ferrous sulfate $(FeSO_4)$ solution

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Both $(A)$ and $(R)$ are correct but $(R)$ is the correct explanation of $(A)$

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(D) The brown ring test is used to detect the presence of nitrate ions $(NO_3^{-})$.In this test,freshly prepared ferrous sulfate $(FeSO_4)$ solution is added to the nitrate solution,followed by the addition of concentrated sulfuric acid $(H_2SO_4)$ along the sides of the test tube.$Fe^{+2}$ ions act as a reducing agent to reduce $NO_3^{-}$ to $NO$ gas.Reaction: $NO_3^{-} + 3Fe^{+2} + 4H^{+} \rightarrow NO + 3Fe^{+3} + 2H_2O$.Then,$NO$ reacts with the remaining $Fe^{+2}$ to form the brown complex: $[Fe(H_2O)_5(NO)]SO_4$.Since $Fe^{+2}$ is required for the reduction of $NO_3^{-}$ and the formation of the complex,$Fe^{+3}$ cannot be used.Thus,both $(A)$ and $(R)$ are correct,and $(R)$ explains why $Fe^{+3}$ is not used (because $Fe^{+2}$ is the necessary reductant).

List-$I$ (Anion)	List-$II$ (Gas evolved on reaction with dil. $H_2SO_4$)
$A. CO_3^{2-}$	$I$. Colourless gas which turns lead acetate paper black
$B. S^{2-}$	$II$. Colourless gas which turns acidified potassium dichromate solution green
$C. SO_3^{2-}$	$III$. Brown fumes which turn acidified $KI$ solution containing starch blue
$D. NO_2^{-}$	$IV$. Colourless gas evolved with brisk effervescence,which turns lime water milky

Assertion $(A) :- Fe^{+3}$ is not used in the brown ring test for $NO_3^{-}$.
Reason $(R) :-$ Reduction of $NO_3^{-}$ takes place during the test.

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Assertion $(A) :- Fe^{+3}$ is not used in the brown ring test for $NO_3^{-}$.Reason $(R) :-$ Reduction of $NO_3^{-}$ takes place during the test.