$$K_1, K_2$$ and $$K_3$$ are the equilibrium constants of the following reactions (I), (II) and (III) respectively: (I) $$\text{N}_2 + 2\text{O}_2 \rightleftharpoons 2\text{NO}_2$$ (II) $$2\text{NO}_2 \rightleftharpoons \text{N}_2 + 2\text{O}_2$$ (III) $$\text{NO}_2 \rightleftharpoons \tfrac{1}{2}\text{N}_2 + \text{O}_2$$. The correct relation from the following is

The three reactions are written below exactly as they appear in the statement:

(I) $$\text{N}_2 + 2\,\text{O}_2 \rightleftharpoons 2\,\text{NO}_2 \qquad K_1$$

(II) $$2\,\text{NO}_2 \rightleftharpoons \text{N}_2 + 2\,\text{O}_2 \qquad K_2$$

(III) $$\text{NO}_2 \rightleftharpoons \tfrac12\,\text{N}_2 + \text{O}_2 \qquad K_3$$

Step 1: Relate $$K_1$$ and $$K_2$$
Reaction (II) is the exact reverse of reaction (I).
Rule: If a reaction is reversed, its equilibrium constant is the reciprocal.
Therefore $$K_2 = \frac{1}{K_1} \qquad -(1)$$

Step 2: Relate $$K_2$$ and $$K_3$$
Compare reactions (II) and (III). Multiplying every coefficient in reaction (III) by 2 gives reaction (II):

$$2\Bigl(\text{NO}_2 \rightleftharpoons \tfrac12\,\text{N}_2 + \text{O}_2\Bigr)\; \Longrightarrow\; 2\,\text{NO}_2 \rightleftharpoons \text{N}_2 + 2\,\text{O}_2$$

Rule: When the stoichiometric coefficients of a balanced reaction are multiplied by a factor $$n$$, the new equilibrium constant becomes the old one raised to the power $$n$$. Here $$n = 2$$, so

$$K_2 = (K_3)^2 \qquad -(2)$$

Step 3: Write all three constants in one chain
From (1): $$K_2 = \dfrac{1}{K_1}\;\;\Longrightarrow\;\;\dfrac{1}{K_2} = K_1$$
From (2): $$K_2 = (K_3)^2\;\;\Longrightarrow\;\;\dfrac{1}{(K_3)^2} = \dfrac{1}{K_2}$$

Combining, we obtain the complete relation:

$$K_1 = \dfrac{1}{K_2} = \dfrac{1}{(K_3)^2}$$

Hence the correct option is:
Option B which is: $$K_1 = \dfrac{1}{K_2} = \dfrac{1}{(K_3)^2}$$