Question

AgCl dissolved in excess $N{H}_3$, $KCN$ and $N{a}_2{S}_2{O}_3$ solutions the complex produce ions.

• A. $\left[Ag\right(N{H}_3{)}_2{]}^{+},\left[Ag\right(CN{)}_2{]}^{-},\left[Ag\right(S_2{O}_3{)}_2{]}^{-3}$
• B. $\left[Ag\right(N{H}_3{)}_2{]}^{+2},\left[Ag\right(CN{)}_2{]}^{-3},\left[A{g}_4\right(S_2{O}_3{)}_2{]}^{+3}$
• C. $\left[Ag\right(N{H}_3{)}_2{]}^{+2},\left[Ag\right(CN{)}_2{]}^{+},\left[A{g}_2\right(S_2{O}_3{)}_2{]}^{-2}$
• D. $\left[Ag\right(N{H}_3{)}_4{]}^{+},\left[Ag\right(CN{)}_4{]}^{-3},\left[Ag\right(S_2{O}_3{)}_2{]}^{-2}$

Answer 1

The correct option is A

$\left[Ag\right(N{H}_3{)}_2{]}^{+},\left[Ag\right(CN{)}_2{]}^{-},\left[Ag\right(S_2{O}_3{)}_2{]}^{-3}$

Co-ordination number of $A{g}^{+}=2$ and based on oxidation states of ligands, charges are indicated. Silver exists in the oxidation state of +1.

$AgCl+2(N{H}_3.H_{2}O)\to \left[Ag\right(N{H}_3{)}_2]Cl+2H_{2}O$

$AgCl+2KCN\to K\left[Ag\right(CN{)}_2]+KCl$

$AgCl+2N{a}_2{S}_2{O}_3\to N{a}_3\left[Ag\right(S_2{O}_3{)}_2]+NaCl$