In k2Cr207, there is cange of state from [Cr* +6- Cr*+3 ] its seems that nfactor will easily be +3..but we must emphasis that here in 2 atoms of Cr so nfactor will be 2×3= 6..

Introduction:
The N-factor (or equivalent factor) of a compound in a redox reaction represents the number of moles of electrons gained or lost by one mole of the compound. It is used to balance redox reactions and calculate the stoichiometry of reactants and products.

Explanation:
To determine the N-factor of K2Cr2O7 in an acidic medium, we need to understand the redox reaction that occurs.

The given compound, K2Cr2O7, contains chromium (Cr) in its +6 oxidation state. In an acidic medium, the Cr(VI) is reduced to Cr(III). The balanced equation for this reaction is:

Cr2O7^2- + 14H^+ + 6e^- --> 2Cr^3+ + 7H2O

Step 1: Identify the change in oxidation state:
In the above equation, the chromium ion (Cr) changes from +6 to +3. This means that each Cr atom gains 3 electrons.

Step 2: Determine the N-factor:
The N-factor is equal to the number of electrons gained or lost per mole of the compound. Since each Cr atom in K2Cr2O7 gains 3 electrons, the N-factor for K2Cr2O7 is 3.

Conclusion:
The N-factor of K2Cr2O7 in an acidic medium is 3. This means that one mole of K2Cr2O7 will gain 3 moles of electrons during a redox reaction.