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n-Factor or Valence Factor - (Concept)

n-Factor or Valence Factor: 
It is very important for both redox reactions and non-redox reactions by which we can obtain the following information:

  • It calculates the molar ratio of the species taking part in reactions that are, reactants. The reciprocal of the n-factor 's ratio of the reactants represents the molar ratio Of the reactants. For example, If A (having n-factor = a) reacts with B (having n-factor = b) then its n-factor's ratio is a: b, so molar ratio of A to B is b: a. 
    It can be represented as follows: 
    \\\mathrm{bA\,\: \: \: \: \: \: \: \: \:\: \: \: \: \: +\: \: \: \: \: \: \: \: \: \: \: \, aB\, \rightarrow \, Product}\\(\mathrm{n-factor=\, a)\:\: \: \: \: \: (n-factor=\, a)}
     
  • Equivalent weight = (Molecular weight) / n -factor
                               or
                              = (Atomic weight) / n -factor

Calculation of n-Factor
Before calculating the n-factor of any of the reactants in a given chemical reaction we must have a clear idea about the type of reaction. The reaction may be any of these types: 

(i) Acid-base or neutralization reaction

(li) Redox reaction 

(iii) Precipitation or double decomposition reaction

  • Acid-Base or Neutralization Reactions:
    As we know that according to the Arrhenius concept, "An acid provides H+ ion(s) while a base provides OH- ion(s) in neutralization these H+ and OH- ion/ions combines together".
    The number of H+ ion(s) and OH- ion(s) represent n-factor for acid and base respectively, that is, basicity and acidity respectively. 
    Example, 
    \\ \mathrm{HX\, \rightarrow \, H^{+}\, +\, Cl^{-} }\\ (\mathrm{n=1)\, that\, is,\, monobasic\, acid} \\\\ \mathrm{H_{2}SO_{4}\, \rightarrow \, 2H^{+}\, +\, SO_{4}^{2-}} \\ \mathrm{(n=2)\, that\, is,\, dibasic\, acid}

 

  • Redox Reactions
    These reactions involve oxidation and reduction simultaneously. Here the exchange of electrons occurs. To find the n-factor for Oxidizing or agent we must find out the change in the oxidation state of these species.

    Example(1): When only one atom undergoes oxidation or reduction.
    \\\mathrm{\overset{+3\times3}{C_{2}O_{4}^{2-}}\rightarrow \overset{+4\times2}{2CO_{2}}}\\\textrm{n=2}\\\mathrm{n-factor=\left | (+4)\, \times\, 2\, -(+3)\, \times\, 2\right |=2}\\\\\mathrm{\overset{+6\times2}{Cr_{2}O_{7}^{2-}}\rightarrow \overset{+3\times2}{2Cr^{3+}}}\\\textrm{n=6}\\\mathrm{n-factor=\left | (+3)\, \times\, 2\, -(+6)\, \times\, 2\right |=6}

    Example(2): For the salt which reacts in such a way that one atom undergoes a change in oxidation state but appears in two product having the same oxidation state. 
    \\\mathrm{\overset{+6\times2}{Cr_{2}O_{7}^{2-}}\rightarrow \overset{+3}{Cr^{3+}}\, +\, \overset{+3}{Cr^{3+}}}\\\mathrm{n-factor=\left | (+6)\, \times\, 2\, -(+3)\, \times\, 2\right |=6}

    Example(3): For the salts which react in such a way that one atom undergoes a change in oxidation state but appear in two product having different oxidation state. 

\\\mathrm{\overset{+7}{3MnO_{4}^{-}}\rightarrow \overset{+2}{2Mn^{2+}}\, +\, \overset{+4}{}Mn^{6+} }\\=\left | 2\times(+2)-2\, \times\,(+7)\right |\, +\, \left | 1\times(+6)-1\, \times\,(+7) \right |\, =\, \left |4-14 \right |+\left | 6-7 \right |=11\\\mathrm{Hence,\, n-factor=\frac{11}{3}}

  • Precipitation or double decomposition Reaction 
    It is the reaction in which there is no change in the oxidation state for any atom, Here n-factor of the salt used in the reaction can be found out by multiplying the oxidation state of the cation or anion by the total number of atoms per molecule of the salt.

    \\\mathrm{BaCl_{2}\, +\, K_{2}SO_{4}\rightarrow BaSO_{4}\, +\, 2KCl}\\(n=2)\: \:\: \: (n=2)
    For BaCl2:
    n-factor = Oxidation state of Ba atom in BaCl2 x number of Ba atoms in one molecule of BaCl2 
    = (+2) x 1 = 2

    For K2SO4:
    n-factor = Oxidation state of K x number of K-atoms in one molecule of K2SO4
    = (+1) x 2 = 2

  • Laws of Equivalence
    According to the law Of equivalence, for each and every reactant and product, 
    Equivalents of each reactant reacted = Equivalents of each product formed.

    Example,
    Suppose the reaction is taking place as follows: 
    P\, +\, Q\rightarrow R\, +\, S 
    According to the law of equivalence, Equivalents of P reacted: 
    = Equivalents of Q reacted 
    = Equivalents of R produced 
    = Equivalents of S produced 

    Equivalents of any substance = (Weight of substance (in g)) / (Equivalent weight)
    = Normality (N) x Volume (V) (In litre) 
    Normality (N) = n-Factor x Molarity (M) 

    Normality and molarity are temperature dependent. As on changing the temperature, the volume of solution changes, so normality and molarity change.

  • POAC (Principle of Atom Conservation Method)

    For any chemical reaction, mole atom of any element remain conserved during the chemical reaction.

Exam Chapter
VITEEE Atomic Structure
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