Br_2 + NaOH xrightarrow{R.T.} Y + ZIf Y gives | vedclass

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(C) The reaction of bromine with cold dilute sodium hydroxide is: $Br_2 + 2NaOH \longrightarrow NaBr + NaBrO + H_2O$.Here,$Y$ is $NaBr$ and $Z$ is $Na

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$Al^{3+}(aq.)$

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(C) The reaction of bromine with cold dilute sodium hydroxide is: $Br_2 + 2NaOH \longrightarrow NaBr + NaBrO + H_2O$.Here,$Y$ is $NaBr$ and $Z$ is $NaBrO$.$NaBr$ gives a yellow precipitate of $AgBr$ with $AgNO_3$.$NaBrO$ (sodium hypobromite) acts as an oxidizing agent.It can oxidize $Cr^{3+}$ to $Cr^{6+}$,$Fe^{2+}$ to $Fe^{3+}$,and $Sn^{2+}$ to $Sn^{4+}$.However,$Al^{3+}$ is already in its highest stable oxidation state $(+3)$,so it cannot be further oxidized by $NaBrO$.

$Br_2 + NaOH \xrightarrow{R.T.} Y + Z$
If $Y$ gives a precipitate with $AgNO_3$,then $Z$ does not undergo a reaction with:

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$Br_2 + NaOH \xrightarrow{R.T.} Y + Z$If $Y$ gives a precipitate with $AgNO_3$,then $Z$ does not undergo a reaction with: