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What is the oxidation number of the central metal atom in the coordination compound {Pt(NH₃)3Cl}Cl?

(a) -1

(b) 0

(c) +1

(d) +2

(e) +3

1 Answer

1 vote

Final answer:

The oxidation number of platinum in the coordination compound {Pt(NH3)3Cl}Cl is +2, based on the charge of the ligands and the overall charge of the compound. However, due to a discrepancy in the question, the provided formula should likely be [Pt(NH3)2Cl2]+ to align with the correct oxidation state of +2.

Step-by-step explanation:

The oxidation number of the central metal atom in the coordination compound {Pt(NH3)3Cl}Cl can be determined by looking at the overall charge and the charges of each ligand. Ammonia (NH3) is a neutral ligand and does not contribute to the oxidation state of the central metal, platinum in this case. However, there is one chloride ion (Cl-) within the coordination sphere, contributing a -1 charge.

The entire coordination compound has a net charge of +1 since it is a cationic complex (with an outside Cl- as a counter ion). Therefore, to find the oxidation state of the platinum (Pt), we start with the overall charge on the coordination sphere and subtract the charge of the chloride ligand: +1 = oxidation state of Pt (x) + (-1), thus the oxidation state of Pt is +2. However, there seems to be a typo in the question presented. The formula given should be [Pt(NH3)2Cl2]+ to get the correct oxidation state of +2 for Pt. The coordination sphere provided in the question does not match the oxidation states provided in the answers (c) +3 or (e) +1. With the data provided, the correct oxidation state should be +2 for Pt in a hypothetical accurate formula [Pt(NH3)2Cl2]+. Please note that the example and the student's question do not seem to align perfectly, and there might be some confusion with the given compound formula.

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