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**Hint:**Using the given chemical reaction first calculate the n-factor of $N{H_3}$ and ${N_2}$. Then find their equivalent weights and subtract them to find your answer.

Formula used:

-Equivalent weight: It is defined as molecular weight divided by the n-factor of the molecule.

$Eq.wt. = Mol.wt./n - factor$ …………..(1)

**Complete step by step answer:**

-To find the value of (${y_1} - {y_2}$), first we need to find the value of ${y_1}$ and ${y_2}$ which are the respective equivalent weights of $N{H_3}$ and${N_2}$.

To find the equivalent weights we need to find the n-factor of both because: equation (1)

$Eq.wt. = Mol.wt./n - factor$

-So, first let’s find out the n-factor of $N{H_3}$ and ${N_2}$, using it we will find their equivalent weights. The reaction is: ${N_2} + 3{H_2} \to 2N{H_3}$.

-For ${N_2}$: Initially the oxidation state of N atoms here is 0. But later it is converted into ${N^{ - 3}}$ in $N{H_3}$. This means that 1 atom of N shares 3 electrons with H atoms and so 2 N atoms will share in total 6 electrons. So, the n-factor of ${N_2}$ is 6.

Molecular weight of ${N_2}$ is ${x_2}$ and equivalent weight is ${y_2}$ (given in the question).

So, using equation (1): ${y_2} = {x_2}/6$ ………...(2)

-For $N{H_3}$: Above we just proved how 2 N atoms or 2 moles of $N{H_3}$ are taking 6 electrons. So, 1 mole of $N{H_3}$ will take 3 electrons. Hence the n-factor of $N{H_3}$ will be 3.

Molecular weight of $N{H_3}$ is ${x_1}$ and its equivalent weight is ${y_1}$ (given in the question).

So, using equation (1): ${y_1} = {x_1}/3$ ………….(3)

-Now use the values of ${y_2}$ and ${y_1}$ from equations (2) and (3) to calculate (${y_1} - {y_2}$).

${y_1} - {y_2}$ = $({x_1}/3) - ({x_2}/6)$

= $(2{x_1} - {x_2})/6$

So, the correct option is: (A) $(2{x_1} - {x_2})/6$

**Note:**The most common mistake we make here is while finding out the n-factor that is the number of electrons being shared. So, while calculating the oxidation states always check whether that state is for 1 atom or more atoms.

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