Correct Answer - b
(A) gives white ppt `NH_(4)CI and NH_(4)OH`
(A) is `AI^(3+)` and ppt is of `AI(OH)_(3)`
`AI(OH)_(3)` is soluble in (B) `rArr` (B) is `NaOH`
(C ) is `NaAIO_(2)` (solium meta aluminate)(A) form dimer indicatting (A) is electrons whichin the possible case of `ICI_(3)`
(A) is thus `AICI_(3)` (acidic due to hydrolysis turn blue lithium red)
(B) is `NaOH` (alkline turns red litmus blue)
`underset((A))(AICI_(3))+underset((B))(3NaOH)rarr underset("White ppt")(AI(OH)_(3))+3NaCI`
`AI(OH)_(3) + underset(("Excess"))(NaOH)-underset("Soluble" (C))(NaAIO_(2))+H_(2)O`
`AICI_(3) +3H_(2)O rarr AI(OH)_(3) +3HCI darr` (fumes)
`AICI_(3) + 3NH_(4)OH overset(NH_(4)CI)rarr AI(OH)_(3) darr` (fumes)
`2AICI_(3) rarr AI_(2)CI_(6)`
