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Last updated date: 05th Dec 2023
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# In the complex $[Pt{{(py)}_{4}}][PtC{{l}_{4}}]$ ,the oxidation number of $Pt$ atom in former and later part of the compound are respectively :(A) $0\text{ and 0}$ (B) $+4\text{ and +2}$ (C) $+2\text{ and +2}$ (D) $0\text{ and +4}$

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Hint : The oxidation number of a complex ion or a complex compound is the charge carried by that specific ion in terms of integers. It results because of the presence of extra electrons, which corresponds to negative oxidation state, or the absence of electrons, which corresponds to the positive oxidation state.

Firstly see the later part which is $[PtC{{l}_{4}}]$ now as we know that the chlorine exhibit $-1$ charge, so the charge due to four chlorine will be $-4$ and $py$ doesn’t exhibit any charge. So the platinum has to counterbalance the negative charge due to chloride ions, so it will do one thing: let the charge on $Pt$ be x. As the compound is neutral, therefore, total charge is $0$ .Chlorine has $-1$ charge.
So we will just do the normal calculation $2x-4=0$ as the whole compound is neutral
So we will get our desired answer which is $x=2$