Question

Ni(CN)42- with Br2 and NaOH

Answer 1

The balanced chemical equation for the reaction is:

$\text{Ni(CN)}_4^{2-} + \text{Br}_2 + 2\text{OH}^- \rightarrow \text{Ni(CN)}_4(\text{OH})_2^{2-} + 2\text{Br}^-$

Answer 2

The reaction involves the oxidation of the central metal ion, Nickel(II), in the complex $\text{Ni(CN)}_4^{2-}$ to a higher oxidation state, Nickel(IV). Bromine ($\text{Br}_2$) acts as the oxidizing agent in the presence of a strong base ($\text{NaOH}$).

1. Formation of the Oxidizing Agent:

In an alkaline medium, bromine disproportionates to form bromide ($\text{Br}^-$) and hypobromite ($\text{BrO}^-$) ions. The hypobromite ion is a strong oxidizing agent.

$\text{Br}_2 + 2\text{OH}^- \rightarrow \text{Br}^- + \text{BrO}^- + \text{H}_2\text{O}$

2. Oxidation of the Nickel Complex:

The $\text{Ni(II)}$ (d$^8$) in $\text{Ni(CN)}_4^{2-}$ is oxidized to $\text{Ni(IV)}$ (d$^6$). Since the reaction occurs in an alkaline medium, hydroxide ions ($\text{OH}^-$) can act as ligands, increasing the coordination number from 4 (square planar) to 6 (octahedral).

• Oxidation half-reaction:

$\text{Ni(CN)}_4^{2-} + 2\text{OH}^- \rightarrow \text{Ni(CN)}_4(\text{OH})_2^{2-} + 2e^-$

(Here, Ni changes oxidation state from +2 to +4)

• Reduction half-reaction:

The hypobromite ion is reduced to bromide ion.

$\text{BrO}^- + \text{H}_2\text{O} + 2e^- \rightarrow \text{Br}^- + 2\text{OH}^-$

3. Overall Reaction:

Adding the oxidation and reduction half-reactions:

$\text{Ni(CN)}_4^{2-} + \text{BrO}^- + \text{H}_2\text{O} \rightarrow \text{Ni(CN)}_4(\text{OH})_2^{2-} + \text{Br}^-$

Now, substitute $\text{BrO}^-$ from the first step ($\text{BrO}^- = \text{Br}_2 + 2\text{OH}^- - \text{Br}^- - \text{H}_2\text{O}$):

$\text{Ni(CN)}_4^{2-} + (\text{Br}_2 + 2\text{OH}^- - \text{Br}^- - \text{H}_2\text{O}) + \text{H}_2\text{O} \rightarrow \text{Ni(CN)}_4(\text{OH})_2^{2-} + \text{Br}^-$

Simplifying the equation:

$\text{Ni(CN)}_4^{2-} + \text{Br}_2 + 2\text{OH}^- \rightarrow \text{Ni(CN)}_4(\text{OH})_2^{2-} + 2\text{Br}^-$

The product $\text{Ni(CN)}_4(\text{OH})_2^{2-}$ is an octahedral Nickel(IV) complex.

Explanation of the solution:

Nickel(II) in $\text{Ni(CN)}_4^{2-}$ is oxidized to Nickel(IV) by bromine in alkaline solution. Bromine disproportionates in $\text{NaOH}$ to form hypobromite ($\text{BrO}^-$), which is the active oxidizing agent. Hydroxide ions from $\text{NaOH}$ also act as ligands, forming an octahedral $\text{Ni(IV)}$ complex, $\text{Ni(CN)}_4(\text{OH})_2^{2-}$, and bromide ions.