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# For the reaction ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]\rightarrow { Fe }^{ 3+ }+{ CO }_{ 2 }+{ NO }_{ 3 }^{ - }$, the n-factor is:(a) 1(b) 11(c) $\cfrac { 5 }{ 3 }$(d) 61

Hint: The n-factor of a substance is also called the valency-factor or the conversion factor. In the reaction given above, ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ is undergoing oxidation. The n-factor for a redox reaction is equal to the number of electrons lost or gained per molecule/per mole of the reactant.

The n-factor of a substance is also called the valency-factor. In the above reaction, the species ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ is undergoing oxidation since the oxidation of iron present in the species is changing from +2 to +3, the oxidation number of carbon present in the species is changing from +2 to +4 and the oxidation number of nitrogen present in the species is changing from -3 to +5. The n-factor for a redox reaction is equal to the number of electrons lost or gained per molecule/per mole of the reactant. The reaction given in the question is called combined reaction since the different atoms present in the same species are all getting oxidised. For such reactions we can find the n-factor by finding the n-factor for the individual oxidation reactions and then adding the n-factors for these individual oxidation reactions.
First we will consider the oxidation of iron. Iron is present in +2 oxidation state in the reactant ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ and it’s getting oxidised to ${ Fe }^{ 3+ }$ as is given in the reaction. We will find the n-factor for this individual oxidation reaction:
${ Fe }^{ 2+ }\rightarrow { Fe }^{ 3+ }+{ e }^{ - }$
From the above reaction it is clear that the n-factor for this individual oxidation reaction is 1.
Now we will consider the oxidation of carbon. Carbon is present in +2 oxidation state in the reactant ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ and it’s getting oxidised to +4 oxidation state in ${ CO }_{ 2 }$ as is given in the reaction. We will find the n-factor for this individual oxidation reaction:
${ CN }^{ - }\rightarrow { CO }_{ 2 }+2{ e }^{ - }$
Since the number of carbon atoms in ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ is six, we will multiply the above equation by 6:
${ 6CN }^{ - }\rightarrow { 6CO }_{ 2 }+12{ e }^{ - }$
From the above reaction it is clear that the n-factor for this individual oxidation reaction is 12.
Now we will consider the oxidation of nitrogen. Nitrogen is present in -3 oxidation state in the reactant ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ and it’s getting oxidised to +5 oxidation state in ${ NO }_{ 3 }^{ - }$ as is given in the reaction. We will find the n-factor for this individual oxidation reaction:
${ CN }^{ - }\rightarrow { NO }_{ 3 }^{ - }+8{ e }^{ - }$
Since the number of nitrogen atoms in ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]$ is six, we will multiply the above equation by 6:
$6{ CN }^{ - }\rightarrow 6{ NO }_{ 3 }^{ - }+48{ e }^{ - }$
From the above reaction it is clear that the n-factor for this individual oxidation reaction is 48.
Now in order to find the n-factor of the reaction ${ K }_{ 4 }[Fe(CN{ ) }_{ 6 }]\rightarrow { Fe }^{ 3+ }+{ CO }_{ 2 }+{ NO }_{ 3 }^{ - }$ we will add the n-factors of the individual oxidation reactions.
Hence 1+12+48=61

Hence the correct answer is (d) 61.

Note: The n-factor for a particular species is not fixed; it can change depending upon the type of reaction, the medium of the reaction i.s. acidic or basic. Also the method of calculating the n-factor for a particular species will vary depending upon the type of reaction example combined reaction, disproportionation reaction, neutralisation reaction etc.