If one mole of original complex is treated with excess $P b (N O_{3})_{2}$ solution, then the number of moles of white precipitate formed will be

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Solution

The correct option is C $0.0$
Since on treatment with silver nitrate solution, a yellow precipitate (which is insoluble in ammonia) is obtained, iodine is present in the ionization sphere and chlorine is present in the coordination sphere.
Since chlorine is not present in the ionization sphere, it will not react with lead nitrate solution.
Hence, no white precipitate will be obtained.
In other words, zero moles of white precipitate will be formed.

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If one mole of original complex is treated with excess Pb(NO3)2 solution, then the number of moles of white precipitate formed will be

If one mole of original complex is treated with excess $P b (N O_{3})_{2}$ solution, then the number of moles of white precipitate formed will be