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Chapter 20: Problem 72

The compound cisplatin, \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2},\) has been studied extensively as an antitumor agent. The reaction for the synthesis of cisplatin is: \(\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)\) Write the electron configuration for platinum ion in cisplatin. Most \(d^{8}\) transition metal ions exhibit square planar geometry. With this and the name in mind, draw the structure of cisplatin.

Short Answer

Expert verified
The electron configuration for the platinum ion (\(\operatorname{Pt}^{2+}\)) is: \[1s^2 \ 2s^2 2p^6 \ 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^9\] The structure of cisplatin (\(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2}\mathrm{Cl}_{2}\)) with square planar geometry is: ``` NH3 | Cl ----- Pt ----- Cl | NH3 ```

Step by step solution

01

Identify the oxidation state of platinum in cisplatin

The compound cisplatin contains Pt (platinum), \(\mathrm{NH}_3\) (ammonia) and \(\mathrm{Cl}_2\) (chlorine) molecules. By looking at the synthesis reaction: \(\mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q)\) We can identify that Pt is originally bonded with 4 Cl ions in the complex \(\mathrm{PtCl}_{4}^{2-}\). By this, we know that platinum is in the oxidation state of +2 or \(\operatorname{Pt}^{2+}\).
02

Write the electron configuration for platinum ion in cisplatin

To find the electron configuration for platinum ion (\(\operatorname{Pt}^{2+}\)), first, write the electron configuration for the neutral Pt atom. Neutral Pt has 78 electrons, and its electron configuration is: \[1s^2 \ 2s^2 2p^6 \ 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^9\] Now, to write the electron configuration for \(\operatorname{Pt}^{2+}\), we should remove 2 electrons from the neutral Pt atom. We always remove electrons from the highest principal quantum number (n). In this case, it's the 6s^2 orbitals. So the electron configuration for \(\operatorname{Pt}^{2+}\) becomes: \[1s^2 \ 2s^2 2p^6 \ 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 4f^{14} 5d^9\]
03

Draw the structure of cisplatin considering its square planar geometry

Since \(\operatorname{Pt}^{2+}\) is a \(d^8\) transition metal ion, it has a square planar geometry. In this structure, the platinum ion is at the center, and two NH3 molecules and two Cl ions surround it. The cisplatin molecule can be represented as Pt(NH3)2Cl2, with cis geometry having the two NH3 molecules adjacent to each other and the two Cl ions adjacent to each other as well. So the structure of cisplatin will look like: ``` NH3 | Cl ----- Pt ----- Cl | NH3 ``` The structure above represents cisplatin with a square planar geometry.

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