Question

In which of the following reaction, oxidation number of N remains unchanged?

Solution

Correct option is

The oxidation number of N in NO2 is +4 and in N2O4 is also +4 (i.e. no change)

SIMILAR QUESTIONS

Q1

To neutralize completely 20ml of 0.1 M phosphorous acid 40ml of KOH was required. What volume of this KOH solution will be required to neutralize 0.66 g of H3PO2?

Q2

10ml of KHC2O4 solution requires 10 ml of NaOH for complete neutralization. 10 ml of same KHC2O4 solution is oxidized by 10 ml of KMnO4 solution in acidic medium. Molarity of KMnO4 solution is

Q3

The molarity, molality and density of a solution are denoted by ω, x and yg ml–1 respectively. If molecular mass of the solute is z, then which of the following relation is correct? 

Q4

KMnO4 oxidizes Xn+ into,  itself changing to Mn2+ in acid solution. 2.68 × 10–3 mol of Xn+ require  mol of KMnO4. The value of n is

Q5

To 1 mole of pyrophosphoric acid (H4P2O7) in aqueous solution 2 mole of NaOH are added. The salt formed is

Q6

What volume of 0.01 M K2Cr2O7 would be required to oxidize Fe (II) in 50ml of 0.03 M solution of ferrous ammonium sulphate in acidic medium?

Q7

100ml of 0.01 M KMnO4 oxidizes 10ml of H2O2 sample in acidic medium. The volume strength of H2O2 sample is

Q8

A sample of mixture of AO and A2O3 in 2 : 1 molar ratio takes 0.015 mole of K2Cr2O7 to oxidize the sample completely to from  in acidic solution. Millimoles of AO and A2O3 in the sample are respectively

Q9

0.01 mole of Fe3O4 is treated with excess of KI solution in presence of dilute H2SO4 whereby iron is reduced to Fe2+ along with liberation of iodine. What volume of 0.1 M Na2S2O3 will be needed to reduce the librated iodine?

Q10

 

KMnO4 acts as a strong oxidizing agent in acidic medium. The reaction that occurs is

What will be the effect on the electrode potential of a half-cell represented as  if the concentration of H+ ion is doubled in solution?