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Which oxidizing agent from the pair $\ce{K_2IrCl_6/K_2PtCl_6}$ is stronger? Explain your answer using their electron configurations.

We write the electron configurations for $\ce{Ir/Ir^{4+}}$ and $\ce{Pt/Pt^{+4}}$:

$$\begin{align} \ce{Ir:\ [Xe] 4f^{14}\ 5d^{7}\ 6s^2 \ &\implies \ Ir^{4+}:\ [Xe] 4f^{14}\ 5d^{5}}\\ \\ \ce{Pt:\ [Xe] 4f^{14}\ 5d^{8}\ 6s^2 \ &\implies \ Pt^{4+}:\ [Xe] 4f^{14}\ 5d^{6}} \end{align}$$

From those, we notice that $\ce{Ir^{4+}}$ has all 5d orbitals half-filled, whereas $\ce{Pt^{4+}}$ has 4 half-filled 5d orbitals and one filled entirely (which is a less stable configuration); so it's much easier for $\ce{Pt^{4+}}$ to get reduced than it is for $\ce{Ir^{4+}}$.

Therefore, the strongest oxidizing agent is $\ce{K_2PtCl_6}$. What's wrong with my solution?

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  • 2
    $\begingroup$ Read wikipedia's article on crystal field theory. If it won't be sufficient, come again. $\endgroup$ – permeakra Jan 23 '16 at 10:42

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