wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The correct descending order of oxidising power of the following is:

A
Cr2O27>MnO4>VO+2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
MnO4>Cr2O27>VO+2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
VO+2>MnO4>Cr2O27
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
MnO4>VO+2>Cr2O27
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
Cr3O27>VO+2>MnO4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A MnO4>Cr2O27>VO+2
The ions in which the central metal atom is present in the highest oxidation state will have the highest oxidizing power. In VO+2, vanadium is present in the +5 oxidation state, while in Cr2O27 ion, Cr is present in the +6 oxidation state. Similarly in the MnO4, Mn is present in the +7 oxidation state.

Oxidizing power depends upon the stability of the reduction product obtained. Since lower oxides of Mn are more stable than that of Cr and V.

So the order of increasing stability of the reduced species is Mn2+>Cr3+>V3+

The correct descending order of oxidizing power is:
MnO4>Cr2O27>VO+2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrochemical series_tackle
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon