The rate of the substitution reaction of \([Co(CN)_5Cl]^3\)\(^-\) with OH\(^-\) to give \([Co(CN)_5(OH)]^3\)\(^-\) follows:
depends on the concentrations of both [Co(CN)\(_5\)Cl]\(^3\)\(^-\) and OH\(^-\)
depends on the concentration of [Co(CN)\(_5\)Cl]\(^3\)\(^-\) only
Step 1: Identifying the nucleophile and the complex.
- The complex \([Co(CN)_5Cl]^{3-}\) contains a chloride ion (Cl\(^-\)) that is substituted by OH\(^-\), which is the nucleophile in the reaction.
- The rate of the reaction will primarily depend on the availability of the complex \([Co(CN)_5Cl]^{3-}\) and the nucleophile OH\(^-\). The more of the complex present, the more collisions there will be between the complex and OH\(^-\), increasing the rate of the substitution.
Step 2: Understanding the mechanism.
In most substitution reactions in coordination compounds, the rate is governed by the concentration of the complex undergoing substitution. For this reaction, OH\(^-\) acts as a nucleophile, but the reaction rate is primarily dependent on the concentration of the complex \([Co(CN)_5Cl]^{3-}\). This is because the substitution occurs when OH\(^-\) attacks the metal center, displacing the chloride ion. The rate law for such a substitution would be first order with respect to \([Co(CN)_5Cl]^{3-}\), meaning the reaction rate increases with the concentration of the complex.
Step 3: Conclusion.
Since the rate of the substitution reaction is dependent on the concentration of the complex \([Co(CN)_5Cl]^{3-}\), the correct answer is (B).
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Courage : Bravery :: Yearning :
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