Reaction Analysis: Platinum Complex Substitution
The reaction involves the substitution of ligands in a Platinum(II) complex. The starting complex is tetraammineplatinum(II), $[Pt(NH_3)_4]^{2+}$. Hydrochloric acid, HCl, provides chloride ions ($Cl^-$).
Ligand Substitution Process
The reaction is:
$ [Pt(NH_3)_4]^{2+} + 2 HCl \rightarrow X + 2 NH_3 $
- The stoichiometry indicates that two ammine ($NH_3$) ligands are replaced by two chloride ($Cl^-$) ligands.
- The neutral complex formed has the formula $[PtCl_2(NH_3)_2]$.
Determining the Geometric Isomer (X)
The complex $[PtCl_2(NH_3)_2]$ is square planar and can exist as two geometric isomers: cis and trans.
- The formation of the trans isomer is generally favored in this reaction due to the trans effect.
- In $[Pt(NH_3)_4]^{2+}$, all ligands are identical. Upon substitution by the first $Cl^-$, a mono-chloro complex $[PtCl(NH_3)_3]^+$ is formed.
- The second substitution involves an incoming $Cl^-$ replacing an $NH_3$. The existing $Cl^-$ ligand directs the incoming $Cl^-$ to the trans position (as $Cl^-$ has a stronger trans effect than $NH_3$).
- This results in the two chloride ligands being positioned opposite each other (trans).
Therefore, X is trans-$[PtCl_2(NH_3)_2]$.
Conclusion
The product identified as X in the reaction $[Pt(NH_3)_4]^{2+} + 2 HCl$ is trans-$[PtCl_2(NH_3)_2]$.
Important Questions from Coordination Chemistry
An aqueous solution of $Co(ClO_4)_2 \cdot 6H_2O$ is light pink in colour. Addition of conc. HCl results in an intense blue coloured solution due to the formation of a new species. The new species among the following is
[Given: Atomic number of Co = 27]- Among the following, the compound with the lowest CO stretching frequency is
The rates of substitution for the following reaction vary with L in the order
- Among the following octahedral complexes, the one that has the highest enthalpy of hydration is
Among the given platinum(II) complexes, the one that is thermally the most unstable is
