The reaction $$A_{2}+B_{2}\rightarrow 2AB$$ follows the mechanism
$$A_{2}k_{1}A+A(fast)\\ \text{     }{_{k_{-1}}}\\ \\A+B_{2}\xrightarrow{k_{2}}AB+B(slow)\\A+B \rightarrow AB(fast)$$
The overall order of the reaction is :

The reaction $$A_2 + B_2 \rightarrow 2AB$$ follows the given mechanism and we need to find the overall order.

We note that the slow step is $$A + B_2 \xrightarrow{k_2} AB + B$$, and the rate law is Rate = $$k_2[A][B_2]$$.

Next, the first step is a fast equilibrium: $$A_2 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} A + A$$. At equilibrium, $$K = \frac{k_1}{k_{-1}} = \frac{[A]^2}{[A_2]}$$, so $$[A] = \sqrt{K[A_2]} = \sqrt{\frac{k_1}{k_{-1}}} \cdot [A_2]^{1/2}$$.

Substituting this expression into the rate law gives Rate = $$k_2 [A][B_2] = k_2 \sqrt{\frac{k_1}{k_{-1}}} [A_2]^{1/2} [B_2]$$ and $$= k'[A_2]^{1/2}[B_2]^1$$.

Therefore the overall order is $$\frac{1}{2} + 1 = 1.5$$.

The correct answer is Option D) 1.5.