Question

${\mathrm{PbI}}_4$ Does Not Exist Because?

Answer 1

${\mathrm{PbI}}_4$ does not exist :

• ${\mathrm{PbI}}_4$ does not exist as the iodine present reduces the Lead$\mathrm{Pb}\left(\mathrm{IV}\right)\mathrm{to}\mathrm{Pb}\left(\mathrm{II}\right)$.
• Here the formation $\mathrm{Pb}\left(\mathrm{II}\right)$ is more stable due to the inert pair effect.
• Also this process $\mathrm{Pb}\left(\mathrm{IV}\right)$acts as a strong oxidizing agent which helps in oxidizing the Iodine to iodine$\left({\mathrm{I}}_2\right)$.
• Since the Iodine is not known as a strong reducing agent to reduce$\mathrm{Pb}\left(\mathrm{II}\right)\mathrm{to}\mathrm{Pb}$, the compound ${\mathrm{Pbl}}_2$ gets formed, instead of ${\mathrm{PbI}}_4$.
• The reaction for the above process can be given as :
• $\underset{\left(\mathrm{Lead}\mathrm{ion}\right)}{{\mathrm{Pb}}^{+4}\left(\mathrm{aq}\right)}+\underset{\left(\mathrm{Iodine}\mathrm{ion}\right)}{4{\mathrm{I}}^{-}\left(\mathrm{aq}\right)}\to \underset{\left(\mathrm{Lead}\mathrm{iodide}\right)}{{\mathrm{PbI}}_2\left(\mathrm{aq}\right)}+\underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2\left(\mathrm{g}\right)}$

Hence, the reason why ${\mathrm{PbI}}_4$ does not exist has been explained above.