Question

in aqueous solution the ni²⁺ ion forms a complex with four cyanide anions. write the formation constant expression for the equilibrium between the hydrated metal ion and the aqueous complex. under that, write the balanced chemical equation for the last step in the formation of the complex. k_f = write the last step: last step:

Explanation:

Step1: Determine the complex formation

The hydrated nickel(II) ion is $\ce{[Ni(H_{2}O)_{6}]^{2+}}$ (we can simplify the hydrated form as $\ce{Ni^{2+}(aq)}$ for the formation constant expression considering the aqua ligands are part of the hydrated ion, but the complex with cyanide: the overall complex is $\ce{[Ni(CN)_{4}]^{2-}}$? Wait, no, $\ce{Ni^{2+}}$ with four $\ce{CN^-}$: the charge is $2 + 4\times(- 1)=- 2$, so the complex is $\ce{[Ni(CN)_{4}]^{2-}}$. The formation reaction from the hydrated metal ion (let's represent the hydrated $\ce{Ni^{2+}}$ as $\ce{Ni^{2+}(aq)}$ for simplicity, considering the water ligands are in excess or part of the aqueous environment) and four $\ce{CN^-}$ ions: $\ce{Ni^{2+}(aq) + 4CN^-(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. The formation constant $K_f$ is the ratio of the concentration of the complex to the product of the concentrations of the reactants, each raised to their stoichiometric coefficients. So $K_f=\frac{[\ce{Ni(CN)_{4}}^{2 - }]}{[\ce{Ni^{2+}}][\ce{CN^-}]^4}$.

Step2: Last step in complex formation

The complex formation of $\ce{[Ni(CN)_{4}]^{2-}}$ occurs in steps. The last step would be the addition of the fourth $\ce{CN^-}$ to the tris(cyano)nickel(II) complex. The previous step is $\ce{[Ni(CN)_{3}]^-(aq) + CN^-(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. Wait, let's check the charges: $\ce{Ni^{2+}}$ + 1 $\ce{CN^-}$: $\ce{[Ni(CN)]^+}$; +2nd $\ce{CN^-}$: $\ce{[Ni(CN)_{2}]}$ (charge 0? No, $\ce{Ni^{2+}}$ + 2 $\ce{CN^-}$: charge 0? Wait, no, $\ce{Ni^{2+}}$ has charge +2, each $\ce{CN^-}$ is - 1. So first step: $\ce{Ni^{2+} + CN^- <=> [Ni(CN)]^+}$ (charge +1). Second step: $\ce{[Ni(CN)]^+ + CN^- <=> [Ni(CN)_{2}]}$ (charge 0). Third step: $\ce{[Ni(CN)_{2}] + CN^- <=> [Ni(CN)_{3}]^-}$ (charge - 1). Fourth step: $\ce{[Ni(CN)_{3}]^- + CN^- <=> [Ni(CN)_{4}]^{2-}}$ (charge - 2). So the last step is the addition of the fourth cyanide ion to the $\ce{[Ni(CN)_{3}]^-}$ complex: $\ce{[Ni(CN)_{3}]^-(aq) + CN^-(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. But maybe the problem considers the hydrated ion as $\ce{[Ni(H_{2}O)_{6}]^{2+}}$, so the last step would be replacing the last water ligand? Wait, no, the complex is with four $\ce{CN^-}$, so the hydrated $\ce{[Ni(H_{2}O)_{6}]^{2+}}$ reacting with $\ce{CN^-}$ in steps. But the problem says "the last step in the formation of the complex" when forming $\ce{[Ni(CN)_{4}]^{2-}}$ from $\ce{Ni^{2+}}$ and four $\ce{CN^-}$. Alternatively, if we consider the overall formation, but the last step is the addition of the fourth $\ce{CN^-}$ to the $\ce{[Ni(CN)_{3}]^-}$ complex. But maybe the problem expects the overall formation reaction's last step as the addition of the fourth $\ce{CN^-}$ to the $\ce{Ni^{2+}}$ with three $\ce{CN^-}$ already bound. So the last step reaction is $\ce{[Ni(CN)_{3}]^-(aq) + CN^-(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. But let's confirm the charge: $\ce{[Ni(CN)_{3}]^-}$ has charge $2 + 3\times(-1)= - 1$, adding $\ce{CN^-}$ (charge - 1) gives a complex with charge $-1+(-1)= - 2$, so $\ce{[Ni(CN)_{4}]^{2-}}$, which matches.

Answer:

For $K_f$: $\boldsymbol{K_f=\frac{[\ce{Ni(CN)_{4}}^{2 - }]}{[\ce{Ni^{2+}}][\ce{CN^-}]^4}}$

For the last step: $\boldsymbol{\ce{[Ni(CN)_{3}]^-(aq) + CN^-(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}}$ (or if we consider the hydrated ion as $\ce{[Ni(H_{2}O)_{6}]^{2+}}$, but the simpler form is as above. Alternatively, if the hydrated ion is represented as $\ce{Ni^{2+}(aq)}$, but the stepwise formation's last step is the addition of the fourth $\ce{CN^-}$ to the tris(cyano) complex.)