Question

A metal nitrate (X) gives a white precipitate with ammonia solution, but the precipitate get dissolved on adding ammonium salt. Lead dioxide and concentrated nitric acid on boiling with a dilute solution of metal nitrate (X) produces a violet-red (or purple) colour solution. Small amount of metal nitrate (X) gives an amethyst-red bead with borax in oxidising flame when cold. The cation of metal nitrate is :

• A. $B{i}^{3+}$
• B. $M{g}^{2+}$
• C. $Z{n}^{2+}$
• D. $M{n}^{2+}$

Answer 1

The correct option is D $M{n}^{2+}$

The cation of metal nitrate is $M{n}^{2+}$ as manganese ion gives an amethyst red bead with borax in oxidising flame, when cold $M{n}^{2+}$ reacts with ammonia to form white precipitate of manganese hydroxide.
$M{n}^{2+}+2N{H}_3+2H_{2}O\rightleftharpoons Mn(OH{)}_2\downarrow (\text{white ppt.})+2N{H}_4^{+}$
On adding ammonium salts, the precipitate dissolves and reaction proceeds towards left.
When manganese nitrate is boiled with lead dioxide and concentrated nitric acid purple colouration of $Mn{O}_4^{-}$ is formed.
$5Pb{O}_2+2M{n}^{2+}+4H^{+}\to 2Mn{O}_4^{-}\left(\text{violet-red or purple}\right)+5P{b}^{2+}+2H_{2}O$