MCQ
$PbI_4$ does not exist because
- Aiodine is not a reactive
- ✓$Pb(IV)$ is oxidizing and $I^-$ is strong reducing agent
- C$Pb(IV)$ is less stable than $Pb(II)$
- D$Pb^{4+}$ is not easily formed
✓
Answer
Correct option: B.
$Pb(IV)$ is oxidizing and $I^-$ is strong reducing agent
b
$PbI_4$ does not exist because the iodine reduces the lead to $Pb\,(II)$ and the $Pb$ oxidizes the iodine to iodine$(I_2)$. Since the iodine is not a strong reducuing agent to reduce $Pb ( II )$ to $Pb$, the compound $PbI_2$ is formed.
$PbI_4$ does not exist because the iodine reduces the lead to $Pb\,(II)$ and the $Pb$ oxidizes the iodine to iodine$(I_2)$. Since the iodine is not a strong reducuing agent to reduce $Pb ( II )$ to $Pb$, the compound $PbI_2$ is formed.
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