The correct order of increasing oxidation number of the transition metal ions for the compounds K₂Cr₂O₇, V₂O₅ and KMnO₄ is

V₂O₅ < K₂Cr₂O₇ < KMnO₄

K₂Cr₂O₇ < KMnO₄ < V₂O₅

KMnO₄ < K₂Cr₂O₇ < V₂O₅

KMnO₄ < V₂O₅ < K₂Cr₂O₇

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bennycephas247

4 years ago

v2......+5
Cr.......+6
Mn.......+7

chinenye12345

1 year ago

The transition elements in the compounds k2Cr2O7,V205 and kMnO4 are Cr,v,MN respectively.

The oxidation number of Cr in k2Cr2O7 is +1*2+X*2+(-2*7)=0
2+2x+-14=0
2x+2-14=0
2x-12=0
2x=12
2x/2=12/2
x=6
Cr=6

V205
x2+(-2*5)=0
x2-10=0
x2/2=10/2
X=5
v=5

kMnO4
+1+x+-2*4=0
+1+x-8=0
X-7=0
X=+7
Mn=+7
A is the correct option that arranges the oxidation numbers in an increasing order.