Acidified chromic acid $+ H_2O_2 \xrightarrow{\text{Org. solvent}} X + Y$ (blue colour). $X$ and $Y$ are:

Which from the following is the general formula of the compound obtained when lanthanoid $(Ln)$ reacts with carbon at elevated temperature?

The oxidising ability of the given anions follows the order:

$KMnO_4$ decomposes on heating at $513 \ K$ to form $O_2$ along with

In the scheme given below,$X$ and $Y$,respectively,are:
$\text{Metal halide} \xrightarrow{\text{aq. NaOH}} \text{White precipitate } (P) + \text{Filtrate } (Q)$
$P \xrightarrow[\text{heat}]{\text{aq. } H_2SO_4, PbO_2 (\text{excess})} X (\text{a coloured species in solution})$
$Q$ $\xrightarrow[\text{warm}]{\text{MnO(OH)}_2, \text{Conc. } H_2SO_4} Y (\text{gives blue-coloration with KI-starch paper})$

When acidified $K_2Cr_2O_7$ reacts with $H_2O_2$,a blue-coloured solution is obtained. This blue colour is due to the formation of: