Question

Why PbF4 exist but PbI4 doesn't

Answer 1

PbI4 does not exist because the iodine reduces the lead to Pb(II) and the Pb oxidizes the iodine to iodine(I2). Since the iodine is not a strong reducuing agent to reduce Pb(II) to Pb, the compound Pbl2 is formed.

Answer 2

$\bold{\mathrm{PbF}_{4}}$ exists but "$\mathrm{Pbl}_{4}$ does not exist" because Iodine reduces Pb or lead to Pb(II).

Explanation:

• On the other hand, Pb oxidizes Iodine or I to $I_{2}$. Iodine is not a strong reducing agent and so cannot reduce Pb(II) to Pb. This leads to the formation of ionic compound $\bold{\mathrm{Pbl}_{2}}$.

• Also $\mathrm{P} \mathrm{b}_{4+}$ is a very strong acid or an oxidant and takes up electrons from the soft base $I^{-}$ to oxidise it and lead to the formation of $I_{2}$.

• Also $\mathrm{Pb}_{4}$ will reduce itself to become $\mathrm{Pb}_{2+}$ leading to the formation of $\mathrm{Pbl}_{2}$.  

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