Correct Answer - A
`K_(4)Fe(CN_(6)) hArr 4 K^(+) + [Fe(CN)_(6)]^(4-)` :. (5 particles)
`Al_(2)(SO_(4))_(3) hArr 2Al^(3+) + 3 SO_(4)^(-2)`:. (5 particles)
`NaCl hArr Na^(+) + Cl^(-)`(2 particles)
`Al(NO_(3))_(3) hArr Al^(3+) + NO_(3)^(-)` (4 particles)
`Na_(2)SO_(4) hArr 2Na^(+) + SO_(4)^(-) -` (3 particles
`" Both" K_(4)Fe (CN)_(6) "and" Al_(2)(SO_(4)) ^(3)` give 5 ions on ionisation.