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 $\ce{[Ni(H_{2}O)_{6}]^{2+}}$ (or we can consider the hydrated ion as $\ce{Ni^{2+}(aq)}$ for simplicity in the formation constant, but the complex with four cyanide is $\ce{[Ni(CN)_{4}]^{2-}}$? Wait, no, nickel(II) with cyanide: the oxidation state of Ni is +2, cyanide is -1, so four cyanide would give a charge of \(2 + 4\times(-1)= -2\)? Wait, no, the formation of the complex: the hydrated metal ion is $\ce{Ni(H_{2}O)_{6}^{2+}}$ (hexaaquanickel(II) ion), and it reacts with cyanide ions to form the tetracyanonickel(II) complex. The overall formation reaction is $\ce{Ni^{2+}(aq) + 4CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. Wait, but the formation constant for the overall reaction? Wait, no, the question is about the equilibrium between the hydrated metal ion and the aqueous complex. Wait, maybe the hydrated metal ion is $\ce{Ni(H_{2}O)_{6}^{2+}}$, but when forming the complex with four CN⁻, the reaction would be $\ce{Ni(H_{2}O)_{6}^{2+}(aq) + 4CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq) + 6H_{2}O(l)}$. But the formation constant expression for a complex formation reaction is the concentration of the complex divided by the product of the concentrations of the metal ion and the ligands, each raised to their stoichiometric coefficients. So for the reaction $\ce{Ni^{2+}(aq) + 4CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$, the formation constant \(K_f=\frac{[\ce{Ni(CN)_{4}^{2-}}]}{[\ce{Ni^{2+}}][\ce{CN^{-}}]^{4}}\). Wait, but if the hydrated metal ion is considered as $\ce{Ni(H_{2}O)_{6}^{2+}}$, then the reaction would be $\ce{Ni(H_{2}O)_{6}^{2+}(aq) + 4CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq) + 6H_{2}O(l)}$, and the formation constant would be \(K_f=\frac{[\ce{Ni(CN)_{4}^{2-}}]}{[\ce{Ni(H_{2}O)_{6}^{2+}}][\ce{CN^{-}}]^{4}}\), but usually, the hydrated metal ion is represented as $\ce{Ni^{2+}(aq)}$ for simplicity in formation constant expressions, assuming the water is the solvent and its concentration is constant (or incorporated into the constant). So first, let's confirm the complex: Ni²⁺ forms a complex with four CN⁻, so the complex is $\ce{[Ni(CN)_{4}]^{2-}}$ (since Ni is +2, 4 CN⁻ are -4, so overall charge is -2? Wait, no: 2 + 4*(-1) = -2? Wait, Ni²⁺ is +2, CN⁻ is -1, so 4 CN⁻ is -4, so the complex charge is 2 - 4 = -2. So the complex is $\ce{[Ni(CN)_{4}]^{2-}}$.

Step2: Write the formation constant expression

The formation reaction is $\ce{Ni^{2+}(aq) + 4CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. The formation constant \(K_f\) is the ratio of the concentration of the complex to the product of the concentrations of the metal ion and the ligands, each raised to their stoichiometric coefficients. So \(K_f=\frac{[\ce{Ni(CN)_{4}^{2-}}]}{[\ce{Ni^{2+}}][\ce{CN^{-}}]^{4}}\).

Step3: Determine the last step in complex formation

The formation of a complex with four ligands (cyanide) occurs in steps. The first step is the addition of one CN⁻: $\ce{Ni^{2+}(aq) + CN^{-}(aq) <=> [Ni(CN)]^{+}(aq)}$, second step: $\ce{[Ni(CN)]^{+}(aq) + CN^{-}(aq) <=> [Ni(CN)_{2}](aq)}$, third step: $\ce{[Ni(CN)_{2}](aq) + CN^{-}(aq) <=> [Ni(CN)_{3}]^{-}(aq)}$, fourth (last) step: $\ce{[Ni(CN)_{3}]^{-}(aq) + CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$. So the last step is the addition of the fourth cyanide ion to the tris(cyano)nickelate(II) ion to form the tetracyanonickelate(II) ion.

Step4: Write the balanced chemical equation for the last step

The last step is the reaction between $\ce{[Ni(CN)_{3}]^{-}(aq)}$ and $\ce{CN^{-}(aq)}$ to form $\ce{[Ni(CN)_{4}]^{2-}(aq)}$. So the balanced equation is $\ce{[Ni(CN)_{3}]^{-}(aq) + CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}$.

Answer:

For the formation constant expression: \(K_f = \frac{[\ce{Ni(CN)_{4}^{2-}}]}{[\ce{Ni^{2+}}][\ce{CN^{-}}]^{4}}\)

For the last step: \(\ce{[Ni(CN)_{3}]^{-}(aq) + CN^{-}(aq) <=> [Ni(CN)_{4}]^{2-}(aq)}\)