Mg_3N_2(s) + H_2O xrightarrow{R.T.} P downarr | vedclass

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(B) The reaction is: $Mg_3N_2(s) + 6H_2O(l) \rightarrow 3Mg(OH)_2(s) \downarrow + 2NH_3(g) \uparrow$.Here,$P$ is $Mg(OH)_2$ and $Q$ is $NH_3$ gas.Exce

Vedclass · Accepted Answer

$Zn^{2+}(aq.)$

Vedclass · Answer

(B) The reaction is: $Mg_3N_2(s) + 6H_2O(l) \rightarrow 3Mg(OH)_2(s) \downarrow + 2NH_3(g) \uparrow$.Here,$P$ is $Mg(OH)_2$ and $Q$ is $NH_3$ gas.Excess $NH_3$ forms coloured complexes with transition metal ions like $Ni^{2+}$,$Cr^{3+}$,and $Cu^{2+}$ due to $d-d$ transitions.However,$Zn^{2+}$ has a $3d^{10}$ electronic configuration (no unpaired electrons),so the complex $[Zn(NH_3)_4]^{2+}$ is colourless.Therefore,the correct option is $B$.

$List-I$	$List-II$
$A$. $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta}$	$I$. $H_2$
$B$. $KMnO_4 + HCl \rightarrow$	$II$. $N_2$
$C$. $Al + NaOH + H_2O \rightarrow$	$III$. $O_2$
$D$. $NaNO_3 \xrightarrow{\Delta}$	$IV$. $Cl_2$

$Mg_3N_2(s) + H_2O \xrightarrow{R.T.} P \downarrow + Q \uparrow$. Excess $Q$ gas does not form a coloured complex with:

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