N Factors Assignment Help
1.6 ‘N’ Factors
In Non Redox Change
(i)n-factor for element: Valency of the element
(ii) For Acids: Acids will be treated as species which furnish H+ ions when dissolved in a solvent. The n factor of an acid is the no. of acidic H+ ions that a molecule of the acid would give when dissolved in a solvent (Basicity).
e.g. for HCl (n = 1), HNO3 (n = 1), H2SO4 (n = 2), H3PO4 (n = 3) and H3PO3 (n = 2)
(iii) For Bases: Bases will be treated as species which furnish OH– ions when dissolved in a solvent. The n factor of a base is the no. of OH– ions that a molecule of the base would give when dissolved in a solvent (Acidity).
e.g. NaOH (n = 1), Ba(OH)2 (n = 2), al(OH)3 (n = 3), etc.
(iv) For Salts: A salts reacting such that no atom of the salt undergoes any change in oxidation state.
e.g. 2AgNO3 + MgCl2 ¾® Mg(NO3)2 + 2AgCl
In this reaction it can be seen that the oxidation state of Ag, N, O, Mg and Cl remains the same even in the product. The n factor for such a salt is the total charge on cation or anion.
In Redox Change
`For oxidizing agent or reducing agent n-factor is the change in oxidation number per mole of the substance.
(i) If we have a salt which react in a fashion that atoms of one of the element are getting oxidized and the atoms of another element are getting reduced and no other element on the reactant side is getting oxidized or reduced, than the n–factor of such a salt can be calculated either by taking the total number of moles of electrons lost or total number of mole of electrons gained by one mole of the salt.
For example, decomposition reaction of KClO3 is represented as
In this reaction, O2– is getting oxidized to O2 and Cl+5 is getting reduced to Cl–1. In each case, 6 mole of electrons are exchanged whether we consider oxidation or reduction.
n-factor of KClO3 considering oxidation = | 3 (–2) – 3 (0)| = 6
or n–factor of KClO3 considering reduction = | 1 ´ (+5) –1 ´ (–1)| = 6.
(ii) Disproportionation reactions in which moles of compound getting oxidsed and reduced are not same i.e. moles of oxidizing agent and reducing agent are not same.
For example,
6Br2 + 12 OH¯
10 Br¯ + 
+ 6H2O
In this reaction, the mole of electrons lost by the oxidation of some of the moles of Br3 are same as the number of mole of electrons gained by the reduction of rest of the moles of Br3. Of the 6 moles of Br2 used, one mole is getting oxidized, loosing 10 electrons (as reducing agent) and 5 moles of Br2 are getting reduced and accepts 10 moles of electrons (as oxidizing agent).
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