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JEE Advanced levelSingle correct optionDifficulty 3 of 3bn220

Red lead is described as lead(II) plumbate(IV), $Pb X_{2} [PbO X_{4}]$ = $Pb X_{3} O X_{4}$ (O = $- 2$ throughout). Which option is correct?

A2 Pb(II) ions and a plumbate(IV) anion of charge $2 -$
B3 Pb(III) ions with no anion
C1 Pb(II) ion and a plumbate anion of charge $2 -$ with Pb(IV)
D2 Pb(II) ions and a plumbate(IV) anion of charge $4 -$

Answer: D

Plumbate(IV):

PbO X_{4} X^{4 -}

since

+ 4 - 8 = - 4

. Two

Pb X^{2 +}

balance it:

2 (+ 2) - 4 = 0

. Total O is 4, so

Pb X_{3} O X_{4}

with average Pb

8/3

.

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