Question

n-factor of NaCN in the change , NaCN$\to$NaCNO, is:

Answer 1

For a redox reaction, n-factor is equal to the number of electrons lost or gained by the atoms in the given compound. Overall charge on cyanide ($C{{N}^{-}}$) and cyanate ($CN{{O}^{-}}$) is -1.

Complete Step-by-Step Solution:
n-factor is a valency factor or conversion factor of a substance. The method to find n-factor for such substances depends upon the type of reaction it undergoes. There are different formulas for finding n-factors for acids, bases, for redox reactions and for reactions that do not involve change in oxidation state of a compound.
To calculate the n-factor of NaCN, first we will need to identify which type of reaction this is.
Let’s see the oxidation states of all atoms in NaCN. Actually NaCN is a salt and is made up of two ions, sodium and cyanide.
- Sodium is positively charged and its oxidation state will be +1 as CN has an overall charge of -1. Now, the Nitrogen atom is in -3 oxidation state and so carbon needs to be in +2 oxidation state because overall charge on cyanide is -1.
- In NaCNO, sodium is positively charged ions and so it is in +1 oxidation state. The negatively charged ion is cyanate ion and it has an overall charge of -1. So, here oxidation will be in its normal oxidation state and that will be -2. Nitrogen will also be in its normal oxidation state which will be +3. So, the oxidation state of the carbon atom should be +4 because only in that case, the overall charge on the ion will be equal to -1.
- So, we can say that this is a redox reaction and the oxidation state of carbon changes from +2 to +4 in the reaction.
- It is very easy in the case of redox reactions to find the n-factor of such a compound.
- n-factor for a compound undergoing redox reaction is equal to the number of electrons lost or gained by any atom in the compound.
- So we can say that only carbon changes its oxidation state in the above reaction that can be shown as below.
${{C}^{+2}}\to {{C}^{+4}}+2{{e}^{-}}$
So, the number of electrons lost by one mole of NaCN will be equal to 2.

So, the n-factor of this reaction is 2.

Note: Do not consider the oxidation state of nitrogen atoms in NaCN equal to +5, this is not possible because one of these two atoms should have a negative oxidation state and nitrogen being more electronegative, should have a negative oxidation state which is -3.