Question

Which one of the following reagents can be used for differentiation of nitrates of $C{u}^{2+}$ and $B{i}^{3+}$?

• A. Excess of water
• B. $N{H}_{4}OH$ (excess)
• C. Excess of $KI$
• D. All of the above
  

Answer 1

Answer: (D)

All of the above

(A) Nitrates of $C{u}^{2+}$ gives no change with excess of water. Nitrates of $B{i}^{3+}$ reacts with excess of water to form white coloration of $BiO\left(N{O}_3\right)$.

$C{u}^{2+}+H_{2}O\left(\text{excess}\right)\to \text{no change}$

$B{i}^{3+}+N{O}_3^{-}+H_{2}O\left(\text{excess}\right)\to BiO\left(N{O}_3\right)\left(\text{white}\right)+2H^{+}$

(B) Nitrates of $C{u}^{2+}$ gives deep blue coloration with excess of ammonium hydroxide while that of $B{i}^{3+}$ gives white color solution.

$C{u}^{2+}+4N{H}_{4}OH\to \left[Cu\right(N{H}_3{)}_4{]}^{-2}\left(\text{deep blue}\right)+4H_{2}O$

$B{i}^{3+}3N{H}_{4}OH\to Bi\left(OH{)}_3\right(\text{white ppt.})+3N{H}_4^{+}$

(C) Excess of $KI$ reacts with $C{u}^{2+}$ to form white colored solution of cupric iodide. $B{i}^{3+}$ reacts with excess of $KI$ to form $(Bi{I}_4{)}^{-}$ orange color solution.

$C{u}^{2+}+4I^{-}\to C{u}_2{I}_2\left(\text{white ppt.}\right)+I_2$

$B{i}^{3+}+4I^{-}\to \left[Bi{I}_4{]}^{-}\right(\text{orange colour solution})$