Home
Class 11
CHEMISTRY
PbCl(4) exists but PbBr(4) and PbI(4) do...

`PbCl_(4)` exists but `PbBr_(4)` and `PbI_(4)` do not because of

A

bromine and iodine are more electronegative these chlorine.

B

iodine and bromine are smaller in size.

C

larger iodine and bromine are able to reduce `Pb^(4+)` to `Pb^(2+)` or `Pb`.

D

the statement is incorrect

Text Solution

Verified by Experts

The correct Answer is:
C
`Pb^(4+)` has higher polarising power and `Br^(-)` and `I^(-)` being larger in size can easily give the electrons to `Pb^(4+)`, i.e. as compared to `Cl^(-),Br^(-)` and `I^(-)` are good reducing agents.
Promotional Banner

Topper's Solved these Questions

  • P BLOCK ELEMENTS ( Group 13 -14)

    A2Z|Exercise Compound Of Carbon Family|52 Videos

  • P BLOCK ELEMENTS ( Group 13 -14)

    A2Z|Exercise Section B - Assertion Reasoning|21 Videos

  • P BLOCK ELEMENTS ( Group 13 -14)

    A2Z|Exercise Compound Of Boron Family|57 Videos

  • MOCK TEST

    A2Z|Exercise Mock Test 2|45 Videos

  • REDOX REACTIONS

    A2Z|Exercise Section D - Chapter End Test|29 Videos

Similar Questions

Explore conceptually related problems

PbF_(4),PbCl_(4) exist but PbBr_(4) and Pbl_(4) do not exist because of -

iv. PbR_4 and PbCl_4 exists , but PbBr_4 and Pbl_4 do not exist because of

PbI_(4) does not exist because:

PbF_(4), PbCl_(4) exist but PbBr_(4) do not exist because of

OF_(2) exists but OF_(4) does not because:

Assertion : PbI_(4) doesn't exist and converts into PbI_(2) and I_(2) spontaneously at room temperature but PbCl_(4) needs heatin to convert into PbCl_(2) and Cl_(2) . Reason : Pb^(2+) is more stable than Pb^(4+) due to inert pair effect.