Which is/are correct statement?

Options

A. In less acidic solution `K_(2)Cr_(2)O_(7)` and `H_(2)O_(2)` gives violet coloured diamagnetic `[CrO(O_(2))(OH)]^(-) ion`.
B. In alkaline `H_(2)O_(2).K_(3)CrO_(8)`(with tetraperoxo species `[Cr(O_(2))_(4)]^(3-)` is formed with `K_(2)Cr_(2)O_(7).`
C. In ammonical solution of `H_(2)O_(2),(NH_(3))_(3)CrO_(4)` is formed with `K_(2)Cr_(2)O_(7)`.
D. `CrO_(4)^(2-)` change to `Cr_(2)O_(7)^(2-)` by oxidation.

Answer

ABC

Explanation

(a) Acidified `K_(2)Cr_(2)O_(7)` solution reacts with `H_(2)O_(2)` solution reacts with `H_(2)O_(2)` to give a deep blue solution due to the formation of `CrO_(5)`.
`Cr_(2)O_(7)^(2-)+2H^(+)+4H_(2)O_(2) to 2CrO_(5)+5H_(2)O`
(B) In alkaline medium with `30% h_(2)O_(2)`. a red -brown `K_(2)CrO_(8)` (diperox) is formed . it is tetra peroxo species `[Cr(O_(2))_(4)]^(3-)` amd thus the `Cr` is in `+V` oxidation state.
(C ) In ammoniacal solution a dark red-brown compound, `(NH_(3))_(3)CrO_(4)`-doperoxo compound with `Cr(IV)` is formed.
(D ) In `CrO_(4)^(2-)` the `Cr` is in its highest `+6` oxidation state. so it can not be further oxidised.