The correct option is A 5, 6, 3, 3
IO−3+aI−+bH+→cH2O+dI2
Step 1 : I−1→I2 (oxidation)IO−3→I2 (reduction)
Step 2: 2IO−3+12H+→I2+6H2O
Step 3:2IO−3+12j++10e→I2+6H2O2I−→I2+2e
Step 4: 2IO−3+12H++10e−→I2+6H2O[2I−→I2+2e]5
Step 5 : 2IO−3+10I−+12H+→6I2+6H2OIO−3+5I−+6H+→3I2+3H2O
On comparing, a = 5, b = 6, c = 3, d = 3