The question is asking to explain why PbCl4 exists, while PbBr4 and PbI4 do NOT

Question

The question is asking to explain why PbCl4 exists, while PbBr4 and PbI4 do NOT exist. Here is what I have found, is this correct?

Upon ionization: PbI4----> Pb^4+ + 4I^-

The I^- wil be oxidized, and reduce Pb^4+ to Pb^2+ and in the process I2 will form. This is because the oxidation potential of I^- is much higher than that of Cl^- (since the oxidation potential of Cl- is relatively low, it does not act as a reducing agent to reduce Pb^4+).

(The exact same reasoning is applied to why PbBr4 does not exist also)

Explanation / Answer

Due to bigger size of these ions, they have more reducing power(i.e. I- reduces Pb(+4) to Pb(+2) and itself oxidises from I- to I2). Similar reason for PbBr4.

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