${N H}_{4} S C N$ can be used to make distinction between ${C u}^{2 +}$ and ${C o}^{2 +}$ .

If the given statement is true enter 1 else enter 0.

Open in App

Solution

${N H}_{4} S C N$ can be used to make distinction between ${C u}^{2 +}$ and ${C o}^{2 +}$ .

Cobalt(II) chloride reacts with ammonium thiocyanate to form blue coloured complex $(N H_{4})_{2} [C o (S C N)_{4}]$

$C o C l_{2} + 4 N H_{4} S C N \to (N H_{4})_{2} [C o (S C N)_{4}] b l u e c o l o u r + 2 N H_{4} C l$

On the other hand Copper(II) chloride reacts with ammonium thiocyanate to form black ppt of $C u (S C N)_{2}$ .

$C u C l_{2} + 2 N H_{4} S C N \to C u (S C N)_{2} b l a c k p p t . ↓ + 2 N H_{4} C l$ .

So, the given statement is true.

Suggest Corrections

Similar questions

{F e}^{3 +}, {C o}^{2 +}

{C u}^{2 +}

{N H}_{4} S C N

{N H}_{4} C l

({N H}_{4})_{2} {S O}_{4}

{N H}_{4} S C N

{F e}^{3 +}

[{F e (S C N)]}^{2 +}

{C u}^{2 +}, {F e}^{3 +}, {Z n}^{2 +}

{C d}^{2 +}

K_{4} [{F e (C N)}_{6}]

N a O H

{A l (O H)}_{3}

{F e (O H)}_{3}

Join BYJU'S Learning Program