For the reaction $H_2+B{r}_2⟶2HBr$ overall order is found to be $\frac{3}{2}$. The rate of reaction can be expressed as:

$\mathrm{aA}+\mathrm{bB}⟶\mathrm{cC}+\mathrm{dD}$

rate of reaction $=-\frac{1d\left[A\right]}{a}=-\frac{1}{dt}\frac{d\left[B\right]}{dt}=\frac{1}{c}\frac{d\left[C\right]}{dt}=\frac{1}{d}\frac{d\left[D\right]}{dt}=k\left[A{]}^x\right[B{]}^y$

Given reaction,

${\mathrm{aH}}_2+{\mathrm{Br}}_2⟶2\mathrm{HBr}$

The rate of reaction can be expressed as.
According to the Rate Equation, the Order of the Reaction is the product of the Concentration of the Reactants.
So the sum of powers in all the 3

The order of reaction is,

$\left[{\mathrm{H}}_2\right]{\left[{\mathrm{Br}}_2\right]}^{1/2}$, ${\left[{\mathrm{H}}_2\right]}^{1/2}\left[{\mathrm{Br}}_2\right]$, ${\left[{\mathrm{H}}_2\right]}^{3/2}{\left[{\mathrm{Br}}_2\right]}^0$

Hence, option D is correct.