The oxidation states of transition metal atoms in K₂Cr₂O₇, KMnO₄ and K₂FeO₄, respectively, are x,y and z. The sum of x,y and z is [numerical value].

K₂Cr₂O₇ + H₂O₂ + H₂SO₄->  + K₂SO₄ + H₂O compound 'X' is ->CrO₅

Oxidation state of chromium = +6.

Potassium permanganate in alkaline medium oxidise lodide to lodate.

$2Mn{O}_4^{-}+H_{2}O+I^{\Theta }\stackrel{}{\to }2Mn{O}_2+2O{H}^{\Theta }+I{O}_3^{\Theta }(A)$            

Compound A is  . Therefore, oxidation state of 1is + 5.

KMnO₄ decomposes upon heating at 513 K and forms K₂MnO₄ and MnO₂.

2KMnO₄  K₂MnO₄ + O₂ + MnO₂

The colour changes due to change in pH not due to its reaction with water.